NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance on Extramarks is the perfect guide for students to prepare for examinations. Physics is an essential part of various competitive examinations like NEET, JEE or even school exams. And hence, to help students with their exam preparations, our subject matter experts at Extramarks have prepared easy and self-explanatory Class 12 Physics Chapter 2 NCERT Solutions. Students can study Physics Chapter 12 by referring to these study materials available for free.

Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 2

The key topics covered under NCERT Solutions Class 12 Physics Chapter 2 are as follows.

3.1	Introduction
3.2	Electrostatic Potential
3.3	Potential Due to a Point Charge
3.4	Potential Due to an Electric Dipole
3.5	Potential Due to a System of Charges
3.6	Equipotential Surfaces
3.6.1	Relation between field and potential
3.7	Potential Energy of a System of Charges
3.8	Potential Energy in an External Field
3.8.1	Potential energy of a single charge
3.8.2	Potential energy of a system of two charges in an external field
3.8.3	Potential energy of a dipole in an external field
3.9	Electrostatics of Conductors
3.10	Dielectrics and Polarisation
3.11	Capacitors and Capacitance
3.12	The Parallel Plate Capacitor
3.13	Effect of Dielectric on Capacitance
3.14	Combination of Capacitors
3.14.1	Capacitors in series
3.14.2	Capacitors in parallel
3.15	Energy Stored in a Capacitor
3.16	Van De Graaff Generator

Students may click on the respective topics to access the NCERT Solutions Class 12 Physics Chapter 2 offered by Extramarks. Here’s a detailed explanation of all subtopics under NCERT Solutions Class 12 Physics Chapter 2.

3.1 Introduction

The electric potential (also called electrostatic potential, electric field potential, potential drop) is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field.

SI unit: volt

In SI base units: V = kg⋅m2⋅s−3⋅A−1

Common symbols: V, φ

3.2 Electrostatic Potential

The electrostatic potential is the pressure that is outside, however conservative. It is the paintings executed via outside force in bringing a price s from a factor r to a degree p. The distinction withinside the capacity strength of prices among preliminary and last points is distinguished.

3.3 Potential Due To A Point Charge

This section of NCERT Solutions Class 12 Physics Chapter 2 describes the charge point. The electric potential of a factor price is V=kQ/r V = k Q / r. Electric potential is a scalar, and the electric subject is a vector. The addition of voltages as numbers offers the voltage because of a mixture of factor charges, while the addition of individual fields as vectors offers the overall electric subject.

3.4 Potential Due To An Electric Dipole

The potential because of a dipole relies upon r (distance among the factor wherein capacity is calculated and the mid-factor of the dipole) and perspective among role vector r and dipole second p. Dipole potential is inversely proportional to the square of r.

3.5 Potential Due To A System Of Charges

Under this section of NCERT Solutions Class 12 Physics Chapter 2, students will learn about the system of charges. For a system of point charges, the overall potential at a point is given through the algebraic sum of the potential for individual fees.

3.6 Equipotential Surfaces

The floor that’s the locus of all points on identical ability is called the equipotential floor. No work is needed to transport a charge from one factor to another on the equipotential floor.

3.6.1 Relation Between Field And Potential

This section of NCERT Solutions Class 12 Physics Chapter 2 describes the relationship between potential and area (E) as differential: the electric powered area is the potential (V) gradient in the x-direction. This may be represented as: Ex=−DV dx E x = − dV dx . Because the test charge is moved in the x-direction, the rate of its potential change is the electrical area’s value.

3.7 Potential Energy Of A System Of Charges

The electric potential energy of a system of factor charges is described as the work required to assemble this system of charges via bringing them near together, as withinside the system from a countless distance. To know more, students may refer to NCERT Solutions Class 12 Physics Chapter 2.

3.8 Potential Energy In An External Field

The electric potential energy of a system of point charges is described as the work required to collect this device of charges with the aid of bringing them together within the system from an infinite distance. The potential energy of the charge q withinside the field is identical to the work achieved in bringing the charge from infinity to the point. Here we word that the outside electric field E and the corresponding potential energy of the system range from point to point withinside the subject.

3.8.1 Potential Energy Of A Single Charge

The potential energy of the charge q in the field is equal to the work done in bringing the charge from infinity to the point. To know more, students may refer to Extramarks NCERT Solutions Class 12 Physics Chapter 2.

3.8.2 Potential Energy Of A System Of Two Charges In An External Field

The potential energy of the charge q withinside the area is the same as the work achieved in bringing the charge from infinity to the point. Here we notice that the outside electric field E and the corresponding potential energy of the system range from point to point withinside the area.

3.8.3 Potential Energy Of A Dipole In An External Field

τ = p × E. This work is stored because of the system’s potential energy. The potential energy U(θ) can then be related to the dipole’s inclination θ. To know more, students may refer to Extramarks NCERT Solutions Class 12 Physics Chapter 2.

3.9 Electrostatics Of Conductors

Conductors comprise mobile charge carriers. Inside a conductor, the electrostatic field is 0. The free charges have allotted themselves in the static situation that the electrical field is 0 anywhere inside. The electrostatic field is 0 inside a conductor.

3.10 Dielectrics And Polarization

The molecule acquires a dipole moment when a dielectric slab is positioned in an electric field. In such circumstances, the dielectric is stated to be polarized. The Electric Polarization of a dielectric substance is the dipole moment per unit volume. P is the symbol for polarization.

3.11 Capacitors And Capacitance

Capacitance is the electric property of a capacitor. It is the degree of a capacitor’s ability to save an electrical charge onto its plates. The unit of capacitance is the Farad (abbreviated to F), named after the British physicist Michael Faraday.

3.12 The Parallel Plate Capacitor

This section of NCERT Solutions Class 12 Physics Chapter 2 describes The Parallel Plate Capacitor. A parallel plate capacitor can store a finite quantity of energy earlier than dielectric breakdown. When parallel plates are connected throughout a battery, the plates are charged, and an electric field is mounted among them; this setup is called the parallel plate capacitor.

3.13 Effect Of Dielectric On Capacitance

The energy of the electrical field is decreased because of the presence of dielectric. If the total charge at the plates is stored constant, then the potential difference is reduced throughout the capacitor plates. