case study questions class 12 physics chapter 2

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CBSE Class 12 Physics Case Study Questions PDF Download

Case Study questions for the Class 12 Physics board exams are available here. You can read the Class 12 Physics Case Study Questions broken down by chapter. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve success on your final exams, practice the following questions.

CBSE (Central Board of Secondary Education) is a renowned educational board in India that designs the curriculum for Class 12 Physics with the goal of promoting a scientific temperament and nurturing critical thinking among students. As part of their Physics examination, CBSE includes case study questions to assess students’ ability to apply theoretical knowledge to real-world scenarios effectively.

Chapter-wise Solved Case Study Questions for Class 12 Physics

Before the exams, students in class 12 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be asked in Physics exams for Grade 12. These questions were created by our highly qualified standard 12 Physics staff based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 12 in understanding the topics.

Best Books for Class 12 Physics Exams

Strictly in accordance with the new term-by-term curriculum for the class 12 board exams to be held in the academic year 2023–2024, which will include multiple choice questions based on new board typologies including stand-alone MCQs and case-based MCQs with an assertion–reason. Included are inquiries from the official CBSE Question Bank that was released in April 2024. What changes have been made to the book: strictly in accordance with the term-by-term syllabus for the board exams that will be held during the 2024 academic year? Chapter- and topic-based Questions with multiple choices that are based on the unique evaluation method used for the Class 12th Physics Final Board Exams.

Key Benefits of Solving CBSE Class 12 Physics Case Study Questions

• Application of Concepts: Case study questions demand the application of theoretical knowledge in practical scenarios, preparing students for real-world challenges and professional pursuits.
• Critical Thinking: By evaluating and analyzing case studies, students develop critical thinking abilities, enabling them to approach complex problems with a logical mindset.
• In-Depth Understanding: Addressing case study questions necessitates a thorough understanding of physics concepts, leading to a more profound comprehension of the subject matter.
• Holistic Evaluation: CBSE adopts case study questions as they provide a holistic evaluation of a student’s aptitude and proficiency in physics, moving beyond rote memorization.
• Preparation for Competitive Exams: Since competitive exams often include similar application-based questions, practicing case study questions equips students for various entrance tests.

How to Approach CBSE Class 12 Physics Case Study Questions

• Read and Analyze Thoroughly: Carefully read the case study to grasp its context and identify the underlying physics principles involved.
• Identify Relevant Concepts: Highlight the physics theories and concepts applicable to the given scenario.
• Create a Systematic Solution: Formulate a step-by-step solution using the identified concepts, explaining each step with clarity.
• Include Diagrams and Charts: If relevant, incorporate diagrams, charts, or graphs to visually represent the situation, aiding better comprehension.
• Double-Check Answers: Always review your answers for accuracy, ensuring that they align with the principles of physics.

Tips for Excelling in CBSE Class 12 Physics

• Conceptual Clarity: Focus on building a strong foundation of physics concepts, as this will enable you to apply them effectively to case study questions.
• Practice Regularly: Dedicate time to solving case study questions regularly, enhancing your proficiency in handling real-world scenarios.
• Seek Guidance: Don’t hesitate to seek guidance from teachers, peers, or online resources to gain additional insights into challenging concepts.
• Time Management: During exams, practice efficient time management to ensure you allocate enough time to each case study question without rushing.
• Stay Positive: Approach case study questions with a positive mindset, embracing them as opportunities to showcase your skills and knowledge.

CBSE Class 12 Physics case study questions play a pivotal role in promoting practical understanding and critical thinking among students. By embracing these questions as opportunities for growth, students can excel in their physics examinations and beyond.

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1569+ Class 12 Physics Case Study Questions PDF Download

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So, you’re in Class 12 and the concept of Physics case study questions is beginning to loom large. Are you feeling a little lost? Fear not! This article will guide you through everything you need to know to conquer these challenging yet rewarding Class 12 Physics Case Study Questions.

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We have provided here Case Study questions for the Class 12 Physics for board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer and solutions are also provided so that you can check the correct answer for each question. Practice these questions to score excellent marks in your final exams.

Physics is more than just formulas and equations, right? It’s a way to interpret the natural world, and case studies provide a perfect opportunity for students to apply theoretical knowledge to real-world situations. So, what’s the importance of case studies in class 12 Physics? Well, they help you to solidify your understanding and enhance your analytical skills, which are invaluable for exams and beyond.

Table of Contents

Case Study-Based Questions for Class 12 Physics

There are total of 14 chapters Electric Charges And Fields, Electrostatic Potential And Capacitance, Current Electricity, Moving Charges And Magnetism, Magnetism And Matter, Electromagnetic Induction, Alternating Current, Electromagnetic Waves, Ray Optics and Optical Instruments, Wave Optics, Dual Nature Of Radiation And Matter, Atoms, Nuclei, Semiconductor Electronics Materials Devices, And Simple Circuits.

• Case Study Based Questions on class 12 Physics Chapter 1 Electric Charges And Fields
• Case Study Based Questions on Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance
• Case Study Based Questions on Class 12 Physics Chapter 3 Current Electricity
• Case Study Based Questions on Class 12 Physics Chapter 4 Moving Charges And Magnetism
• Case Study Based Questions on Class 12 Physics Chapter 5 Magnetism And Matter
• Case Study Based Questions on Class 12 Physics Chapter 6 Electromagnetic Induction
• Case Study Based Questions on Class 12 Physics Chapter 7 Alternating Current
• Case Study Based Questions on Class 12 Physics Chapter 8 Electromagnetic waves
• Case Study Based Questions on Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
• Case Study Based Questions on class 12 Physics Chapter 10 Wave Optics
• Case Study Based Questions on class 12 Physics Chapter 11 Dual Nature of Matter and Radiation
• Case Study Based Questions on class 12 Physics Chapter 12 Atoms
• Case Study Based Questions on class 12 Physics Chapter 13 Nuclei
• Case Study Based Questions on class 12 Physics Chapter 14 Semiconductor Electronics
• Class 12 Chemistry Case Study Questions
• Class 12 Biology Case Study Questions
• Class 12 Maths Case Study Questions

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy-to-learn solutions.

How to Approach Physics Case Study Questions

Ah, the million-dollar question! How do you tackle these daunting case study questions? Let’s dive in.

Effective Reading Techniques

A vital part of cracking case study questions is understanding the problem correctly. Don’t rush through the question, take your time to fully grasp the scenario, and don’t skip over the details – they can be critical!

Conceptual Analysis

Once you’ve understood the problem, it’s time to figure out which physics principles apply. Remember, your foundational knowledge is your greatest weapon here.

Solving Strategies

Next, put pen to paper and start solving! Follow a systematic approach, check your calculations, and make sure your answer makes sense in the context of the problem.

Best Books for Class 12 Physics Exams

Class 12 Physics Syllabus 2024

Chapter–1: Electric Charges and Fields

Electric charges, Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle, and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in a uniform electric field.

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell (field inside and outside).

Chapter–2: Electrostatic Potential and Capacitance

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only).

Chapter–3: Current Electricity

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s rules, Wheatstone bridge.

Chapter–4: Moving Charges and Magnetism

Concept of magnetic field, Oersted’s experiment.

Biot – Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields.

Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometerits current sensitivity and conversion to ammeter and voltmeter.

Chapter–5: Magnetism and Matter

Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines.

Magnetic properties of materials- Para-, dia- and ferro – magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties.

Chapter–6: Electromagnetic Induction

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction.

Chapter–7: Alternating Current

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer.

Chapter–8: Electromagnetic Waves

Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Chapter–9: Ray Optics and Optical Instruments

Ray Optics:  Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism.

Optical instruments:  Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics

Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only).

Chapter–11: Dual Nature of Radiation and Matter

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

Experimental study of photoelectric effect

Matter waves-wave nature of particles, de-Broglie relation.

Chapter–12: Atoms

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in his orbit, of hydrogen line spectra (qualitative treatment only).

Chapter–13: Nuclei

Composition and size of nucleus, nuclear force

Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction

Semiconductor diode – I-V characteristics in forward and reverse bias, application of junction diode -diode as a rectifier.

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You can click on the chapter-wise Case Study links given above and read the Class 12th Physics Case Study questions and answers provided by us. If you face any issues or have any questions please put your questions in the comments section below. Our teachers will be happy to provide you with answers.

FAQs Class 12 Physics Case Studies Questions

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studyrate.in is the best website for Class 12 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

What are Physics case study questions?

Physics case study questions involve applying physics principles to solve real-world scenarios or problems.

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Strengthen your basic concepts, practice solving a variety of problems, and make use of resources like the Class 12 Physics Case Study Questions PDF.

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CBSE 12th Standard Physics Subject Electrostatic Potential And Capacitance Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

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Cbse 12th standard physics subject electrostatic potential and capacitance case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

(ii) The signs of charges Q 1 and Q 2 respectively are

(iii) Which of the two charges Q 1 and Q 2 is greater in magnitude?

(iv) Which of the following statement is not true?

(v) Positive and negative point charges of equal magnitude are kept at  \(\left(0,0, \frac{a}{2}\right)\)  and  \(\left(0,0, \frac{-a}{2}\right)\)  respectively. The work done by the electric field when another positive point charge is moved from (-a, 0, 0) to (0, a, 0) is

(ii) Nature of equipotential surface for a point charge is

(iii) A spherical equipotential surface is not possible

(iv) The work done in carrying a charge q once round a circle of radius a with a charge Q at its centre is

(v) The work done to move a unit charge along an equipotential surface from P to Q

(ii) How much charge should be placed on a capacitance of 25 pF to raise its potential to l0 5  V?

(iii) Dimensions of capacitance is

(iv) Metallic sphere of radius R is charged to potential V. Then charge q is proportional to

(v) If 64 identical spheres of charge q and capacitance C each are combined to form a large sphere. The charge and capacitance of the large sphere is

(ii) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. The separation between the plates is now reduced half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in second case.

(iii) A dielectric introduced between the plates of a parallel plate condenser

(iv) A parallel plate capacitor of capacitance 1 pF has separation between the plates is d. When the distance of separation becomes 2d and wax of dielectric constant x is inserted in it the capacitance becomes 2 pF. What is the value of x?

(v) A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is

(ii) When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance

(iii) Which of the following is a dielectric?

(iv) For a polar molecule, which of the following statements is true?

(v) When a comb rubbed with dry hair attracts pieces of paper. This is because the

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Cbse 12th standard physics subject electrostatic potential and capacitance case study questions 2021 answer keys.

(i) (b): Here  \(C-50 p F-50 \times 10^{-12} F, V=10^{4} V\) \(R=\frac{1}{4 \pi \varepsilon_{0}} \cdot C=9 \times 10^{9} \mathrm{mF}^{-1} \times 50 \times 10^{-12} \mathrm{~F}\) = 45 x 10 -2 m = 45 cm (ii) (d) : As  \(q=C V=25 \times 10^{-12} \times 10^{5}=2.5 \mu \mathrm{C}\) (iii) (c) (iv) (c): As charge  \(q=C V=\left(4 \pi \varepsilon_{0} R\right) V\) \(\therefore\)   q depends on both V and R. (v) (c): 64 drops have formed a single mop of radius R. Volume oflarge sphere = 64 x Volume of small sphere \(\therefore \frac{4}{3} \pi R^{3}=64 \frac{4}{3} \pi r^{3} \Rightarrow R=4 r \text { and } Q_{\text {total }}=64 q\) \(C^{\prime}=4 \pi \varepsilon_{0} R \Rightarrow C^{\prime}=\left(4 \pi \varepsilon_{0}\right) \cdot 4 r \Rightarrow C^{\prime}=4 C\)

(i) (b):  \(k=\frac{\text { Capacitance with dielectric }}{\text { Capdcitance without dielectric }}\)   \(=\frac{80 \mu \mathrm{F}}{4 \mu \mathrm{F}}=20\) (ii) (c): Capacitance of the capacitor with air between plates \(C^{\prime}=\frac{\varepsilon_{0} A}{d}=8 \mathrm{pF}\) With the capacitor is filled with dielectric (k = 5) between its plates and the distance between the plates is reduced by half, capacitance become  \(C=\frac{\varepsilon_{0} k A}{d / 2}=\frac{\varepsilon_{0} \times 5 \times A}{d / 2}=10 C^{\prime}=10 \times 8=80 \mathrm{pF}\) (iii) (d):  If a dielectric medium of dielectric constant K is filled completely between the plates then capacitance increases by K times . (iv) (b):  \(C=\frac{\varepsilon_{0} A}{d}=1 \mathrm{pF}\)      ....(i) \(C^{\prime}=\frac{x \varepsilon_{0} A}{(2 d)}=2 \mathrm{pF}\)    ....(ii) Divide (ii) by (i), x/2 = 2/1  \(\Rightarrow\)  x = 4 (v) (c) : As capacitance  \(C_{o}=\frac{\varepsilon_{o} A}{d}\) \(\therefore\)  After inserting copper plate,  \(C=\frac{\varepsilon_{o} A}{d-b}\)

(i) (d): In polar molecule the centres of positive and negative charges are separated even when there is no external field. Such molecule have a permanent dipole moment. Ionic molecule like HCl is an example of polar molecule. (ii) (c): As, \(F_{m}=\frac{F_{o}}{K}\) \(\therefore\)  The maximum force decreases by K times. (iii) (b) (iv) (b) : A polar molecule is one in which the centre of gravity for positive and negative charges are separated. (v) (a)

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• Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Chapter 2

Important Practice Problems for CBSE Class 12 Physics Chapter 2: Free PDF Download

Students of class 12 can find the important questions of Chapter 2 physics class 12 provided in a PDF format here. Scoring good marks in class 12 board exams is extremely important for every student. Hence, the important questions for class 12 physics chapter 2 - Electrostatic Potential and capacitance is made available to the students so that they can make a quick revision of all the vital topics and the important questions of chapter 2 physics class 12 and confidently ace the board exams. The physics class 12 chapter 2 important questions PDF can be easily downloaded from vedantu.com and students can refer to them anytime.

Download CBSE Class 12 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 12 Physics Important Questions for other chapters:

Related Chapters

Study Important Questions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

Very Short Answer Question (1 Mark Questions)

1. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans: The electric field that is present inside a dielectric decreases when it is placed in an external electric field because of polarisation as it creates an internal electric field which is opposite to the external electric field inside a dielectric due to which the net electric field gets reduced.

2. What is the work done in moving a $2\mu C$ point change from corner A to corner B of a square ABCD when a $10\mu C$ charge exists at the centre of the square?

Ans: Point A & B are at the same distance from point O.

\[{{V}_{A}}={{V}_{B}}\]

$\Rightarrow \text{ Work done }=\text{0}$

3. Force of attraction between two point electric charges placed at a distance \[d\] in a medium is \[F\]. What distance apart should these be kept in the same medium, so that force between them becomes\[\frac{F}{3}\]?

Ans: If two point charges are \[{{q}_{1}}\] and \[{{q}_{2}}\] separated by distance \[d\], it can be expressed as:

\[F=\frac{K{{q}_{1}}{{q}_{2}}}{{{d}^{2}}}\]

Suppose if force becomes \[\frac{F}{3}\] let the distance be \[x\]

\[\frac{F}{3}=\frac{K{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}\]

\[\Rightarrow \frac{K{{q}_{1}}{{q}_{2}}}{3{{d}^{2}}}=\frac{K{{q}_{1}}{{q}_{2}}}{{{x}^{2}}}\]

\[\Rightarrow {{x}^{2}}=3{{d}^{2}}\]

\[\Rightarrow x=\sqrt{3}d\]

4. The distance of the field point on the equatorial plane of a small electric dipole is halved. By what factors will the electric field due to the dipole changes?

Ans: The formula \[E\propto \frac{1}{{{r}^{3}}}\] gives the relation between electric field and distance. 

\[\therefore E\propto \frac{1}{{{\left( \frac{r}{2} \right)}^{3}}}\Rightarrow E\propto \frac{8}{{{r}^{3}}}\]

Therefore, the electric field becomes eight times larger.

5. The Plates of a charged capacitor are connected by a voltmeter. If the plates of the capacitor are moved further apart, what will be the effect on the reading of the voltmeter?

Ans: The relation between capacitance, area, distance and dielectric constant is \[C=\frac{A{{\in }_{0}}}{d}\Rightarrow C\propto \frac{1}{d}\]

Hence, if distance increases, capacitance decreases.

Since \[V=\frac{Q}{C}\] and charge on the capacitor is constant.

Hence reading of the voltmeter increases.

6. What happens to the capacitance of a capacitor when a dielectric slab is placed between its plates?

Ans: The introduction of dielectric in a capacitor will reduce the effective charge on plate and therefore will increase the capacitance.

Short Answer Questions (2 Marks Questions)

1. Show mathematically that the potential at a point on the equatorial line of an electric dipole is Zero?

Ans:  Electric potential at point on the equatorial line of an electric dipole can be expressed mathematically as:

\[V={{V}_{PA}}+{{V}_{PB}}\]

\[\Rightarrow V=\frac{K\left( -q \right)}{r}+\frac{K\left( +q \right)}{r}\]

\[\Rightarrow V=0\]

2. A parallel plate capacitor with air between the plates has a capacitance of \[8pF\]\[\left( 1pF={{10}^{-12}}F \right)\]. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: For air, capacitance can be expressed as, \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

\[{{C}_{0}}=8\times pF=8\times {{10}^{-12}}F\]

Now \[{d}'=\frac{d}{2}\] and \[K=6\]

\[\Rightarrow {C}'=\frac{A{{C}_{0}}}{{{d}'}}\times 2\times K\]

\[\Rightarrow {C}'=8\times {{10}^{-12}}\times 2\times 6\]

\[\Rightarrow {C}'=96\times {{10}^{-12}}pF\]

3. Draw one equipotential surfaces (1) Due to uniform electric field (2) For a point charge \[\left( q<0 \right)\]?

Ans: The diagrams are given as:

4. If the amount of electric flux entering and leaving a closed surface are \[{{\phi }_{1}}\] and \[{{\phi }_{2}}\] respectively. What is the electric charge inside the surface?

Ans: Net flux can be given as \[={{\phi }_{2}}-{{\phi }_{1}}\]

Since \[\phi =\frac{q}{{{\in }_{0}}}\]

\[\Rightarrow Q=\left( {{\phi }_{2}}-{{\phi }_{1}} \right){{\in }_{0}}\]

The electric charge inside the surface can be given as:

\[\Rightarrow Q=\phi {{\in }_{0}}\]

5. A stream of electrons travelling with speed \[v\,m/s\] at right angles to a uniform electric field E is deflected in a circular path of radius \[r\]. Prove that \[\frac{e}{m}=\frac{{{v}^{2}}}{rE}\]?

Ans: The path of the electron that is travelling with velocity \[v\,m/s\] at right angle of \[\overline{E}\] is of circular shape.

It requires a centripetal in nature, \[F=\frac{m{{v}^{2}}}{r}\]

It is provided by an electrostatic force \[F=eE\]

\[\Rightarrow eE=\frac{m{{v}^{2}}}{r}\]

\[\Rightarrow \frac{e}{m}=\frac{{{v}^{2}}}{rE}\]

6. The distance between the plates of a parallel plate capacitor is \[d\]. A metal plate of thickness \[\left( \frac{d}{2} \right)\] is placed between the plates. What will be the effect on the capacitance?

Ans: For air \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

Thickness \[t=\frac{d}{2}\] only when \[k=\infty \]

\[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

\[{{C}_{metal}}=\frac{A{{\in }_{0}}}{\left( d-t \right)}\]

\[\Rightarrow {{C}_{metal}}=\frac{A{{\in }_{0}}}{\left( d-\frac{d}{2} \right)}\]

\[\Rightarrow {{C}_{metal}}=2A{{\in }_{0}}\]

\[\Rightarrow {{C}_{metal}}=2{{C}_{0}}\]

Hence, capacitance will double.

7. Two charges \[2\mu C\] and \[-2\mu C\] are placed at points A and B \[6cm\] apart.

(a) Identify an equipotential surface of the system.

Ans: The situation is represented in the adjoining figure.

An equipotential surface is the plane on which the total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

(b) What is the direction of the electric field at every point on this surface?

Ans: The direction of the electric field at every point on this surface Is normal to the plane in the direction of AB.

8. In a Van de Graaff type generator a spherical metal shell is to be a 

\[15\times {{10}^{6}}V\]electrode. The dielectric strength of the gas surrounding the electrode is \[5\times {{10}^{7}}V{{m}^{-1}}\]. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Ans: Given that,

Potential difference is given as, \[V=15\times {{10}^{6}}V\]

Dielectric strength of the surrounding gas\[=5\times {{10}^{7}}V{{m}^{-1}}\]

Electric field intensity is given as, E = Dielectric strength = \[5\times {{10}^{7}}V{{m}^{-1}}\]

Minimum radius of the spherical shell required for the purpose is given by,

\[r=\frac{V}{E}\]

\[\Rightarrow r=\frac{15\times {{10}^{6}}}{5\times {{10}^{7}}}\]

\[\Rightarrow r=0.3m=30cm\]

Therefore, the minimum radius of the spherical shell required is \[30cm\].

9. A small sphere of radius \[{{r}_{1}}\] and charge \[{{q}_{1}}\] is enclosed by a spherical shell of radius \[{{r}_{2}}\] and charge \[{{q}_{2}}\]. Show that if \[{{q}_{1}}\] is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge \[{{q}_{2}}\] on the shell is.)

Ans: According to the statement of Gauss’s law, the electric field between a sphere and a shell is determined by the charge \[{{q}_{1}}\] on a small sphere. Therefore, the potential difference, \[V\], between the sphere and the shell is independent of charge \[{{q}_{2}}\]. For positive charge \[{{q}_{1}}\], potential difference \[V\]is always positive.

10. Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

Ans: Equidistant planes which are parallel to the x-y plane are the equipotential surfaces.

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

Ans: Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) a single positive charge at the origin, and

Ans: Concentric spheres centred at the origin are equipotential surfaces.

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans: A periodically changing shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Short Answer Questions (3 Marks Questions)

1. Two dielectric slabs of dielectric constant \[{{K}_{1}}\] and \[{{K}_{2}}\] are filled in between the two plates, each of area \[A\], of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor? Area of each plate \[\frac{A}{2}\].

Ans: In the question, the two capacitors are in parallel

Net Capacitance, \[C={{C}_{1}}+{{C}_{2}}\] 

\[{{C}_{1}}=\frac{{{K}_{1}}{{\in }_{0}}\left( \frac{A}{2} \right)}{d}=\frac{{{K}_{1}}{{\in }_{0}}A}{2d}\]

\[{{C}_{2}}=\frac{{{K}_{2}}{{\in }_{0}}\left( \frac{A}{2} \right)}{d}=\frac{{{K}_{2}}{{\in }_{0}}A}{2d}\]

\[\Rightarrow C=\frac{{{K}_{1}}{{\in }_{0}}A}{2d}+\frac{{{K}_{2}}{{\in }_{0}}A}{2d}\]

\[\Rightarrow C=\frac{{{\in }_{0}}A}{2d}\left( {{K}_{1}}+{{K}_{2}} \right)\], which is the required capacitance.

2. Prove that the energy stored in a parallel plate capacitor is given by \[\frac{1}{2}C{{V}^{2}}\].

Ans: Let us suppose a capacitor is connected to a battery and it supplies a small amount of change \[dq\] at constant potential \[V\], then small amount of work done by the battery is given by

\[\Rightarrow dw=\frac{qc}{dq}\] …… (Since, \[q=CV\])

Total work done where the capacitor is fully changed to \[q\].

\[\int{dw=W}=\int_{0}^{q}{\frac{q}{c}dq}\]

\[\Rightarrow W=\int_{0}^{q}{qdq}\]

\[\Rightarrow W=\frac{{{q}^{2}}}{2C}=\frac{{{C}^{2}}{{V}^{2}}}{2C}\]

\[\Rightarrow W=\frac{1}{2}C{{V}^{2}}\]

This work done is stored in the capacitor in the form of electrostatic potential energy.

\[W=U=\frac{1}{2}C{{V}^{2}}\]

Hence proved.

3. State Gauss’s Theorem in electrostatics? Using this theorem, define an expression for the field intensity due to an infinite plane sheet of change of charge density \[\sigma C/{{m}^{2}}\].

Ans: Gauss’s Theorem has a statement that electric flux through a closed surface enclosing a charge \[q\] in vacuum is \[\frac{1}{{{\in }_{0}}}\] times the magnitude of the charge enclosed is \[\phi =\frac{q}{{{\in }_{0}}}\]

Let us consider a charge distributed over an infinite sheet of area S having surface change density\[\sigma C/{{m}^{2}}\]

To enclose the charge on sheet an imaginary Gaussian surface cylindrical in shape can be assumed and it is divided into three sections \[{{S}_{1}}\], \[{{S}_{2}}\] and \[{{S}_{3}}\].

According to Gauss’s theorem we can infer,

\[\phi =\frac{q}{{{\in }_{0}}}=\frac{\sigma S}{{{\in }_{0}}}\] …… (1)

We know that, \[\phi =\int{E\cdot ds}\]

For the given surface \[\phi =\int\limits_{{{S}_{1}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}+\int\limits_{{{S}_{2}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}+\int\limits_{{{S}_{3}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}\]

\[\Rightarrow \phi =\int\limits_{{{S}_{1}}}{E\cdot ds}\cos {{0}^{\circ }}+\int\limits_{{{S}_{3}}}{E\cdot ds}\cos {{0}^{\circ }}\] …… \[\left( \because \theta ={{90}^{\circ }}\,for\,{{S}_{2}}\,so\,\int\limits_{{{S}_{2}}}{\overrightarrow{E}\cdot \overrightarrow{ds}}=0 \right)\]

\[\Rightarrow \phi =E\int\limits_{{{S}_{1}}}{ds}+E\int\limits_{{{S}_{3}}}{ds}\]

\[\Rightarrow \phi =E\left[ \int_{0}^{S}{ds}+\int_{0}^{S}{ds} \right]\]

\[\Rightarrow \phi =E\cdot 2S\] ……(2)

From both equation (1) & (2)

\[E=\frac{\sigma }{2{{\in }_{0}}}\]

\[\Rightarrow E\cdot 2S=\frac{\sigma S}{{{\in }_{0}}}\], which is the required field intensity.

4. Derive an expression for the total work done in rotating an electric dipole through an angle \[\theta \] in a uniform electric field? \[E=1.5\times {{10}^{-8}}J\]

Ans: We know \[\tau =PE\sin \theta \]

If an electric dipole is rotated through an angle \[\theta \] against the torque, then small amount of work done is

\[dw=\tau d\theta =PE\sin \theta d\theta \]

For rotating through an angle \[\theta \] from \[{{90}^{\circ }}\]

\[W=\int_{{{90}^{\circ }}}^{\theta }{{}}PE\sin \theta \]

5. If \[{{C}_{1}}=3pF\] and \[{{C}_{2}}=2pF\], calculate the equivalent capacitance of the given network between A and B?

Ans: Since \[{{C}_{1}}\], \[{{C}_{2}}\]  and \[{{C}_{1}}\] are in series

\[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]

\[\Rightarrow \frac{1}{C}=\frac{1}{3}+\frac{1}{2}+\frac{1}{3}\]

\[\Rightarrow \frac{1}{C}=\frac{2+3+2}{6}\]

\[\Rightarrow C=\frac{6}{7}pF\]

\[{{C}_{2}}\] and C are in series

\[\Rightarrow {C}'=\frac{2}{1}+\frac{6}{7}=\frac{14+6}{7}=\frac{20}{7}pF\]

\[{{C}_{1}},{C}'\] and \[{{C}_{1}}\] are in series

\[\frac{1}{{{C}_{net}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}'}}+\frac{1}{{{C}_{1}}}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{1}{3}+\frac{7}{20}+\frac{1}{3}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{20+21+20}{60}=\frac{61}{60}\]

\[\Rightarrow \frac{1}{{{C}_{net}}}=\frac{61}{60}pF\Rightarrow {{C}_{net}}=\frac{60}{61}pF\], which is the net capacitance.

6. Prove that energy stored per unit volume in a capacitor is given by \[\frac{1}{2}{{\in }_{0}}{{E}^{2}}\], where E is the electric field of the capacitor.

Ans: We know capacitance of a parallel plate capacitor, \[C=\frac{A{{\in }_{0}}}{d}\], electric field in between the plates 

Energy stored per unit volume\[=\frac{Energy\,stored}{Volume}\]

Energy stored per unit volume\[=\frac{\frac{1}{2}\frac{{{q}^{2}}}{C}}{Ad}\] (Volume of the capacitor = Ad)

\[=\frac{\frac{1}{2}\times \frac{{{\left( \sigma A \right)}^{2}}}{\frac{A{{\in }_{0}}}{d}}}{Ad}=\frac{\frac{1}{2}\times {{E}^{2}}{{\in }^{2}}{{A}^{2}}d}{{{A}^{2}}{{\in }_{0}}d}\]

\[\Rightarrow \]Energy stored/volume\[=\frac{1}{2}{{\in }_{0}}{{E}^{2}}\]. 

Therefore, proved.

7. Keeping the voltage of the charging source constant. What would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by \[10%\]?

Ans: We have the relation, \[U=\frac{1}{2}C{{V}^{2}}\]

For parallel plate \[U=\frac{1}{2}\frac{A{{\in }_{0}}}{d}{{V}^{2}}\]

When \[{d}'=d-\frac{10}{100}\times d=0.9d\]

Then \[{U}'=\frac{1}{2}\frac{A{{\in }_{0}}}{0.9d}{{V}^{2}}\]

Change in energy\[={U}'-U=\frac{1}{2}\frac{A{{\in }_{0}}}{d}\left( \frac{1}{0.9}-1 \right)\]

\[={U}'-U=u\left( \frac{0.1}{0.9} \right)=\frac{U}{9}\]

\[%change=\frac{{U}'-U}{U}\times 100%=\frac{U}{9}\times \frac{1}{U}\times 100%=\frac{100}{9}%\]

\[\Rightarrow \% change=11.1%\]

8. Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of \[1.0cm\] as shown in figure.

Surface A is given a positive potential of \[10V\] and the outer surface of B is earthed.

(a) What is the magnitude and direction of the uniform electric field between point Y and Z? What is the work done in moving a charge of \[20\mu C\] from point X to Y?

Ans:   Since \[E=\frac{dv}{dr}=\frac{10V}{1\times {{10}^{-2}}}=1000\frac{V}{M}\]

Since surface A is an equipotential surface i.e., \[\Delta V=0\]

$\therefore$ Work done in moving a charge of $20 \mu C$ from X to Y = Zero.

(b) Can we have non-zero electric potential in space, where electric field strength is zero?

Ans:  Since surface A is an equipotential surface i.e., \[\Delta V=0\]

$\therefore$ Work done from X to Y = Zero.

\[E=\frac{-dv}{dr}\] if \[E=0\]

\[\frac{dv}{dr}=0\Rightarrow dv=0\] or V=constant (non-zero)

So, we can have non-zero electric potential, where the electric field is zero.

9. A regular hexagon of side \[10cm\] has a charge \[5\mu C\] at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure shows six equal numbers of charges \[q\], at the vertices of a regular hexagon.

Charge, \[q=5\mu C=5\times {{10}^{-5}}C\]

Side of the hexagon, \[AB=BC=CD=DE=EF=FA=10cm\]

Distance of each vertex from center \[O\], \[d=10cm\]

Electric potential at point \[O\],

\[V=\frac{6\times q}{4\pi {{\in }_{0}}d}\]

\[{{\in }_{0}}\]permittivity of free space

\[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

\[V=\frac{6\times 9\times {{10}^{9}}\times 5\times {{10}^{-5}}}{0.1}\]

\[\Rightarrow V=2.7\times {{10}^{6}}V\]

Therefore, the potential at the centre of the hexagon is \[2.7\times {{10}^{6}}V\].

10. A spherical conductor of radius \[12cm\] has a charge of \[1.6\times {{10}^{-7}}C\] distributed uniformly on its surface. What is the electric field? 

(a) Inside the sphere.

Ans: Radius of the spherical conductor is given as, \[r=12cm=0.12m\]

Charge is uniformly distributed over the conductor and can be given as, \[q=1.6\times {{10}^{-7}}C\]

Electric field inside a spherical conductor is zero. This is because if there is a field inside the conductor, then charges will move to neutralize it.

(b) Just outside the sphere  

Ans: Electric field E just outside the conductor is given by the relation,

\[E=\frac{q}{4\pi {{\in }_{0}}{{r}^{2}}}\]

\[{{\in }_{0}}=\]permittivity of free space

\[\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{-9}}}{{{\left( 0.12 \right)}^{2}}}\]

\[\Rightarrow E={{10}^{5}}N{{C}^{-1}}\]

Therefore, the electric field just outside the sphere \[={{10}^{5}}N{{C}^{-1}}\]

(c) At a point \[18cm\] from the centre of the sphere?

Distance of the point from the centre, \[d=18cm=0.18m\]

\[E=\frac{q}{4\pi {{\in }_{0}}{{d}^{2}}}\]

 \[\Rightarrow E=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-7}}}{\left( 18\times {{10}^{-2}} \right)}\]

 \[\Rightarrow E=4.4\times {{10}^{4}}N/C\]

Therefore, the electric field at a point 18 cm from the centre of the sphere is \[4.4\times {{10}^{4}}N/C\].

11. Three capacitors each of capacitance \[9pF\] are connected in series.

(a) What is the total capacitance of the combination

Ans: Capacitance of each of the three capacitors, \[C=9pF\]

Equivalent capacitance \[{C}'\] of the combination of the capacitors is given by the relation,

\[\frac{1}{C}=\frac{1}{{{C}'}}+\frac{1}{{{C}'}}+\frac{1}{{{C}'}}\]

\[\Rightarrow \frac{1}{C}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}\]

\[\Rightarrow C=3\mu F\]

Therefore, the total capacitance of the combination is \[3\mu F\].

(b) What is the potential difference across each capacitor if the combination is connected to a \[120V\] supply?

Ans:   Supply voltage is given as, \[V=100V\]

Potential difference \[V\]  across each capacitor is equal to one-third of the supply voltage.

\[{V}'=\frac{V}{3}=\frac{120}{3}=40V\]

Therefore, the potential difference across each capacitor is \[40V\].

12. Three capacitors of each capacitance \[2pF\] , \[3pF\] and \[4pF\]  are connected in parallel.

(a) What is the total capacitance of the combination?

Ans: Capacitance of the given capacitors are

\[{{C}_{1}}=2pF\]

\[{{C}_{2}}=3pF\]

\[{{C}_{3}}=4pF\]

For the parallel combination of the capacitors, equivalent capacitor \[{C}'\] is given by the algebraic sum,

\[{C}'=\left( 2+3+4 \right)pF\]

Therefore, the total capacitance of the combination is \[9pF\].

(b) Determine the charge on each capacitor if the combination is connected to a \[100V\] supply.

Ans:  Supply voltage, \[V=100V\]

The voltage through all the three capacitors is same as \[V=100V\] 

Charge on a capacitor of capacitance \[C\] and potential difference \[V\] is given by the relation,

\[q=CV\] …… (1)

For \[C=2pF\],

Charge\[=VC=100\times 2=200pC=2\times {{10}^{-10}}C\]

For \[C=3pF\], 

Charge\[=VC=100\times 3=300pC=3\times {{10}^{-10}}C\]

For \[C=4pF\], 

Charge\[=VC=100\times 4=400pC=4\times {{10}^{-10}}C\]

13. In a parallel plate capacitor with air between the plates, each plate has an area and the distance between the plates is \[3mm\]. Calculate the capacitance of the capacitor. If this capacitor is connected to a \[100V\] supply, what is the charge on each plate of the capacitor?

Ans: Area of each plate of the parallel plate capacitor, \[A=6\times {{10}^{-3}}{{m}^{2}}\]

Distance between the plates, \[d=3mm=3\times {{10}^{-3}}m\]

Supply voltage, \[V=100V\]

Capacitance \[C\] of a parallel plate capacitor is given by,

\[C=\frac{A{{\in }_{0}}}{d}\]

\[\Rightarrow {{\in }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{m}^{-2}}{{C}^{-2}}\]

\[\Rightarrow C=\frac{8.854\times {{10}^{-12}}\times 6\times {{10}^{-3}}}{3\times {{10}^{-3}}}\]

\[\Rightarrow C=17.71\times {{10}^{-12}}F\Rightarrow C=17.71pF\]

Potential \[V\] is related with the charge \[q\] and capacitance \[C\] as

\[V=\frac{q}{C}\Rightarrow q=VC\Rightarrow q=1.771\times {{10}^{-9}}C\]

\[\Rightarrow q=100\times 17.71\times {{10}^{-12}}C=17.71pF\]

Therefore, the capacitance of the capacitor is \[17.71pF\] and charge on each plate is \[1.771\times {{10}^{-9}}\].

14. Explain what would happen if in the capacitor given in Exercise 2.8, a \[3mm\] thick mica sheet (of dielectric constant \[=6\]) were inserted between the plates,

(a) While the voltage supply remained connected.

Ans: Dielectric constant of the mica sheet is given in the question as, \[k=6\]

Initial capacitance is provided as, \[C=1.771\times {{10}^{-11}}\]

New capacitance is given as, \[{C}'=kC=6\times 1.771\times {{10}^{-11}}=106pF\]

Supply voltage is given as, \[V=100V\]

New charge \[{q}'={C}'V=6\times 1.7717\times {{10}^{-9}}=1.06\times {{10}^{-8}}Cs\]

Potential across the plates remains \[100V\].

(b). After the supply was disconnected.

Ans: Dielectric constant is given in the question as, \[k=6\]

Initial capacitance, \[C=1.771\times {{10}^{-11}}F\]

New capacitance is given as \[{C}'=kC=6\times 1.771\times {{10}^{-11}}F=106pF\]

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge \[=1.771\times {{10}^{-9}}C\]

Potential across the plates is given by,

\[V=\frac{q}{{{C}'}}\]

\[\Rightarrow V=\frac{1.771\times {{10}^{-9}}}{106\times {{10}^{-12}}}\]

\[\Rightarrow V=16.7V\]

15. A \[12pF\] capacitor is connected to a \[50V\] battery. How much electrostatic energy is stored in the capacitor?

Ans: Capacitor of the capacitance is given as, \[C=12pF=12\times {{10}^{-12}}F\]

Potential difference is provided as, \[V=50V\]

Electrostatic energy stored in the capacitor is given by the relation,

\[{{E}_{B}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}{{\left( BZ \right)}^{2}}}\]

\[\cos \theta =\frac{0.10}{0.18}=\frac{5}{9}=0.5556\]

\[\theta ={{\cos }^{-1}}0.5556={{56.25}^{\circ }}\]

\[\Rightarrow 2\theta ={{112.5}^{\circ }}\]

\[\Rightarrow \cos 2\theta =-0.38\]

\[\overline{{{E}_{2}}}-\overline{{{E}_{1}}}=\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{2{{\in }_{0}}}\hat{n}=\frac{\sigma }{{{\in }_{0}}}\hat{n}\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{\lambda L}{{{\in }_{0}}}\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{1.06\times {{10}^{-10}}}-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=21.73\times {{10}^{-10}}J-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=13.58eV-27.2eV\]

\[\Rightarrow E\left( 2\pi dL \right)=13.58eV\]

\[E=\frac{1}{2}C{{V}^{2}}\]

\[\Rightarrow E=\frac{1}{2}\times 12\times {{10}^{-12}}\times {{\left( 50 \right)}^{2}}\]

\[\Rightarrow E=1.5\times {{10}^{-8}}J\]

Therefore, the electrostatic energy stored in the capacitor is \[1.5\times {{10}^{-8}}J\].

16. A \[600pF\]capacitor is charged by a \[200V\] supply. It is then disconnected from the supply and is connected to another uncharged \[600pF\] capacitor. How much electrostatic energy is lost in the process?

Capacitor of the capacitance, \[C=600pF\] 

Potential difference, \[V=200V\]

Electrostatic energy stored in the capacitor is given by,

\[\Rightarrow E=\frac{1}{2}\times \left( 600\times {{10}^{-12}} \right)\times {{\left( 200 \right)}^{2}}\]

\[\Rightarrow E=1.2\times {{10}^{-5}}J\]

If supply is disconnected from the capacitor and another capacitor of capacitance \[C=600pF\] is connected to it, then equivalent capacitance \[{C}'\] of the combination is given by,

\[\frac{1}{{{C}'}}=\frac{1}{C}+\frac{1}{C}\]

\[\Rightarrow \frac{1}{{{C}'}}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}\]

\[\Rightarrow {C}'=300pF\]

New electrostatic energy can be calculated as

\[{E}'=\frac{1}{2}{C}'\times {{V}^{2}}\]

\[{E}'=\frac{1}{2}\times 300\times {{\left( 200 \right)}^{2}}\]

\[{E}'=0.6\times {{10}^{-5}}J\]

Loss in electrostatic energy can be given as \[=E-{E}'\]

\[\Rightarrow E-{E}'=1.2\times {{10}^{-5}}-0.6\times {{10}^{-5}}\]

\[\Rightarrow E-{E}'=0.6\times {{10}^{-5}}\]

\[\Rightarrow E-{E}'=6\times {{10}^{-6}}J\]

Therefore, the electrostatic energy lost in the process is \[6\times {{10}^{-6}}J\].

17. A spherical conducting shell of inner radius \[{{r}_{1}}\] and outer radius \[{{r}_{2}}\] has a charge

(a) A charge \[q\] is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

Ans: Charge placed at the centre of a shell is \[+q\]. Hence a charge of magnitude \[-q\] will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is \[-q\].

Surface charge density at the inner surface of the shell can be given by the relation,

\[{{\sigma }_{1}}=\frac{Total\,Charge}{Outer\,Surface\,Area}=\frac{-q}{4\pi {{r}_{1}}^{2}}\]

\[{{\sigma }_{2}}=\frac{Total\,Charge}{Outer\,Surface\,Area}=\frac{-q}{4\pi {{r}_{2}}^{2}}\]

A charge of \[+q\] is induced on the outer surface of the shell. A charge of magnitude \[Q\] is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is \[Q+q\]. Surface charge density at the outer surface of the shell.

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but any irregular shape? Explain.

Ans:   Yes, the electric field intensity which is inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Now, take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop will be zero due to the field inside the conductor being zero. Therefore, the electric field is zero, whatever the shape.

18. If one of the two electrons of a \[{{H}_{2}}\] molecule is removed, we get a hydrogen molecular ion \[{{H}_{2}}^{+}\]. In the ground state of an \[{{H}_{2}}^{+}\], the two protons are separated by roughly \[1.5{{A}^{\circ }}\], and the electron is roughly \[1{{A}^{\circ }}\] from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Ans: The system of two protons and one electron is represented in the given figure:

Charge on proton 1, \[{{q}_{1}}=1.6\times {{10}^{-19}}C\]

Charge on proton 2, \[{{q}_{2}}=1.6\times {{10}^{-19}}C\]

Charge on electron, \[{{q}_{3}}=-1.6\times {{10}^{-19}}C\]

Distance between protons 1 and 2, \[{{d}_{1}}=1.5\times {{10}^{-10}}m\]

Distance between proton 1 and electron, \[{{d}_{2}}=1\times {{10}^{-10}}m\]

Distance between protons 2 and electron, \[{{d}_{3}}=1\times {{10}^{-10}}m\]

The potential energy at infinity is zero.

Potential energy of the system,

\[V=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}+\frac{{{q}_{1}}{{q}_{3}}}{4\pi {{\in }_{0}}{{d}_{3}}}+\frac{{{q}_{3}}{{q}_{1}}}{4\pi {{\in }_{0}}{{d}_{2}}}\]

Substituting \[\frac{1}{4\pi {{\in }_{0}}d}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[V=\frac{9\times {{10}^{9}}\times {{10}^{-19}}}{{{10}^{-10}}}\left[ -{{\left( 16 \right)}^{2}}+\frac{{{\left( 1.6 \right)}^{2}}}{1.5}-{{16}^{2}} \right]\]

$\Rightarrow V=-30.7\times {{10}^{-19}}J $

$\Rightarrow V=-19.2eV$

Therefore, the potential energy of the system is \[-19.2eV\].

19. Two charged conducting spheres of radii \[a\] and \[b\] are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions?

Ans: Let \[a\] be the radius of a sphere \[A\], \[{{Q}_{A}}\] be the charge on the sphere, and \[{{C}_{A}}\] be the capacitance of the sphere. Let \[b\] be the radius of a sphere \[B\], \[{{Q}_{B}}\] be the charge on the sphere, and \[{{C}_{B}}\] be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential \[V\] will become equal.

Let \[{{E}_{A}}\] be the electric field of sphere \[A\] and \[{{E}_{B}}\] be the electric field of sphere \[B\]. Therefore, their ratio,

\[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{4\pi {{\in }_{0}}{{a}_{2}}}\times \frac{{{b}^{2}}4\pi {{\in }_{0}}}{{{Q}_{B}}}\]

\[\Rightarrow \frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{{{Q}_{B}}}\times \frac{{{b}^{2}}}{{{a}_{2}}}\]

However, 

\[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{{{C}_{A}}V}{{{C}_{B}}V}\]

\[\frac{{{C}_{A}}}{{{C}_{B}}}=\frac{a}{b}\]

\[\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{a}{b}\]

Putting the value of (2) in (1), we obtain

\[\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{a}{b}\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{b}{a}\]

Therefore, the ratio of the electric field at the surface is \[\frac{b}{a}\].

20. What is the area of the plates of a \[2F\] parallel plate capacitor, given that the separation between the plates is \[0.5cm\]? (You will realize from your answer why ordinary capacitors are in the range \[\mu F\] or less. However, electrolytic capacitors do have a much larger capacitance \[0.1F\] because of very minute separation between the electrolytic capacitors.)

Ans: Capacitance of a parallel capacitor, \[V=2F\]

Distance between the two plates, \[d=0.5cm=0.5\times {{10}^{-2}}m\]. Capacitance of a parallel plate capacitor is given by the relation,

\[{{\in }_{0}}=\]permittivity of free space\[=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{2}}\]

\[A=\frac{2\times 0.5\times {{10}^{-2}}}{8.85\times {{10}^{-12}}}\]

\[\Rightarrow A=1130k{{m}^{2}}\]

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of \[\mu F\].

21. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36).

(Image will be uploaded soon)

Show that the capacitance of a spherical capacitor is given by\[C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\] where \[{{r}_{1}}\] and \[{{r}_{2}}\] re the radii of outer and inner spheres, respectively.

Radius of the outer shell is given in the question as\[={{r}_{1}}\]

Radius of the inner shell is given as\[={{r}_{2}}\]

The inner surface of the outer shell's charge can be given as \[+Q\].

\[V=\frac{Q}{4\pi {{\in }_{0}}{{r}_{2}}}-\frac{Q}{4\pi {{\in }_{0}}{{r}_{1}}}\]

The outer surface of the inner shell has induced charge \[-Q\]. Potential difference between the two shells is given by,

\[V=\frac{Q}{4\pi {{\in }_{0}}}\left[ \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right]\]

\[\Rightarrow V=\frac{Q\left( {{r}_{1}}-{{r}_{2}} \right)}{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}\]

Capacitance of the given system is given by,

\[C=\frac{Charge\,\left( Q \right)}{Potential\,Difference\,\left( V \right)}\]

\[\Rightarrow C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]

22. A cylindrical capacitor has two co-axial cylinders of length \[15cm\] and radii \[1.5cm\] and \[1.4cm\]. The outer cylinder is earthed and the inner cylinder is given a charge of \[3.5\mu C\]. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Length of a co-axial cylinder is given as, \[l=15cm=0.15m\]

Radius of the outer cylinder is given as, \[{{r}_{1}}=1.5cm=0.015m\]

Radius of the inner cylinder is given as, \[{{r}_{2}}=1.4cm=0.014m\]

Charge on the inner cylinder, \[q=3.5\mu C=3.5\times {{10}^{-6}}C\]

Capacitance of a co-axial cylinder of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] is given by the relation,

\[C=\frac{2\pi {{\in }_{0}}l}{{{\log }_{2}}\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}\]

\[{{\in }_{0}}=\]permittivity of free space\[=8.85\times {{10}^{-12}}{{N}^{-1}}{{m}^{-2}}{{C}^{2}}\]

\[C=\frac{2\pi \times 8.85\times {{10}^{-12}}\times 0.15}{2.3026{{\log }_{10}}\left( \frac{0.15}{0.14} \right)}\]

\[\Rightarrow C=\frac{2\pi \times 8.85\times {{10}^{-12}}\times 0.15}{2.3026\times 0.0299}=1.2\times {{10}^{-10}}F\]

Potential difference of the inner cylinder is given by,

\[V=\frac{Q}{C}\]

\[\Rightarrow V=\frac{3.5\times {{10}^{-6}}}{1.2\times {{10}^{-10}}}=2.92\times {{10}^{4}}V\]

23. A parallel plate capacitor is to be designed with a voltage rating \[1kV\], using a material of dielectric constant \[3\]and dielectric strength about \[{{10}^{7}}V{{m}^{-1}}\]. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation). For safety, we should like the field never to exceed, say \[10%\] of the dielectric strength. What minimum area of the plates is required to have a capacitance of \[50pF\]?

Potential rating of a parallel plate capacitor, \[V=1kV=1000V\]

Dielectric constant of a material, \[{{\in }_{r}}=3\] Dielectric strength\[={{10}^{7}}V/m\]

For safety, the field intensity never exceeds 10%of the dielectric strength. Hence, electric field intensity, \[E=10%\] of \[{{10}^{7}}={{10}^{6}}V/m\]

Capacitance of the parallel plate capacitor, $C=50pF=50\times10^{-12}F$

Distance between the plates is given by,

\[d=\frac{V}{E}\]

\[\Rightarrow d=\frac{1000}{{{10}^{6}}}\]

\[\Rightarrow d={{10}^{-3}}m\]

Capacitance is given by the relation,

\[C=\frac{{{\in }_{0}}{{\in }_{1}}A}{d}\]

\[A=\]area of each plate

\[A=\frac{CD}{{{\in }_{0}}{{\in }_{1}}}\]

\[\Rightarrow A=\frac{50\times {{10}^{-12}}\times {{10}^{-3}}}{8.85\times {{10}^{-12}}\times 3}=19c{{m}^{2}}\]

Hence, the area of each plate is about \[19c{{m}^{2}}\].

Long Answer Questions (5 Mark Questions)

(a) Define dielectric constant in terms of the capacitance of a capacitor. On what factor does the capacitance of a parallel plate capacitor with dielectric depend?

Ans: Dielectric constant can be expressed as the ratio of capacitance of a capacitor when the dielectric is filled in between the plates to the capacitance of a capacitor when there is vacuum in between the plates.

In \[K=\frac{{{C}_{m}}}{{{C}_{o}}}=\frac{\text{Capacitance of a capacitor when dielectric is in between the plates}}{\text{Capacitance of a capacitor with vacuum in between the plates}}\]

Capacitance of a parallel plate capacitor with dielectric depends on the following factors

\[{{C}_{m}}=\frac{KA{{\in }_{0}}}{d}\]

Area of the plates

Distance between the plates

Dielectric constant of the dielectric between the plates.

(b) Find the ratio of the potential differences that must be applied across the 

(i) Parallel

(ii) Series combination of two identical capacitors so that the energy stored in the two cases becomes the same.

Ans: Let the capacitance of each capacitor be \[C\]

\[{{C}_{p}}=\frac{C\times C}{C+C}=\frac{C}{2}\] …… (in series)

Let \[{{V}_{p}}\] and \[{{V}_{s}}\] be the values of potential difference

Thus \[{{U}_{p}}=\frac{1}{2}\times {{C}_{p}}\times {{V}_{p}}^{2}=\frac{1}{2}\times 2C\times {{V}_{p}}^{2}=C{{V}_{p}}^{2}\]

\[{{U}_{s}}=\frac{1}{2}\times {{C}_{s}}\times {{V}_{s}}^{2}=\frac{1}{2}\times \frac{C}{2}\times {{V}_{s}}^{2}=\frac{C{{V}_{s}}^{2}}{4}\]

But \[{{U}_{p}}={{U}_{s}}\] …… (given)

\[C{{V}_{p}}^{2}=\frac{C{{V}_{s}}^{2}}{4}\]

\[\Rightarrow \frac{{{V}_{p}}^{2}}{{{V}_{s}}^{2}}=\frac{1}{4}\]

\[\Rightarrow {{V}_{p}}:{{V}_{s}}=1:2\]

(a) An air-filled capacitor is given a charge of \[2\mu C\] raising its potential to \[200V\]. If on inserting a dielectric medium, its potential falls to \[50V\], what is the dielectric constant of the medium?

Ans: \[K=\frac{V}{{{V}'}}=\frac{200}{50}=4\] (where \[V=200V\] for air filled capacitor \[{V}'=50\] after insertion of a dielectric)

(b) A conducting slab of thickness \[t\] is introduced without touching between plates of a parallel plate capacitor separated by a distance d , \[t<d\] . Derive an expression for the capacitance of a capacitor?

Ans: For a parallel plate capacitor when air/vacuum is in between the plates \[{{C}_{0}}=\frac{A{{\in }_{0}}}{d}\]

Since the electric field inside a conducting stab is zero, hence the electric field exists only between the space \[\left( d-t \right)\Rightarrow V={{E}_{0}}\left( d-t \right)\]

Where \[{{E}_{0}}\] is the electric field between the plates.

And \[{{E}_{0}}=\frac{\sigma }{{{\in }_{0}}}=\frac{q}{A{{\in }_{0}}}\]

Where \[A\] is the area of each plate

\[\Rightarrow V=\frac{q}{A{{\in }_{0}}}\left( d-t \right)\]

Hence capacitor of a parallel plate capacitor

\[C=\frac{q}{V}=\frac{q}{q\left( d-t \right)}A{{\in }_{0}}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d-t}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d\left( 1-\frac{t}{d} \right)}\]

\[\Rightarrow C=\frac{{{C}_{0}}}{\left( 1-\frac{t}{d} \right)}\]

3. Figure (a) and (b) shows the field lines of a single positive and negative changes respectively

(a) Give the signs of the potential: \[{{V}_{P}}-{{V}_{Q}}\] and \[{{V}_{B}}-{{V}_{A}}\]

Ans: We know the relation as \[V\propto \frac{1}{r}\]

\[{{V}_{P}}>{{V}_{Q}}\Rightarrow {{V}_{P}}-{{V}_{Q}}\,=\,Positive\]

\[{{V}_{A}}<{{V}_{B}}\Rightarrow {{V}_{B}}-{{V}_{A}}\,=\,Positive\]

Because \[{{V}_{B}}\] is less negative than \[{{V}_{A}}\].

(b) Give the sign of the work done by the field in moving a small positive change from \[Q\] to \[P\].

Ans: In moving a positive change form \[Q\] to \[P\] work has to be done against the electric field so it is negative, and the sign is, hence, negative.

(c) Give the sign of the work done by the field in moving a small negative change from \[B\] to \[A\].

Ans:  In moving a negative change form \[B\] to \[A\] work is done along the same direction of the field so it is positive, and the sign is, hence, positive.

4. With the help of a labelled diagram, explain the principle, construction and working of a Van de Graff generator. Mention its applications.

Ans: Van de Graff generator can be expressed as a device which is capable of producing a high potential of the order of million volts.

Principle: 

The charge always resides on the outer surface of the hollow conductor.

The electric discharge in air or gas takes place readily at the pointed ends of the conductors.

Construction of Van de Graff Generator: 

It consists of a large hollow metallic sphere \[S\] mounted on two insulating columns A and B and an endless belt made up of rubber which is running over two pulleys \[{{P}_{1}}\] and \[{{P}_{2}}\] with the help of an electric motor. \[{{B}_{1}}\] and \[{{B}_{2}}\] are two sharp metallic brushes. The lower brush \[{{B}_{1}}\] is given a positive potential by high tension battery and is called a spray brush while the upper brush \[{{B}_{2}}\] connected to the inner part of the sphere \[S\].

Working: 

When brush \[{{B}_{1}}\] is given a higher positive potential then it produces ions, due to the action of sharp points. Thus, the positive ions so produced get sprayed on the belt due to repulsion between positive ions and the positive charge on brush \[{{B}_{1}}\]. Then it is carried upward by the moving belt. The pointed end of \[{{B}_{2}}\] just touches the belt collects the positive change and make it move to the outer surface of the sphere S. This process continues and the potential of the shell rises to several million volts.

Applications: 

Particles like protons, deuterons, \[\alpha \]-particles etc are accelerated to high speeds and energies.

5. Two charges \[5\times {{10}^{-8}}C\] and \[-3\times {{10}^{-8}}C\] are located \[16cm\] apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Ans: There are two charges which are given as,

\[{{q}_{1}}=5\times {{10}^{-8}}C\]

\[{{q}_{2}}=-3\times {{10}^{-8}}C\]

Distance between the two charges, \[d=16cm=0.16m\]

Consider a point \[P\] on the line joining the two charges, as shown in the given figure.

\[r=\] distance of point \[P\] from charge \[{{q}_{1}}\]

Let the electric potential \[V\] at point \[P\] be zero.

Potential at point \[P\] is the sum of potentials caused by charges \[{{q}_{1}}\] and \[{{q}_{2}}\] respectively.

\[V=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}r}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( d-r \right)}\] …… (1)

For \[V=0\], equation (1) reduces to

\[\frac{{{q}_{1}}}{4\pi {{\in }_{0}}r}=-\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( d-r \right)}\]

\[\Rightarrow \frac{{{q}_{1}}}{r}=-\frac{{{q}_{2}}}{\left( d-r \right)}\]

\[\Rightarrow \frac{5\times {{10}^{-8}}}{r}=\frac{\left( -3\times {{10}^{-8}} \right)}{\left( 0.16-r \right)}\]

\[\Rightarrow \frac{0.16}{r}=\frac{8}{5}\]

\[\Rightarrow r=0.1m=10cm\]

Therefore, the potential is zero at a distance of \[10cm\] from the positive charge between the charges.

Suppose point \[P\] is outside the system of two charges where potential is zero, as shown in the given figure.

For this arrangement, potential is given

 \[V=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}s}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( s-d \right)}\]…… (2)

For \[V=0\], equation (2) reduces to

\[\frac{{{q}_{1}}}{4\pi {{\in }_{0}}s}=-\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( s-d \right)}\]

\[\Rightarrow \frac{{{q}_{1}}}{s}=-\frac{{{q}_{2}}}{\left( s-d \right)}\]

\[\Rightarrow \frac{5\times {{10}^{-8}}}{s}=\frac{\left( -3\times {{10}^{-8}} \right)}{\left( s-0.16 \right)}\]

\[\Rightarrow 1-\frac{0.16}{s}=\frac{3}{5}\]

\[\Rightarrow \frac{0.16}{s}=\frac{2}{5}\]

\[\Rightarrow s=0.4m=40cm\]

Therefore, the potential is zero at a distance of \[40cm\] from the positive charge outside the system of charges.

6. A parallel plate capacitor with air between the plates has a capacitance of \[8pF\], \[\left( 1pF={{10}^{-12}}F \right)\]. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: Capacitance between the parallel plates of the capacitor, \[C=8pF\]

Initially, the distance between the parallel plates was \[d\] and it was filled with air. Dielectric constant of air, \[k=1\]

Capacitance, \[C\] is given by the formula,

\[C=\frac{kA{{\in }_{0}}}{d}\]

\[\Rightarrow C=\frac{A{{\in }_{0}}}{d}\] ……(1)

\[A=\] Area of each plate

\[{{\in }_{0}}=\] Permittivity of free space

If distance between the plates is reduced to half, then the new distance can be given as, \[{{d}^{-TM}}=\frac{d}{2}\]. Dielectric constant of the substance filled in between the plates, \[{k}'=6\]. 

Therefore, capacitance of the capacitor becomes

\[{C}'=\frac{{k}'A{{\in }_{0}}}{d}=\frac{6A{{\in }_{0}}}{\frac{d}{2}}\] …… (2)

Taking the ratios of equation \[1\] and \[2\], we obtain

\[{C}'=2\times 6C\]

\[\Rightarrow {C}'=12C\]

\[\Rightarrow {C}'=12\times 8=96pF\]

Therefore, the capacitance between the plates is \[96pF\].

7. A charge of \[8mC\] is located at the origin. Calculate the work done in taking a small charge of \[-2\times {{10}^{-9}}C\] from a point \[P\left( 0,0,3 \right)cm\] to a point \[Q\left( 0,4,0 \right)cm\], via a point \[R\left( 0,6,9 \right)cm\] .

Ans: Charge located at the origin is given as, \[q=8mC=8\times {{10}^{3}}C\].

Magnitude of a small charge, which is taken from a point \[P\] to point \[R\] to point \[Q\].

\[{{q}_{1}}=-2\times {{10}^{-9}}C\]

All the points are represented in the figure below.

Point \[P\] is at a distance, \[{{d}_{1}}=3cm\] from the origin along z-axis. Point \[Q\] is at a distance, \[{{d}_{2}}=4cm\], from the origin along y-axis.

Potential at point \[P\]can be given as, \[{{V}_{1}}=\frac{q}{4\pi {{\in }_{0}}{{d}_{1}}}\]

Potential at point \[Q\], \[{{V}_{2}}=\frac{q}{4\pi {{\in }_{0}}{{d}_{2}}}\]

Work done \[W\] by the electrostatic force is independent of the path.

\[W={{q}_{1}}\left( {{V}_{2}}-{{V}_{1}} \right)\]

\[\Rightarrow W={{q}_{1}}\left[ \frac{q}{4\pi {{\in }_{0}}{{d}_{2}}}-\frac{q}{4\pi {{\in }_{0}}{{d}_{1}}} \right]\]

\[\Rightarrow W=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}}\left[ \frac{1}{{{d}_{2}}}-\frac{1}{{{d}_{1}}} \right]\] …… (1)

Where, 

\[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[\Rightarrow W=9\times {{10}^{9}}\times 8\times {{10}^{-3}}\times \left( -2\times {{10}^{-9}} \right)\left[ \frac{1}{0.04}-\frac{1}{0.03} \right]\]

\[\Rightarrow W=-144\times {{10}^{-3}}\times \left( \frac{-25}{3} \right)\]

\[\Rightarrow W=1.27J\]

Therefore, work done during the process is \[1.27J\].

8. A cube of side \[b\]has a charge \[q\] at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Ans: Length of the side of a cube is given as \[=b\]

Charge at each of its vertices\[=q\]

A cube of side \[b\] is shown in the following figure.

\[d=\]Diagonal of one of the six faces of the cube

\[\Rightarrow {{d}^{2}}=\sqrt{{{b}^{2}}+{{b}^{2}}}=\sqrt{2{{b}^{2}}}\]

\[\Rightarrow d=b\sqrt{2}\]

\[l=\] length of the diagonal of the cube

\[{{l}^{2}}=\sqrt{{{d}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{{{\left( \sqrt{2}b \right)}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{2{{b}^{2}}+{{b}^{2}}}\]

\[\Rightarrow {{l}^{2}}=\sqrt{3{{b}^{2}}}\]

\[\Rightarrow l=b\sqrt{3}\]

\[\Rightarrow r=\frac{l}{2}=\frac{b\sqrt{2}}{2}\] is the difference between the centre of the cube and one of the eight vertices.

The electric potential \[V\] at the centre of the cube is due to the presence of eight charges at the vertices.

\[V=\frac{8q}{4\pi {{\in }_{0}}}\]

\[\Rightarrow V=\frac{8q}{4\pi {{\in }_{0}}\left( \frac{b\sqrt{3}}{2} \right)}\]

\[\Rightarrow V=\frac{4q}{\sqrt{3}\pi {{\in }_{0}}b}\]

Therefore, the potential at the centre of the cube is \[\frac{4q}{\sqrt{3}\pi {{\in }_{0}}b}\].

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the cube. 

Therefore, the electric field is zero at the centre.

9. Two tiny spheres carrying charges \[1.5\mu C\] and \[2.5\mu C\] are located \[30cm\] apart. Find the potential and electric field:

(a) At the mid-point of the line joining the two charges, and

Ans: The charges are depicted as:

Two charges placed at points \[A\] and \[B\] are depicted in the provided figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at \[A\], \[{{q}_{1}}=1.5\mu C\]

Magnitude of charge located at \[B\], \[{{q}_{2}}=2.5\mu C\]

Distance between the two charges, \[d=30cm=0.3m\]

Let \[{{V}_{1}}\]and  \[{{E}_{1}}\] are electric potential and electric field respectively at O.

\[{{V}_{1}}=\] Potential due to charge at \[A+\]Potential due to charge at \[B\]

\[{{V}_{1}}=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}+\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}=\frac{1}{4\pi {{\in }_{0}}\left( \frac{d}{2} \right)}\left( {{q}_{1}}+{{q}_{2}} \right)\]

Where, \[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{C}^{2}}{{m}^{-2}}\]

\[{{V}_{1}}=\frac{9\times {{10}^{9}}\times {{10}^{-5}}}{\left( \frac{0.30}{2} \right)}\left( 2.5+1.5 \right)\]

\[{{V}_{1}}=2.4\times {{10}^{5}}V\]

\[{{E}_{1}}=\] Electric field due to \[{{q}_{2}}\] \[-\]Electric field due to \[{{q}_{1}}\]

\[\Rightarrow {{E}_{1}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}{{\left( \frac{d}{2} \right)}^{2}}}-\frac{{{q}_{1}}}{4\pi {{\in }_{0}}{{\left( \frac{d}{2} \right)}^{2}}}\]

\[\Rightarrow {{E}_{1}}=\frac{9\times {{10}^{9}}}{{{\left( \frac{0.30}{2} \right)}^{2}}}\times {{10}^{6}}\times \left( 2.5-1.5 \right)\]

\[\Rightarrow {{E}_{1}}=4\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at mid-point is \[2.4\times {{10}^{5}}V\] and the electric field at mid-point is \[4\times {{10}^{5}}V{{m}^{-1}}\]. The field is directed from the larger charge to the smaller charge

(b) At a point \[10cm\] from this midpoint in a plane normal to the line passing through the mid-point.

Ans: Consider a point \[Z\]such that the normal distance \[OZ=10cm=0.1m\] as shown in the following figure.

\[{{V}_{2}}\] and \[{{E}_{2}}\] are the electric potential and electric field respectively at \[Z\]. It can be observed from the figure that distance,

\[BZ=AZ=\sqrt{{{\left( 0.1 \right)}^{2}}+{{\left( 0.15 \right)}^{2}}}=0.18m\]

\[{{V}_{2}}=\] Electric potential due to \[A+\]Electric potential due to point \[B\]

\[\Rightarrow {{V}_{2}}=\frac{{{q}_{2}}}{4\pi {{\in }_{0}}\left( AZ \right)}+\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( BZ \right)}\]

\[\Rightarrow {{V}_{2}}=\frac{9\times {{10}^{9}}\times {{10}^{-5}}}{0.18}\left( 1.5+2.5 \right)\]

\[\Rightarrow {{V}_{2}}=2\times {{10}^{5}}V\]

Electric field due to \[{{q}_{1}}\] at \[Z\],

\[{{E}_{A}}=\frac{{{q}_{1}}}{4\pi {{\in }_{0}}\left( AZ \right)}\]

\[\Rightarrow {{E}_{A}}=\frac{9\times {{10}^{9}}\times 1.5\times {{10}^{-5}}}{0.18}\]

\[\Rightarrow {{E}_{A}}=0.416\times {{10}^{6}}V/m\]

Electric field due to \[{{q}_{2}}\] at \[Z\],

\[\Rightarrow {{E}_{B}}=\frac{9\times {{10}^{9}}\times 2.5\times {{10}^{-5}}}{{{\left( 0.18 \right)}^{2}}}\]

\[\Rightarrow {{E}_{B}}=0.69\times {{10}^{6}}V/m\]

The resultant field intensity at \[Z\]

\[E=\sqrt{{{E}^{2}}_{A}+{{E}^{2}}_{B}+2{{E}_{A}}{{E}_{B}}\cos 2\theta }\]

Where, \[2\theta \] is the angle, \[\angle AZB\]

From the given figure, we obtain

\[\Rightarrow \theta ={{\cos }^{-1}}0.5556={{56.25}^{\circ }}\]

\[E=\sqrt{{{\left( 0.416\times {{10}^{6}} \right)}^{2}}\times 2\times 0.416\times 0.69\times {{10}^{12}}\times \left( -0.38 \right)}\]

\[\Rightarrow E=6.6\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at a point \[10cm\] (perpendicular to the mid-point) is $ 2.0 \times {{10}^{5}} V$ and the electric field is \[6.6\times {{10}^{5}}V{{m}^{-1}}\].

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by \[\left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)\cdot \hat{n}=\frac{\sigma }{{{\in }_{0}}}\]. Where \[\hat{n}\] is a unit vector normal to the surface at a point and \[\sigma \] is the surface charge density at that point. (The direction of \[\hat{n}\] is from side \[1\]to side \[2\]). Hence show that just outside a conductor, the electric field is \[\frac{{\hat{n}}}{{{\in }_{0}}}\].

Ans: Electric field on one side of a charged body is \[{{E}_{1}}\] and electric field on the other side of the same body is \[{{E}_{2}}\]. If an infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

\[\overline{{{E}_{1}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

\[\hat{n}=\] Unit vector normal to the surface at a point 

\[\sigma =\]Surface charge density at that point 

Electric field due to the other surface of the charge body,

\[\overline{{{E}_{2}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

Electric field at any point due to the two surfaces,

$\left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)=\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{2{{\in }_{0}}}\hat{n}+\frac{\sigma }{{{\in }_{0}}}\hat{n} $

$\Rightarrow \left( \overline{{{E}_{2}}}-\overline{{{E}_{1}}} \right)\cdot \hat{n}=\frac{\sigma }{{{\in }_{0}}} $

Since inside a closed conductor, \[\overline{{{E}_{1}}}=0\],

\[\overline{E}=\overline{{{E}_{2}}}=-\frac{\sigma }{2{{\in }_{0}}}\hat{n}\]

Therefore, the electric field just outside the conductor is \[\frac{\sigma }{{{\in }_{0}}}\hat{n}\].

(b) Show that the tangential component of an electrostatic field is continuous from one side of a charged surface to another. 

Ans: When a charge particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of an electrostatic field is continuous from one side of a charged surface to the other.

11. A long charged cylinder of linear charge density \[\lambda \] is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans: Charge density of the long, charged cylinder of length \[L\] and radius \[r\] is \[\lambda \].

Another cylinder of the same length surrounds the previous cylinder. The radius of this cylinder is \[R\].

Let \[E\] be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gauss's theorem as, \[\phi =E\left( 2\pi d \right)L\]

\[d=\]Distance of a point from the common axis of the cylinders. Let \[q\]be the total charge on the cylinder.

It can be written as

\[\Rightarrow \phi =E\left( 2\pi d \right)L=\frac{q}{{{\in }_{0}}}\]

\[q=\] Charge on the inner sphere of the outer cylinder

\[\Rightarrow E\left( 2\pi d \right)L=\frac{\lambda L}{{{\in }_{0}}}\]

\[\Rightarrow E=\frac{\lambda }{2\pi {{\in }_{0}}d}\]

Therefore, the electric field in the space between the two cylinders is \[\frac{\lambda }{2\pi {{\in }_{0}}d}\].

12. In a hydrogen atom, the electron and proton are bound at a distance of about \[0.53{{A}^{\circ }}\].

(a) Estimate the potential energy of the system in \[eV\], taking the zero of the potential energy at infinite separation of the electron from the proton.

Ans: The distance between the electron-proton of a hydrogen atom is given as, \[d=0.53\overset{{}^\circ }{\mathop{A}}\,\].

Charge on an electron, \[{{q}_{1}}=1.6\times {{10}^{-19}}C\]

Charge on a proton, \[{{q}_{2}}=+1.6\times {{10}^{-19}}C\]

Potential at infinity is zero.

\[=0-\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}d}\]

Potential energy of the system, \[p-e=\] Potential energy at infinity – Potential energy t distance, \[d\]

\[{{\in }_{0}}\] is the permittivity of free space

We can further express,

\[\Rightarrow \frac{1}{4\pi {{\in }_{0}}}=0-\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{0.53\times {{10}^{10}}}\]

\[Potential\,Energy=\frac{-43.7\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\]

\[\Rightarrow Potential\,Energy=-27.2eV\]

Therefore, the potential energy of the system is \[-27.2eV\].

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)

Ans: Kinetic energy can be given as half times the magnitude of potential energy.

Kinetic Energy\[=\frac{1}{2}\times \left( -27.2eV \right)=13.6eV\]

Total energy \[=13.6eV\]

Therefore, the minimum work required to free the electron is \[13.6eV\].

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at \[1.06\overset{{}^\circ }{\mathop{A}}\,\] separation?

Ans: When zero of potential energy is taken, \[{{d}_{1}}=1.06\overset{{}^\circ }{\mathop{A}}\,\]

$\therefore$ Potential energy of the system = Potential energy at \[{{d}_{1}}\]\[-\]Potential energy at \[d\].

\[Potential\,Energy\,of\,the\,\text{ }system=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\in }_{0}}{{d}_{1}}}-27.2eV\]

\[\Rightarrow Potential\,Energy\,of\,the\,\text{ }system=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{1.06\times {{10}^{-10}}}-27.2eV\]

\[\Rightarrow 21.73\times {{10}^{-10}}J-27.2eV\]

\[\Rightarrow 13.58eV-27.2eV\]

\[\Rightarrow 13.6eV\]

13. Two charges \[-q\] and \[+q\] are located at points \[\left( 0,0,-a \right)\] and \[\left( 0,0,a \right)\] , respectively.

(a) What is the electrostatic potential at these points?

Ans:  In this situation, charge \[-q\] is located at \[\left( 0,0,-a \right)\] and charge \[+q\] is located at \[\left( 0,0,a \right)\]. Therefore, they form a dipole. Point \[\left( 0,0,z \right)\] is on the axis of this dipole and point \[\left( x,y,0 \right)\] is normal to the axis of the dipole. Hence, the electrostatic potential at point \[\left( x,y,0 \right)\] is zero. Electrostatic potential at point \[\left( 0,0,z \right)\] is given by,

\[V=\frac{1}{4\pi {{\in }_{0}}}\left( \frac{q}{z-a} \right)+\frac{1}{4\pi {{\in }_{0}}}\]

\[\Rightarrow V=\frac{q\left( z+a-z+a \right)}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[\Rightarrow V=\frac{2qa}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[\Rightarrow V=\frac{p}{4\pi {{\in }_{0}}\left( {{z}^{2}}-{{a}^{2}} \right)}\]

\[p=\]Dipole moment of the system of two charges \[=2qa\]

(b) Obtain the dependence of potential on the distance \[r\] of a point from the origin when \[\frac{r}{a}>>1\].

Ans: Distance \[r\] is much greater than half of the distance between the two charges. Therefore, the potential \[\left( V \right)\] at a distance \[r\] is inversely proportional to the square of the distance.

i.e., \[4\propto \frac{1}{{{r}^{2}}}\]

(c) How much work is done in moving a small test charge from the point \[1.06{{A}^{\circ }}\] to \[\left( -7,0,0 \right)\] along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Ans:   Zero

The answer does not change if the path of the test charge is not along the x-axis.

A test charge is moved from point \[\left( 5,0,0 \right)\] to point \[\left( -7,0,0 \right)\] along the x-axis. Electrostatic potential \[{{V}_{1}}\] at point \[\left( 5,0,0 \right)\] is given by,

\[{{V}_{1}}=\frac{-q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 5-0 \right)}^{2}}-{{\left( -a \right)}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 5-0 \right)}^{2}}-{{\left( -a \right)}^{2}}}}\]

\[\Rightarrow {{V}_{1}}=\frac{-q}{4\pi {{\in }_{0}}\sqrt{25+{{a}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}\sqrt{25+{{a}^{2}}}}\]

\[\Rightarrow {{V}_{1}}=0\]

Electrostatic potential, \[{{V}_{2}}\] at point \[\left( -7,0,0 \right)\] is given by,

\[{{V}_{2}}=\frac{-q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( 7 \right)}^{2}}-{{\left( -a \right)}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}}\frac{1}{\sqrt{{{\left( -7 \right)}^{2}}-{{\left( -a \right)}^{2}}}}\]

\[\Rightarrow {{V}_{2}}=\frac{-q}{4\pi {{\in }_{0}}\sqrt{49+{{a}^{2}}}}+\frac{q}{4\pi {{\in }_{0}}\sqrt{49+{{a}^{2}}}}\]

\[\Rightarrow {{V}_{2}}=0\]

Therefore, no work is done in moving a small test charge from point \[\left( 5,0,0 \right)\] to point \[\left( -7,0,0 \right)\] along the x-axis.

The answer will not change because work done by the electrostatic field in moving a test charge between the two points is not dependent of the path connecting the two points.

14. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on \[r\] for \[\frac{r}{a}>>1\] and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge)

Ans: Four charges of the same magnitude are placed at points \[X\], \[Y\] and \[Z\] respectively, as depicted in the following figure.

A point is located at \[P\], which is \[r\] distance away from point \[Y\]. The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges

Charge \[+q\] placed at point \[X\]

Charge \[-2q\] placed at point \[Y\]

Charge \[+q\] placed at point \[Z\]

\[XY=YZ=a\]

Electrostatic potential that is caused by the system of three charges at point \[P\] can be given by,

\[V=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{q}{XP}-\frac{2q}{YP}+\frac{q}{ZP} \right]\]

\[\Rightarrow V=\frac{1}{4\pi {{\in }_{0}}}\left[ \frac{q}{r+a}-\frac{2q}{r}+\frac{q}{r-a} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{r\left( r-a \right)-2\left( r+a \right)\left( r-a \right)+r\left( r+a \right)}{r\left( r+a \right)\left( r-a \right)} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{{{r}^{2}}-ra-2{{r}^{2}}+2{{a}^{2}}+{{r}^{2}}+ra}{r\left( {{r}^{2}}-{{a}^{2}} \right)} \right]\]

\[\Rightarrow V=\frac{q}{4\pi {{\in }_{0}}}\left[ \frac{2{{a}^{2}}}{r\left( {{r}^{2}}-{{a}^{2}} \right)} \right]\]

\[\Rightarrow V=\frac{2q{{a}^{2}}}{4\pi {{\in }_{0}}{{r}^{3}}\left( 1-\frac{{{a}^{2}}}{{{r}^{2}}} \right)}\]

Since \[\frac{r}{a}>>1\]

\[\frac{{{r}^{2}}}{{{a}^{2}}}\] is taken as negligible.

\[V=\frac{2q{{a}^{2}}}{4\pi {{\in }_{0}}{{r}^{3}}}\]

It can be inferred that potential, \[V\propto \frac{1}{{{r}^{3}}}\]

But it is known that for a dipole, \[V\propto \frac{1}{{{r}^{2}}}\]

And, for a monopole, \[V\propto \frac{1}{r}\]

15. An electrical technician requires a potential difference of $2\mu F$ in a circuit across a potential difference of  \[1kV\]. A large number of  $2\mu F$ capacitors are available to him, each of which can withstand a potential difference of not more than $400V$. Suggest a possible arrangement that requires the minimum number of capacitors.

Ans: Total required capacitance is given as, $C=2\mu F$ 

Potential difference is given as, $V=1kV=1000V$

Capacitance of each capacitor, ${{C}_{1}}=1\mu F$

Each capacitor can hold a potential difference,

 ${{V}_{1}}=400V$

$\frac{1000}{400}=2.5$

Let us suppose a number of capacitors are connected in series and these series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be $1000V$ and the potential difference across each capacitor must be $400V$. Hence, the number of capacitors in each row is given as

Therefore, there are three capacitors in each row. 

Capacitance of each row$=\frac{1}{1+1+1}=\frac{1}{3}\mu F$

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....\text{n terms = }\frac{n}{3}$

However, capacitance of the circuit is given as $2\mu F$.

$\therefore \frac{n}{3}=2$

Let there be $n$ rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as

Hence, $6$ rows of three capacitors are present in the circuit. A minimum of $6\times 3$ i.e.,$18$ capacitors are required for the given arrangement.

16. Obtain the equivalent capacitance of the network in Fig. 2.35. For a $300V$ supply, determine the charge and voltage across each capacitor.

Ans: Capacitance of capacitor ${{C}_{1}}$ is given as $100pF$.

Capacitance of capacitor ${{C}_{2}}$ is given as $200pF$.

Capacitance of capacitor ${{C}_{3}}$ is given as $200pF$.

Capacitance of capacitor ${{C}_{4}}$ is given as $200pF$.

Supply potential, $V=300V$

Capacitors ${{C}_{2}}$ and ${{C}_{3}}$ are connected in series. Let their equivalent capacitance be C 1

$\therefore \frac{1}{{{C}^{1}}}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\Rightarrow {{C}^{1}}=100pF$

Capacitors ${{C}^{1}}$ and $C'$ are parallel. Let their equivalent capacitance be ${{C}^{n}}$

$\therefore {{C}^{n}}={{C}^{1}}+{{C}_{1}}=100+100=200pF$

${{C}^{n}}$ and ${{C}_{4}}$ are connected in series. Let their equivalent capacitance be $C$.

$\therefore \frac{1}{C}=\frac{1}{{{C}^{n}}}+\frac{1}{{{C}_{4}}}=\frac{1}{200}+\frac{1}{100}=\frac{2+1}{200}\Rightarrow C=\frac{200}{3}pF$

Hence, the equivalent capacitance of the circuit is $\frac{200}{3}pF$.

 Potential difference across

${{C}^{n}}={V}''$

Potential difference across ${{C}_{4}}={{V}_{4}}$

$\therefore {{V}^{5}}+{{V}_{4}}=V=300V$

Charge on ${{C}_{1}}$ is given by,

${{Q}_{4}}=CV=\frac{200}{3}\times {{10}^{-12}}\times 300=2\times {{10}^{-8}}C$

$\therefore {{V}_{4}}=\frac{{{Q}_{4}}}{{{C}_{4}}}=\frac{2\times {{10}^{-8}}}{100\times {{10}^{-12}}}=200V$

$\therefore$ Voltage across ${{C}_{1}}$ is given by,

${{V}_{1}}=V-{{V}_{4}}=300-200=100V$

Hence, the potential difference, ${{V}_{1}}$ across ${{C}_{1}}$ is $100V$. Charge on ${{C}_{1}}$ is given by, 

${{Q}_{1}}={{C}_{1}}{{V}_{1}}=100\times {{10}^{-12}}\times 100={{10}^{-8}}$

Now,  ${{C}_{2}}$ and ${{C}_{3}}$ having the same capacitance have a potential difference of $100V$together. Since ${{C}_{2}}$ and ${{C}_{3}}$ are in series, the potential difference across ${{C}_{2}}$ and ${{C}_{3}}$ is given by,

${{V}_{2}}={{V}_{3}}=50V$

Therefore, charge on ${{C}_{2}}$ is given by,

${{Q}_{2}}={{C}_{2}}{{V}_{2}}=200\times {{10}^{-12}}\times 50={{10}^{-8}}C$

And charge on ${{C}_{3}}$ is given by,

${{Q}_{3}}={{C}_{3}}{{V}_{3}}=200\times {{10}^{-12}}\times 50={{10}^{-8}}C$

Therefore, the equivalent capacitance of the given circuit is  $\frac{200}{3}\text{pF with}$

${{Q}_{1}}={{10}^{-8}}C{{V}_{1}}=100V$

${{Q}_{2}}={{10}^{-8}}C{{V}_{2}}=50V$

${{Q}_{3}}={{10}^{-8}}C{{V}_{3}}=50V$

${{Q}_{4}}=2\times {{10}^{-8}}C{{V}_{4}}=200V$

17. The plates of a parallel plate capacitor have an area of $90c{{m}^{2}}$ each and are separated by $2.5mm$. The capacitor is charged by connecting it to a $400V$ supply.

(a) How much electrostatic energy is stored by the capacitor?

Ans: Area of the plates of a parallel plate capacitor is given as, $A=90c{{m}^{2}}=90\times {{10}^{-4}}{{m}^{2}}$

Distance between the plates, $d=2.5mm=2.5\times {{10}^{-3}}m$

Potential difference across the plates, $V=400V$

Capacitance of the capacitor is given by the relation,

$C=\frac{A{{\varepsilon }_{0}}}{d}$

Electrostatic energy stored in the capacitor is given by the relation, ${{E}_{1}}=\frac{1}{2}C{{V}^{2}}$

$=\frac{1}{2}\frac{{{\varepsilon }_{0}}A}{d}{{V}^{2}}$

${{\varepsilon }_{0}}=$ Permittivity of free space$=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\therefore {{E}_{1}}=\frac{1\times 8.85\times {{10}^{-12}}\times 90\times {{10}^{-4}}\times {{\left( 400 \right)}^{2}}}{2\times 2.5\times {{10}^{-3}}}=2.55\times {{10}^{-6}}J$

Hence, the electrostatic energy stored by the capacitor is $2.55\times {{10}^{-6}}J$.

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

Ans: Volume of the given capacitor,

${{V}^{1}}=A\times d=90\times {{10}^{-4}}\times 25\times {{10}^{-3}}=2.25\times {{10}^{-4}}{{m}^{3}}$

Energy stored in the capacitor per unit volume is given by,

$u=\frac{{{E}_{1}}}{{{V}_{1}}}=\frac{2.25\times {{10}^{-6}}}{2.25\times {{10}^{-4}}}=0.113J{{m}^{-3}}$

Again, $u=\frac{{{E}_{1}}}{{{V}_{1}}}$

$\frac{V}{d}=$Electric intensity$=E$

$\therefore u=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$, which is the required energy stored.

18. A $4\mu F$ capacitor is charged by a $200V$ supply. It is then disconnected from the supply, and is connected to another uncharged $2\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Capacitance of a charged capacitor,  ${{C}_{1}}=4\mu F=4\times {{10}^{-5}}F$

Supply voltage, ${{V}_{1}}=200V$

Electrostatic energy stored in ${{C}_{1}}$ is given by,

${{E}_{1}}=\frac{1}{2}{{C}_{1}}{{V}_{1}}^{2}=\frac{1}{2}\left( 4\times {{10}^{-5}} \right)\times {{\left( 200 \right)}^{2}}=8\times {{10}^{-2}}J$

Capacitance of an uncharged capacitor, ${{C}_{2}}=2\mu F=2\times {{10}^{-6}}F$

When ${{C}_{2}}$ is connected to the circuit, the potential acquired by it is ${{V}_{2}}$,

$\therefore {{V}_{2}}\left( {{C}_{1}}+{{C}_{2}} \right)={{C}_{1}}{{V}_{1}}$

$\Rightarrow {{V}_{2}}\left( 4+2 \right)\times {{10}^{-5}}=4\times {{10}^{-6}}\times 200$

$\Rightarrow {{V}_{2}}=\frac{400}{3}V$

According to the conservation of charge, initial charge on capacitor ${{C}_{1}}$ is equal to the final charge on capacitors ${{C}_{1}}$ and ${{C}_{2}}$.

Electrostatic energy for the combination of two capacitors is given by,

\[{{E}_{2}}=\frac{1}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{V}_{2}}^{1}=\frac{1}{2}\left( 2+4 \right)\times {{10}^{-5}}\times {{\left( \frac{400}{3} \right)}^{2}}=5.33\times {{10}^{-2}}J\]

Hence, the amount of electrostatic energy lost by the capacitor ${{C}_{1}}$ is \[5.33\times {{10}^{-2}}J\]. 

19. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2}QE$, where $Q$ is the charge on the capacitor, and $E$ is the magnitude of the electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

Ans: Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of$\Delta x$. Hence, work done by the force to do so $=F\Delta x$

As a result, the potential energy of the capacitor increases by an amount given as$uA\Delta x$.

$u=$Energy density

$A=$Area of each plate

$d=$ Distance between the plate

$V=$Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

$F\Delta x=uA\Delta x$

$\Rightarrow F=uA=\left( \frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}} \right)A$

Electric intensity is given by,

$E=\frac{V}{d}$

$\Rightarrow F=\frac{1}{2}{{\varepsilon }_{0}}\left( \frac{V}{d} \right)EA=\frac{1}{2}\left( {{\varepsilon }_{0}}A\frac{V}{d} \right)E$

However, capacitance can be given as, $C=\frac{{{\varepsilon }_{0}}A}{d}$

$\therefore F=\frac{1}{2}\left( CV \right)E$

Charge on the capacitor is given by,

$Q=CV\Rightarrow F=\frac{1}{2}QE$

The physical origin of the factor, $\frac{1}{2}$, in the force formula lies in the fact that just outside the conductor, the field is $E$ and inside it is zero. Hence, it is the average value, $\frac{E}{2}$, of the field that contributes to the force.

20. A spherical capacitor has an inner sphere of radius \[12cm\] and an outer sphere of radius \[13cm\]. The outer sphere is earthed and the inner sphere is given a charge of \[2.5\mu C\]. The space between the concentric spheres is filled with a liquid of dielectric constant \[32\]

(a) Determine the capacitance of the capacitor.

Radius of the inner sphere, \[{{r}_{2}}=12cm=0.12m\]

Radius of the outer sphere, \[{{r}_{1}}=13cm=0.13m\]

Charge on the inner sphere, \[q=2.5\mu C=2.5\times {{10}^{-5}}\]

Dielectric constant of a liquid, \[{{\in }_{r}}=32\]

\[C=\frac{4\pi {{\in }_{0}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}\]

\[{{\in }_{0}}=\]Permittivity of free space \[=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\]

\[V=\frac{1}{4\pi {{\in }_{0}}}9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

\[C=\frac{32\times 0.12\times 0.13}{9\times {{10}^{9}}\times \left( 0.13-0.12 \right)}\Rightarrow C=5.5\times {{10}^{-9}}F\]

Hence, the capacitance of the capacitor is approximately \[5.5\times {{10}^{-9}}F\].

(b) What is the potential of the inner sphere?

Ans: Potential of the inner sphere is given as below

\[\frac{q}{C}=\frac{2.25\times {{10}^{-6}}}{5.5\times {{10}^{-9}}}=4.5\times {{10}^{2}}V\]

Hence, the potential of the inner sphere is \[4.5\times {{10}^{2}}V\].

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius \[12cm\]. Explain why the latter is much smaller.

Ans:  Given that,

Radius of an isolated sphere given, \[r=12\times {{10}^{-2}}m\]

Capacitance of the sphere is given by the relation,

\[{C}'=4\pi {{\in }_{0}}r\]

\[\Rightarrow {C}'=4\pi \times 8.85\times {{10}^{-12}}\times 12\times {{10}^{-12}}\]

\[\Rightarrow {C}'=1.33\times {{10}^{-11}}F\]

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

21. Answer carefully

(a) wo large conducting spheres carrying charges \[{{Q}_{1}}\] and \[{{Q}_{2}}\] are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \[\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}\], where \[r\] is the distance between their centres?

Ans: The force between two conducting spheres is not exactly the same given by the expression \[\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\in }_{0}}{{r}^{2}}}\], because there is a non-uniform charge distribution on the sphere.

(b) If Coulomb’s law involved \[\frac{1}{{{r}^{3}}}\] dependence (instead of \[\frac{1}{{{r}^{2}}}\]), would Gauss’s law be still true?

Ans: Gauss’s law will not hold true, if Coulomb’s law involved \[\frac{1}{{{r}^{3}}}\] dependence, instead of \[\frac{1}{{{r}^{2}}}\] on \[r\].

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

Ans: Yes, if a small test charge is released at rest at a point in an electric field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

Ans: Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Ans: No Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) What meaning would you give to the capacitance of a single conductor?

Ans: The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(g) Guess a possible reason why water has a much greater dielectric constant \[=80\] than say, mica \[=6\].

Ans:   Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

22. Answer the following

a) The top of the atmosphere is at about \[400kV\] with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about \[100V{{m}^{-1}}\]. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside).

Ans: We do not get an electric shock as we step out of our house because the original equipotential surfaces of open-air changes, thus maintaining our body and the ground at the same potential.

b) A man fixes outside his house one evening a two-metre-high insulating slab carrying on its top a large aluminium sheet of area \[1{{m}^{2}}\]. Will he get an electric shock if he touches the metal sheet next morning?

Ans: Yes, the man will get an electric shock when he touches the metal slab. The steady discharging current in the atmosphere charges up the aluminium sheet. Resulting in a gradual rise in voltage. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

c) The discharging current in the atmosphere due to small conductivity of air is known to be \[1800A\] on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

Ans: The occurrence of thunderstorms and lightning charges the atmosphere continuously. Thus, even with the presence of discharging current of \[1800A\], the atmosphere maintains its electrical neutrality.

d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning strike? (Hint: The earth has an electric field of about \[100V{{m}^{-1}}\] at its surface in the downward direction, corresponding to a surface charge density\[=-{{10}^{-9}}C{{m}^{-2}}\]. Due to the slight conductivity of the atmosphere up to about \[50km\] (beyond which is a good conductor), about \[1800C\] is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Ans: During lightning and thunderstorms, light energy, heat energy, and sound energy are emitted to the atmosphere.

Class 12 Physics Chapter 2 Important Questions

Chapter 12 - electrostatic potential and capacitance, electrostatic potential .

When a unit positive charge is moved from one point to the other in an electric field, then the amount of work done to move that charge against an electrostatic force with zero acceleration is known as electrostatic potential.

The formula of electrostatic potential is:-

 Electrostatic potential (V) = Work done (W)/Charge(q) 

Volt (V) is the SI unit of electrostatic potential.

When a charge of 1 coulomb is moved is an electric field against an electrostatic force, then the amount of work done is 1J, hence the electrostatic potential will be 1JC-1.

Electrostatic Potential Difference 

When a unit positive charge is moved from say point A to another point say point B against an electrostatic force in an electric field, then the electrostatic potential difference between point A and point B is the amount of work done to move the charge from point A to point B with zero acceleration.

The formula of electrostatic potential difference is:-

              \[ V_{B}-V_{A}=\frac{W_{AB}}{q}\]

Another formula of electrostatic potential difference is:-

              \[ V_{B}-V_{A}=\int_{B}^{A}E.dl\]

The potential difference point A and point B can be found with the line integral of the electric field from point A to point B.

When a point charge q is at any point P and a distance r, then the electrostatic potential is computed:-

              \[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{q}{r}\]

If the point charge is positive, then the electric potential at that point will be positive. Whereas, when the p[oint charge is negative, then the electric potential at that point will be negative.

A positive charge travels from higher potential to lower potential when placed in an electric field. Whereas, negative potential travels from lower potential to higher potential when placed in an electric field.

The Formula of Electrostatic Potential Due to an Electric Dipole

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The formula of electrostatic potential due to an electric dipole at point P is:-

                \[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{pcos\Theta }{r^{2}}\]

Where, is the angle between p and r.

The Formula of Electrostatic Potential of a Thin Charged Spherical Shell

The formula of the electrostatic potential at a point P inside a thin charged spherical shell that carries charge q and has radius R is:-

\[V=\frac{1}{4\pi\varepsilon _{o}}.\frac{q}{R}\]

The formula of the electrostatic potential at a point P on the surface of a thin charged spherical shell that carries charge q and has radius R is:-

\[V = \frac{1}{4 \pi \epsilon_{0}} . \frac{q}{R}\]

The formula of the electrostatic potential at a point P outside the thin charged spherical shell that carries charge q and the distance between the point P outside the shell and the centre of the shell is ‘r’  is:-

\[V = \frac{1}{4 \pi \epsilon_{0}} . \frac{q}{r}\], where r is greater than R.

Electrostatic Potential Energy

When a charge q is moved from one point to another, the work done by the charge q is stored as electrical potential energy.

The formula of electrostatic potential energy in a system having two charges namely q1 and q2 is:-

\[U = \frac{1}{4 \pi \epsilon_{0}} . \frac{q_{1}q_{2}}{r}\]

These are the important concepts and formulas that are imperative to the students for solving the electrostatic potential and capacitance class 12 important questions. They can also refer to the NCERT question solutions provided by the expert teachers of Vedantu so that grasping of the conceptual part of Physics becomes easy for them.

Important Related Links for CBSE Class 12 Physics 

Mastering the important questions for CBSE Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance is crucial for success in the upcoming 2024-25 exams. By thoroughly understanding the concepts and solving these questions, students can strengthen their knowledge, improve problem-solving skills, and gain confidence. These questions cover various aspects of electrostatic potential and capacitance, allowing students to deepen their understanding of the topic. With dedicated practice and thorough preparation using these important questions, students can aim for excellent results and excel in their Physics examinations. So, buckle up and embark on this learning journey to achieve academic excellence.

FAQs on Electrostatic Potential and Capacitance Important Questions for CBSE Class 12 Physics Chapter 2

1. How is Electrostatic Potential used for?

Electrostatic potential is used to describe the electric potential energy per unit charge at a point in space. It is a fundamental concept in electrostatics and is used to analyse and understand the behavior of electric fields and charges. Electrostatic potential is used in a wide range of applications, including in the design of electrical devices, the study of biological systems, and the analysis of atmospheric phenomena. It is also used to calculate the work done by electric fields, and to describe the behavior of conductors, dielectrics, and other materials in the presence of electric fields.

2. What is Capacitance used for?

Capacitance is used for storing electrical energy in an electric field. It is commonly used in electronic circuits to store and release electrical energy as needed. Capacitors, which are devices that have capacitance, are used in a wide range of applications, including power supplies, audio filters, timing circuits, and many more. Capacitance also plays a crucial role in many technologies, including telecommunications, computing, and electric vehicles.

3. Can I download the Important Questions for Class 12 Physics Chapter 2 for free?

Yes, you can download the Important Questions for Class 12 Physics Chapter 2 for free from Vedantu. The Important Questions for this chapter are provided in a PDF file on Vedantu. You can refer to these notes online or download them and practice offline. All you need for downloading the Important Questions of Class 12 Physics Chapter 2, Electrostatic Potential and Capacitance, is an internet connection and a digital screen. Also, you can print a hardcopy of these Important Questions for your convenience.

4. What are the uses of Electrostatic Potential and Capacitance according to NCERT for Class 12 Physics Chapter 2?

According to NCERT for Class 12 Physics Chapter 2, the uses of electrostatic potential and capacitance are:

Capacitors: Capacitance is used in capacitors, which are devices used to store electrical energy in an electric field. Capacitors are used in a wide range of applications, including power supplies, audio filters, timing circuits, and many more.

Van de Graaff generator: Electrostatic potential is used in the Van de Graaff generator, which is a device that generates high voltages using electrostatics. This device is used in various scientific experiments, such as particle accelerators, and also in educational demonstrations.

Cathode Ray Oscilloscope: Electrostatic potential is also used in Cathode Ray Oscilloscope (CRO), which is an instrument used to display and analyze electronic signals. The CRO uses electrostatic deflection to move the electron beam across the screen.

Lightning rods: Electrostatic potential is used in lightning rods, which are devices used to protect buildings and structures from lightning strikes. The lightning rod works by creating a path of low resistance for the lightning to follow, thus preventing damage to the structure.

Overall, Electrostatic potential is used in Van de Graaff generators, Cathode Ray Oscilloscopes, and lightning rods. Capacitance is used in capacitors for storing electrical energy in electronic circuits and in a wide range of applications.

5. How many chapters are there in Electrostatic Potential and Capacitance Class 12 Physics?

There is one chapter on Electrostatic Potential and Capacitance in Class 12 Physics. The chapter covers topics such as electric potential, potential difference, capacitance, capacitors, and their applications in electronic circuits and other technologies.

CBSE Class 12 Physics Important Questions

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• Chapter 2: Electrostatic Potential Capacitance

Important Questions for Class 12 - Physics - Chapter 2 - Electrostatic Potential and Capacitance

If an external force does work to move a body from one point to another while opposing a force, such as a gravitational force or spring force, the work is converted into the potential energy of the body. When the external force is no longer applied, the body will start moving, gaining kinetic energy while losing an equal amount of potential energy. The sum of kinetic and potential energy is preserved in this process. These types of forces are called conservative forces, and spring force and gravitational force are examples of such forces.

The Coulomb force between two stationary charges is conservative. This is unsurprising, as both the Coulomb force and gravitational force exhibit an inverse-square dependence on distance and differ mainly in their proportionality constants (charge in Coulomb’s law versus mass in the gravitational law). Therefore, similar to how we can define the potential energy of a mass in a gravitational field, we can also define the electrostatic potential energy of a charge in an electrostatic field.

To define the potential energy of a test charge q, we consider the work done on the charge q. The work is proportional to q; the value of force at any point is equal to qE, where E is the electric field at that point due to the given charge configuration. Therefore, it is convenient to divide the work by the magnitude of the charge q, resulting in a quantity that is independent of q. In other words, the work done per unit charge is a characteristic of the electric field associated with the charge configuration.

We are now considering the potential energy of a charge or charges in an external field rather than the field produced by the charge itself. The external field, denoted as E, is produced by sources outside of the charge being considered and is not produced by the charge itself. In many cases, the sources of the external field are unknown or unspecified. However, the electric field E or the electrostatic potential V caused by these external sources is specified. It is assumed that the charge q does not significantly alter the external sources that produce the field. This assumption is valid if the charge q is very small or if the external sources are fixed in place by other forces. Even if q is not infinitesimally small, it may still be assumed to have no significant impact on the external sources if there are strong sources far away that produce a finite field E in the region of interest. This is because the influence of q on external sources may be negligible in comparison to strong sources.

It is important to reiterate that our focus is on determining the potential energy of a specific charge q (or system of charges) in an external field rather than the potential energy of the sources that produce the external field. The external electric field E and the associated external potential V can vary from one point to another. The potential V at point P is defined as the work required to bring a unit positive charge from infinity to point P. (We maintain the convention of taking the potential at infinity to be zero.) Therefore, the work required to bring a charge q from infinity to the point P in an external field is qV, and this work is stored as the potential energy of the charge q.

Important Questions for Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance are provided here. Students must go through these questions and solve them to prepare for their Physics papers. They can also refer to these questions for quick revision. These questions are more likely to be asked in the exam; hence, students must practise them thoroughly.

Very Short Answer Type Questions

1. Consider two conducting spheres of radii R 1 and R 2 with R 1 > R 2 . If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

In this case, V 1 = V 2

It is given that R 1 > R 2

Therefore, the bigger sphere has more electric charge than the smaller sphere. Then the charge densities

The charge density of a smaller sphere is greater than that of the larger sphere.

2) Do free electrons travel to regions of higher potential or lower potential?

Solution: The free electrons experience an electrostatic force in the opposite direction of the electric field due to their negative charge. Since the electric field always moves from higher potential to lower potential, the electrostatic force and the direction of travel of the electrons is from the region of lower potential to the region of higher potential.

3) Can there be a potential difference between two adjacent conductors carrying the same charge?

From the mathematical relation Q = CV

So, the potential difference across a conductor depends not only on the charge present but also on the geometry of the conductor, which influences its capacitance. As a result, two adjacent conductors with the same charge but different dimensions may have different potential differences.

4) Can the potential function have a maximum or minimum in free space?

The potential function does not have a maximum or minimum value in free space because there is no atmosphere present to cause electric discharge or potential leakage around the conductor.

5) A test charge q is made to move in an electric field of a point charge Q along the two different closed paths. The first path has sections along and perpendicular to the lines of an electric field. The second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

The work done by an electrostatic force is conservative, which means that the work done in a closed loop is always zero, regardless of the path taken within the loop.

Short Answer Type Questions

1) Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Assuming a contradicting statement, the potential within a closed equipotential surface is not uniform. Let the potential just inside the surface be different to that on the surface having a potential gradient (dV/dr). Therefore, an electric field is created, which is represented by the following:

As a result, field lines either point inward or outward from the surface. These lines cannot originate from the surface itself, as it is equipotential. Field lines can only be formed when they have a charge at the other end within the surface. This conclusion contradicts the initial assumption. Therefore, it must be the case that the entire volume inside is equipotential.

2) A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected, and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, the charge stored, and the voltage will increase, decrease or remain constant.

When a capacitor is connected to a DC source, it becomes charged. When the battery is disconnected, no more charge can flow into the capacitor. Removing the dielectric material from the capacitor will cause its capacitance to decrease.

Imagine that a charged body, designated as A, is placed near an insulated, uncharged conductor, designated as B. A has a positive charge. The setup is shown in the accompanying figure.

The uncharged conductor is situated between the charged conductor and infinity, causing the potential to decrease from body A to infinity. As a result, the potential of the uncharged body varies between the potential of A and infinity.

4) Calculate the potential energy of a point charge -q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E as a function of axial distance z from the centre of the ring. Looking at the graph, can you see what would happen if -q is displaced slightly from the centre of the ring (along the axis)?

The potential energy (U) of a point charge q at a potential V is given by U=qV. In this case, a negatively charged particle is located on the axis of a ring with charge Q. The ring has a radius of a, and the electric potential at a distance x from the centre of the ring along the axis can be calculated.

The graph in Fig. shows how the potential energy changes as a function of axial distance z from the centre of the ring. If a charge of -q is slightly displaced from the centre of the ring along the axis and released, it will undergo oscillations. However, it is not possible to determine the nature of these oscillations simply by examining the graph.

5) Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

Let’s assume a ring of radius R possessing charge +Q that is evenly distributed. A point P at length z passes through the centre O, and it is perpendicular to the ring’s plane.

Next, assume an element of the ring at point S (length di) possessing a charge dq and SP is equivalent to r.

Consequently, the potential energy due to element r and the potential energy due to element dl at p

Charge on the 2πR length of ring is equal to Q.

By integrating over the ring potential V p at point P ,

Therefore, the potential caused by element di at point P is

By integrating over the ring the potential V p at point P, we get

Long Answer Type Questions

1) Find the equation of the equipotentials for an infinite cylinder of radius r 0 , carrying charge of linear density λ.

Take the case of a cylindrical Gaussian dotted surface (S) at a length r from the cylinder’s centre of radius r 0 of infinite length. The electric field lines are perpendicular and radial to the surface.

Consider electric field intensity on the given Gaussian surface at point P is E, and net charge q on the cylinder be q=λl

In this scenario, ‘ε 0 ’ is the free space’s permittivity, ‘l’ is the length of the cylinder, ‘q’ is the charge bounded within the surface, and ‘S’ is the area vector perpendicular to the surface. ‘2π rl’ is the area of the curved cylindrical surface.

When r 0 is the infinite cylinder’s radius,

Therefore, equipotential surfaces are the coaxial curved surfaces of cylinders with the given cylinder of radius r.

2) Two point charges of magnitude +q and -q are placed at (-d/2, 0,0) and (d/2, 0,0), respectively. Find the equation of the equipotential surface where the potential is zero.

The potential caused by charges -q and +q are zero in between the line connecting the two charges -q and +q. Consider zero potential is at point S. Then the equipotential surface will go through S and normal to the line connecting the two charges (AB).

Therefore, the total potential at point P = 0

Thus, the equipotential surface will be normal to the x-axis going through x = 0 (origin in y-z plane).

3) A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V -1 . A similar capacitor with no dielectric is charged U 0 = 78 V. It is then connected to the unchanged capacitor with the dielectric. Find the final voltage on the capacitors.

After connecting the two capacitors, let’s consider the final voltage to be ‘V’. When the capacitance of the capacitor with dielectric is C, the charge on the capacitor is q 1 (equal to CV). Let’s also assume that the other capacitor with dielectric possesses εC capacitance.

So, the charge on it is

SInce ε = αV, q 2 = ε = αV

The initial charge on the given capacitor (without the dielectric) that was electrically charged is

From the law of conservation charges,

Since V is positive,

Thus, the charge ‘q’ transferred to the disc is

Then, the force acting on the disc is

When the disc is lifted,

(b) Repeat the above exercise for a proton which is made of two up and one down quarks.

(a) Construct a rough figure.

The charge on an up-quark is

The charge on a down-quark is

The length between each quark is

The potential energy of the charge system is expressed by,

By comparing the resulted energy with the mass of the neutron (939 MeV), the potential energy is,

The length between each quark is 10 -15 m. The potential energy of the charge system is given by

Substitute the corresponding values, and then we get

Before making contact, the charges of the two spheres are

If the two spheres are made to touch one another, their charges are shared till their potentials become the same value, V 1 = V 2

Since there is no loss of charge in the phenomenon

If K 1 is closed and K 2 is open, the capacitors C 2 and C 1 will charge, and potential arises across them V 2 and V 1, respectively, which will be equivalent to the potential of the 9V battery.

From the equations (i) and (ii)

Therefore, the charges on each capacitor, q 1 = q 2 = 18μc

If k 1 is open and k 2 is closed off, then the charge q 2 will be infused in C 2 and C 3 . Let’s consider it as q 2 and q 3

Since C 2 and C 3 are connected parallelly, their potentials stay the same.

Thus, the potential on C 3 and C 2 are 3 volts each.

Take a point P on the axis normal to the disc’s plane and length ‘x’ from the centre O as represented in the diagram. Then, take a ring of radius r with thickness dr on the disc with radius R, as represented in the above diagram. Also, consider the charge on the ring is dq. Then the potential dV due to ring at point P will be

‘dq’ is the charge on the ring.

Since dr is very small, dr 2 is practically negligible.

Thus, the potential due to the charged disc is

Let’s consider the potential at any random point P(x, y, z) is zero, then

This is the sphere’s equation with centre (a, b, c) as the needed point is on the z-axis.

Therefore, it is in an unstable equilibrium.

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Class 12 Physics Case Study Questions

Please refer to the chapter-wise Class 12 Physics Case Study Questions given below. These questions are expected to come in your class 12 Physics board exams. Our expert teachers have developed these Case Study questions and answers based on the latest examination guidelines and updated syllabus issued by CBSE, NCERT, and KVS. These will help you to understand all topics too and bet way to write answers in your exams. Revise these solved problems prior to your exams to score better marks in Class 12 exams.

Case Study Questions Class 12 Physics

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Chapter-wise Solved Case Study Questions for Class 12 Physics

Class 12 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Physics examinations. Our expert faculty for standard 12 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy to learn solutions.

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Important Questions Class 12 Physics Chapter 2

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Important Questions for CBSE Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance

These Important Questions in Class 12 Physics Chapter 2 will teach students about electrostatic potential and capacitance. These Chapter 2 Class 12 Physics Important Questions will prepare students for their exams. After studying these Physics Class 12 chapter 2 Important Questions, students will be able to solve many CBSE sample papers.

CBSE Class 12 Physics Chapter-2 Important Questions

Study important questions for class 12 physics chapter 2 – electrostatic potential and capacitance.

Electrostatic Potential and Capacitance is an important chapter and going through these important question notes can help you understand the exam pattern. These Class 12 Physics Chapter 2 Important Questions contain important formulas and CBSE extra questions that can help you test student’s understanding. In addition, these notes are crafted as per the prescribed CBSE syllabus and CBSE revision notes .

Short answer questions

Q.1. Why does a dielectric’s internal electric field weaken when exposed to an external electric field?

Ans. The polarisation causes an internal electric field that is opposite to the external electric field inside a dielectric, which causes the net electric field to decrease when the dielectric is exposed to an external electric field.

Q.2. A 10 cm square with a 500 C charge in the centre. Find the amount of effort required to move a charge of 10 C between two square spots that are diagonally opposed.

Ans. Since these two spots on the square will be equipotential, the work required to move a charge of 10C between them will be zero.

Q.3. When a 10 C charge is present in the square’s centre, how much work is required to move a 2 C point shift from corner A to corner B?

Ans. Points A and B are equally distant from point O. Hence the work done is equal to zero.

Therefore, the work done=0

Q.4. What physical effort is expended when an electric dipole’s equatorial axis is traversed by a test charge q over a distance of 1 cm?

Ans. Given that the equatorial axis potential is V = 0

∴W = qV = 0,

Q.5. A voltmeter connects the plates of a charged capacitor. What will happen when the capacitor’s plates are separated further by voltmeter reading?

Ans. Capacitance, area, distance, and dielectric constant are all related.

Hence, the capacitance will decrease if the distance increases.

Since V=QC and there is a constant charge on the capacitor,

Therefore, the voltmeter reading will increase.

Q.6. A hollow metal sphere with a 10 cm radius is charged to a surface potential of 5 V. What potential exists in the sphere’s centre?

Ans. The potential in the centre of a hollow metal sphere will be 5 V because it functions as an equipotential surface.

Q.7. Why must every point of an empty charged conductor’s electrostatic potential be the same?

Ans. Since there is no electric field within the hollow-charged conductor, there is no effort expended in moving the test charge. The electrostatic potential is hence constant throughout a hollow-charged conductor.

Q.8. Imagine a parallel plate capacitor having air in between the plates has a capacitance of  8pF ( 1pF = 10 − 12 F ) . What would the capacitance be if the space between the plates is halved and the space between them is taken over with a substance of dielectric constant 6?

Ans. For air, capacitance can be written as C0=A0d

C0 = 8pF=810-12F

Now, d’=d2 and k=6

C’=810-1226

C’=9610-12pF

Q.9. If there are 1 and 2 volts of electric flux entering and exiting a closed surface, respectively, what is the surface’s internal electric charge?

Ans. Net flux can be represented as 2 -1

since  = q0

Q= (2 -1) 0

The electric charge that is present on the surface can be represented as

Long Answer Questions

Q.10. (i) Is it possible for two equipotential surfaces to intersect? Give reasons. 

(ii) At positions A (0, 0, -a) and B (0, 0, +a), respectively, are two charges, -q and +q. In transferring a test charge from location P (7, 0, 0) to point Q (-3, 0, 0), how much work is involved?

Ans. (i) This is incorrect because if they connect, the electric field there would be pointing in two distinct directions. If they do, there will be two potential values present at the junction. As a result, two equipotential surfaces cannot intersect because this is not conceivable.

(ii) Since V = 0 at every point on the dipole’s equatorial line, which includes points P and Q, there will be no work done. Additionally, since any force of a charge is perpendicular to the equatorial line, no work is done.

Q.11. Each vertex of a regular hexagon having 10 cm of side length carries a charge of.

Calculate the potential in the hexagon’s centre.

Ans: The given illustration depicts six identical charges named q at the vertices of a regular hexagon.

In the case, Charge, q = 5C = 510-5 C

AB=BC=CD=DE=EF=FA = 10cm

Each vertex’s distance from the hexagon’s centre, O, d, is 10 cm, and the electric potential at point O is 10 cm.

0 is the permittivity of free space,

140 = 9109NC-2m-2

V = 69109510-50.1

V = 2.7106V

Hence, 2.7106V is the potential at the centre of the hexagon.

Q.12. A spherical conductor with a radius of 12 cm has a 1.6 107 C charge evenly dispersed over its surface. What will the electric field be in the following cases?

• Inside the sphere
• Just outside the sphere
• At a location 18 cm from the sphere’s centre

Ans. (a) Radius of the spherical conductor can be represented as r = 12cm = 0.12m

The conductor is evenly covered with charge, which can be applied as,

q = 1.610-7C

There is no electric field inside of a spherical conductor. This is due to the fact that charges will flow to neutralise any fields present inside the conductor.

(b) The electric field E present just outside the conductor can be represented by the following relation,

140 = 9109NC-1m-2

E =1.610-7910-9(0.12)2

E = 105 NC-1

Hence, the electric field that will exist just outside the sphere will be 105 NC-1.

(c) The distance between the point from the centre, d = 18cm = 0.18m

E= 91091.610-7(1810-2)

E = 4.4104 N/C

Hence, the electric field having an 18 cm distance from the centre of the sphere will be 4.4104 N/C.

Q.13. A constant electric field holds an electric dipole in place.

(i) Demonstrate that there is no net force operating on it.

(ii) The dipole is parallel to the field in alignment.

Find the work that was done to rotate it via a 180° angle.

Ans. (i) Force resulting from charge -q is -qE acting on point A

Point B is exposed to a force from charge +q. is + qE.

The net force acting on equals -qE plus qE, or 0. (zero)

Hence, the net force acting on an electric dipole held in a uniform electric field is zero.

(ii) W = -pE(cos 02 — cos θ2)

W = -pE(cos 180° – cos 0°)

=> W = -pE(-1 – (1)) = +2pE

Q.14. A 12 V battery is linked to three identical capacitors C1, C2, and C3, with capacitances of 6 F each.

Answer the following questions. 

(i) What charge will be present on each capacitor? 

(ii) What will the equivalent capacitance of the network be?

(iii) What will the energy stored in the network of capacitors be?

Ans: (i) C1 and C2 is series, make C4=3µF

Using the equation 1C4=1C1+1c2

(i) the potential that is available in C4 and C3

Charge in C3=Q3=C3V

=610-612=72C

Charge in C4=Q4=C4V

310-612=36C

∴Here the charge on CC1 and C2 will be the same, i.e., 36C

(ii) C4 and C4 are parallel to the source

∴Ceq=3+6=9F

(iii) Energy stored=12Ceq V²

=12(910-6)122

=64810-6=6.4810-4 joule.

Class 12 Physics Chapter 2 Important Questions

Q.15. A battery supplies power to a parallel plate capacitor. After some time, the battery is removed, and a dielectric slab is put between the plates with a thickness equal to the plate separation. Find the following:

(i) the capacitance of the capacitor

(ii) the electric field in between the plates

(iii) how the energy stored in the capacitor will be affected

  Ans. Take C as the capacitance and take V to be the potential difference.

Q = CV will be the charge on the capacitor plates

E = Vd will be the energy stored between the plates

En = Q22C or 12CV²

When the battery has disconnected, the dielectric (k) will be introduced

Then, the new volts of charge will be Q’= Q Capacitance C’= KC

Potential V’= QKC = VK

(i) New Capacitance is K times its original.

The new electric field E will be = V’d=VKd=EKi.e., 1K times the original field.

New energy = Q22C’ = Q22KC = 1K(En), i.e., 1K time the original energy

Q.16. A parallel plate capacitor is charged to a potential difference V. Each plate has an area of A and a distance of d. The battery that is used for charging is still attached. Between the plates is now a dielectric slab with the dimensions d, where k is the dielectric constant. What alterations, if any, will be made to

(i) the charge on plates?

(ii) the electric field intensity that would exist between the plates?

(iii) the capacitance of the capacitor?

Ans. The plate area of a parallel plate capacitor’s either plate equals A.

Potential difference between the plates equals V, and the distance between the plates equals d.

∴ Initially capacitance, C = 0Ad

Charge on the plate, Q – CV

When a dielectric slab of thickness denoted with “d” and dielectric constant denoted with “k” is placed between the plates, the potential difference between the plates remains unchanged (V’ = V) as long as the battery is connected throughout.

(i) New charge on the plates, Q’= C’V’= kCV = kQ

Hence, the charge will change to k times its original value.

(ii) Electric field intensity existing between the plates,

E’ = V’d=Vd = E

Hence, no change in the electric field intensity existing between the place of the capacitor.

(iii) The new capacitance of the capacitor will be

Q.17. Imagine there are two charges, 2 C and 2 C, that are placed 6 cm apart from points A and B. Find the following.

(a) The equipotential surface of the system

(b) The electric field direction that exists at every point on the given surface

Ans. (a) This situation can be represented in the figure given below.

The plane where the total potential is zero everywhere is referred to as an equipotential surface. This plane is typical for line AB. Because the magnitude of the charges is the same throughout line AB, the plane is situated at its midpoint.

(b) Every point on this surface has an electric field that is normal to the plane and points in the direction of AB.

Q.18. Charge exists in a spherical conducting shell with an inner radius r1 and outer radius r2.

(a) In the middle of the shell is a charge q. What will the surface charge density of the shell’s interior and exterior surfaces be?

(b) Even if the shell is not spherical, but has any irregular shape, is the electric field within a cavity (with no charge) zero? Explain.

Ans. (a) A shell’s centre charge is designated as +q. So, the shell’s inner surface will generate a charge of size -q. As a result, the inner shell’s inner surface has a total charge of -q. The relation can be used to calculate the surface charge density at the shell’s inner surface.

1 = Total ChargeOuter Surface Area = -q4r12

2=Total ChargeOuter Surface Area = -q4r22

On the outside of the shell, a charge of +q is induced. The exterior of the shell is charged with a charge of Q magnitude. Therefore, the total charge on the shell’s exterior is Q+q.

(b) Yes, even though the electric field intensity existing inside a cavity is zetrot The shell has an uneven shape and is not spherical. Take a closed loop now, like so just a portion of it, along a field line, is within the hollow, and the remainder is inside the conductor. The field’s net effort in transporting a test charge over a closed loop. Due to the zero field inside the conductor, the loop will also be zero. Therefore, in whatever shape, the electric field is zero.

Q. A 50 V battery is coupled with a 12 pF capacitor. How much electrostatic energy does the capacitor have stored? Find the charge stored and the potential difference across each capacitor if a second 6 pF capacitor is placed in series with it and the same battery is connected across the combination. 

Ans. The 12 pF capacitor’s capacity stores energy.

(i) U = 12CV²

= 12(1210-12)5050 J                (Given V = 50)

= 1.510-8 J

(ii) C = Equivalent capacitance of 12 pF and 6 pF in series can be represented by

1C=112+16=1+212

∴ The charge that will be stored across each capacitor will be

q=CV=(410-12)50 C=210-10 C

The charge that will be present on each capacitor 12 pF, and 6 pF is 210-10 C

∴ The potential difference existing across capacitor C1

V1=qC1=(210-12)1210-12 Volt=503V

The potential difference existing across capacitor C2

V2=qC2=(210-10)610-12 Volt=1003V

Q.19. There is a connection between two charged conducting spheres with radii a and b by a wire to another. How many electric fields are there on the surfaces of the

two spheres? Use the outcome to demonstrate why the charge density on the conductor’s sharp and pointed ends is higher than its flatter ones.

Ans. Imagine “a” to be the sphere A’s radius, QA represents the sphere’s charge and CA represent the sphere’s capacitance. If “b” is a sphere B’s radius and Q B is the sphere’s charge, let CB represent the sphere’s capacitance. The potential V of the two spheres will equalize since they are joined together by a wire.

Let EB and EA represent the electric fields of spheres B and A, respectively. Consequently, their ratio,

EAEB= QA40a2b240QB

EAEB=QAQBb2a2

QAQB=CAVCBV

When we put the value obtained together, we get

EAEB = abb2a2 = ba

Hence, ba will be the electric field’s ratio at the surface.

Q.20. Given that there is a 0.5cm gap between the plates of a 2F parallel plate capacitor, what is the area of the plates? (You’ll see from your response why standard capacitors have a range of F or less. However, due to the extremely thin spacing between the electrolytic capacitors, electrolytic capacitors do have a significantly higher capacitance of 0.1F.)

Ans. The capacitance of a parallel capacitor can be written as V = 2F.

The gap between the two plates will be written as d = 0.5cm = 0.510-2m.

The Capacitance of a parallel plate capacitor can be represented by

0 = permittivity of free space = 8.8510-12C²N-1m²

A = 20.510-28.8510-12

A = 1130 km2

Therefore, the plate’s area is too big. The capacitance of the F range is taken to avoid this situation.

Q.21. (a) Give a definition of the dielectric constant with respect to the capacitance of a capacitor. What factor influences a parallel plate capacitor’s dielectric capacitance?

(b) To make the energy stored in the two cases equal, determine the ratio of the potential variations that must be applied across the

(i) parallel

(ii) series combination of two identical capacitors

Ans. Dielectric constant means the ratio of a capacitor’s capacitance in the case of the dielectric being filled between the plates with a capacitor’s capacitance in case of a vacuum present between the plates.

In K = CmCo = Capacitance of a capacitor when dielectric is in between the platesCapacitance of a capacitor with vacuum in between the plates

The following variables affect the capacitance of a parallel plate capacitor with a dielectric.

Imagine the C to be the capacitance of each capacitor

CP = CCC+C = C2          ………….(in series)

Let the value of potential difference be  VP and Vs

Hence, UP = 12CPVP2 = 122CVP2= CVP2

US = 12CSVS2 = 12C2VS2 = CVS24

However, UP= US is already given,

CVP2 = CVS24

VP2VS2 = 14

VP: VS = 1:2

Q.22. Explain the construction, working, and the principle of a Van de Graaff generator using a labelled diagram. Also, mention its uses.

Ans. Van de Graaff generators are devices that can generate enormous potentials of the order of millions of volts.

Principle: In a hollow conductor, the charge is always located on the outer surface.

At the pointed ends of the conductors, the electric discharge in air or gas occurs easily.

How is it constructed?

It is made up of a big metallic hollow spherical S set on two insulating columns.

A and B are connected by a rubber belt that runs over two pulleys with the aid of an electric motor, P1 and P2, in an unending fashion. B1 and B2 are two pointed metallic brushes. High tension battery provides a positive voltage to the lower brush B1.

Spray brush, with the higher brush B2 attached to the S-sphere’s inner portion.

Brush B1 produces ions when its positive potential is increased because of the impact of sharp edges. The resulting positive ions are then sprayed on the belt as a result of electrostatic attraction between positive ions and brush B1’s positive charge.

It is then raised by the belt as it moves. B2’s pointy end barely touches

The belt gathers the favourable change and moves it to the belt’s outer surface S sphere. The potential of the shell increases to millions of volts as a result of this ongoing process.

Applications:

Protons, deuterons, and other particles, among others, are accelerated to very high speeds and energy.

Chapter 12 – Electrostatic Potential and Capacitance

These Class 12 Physics Chapter 2 Important Questions include the Important Questions mentioned in NCERT books. After going through these important questions, you will know all about electric charges and fields and will gain confidence to solve CBSE past years’ question papers .

Electrostatic Potential

A electrostatic potential is the work required to shift a unit of positive charge from one location to another in an electric field when that charge is up against an electrostatic force with zero acceleration.

The electrostatic potential equation is:

Work done (W)/Charge = Electrostatic potential (V) (q). The SI unit for electrostatic potential is the volt (V).

The electrostatic potential is equal to 1JC-1 when a charge of 1 coulomb is pushed by an electric field in opposition to an electrostatic force.

Electrostatic Potential Difference

The electrostatic potential difference between points A and B is the amount of effort required to transport a unit positive charge from point A to point B with no acceleration when doing so in the presence of an electrostatic force and an electric field.

The electrostatic potential difference is calculated as follows.

The following is another electrostatic potential difference formula.

VB-VA= bA Edl

The line integral of the electric field from point A to point B can be used to determine the potential difference between those two points.

The electrostatic potential is calculated when a point charge q is present at any location P and a distance r as follows.

The electric potential at that location will be positive if the point charge is positive. In contrast, the electric potential at a point will be negative if the point charge is negative.

In an electric field, a positive charge moves from a higher potential to a lower potential. In contrast, when placed in an electric field, negative potential moves from lower potential to greater potential.

The Formula of Electrostatic Potential Due to an Electric Dipole

The formula of electrostatic potential caused to an electric dipole at a point S is given below.

V=140scosr2

The Formula of Electrostatic Potential of a Thin Charged Spherical Shell

The electrostatic potential at a point P inside a thin, charged, spherical shell of radius R and charge q have the following formula.

The electrostatic potential at a point S on the surface of a thin charged spherical shell with charge q and radius R has the following formula.

The electrostatic potential at a point S outside of a thin, charged spherical shell carrying charge q and the distance r between that point and the shell’s centre is calculated as follows.

In the case where r is larger than R, V = 140qr

Electrostatic Potential Energy

The work that a charge q performs when it is transported from one location to another is saved as electrical potential energy.

The following equation describes the electrostatic potential energy in a system with two charges, q1 and q2.

U = 140q1q2r

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Four capacitors are connected as shown in the figure. What is the equivalent capacitances between A and B?

According to the given figure, capacitance in parallel is 12 ?

According to the given figure, capacitance in parallel is 12 ¼F and 12 ¼F. Therefore, C p = 12 + 12 = 24 ¼F

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Faqs (frequently asked questions), 1. what variables affect a parallel plate capacitor's capacitance.

• Area of plates
• The difference between the plates’ spacing and
• The nature of dielectric that exists between them

2. A hollow metal sphere with a 5 cm radius is charged to have a 10 V potential on its surface. What potential exists in the sphere's centre?

3. what will the geometrical shape be if an equipotential surface is caused by a single isolated charge.

The geometrical shape created for an equipotential surface caused by a single isolated charge is concentric circles.

4. In the presence of an external electric field E, write a relation for the polarisation P of a dielectric substance.

5. how much work is involved in moving a point charge across an r-radius circular arc where another point charge is situated.

The work done in this case will be zero as it is an equipotential surface.

6. Justify the following statement. “For any charge configuration, equipotential surface through a point is normal to the electric field”.

An equipotential surface requires no work to move a charge over it. Hence, every point on the surface will be normal to the electric field.

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NCERT Solutions For Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT Solutions For Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance are given below. Get complete solutions for the 2024 board exams.

February 9, 2024

Table of Contents

NCERT Solutions For Class 12 Physics Chapter 2

NCERT Solutions for class 12 Physics Chapter 2 Electrostatic Potential and Capacitance is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class 12 in NCERT textbook, also do read theory of this Chapter 2 Electrostatic Potential and Capacitance while going before solving the NCERT questions. You can download and share NCERT Solutions of Class 12 Physics from Physics Wallah.

NCERT Solutions for class 12 Physics Chapter 1

Answer The Following Question Answers of class 12 physics Chapter 2 – Electrostatic Potential and Capacitance:

Question 1. Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution : Given,

Two charges q A  = 5 x 10 -8  C and q B  = -3×10 -8  C

Distance between two charges, r = 16 cm = 0.16 m

Consider a point O on the line joining two charges where the electric potential is zero due to two charges.

Question 2. A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.

Solution : Let O be the center of the hexagon. It contains the charges at all its 6 vertices, each charge = + 5 μC = 5×10 -6  C. The side of the hexagon is 10 cm = 0.1 m

It follows that the point O, when joined to the two ends of a side of the hexagon forms an equilateral triangle Electric potential at O due to one charge,

Question 3. Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this surface?

(a) For the given system of two charges, the equipotential surface will be a plane normal to the line AB joining the two charges and passing through its mid-point O. On any point on this plane, the potential is zero.

(b) The electric field is in a direction from the point A to point B i.e. from the positive charge to the negative charge and normal to the equipotential surface.

Question 4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field

(a) Inside the sphere

(b) Just outside the sphere

(c) At a point 18 cm from the centre of the sphere?

Solution : Given, q = 1.6 x 10 -7  C

Radius of the sphere, r = 12 cm = 0.12 m

(a)  Inside the sphere: The charge on a conductor resides on its outer surface. Therefore, electric field inside the sphere is zero.

(b)  Just outside the sphere: For a point on the charged spherical conductor or outside it, the charge may be assumed to be concentrated at its center.

Question 5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Solution : Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF = 8×10 –12  F

Question 6. Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Solution : Given, C 1  = C 2  = C 3  = 9 pF = 9 x 10 -12  F

V = 120 volt

Question 7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution : Given, C 1  = 2 pF

C 2  = 3 pF

C 3  = 4 pF , V = 100 volt

• Total capacitance of the  parallel  combination is

C = C 1  + C 2  + C 3  = 2 + 3 + 4 = 9 pF

• Let q 1  , q 2  and q 3  be that charges on the capacitor C 1 ,  C 2  and C 3  respectively.

In the parallel combination the potential difference across  each  capacitor will be equal to the supply voltage i.e., 100 V

⇒ q 1  = C 1 V = 2 x 10 -12 ×100 = 2× 10 -10  C

⇒ q 2  = C 2 V = 3 x 10 -12 ×100 = 3× 10 -10  C

⇒ q 3  = C 3 V = 4 x 10 -12 ×100 = 4× 10 -10  C

Question 8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Question 9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Solution : (a) When the voltage supply remains connected:

The capacitance of the capacitor will become K times.

Therefore, C’ = kC

Where k = dielectric constant = 6×17.7pF = 106.2 pF

The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V

The charge on the capacitor,

q’ = C’V = 160.2 x 10 -12  x 100

= 1.602 x 10 -8  C

(b) After the voltage supply is disconnected:

As calculated above, the capacitance of the capacitor, C’ = 106.2 pF

The potential difference will decrease on introducing mica sheet by a factor of K,

Question 10. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Solution : Given, C = 12 pF = 12 x 10 -12  F

The electrostatic energy stored in the capacitor,

W = (½) CV 2  = (½) × 12 × 10 -12 × (50) 2  = 1.5 × 10 -8  J

Question 11. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution : Given, C 1  = 600 pF = 600 x 10 -12  F

V 1  = 200 V

Energy stored in the capacitor,

U 1  = (½) C 1  (V 1 ) 2  = (½)×600× 10 -12 × (200) 2

= 12×10 -6  J

When this charged capacitor is connected to another uncharged capacitor C 2  ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.

Total charge on the two capacitors,

q = C 1 V 1  + C 2 V 2  = 600 × 10 -12 × 200 + 0

= 12 ×10 -8  C

Total capacitance of the two capacitors,

C = C 1  + C 2  = 600 pF + 600 pF

= 1200 x 10 -12  F

ADDITIONAL EXERCISES:

Question 12. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

Charge at origin (O), q 0  = 8×10 -3  C

The charge (q 1  = -2 Nc) is moving through the given points, P(0, 0, 3)cm, R(0, 6, 9)cm and Q(0, 4, 0)cm respectively.

The picture below represents above situation,

Question 13. A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Solution : Given, side of the cube = b units

Charge at each vertices = q C

Question 14. Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Question 15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Question 16. 

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Solution : (A)Electric field on one side of the charged is body is E 1  and the electric field on the other side of the same body be E 2 . If infinite plane charged body has uniform thickness then the electric field due to one surface of the body is given by,

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

Question 17. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Charge density of the long-charged cylinder of length L and radius r is λ. Another same type of cylinder with radius R surrounded it.

Let E is the electric field produced in the space between the two cylinders.

Electric flux through a Gaussian surface is given by the Gaussian theorem as,

Φ = E(2πd)L

Where, d = distance of a point from common axis of the cylinders.

Let q be the total charge on the cylinder,

Question 18. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

The distance between electron-proton of hydrogen atom = 0.53 Å = 0.53× 10 -10 m

Charge on electron, q 1  = – 1.6×10 -19  C

Charge on proton, q 2  = 1.6×10 -19  C

(a)Potential energy at infinity is Zero.

Potential energy of a system,

Question 19.

Question 20. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Solution : Let a be the radius of the sphere A, Q A  be the charge on the sphere, and C A  be the capacitance of the sphere. Let b be the radius of the sphere B, Q B  be the charge on the sphere, and C B  be the capacitance of the sphere. Since, the two spheres are connected with wire, their potential V is equal.

Question 21. Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.

(a) What is the electrostatic potential at the points?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

B. Given r/a>>1, which implies r>>a

the distance of point where potential is to be obtained is much greater than half of the distance between the two charges.

Hence, the potential (V) at a distance r is inversely proportional to square of the distance, i.e. V ∞1/r 2

Question 22. Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

The given charges of same magnitude placed at points X, Y, and Z respectively, forms an electric quadrupole.

Where, charge + q is at point X, charge -2q is at point Y, and charge + q is at point Z.

The point P is at a distance r from point Y.

Here, XY = YZ = a

So, YP = r, PX = r + a, PZ = r-a.

The electrostatic potential due to the system of three charges at point P is given by,

Question 23. An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Solution : Potential difference across the circuit = 1kV = 1000V

Capacitance of each capacitor = 1 μF

Potential difference each capacitor can withstand = 400V

Capacitance required across the circuit = 2 μF

Assume n number of capacitors are connected in series and further m number of such series circuits are connected in parallel to each other.

As the potential difference in the circuit is 1000V so the potential difference across each row of n capacitors is 1000V, as the potential difference each capacitor can withstand is 400V,

Therefore, 400V × n = 1000V

⇒ n = 1000V/400V = 2.5~3capacitors in each row.

Question 24. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Question 25. Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

Question 26. The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Question 27. A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

∴ U = 5.33×10 -2  J

Thus, the electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation = 8×10 -2  J-5.33×10 -2  J = 6.67×10 -2 J

Question 28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½.

Question 29.

Question 30. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Radius of the outer shell (r 1 ) = 13cm = 0.13m

Radius of the inner shell (r 2 ) = 12cm = 0.12m

Charge on the outer surface of the inner shell = 2.5 μC = 2.5×10 -6 C

Dielectric constant of liquid = 32

Since, Potential difference between the two shells,

Question 31. Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1 Q2/4π∈ o r 2, where r is the distance between their centres?

(b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Solution : (a)

(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of1/r2, on r.

If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.

(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

Question 32. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Solution : Radius of outer cylinder(R) = 1.5cm = 0.015m

Radius of inner cylinder(r) = 1.4cm = 0.014m

Charge on the inner cylinder(q) = 3.5µC =  3.5×10 -6 C

Question 33. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Solution : Potential difference of a parallel plate capacitor(V) = 1kV = 1000V

Dielectric constant of a material(ϵ r ) = 3

Dielectric strength = 10 7 V/m

Electric field intensity(E) = 10%of 10 7

⇒ E = 10 6 V/m

(since, the field intensity never exceeds 10% of the dielectric strength)

Question 34. Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Solution : (a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.

(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) Concentric spheres centered at the origin are equipotential surfaces.

(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Question 35. In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Solution : Potential difference, V = 15 × 106 V

Dielectric strength of the surrounding gas = 5 × 107 V/m

Electric field intensity, E = Dielectric strength = 5 × 107 V/m

Minimum radius of the spherical shell required for the purpose is given by,

Hence, the minimum radius of the spherical shell required is 30 cm.

Question 36. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

Solution : According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.

Question 37. Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)

Solution : (a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.

NCERT Solutions For Class 12 Physics Chapter 3 Current Electricity

NCERT Solutions for Class 7 Social Science Civics Chapter 1 On Equality

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CBSE Class 12 Physics Chapter 2 Extra Questions

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CBSE Class 12 Physics Chapter 2 Extra Questions. We know Physics is tough subject within the consortium of science subjects physics is an important subject. But if you want to make career in these fields like IT Consultant, Lab Technician, Laser Engineer, Optical Engineer etc. You need to have strong fundamentals in physics to crack the exam .  myCBSEguide has just released Chapter Wise Question Answers for class 12 Physics. There chapter wise Extra Questions with complete solutions are available for download in  myCBSEguide   website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Physics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in  CBSE Class 12 Physics syllabus  and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Physics Extra Questions

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Class 12 Physics Electrostatic Potential and Capacitance Practice Questions

Correct formula for capacitance in terms of area A and distance d, of a parallel plate capacitor in vacuum is

• {tex}\epsilon_0\frac{A}{d^2}{/tex}
• {tex}\epsilon_0\frac{A^2}{d}{/tex}
• {tex}\epsilon_0\frac{d}{A}{/tex}
• {tex}\epsilon_0\frac{A}{d}{/tex}

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V by a battery. The battery is then disconnected and the space between the plates of capacitor C is completely filled with a material of dielectric constant K = 3. The potential difference across the capacitors now becomes

• {tex}{2V \over 5}{/tex}
• {tex}{3V \over 6}{/tex}
• {tex}{V \over 4}{/tex}
• {tex}{3V \over 5}{/tex}
• Decreasing the battery potential
• Increasing the area of overlapping of the plates
• Decreasing the distance between the plates
• Placing a dielectric between the plates
• 1 {tex}\Omega {/tex}

What is the geometrical shape of equipotential surfaces due to a single isolated charge?

Draw equipotential surfaces due to a single point charge.

Two charges 2 µC and -2 µC are placed at points A and B, 5 cm apart. Depict an equipotential surface of the system.

What are the dimensions of capacitance?

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor, but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

Electric charge is distributed uniformly on the surface of a spherical rubber balloon. Show how the value of electric intensity and potential vary

• on the surface,

A {tex}5\mu F{/tex} capacitor is charged by a 100 V supply. The supply is then disconnected and the charged capacitor is connected to another uncharged {tex}3\mu F{/tex} capacitor. How much electrostatic energy of the first capacitor is lost in the process of attaining the steady situation?

• Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
• Two charged spherical conductors of radii R 1 and R 2 when connected by a conducting plate respectively. Find the ratio of their surface charge densities in terms of their radii.
• conductor and
• dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility.
• the force on the charge at the centre of shell and at the point A,
• the electric flux through the shell.

Class – 12 Physics (Electrostatics Potential and Capacitance) Answers

• {tex}\epsilon_0\frac{A}{d}{/tex} Explanation:  The capacitance of a parallel plate capacitor is given by {tex}C = \frac{{{\varepsilon _0}A}}{d}{/tex}
• {tex}{3V \over 5}{/tex} Explanation:  The charges on the capacitors after being charged to a potential V are Q1 = CV; Q2 = 2CV. After being filled with a material of dielectric K = 3 the capacitor which initially had a capacitance C has now the capacitance KC = 3C. The common potential {tex}{V_1} = \frac{{{\text{Total charge}}}}{{{\text{Total capacitance}}}} = \frac{{{Q_1} + {Q_2}}}{{3C + 2C}}{/tex} {tex} = \frac{{CV + 2CV}}{{5C}} = \frac{3}{5}V{/tex}
• Decreasing the battery potential Explanation:  An electroscope is a device which measures the potential difference. If it is connected in parallel to the capacitor, the potential across it will be equal to the potential across the capacitor, which is equal to the potential across the battery. On decreasing the battery potential, the potential difference across the electroscope reduces and hence the reading reduces. While the capacitor is connected to the battery, Placing a dielectric between the plates, or decreasing the distance between the plates or increasing the area of the plates will not change the potential difference across it; since it will always remain equal to the potential difference maintained by the battery. In the cases B, C and D, The capacitance of the capacitor, however increases; but this increase happens due to increase in the charge stored in the capacitor while the potential remains constant.
• infinite Explanation: Capacitor does not allow DC to pass through it. The effective capacitance or the capacitive reactance {tex}{X_C} = \frac{1}{{C\omega }}{/tex} , where ω is the frequency of voltage source. Since DC current is a constant current, its frequency is zero. The capacitive reactance is therefore infinity.
• voltage Explanation: Voltage is an electrical potential difference, the difference in electric potential between two places. The unit for electrical potential difference, or voltage, is the volt.
• {tex}\left[ {{M^{ – 1}}{L^{ – 2}}{A^2}{T^4}} \right]{/tex}
• On the surface E = constant, V = constant. Inside the surface, E = 0, V = constant = potential on surface. Outside the balloon. {tex}E \propto \frac{1}{{{r^2}}}{/tex} and {tex}V \propto \frac{1}{r}{/tex} where r is the distance of the point from the centre of the balloon.
• Surface charge density of a spherically charged body is given by {tex}\sigma = \frac { q } { 4 \pi R ^ { 2 } }{/tex} After connecting both the conductors, their potentials will become equal. {tex}\Rightarrow \frac { K q _ { 1 } } { R _ { \mathrm { l } } } = \frac { K q _ { 2 } } { R _ { 2 } }{/tex} [ For a spherically charged conductor with charge q potential is given by, {tex}V = \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { q } { R } \text { or } V = \frac { K q } { R }{/tex} ] {tex}\therefore \frac { q _ { 1 } } { q _ { 2 } } = \frac { R _ { 1 } } { R _ { 2 } } ~Hence~ \frac { \sigma _ { 1 } } { \sigma _ { 2 } } = \frac { q _ { 1 } / 4 \pi R _ { 1 } ^ { 2 } } { q _ { 2 } / 4 \pi R _ { 2 } ^ { 2 } }{/tex} {tex}= \frac { q _ { 1 } } { q _ { 2 } } \left( \frac { R _ { 2 } } { R _ { 1 } } \right) ^ { 2 } = \frac {R_1}{R_2}× \left (\frac {R_2}{R_1} \right)^{2}= \frac { R _ { 2 } } { R _ { 1 } }{/tex}
• When a conductor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static situation is achieved, i.e. when the two fields cancels each other, then the net electrostatic field in the conductor becomes zero.
• At point C, inside the shell. The electric field inside a spherical shell is zero. Thus, the force experienced by the charge Q/2 kept at the centre C of the shell will also be zero. {tex}\because F _ { C } = q E \quad \left( E _ { \text { nside the shell } } = 0\right.{/tex} ) {tex}\therefore F _ { C } = 0{/tex} At point ‘A’, magnitude of the force experienced by charge +2Q there, IF A = 2Q × {tex}\left( \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \frac { 3 Q / 2 } { x ^ { 2 } } \right){/tex} (since 3Q/2 is the net charge of the shell) {tex}\therefore F = \frac { 3 Q ^ { 2 } } { 4 \pi \varepsilon _ { 0 } x ^ { 2 } }{/tex} , direction of this force is away from shell
• Electric flux through the shell, {tex}\phi = \frac { 1 } { \varepsilon _ { 0 } } {/tex} × magnitude of the net charge enclosed by the shell. (In this case there is no effect of the charge which lies above the shell) {tex}\Rightarrow \quad \phi = 1 / \varepsilon _ { 0 } \times Q / 2 = Q / 2 \varepsilon _ { 0 }{/tex}

Chapter Wise Extra Questions of Class 12 Physics Part I & Part II

• Electric Charges and Fields
• Electrostatic Potential and Capacitance
• Current Electricity
• Moving Charges and Magnetism
• Magnetism and Matter
• Electromagnetic Induction
• Alternating Current
• Electromagnetic Waves
• Ray Optics and Optical
• Wave Optics
• Dual Nature of Radiation and Matter
• Electronic Devices

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Physics Case Study for Class 12

While preparing for the board exams, students are being judged on different levels of skills, such as writing, reading, etc. Physics Case Study for Class 12 questions are one of them that helps in assessing critical thinking.

The Central Board of Secondary Education will be asking the case study questions in the Class 12 board examination. Therefore, here on this page, we have provided the Physics Case Study for Class 12 at free of cost. Our subject matter experts have prepared Case Study questions so that Apart from the basic standard questions, students can have a variety of problems to solve. 

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Download Chapter Wise Physics Case Study for Class 12 Question and Answers PDF 

There are lots of chapters in class 12 Physics from which Class 12 Case Study Questions can be framed. Going through such types of questions help the students to assess their understanding level in the topics discussed in NCERT Class 12 Physics Books. By practicing the Class 12 Case Study Questions for Physics students will be very confident to ace the board exam. Also the Physics case study will be very useful for the NEET exam preparation.

Doing a regular practice of Class 12 Physics Case Study questions is a great way to score higher marks in the board exams as it will help students to develop a grip on the concepts. 

Here our subject experts have crafted the Chapter Wise Case Study For Class 12. Download Chapter Wise CBSE Case Study Class 12 Question and Answers PDF from the below given links.

Case Study Questions Class 12 Chapter 1 Electric Charges and Fields

Case Study Questions Class 12 Chapter 2 Electrostatic Potential And Capacitance

Case Study Questions Class 12 Chapter 3 Current Electricity

Case Study Questions Class 12 Chapter 4 Moving Charges And Magnetism

Case Study Questions Class 12 Chapter 5 Magnetism And Matter

Case Study Questions Class 12 Chapter 6 Electromagnetic Induction

Case Study Questions Class 12 Chapter 7 Alternating Current

Case Study Questions Class 12 Chapter 8 Electromagnetic Waves

Case Study Questions Class 12 Chapter 9 Ray Optics & Optical Instruments

Case Study Questions Class 12 Chapter 10 Wave Optics

Case Study Questions Class 12 Chapter 11 Dual Nature Radiation & Matter

Case Study Questions Class 12 Chapter 12 Atoms

Case Study Questions Class 12 Chapter 13 Nuclei

Case Study Questions Class 12 Chapter 14 Semiconductor Electronics - Materials, Devices & Simple Circuits

Case study types of questions are generally descriptive that helps to gather more information easily so, it is kinda easy to answer. However, our subject matter experts have given the solutions of all the Physics Case Study for Class 12 Physics questions.

Passage Based Class 12 Physics Case Study Questions in PDF

CBSE Class 12 Case studies are known as Passage Based Questions. These types of problems usually contain a short/long paragraph with 4 to 5 questions. 

Students can easily solve Passage Based Class 12 Case Study Questions by reading those passages. By reading the passage students will get the exact idea of what should be the answers. Because the passage already contains some vital information or data. However a better understanding of the basic concepts that can be learned from the NCERT Class 12 Textbooks will aid in solving the Case based questions or passage based questions.

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• Go to the navigation menu that look like this
• Now, click on CBSE and then Case Study respectively
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• Read each line of paragraph carefully and pay attention to the given data/numbers. Often questions are framed according to the highlighted data of the passage.
• Since case study questions are often framed in Multiple choice questions, students should have the knowledge of elimination methods in MCQs.
• Having a good understanding of the topics that are discussed in CBSE Class 12 Books are ideal to Solve Case Study Based Questions of Class 12.

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• In Exam Preparation:  Those who go through the Physics Case Study for Class 12 will find support during the exam preparation as case based questions are also asked in the CBSE Class 12 Board examination. 
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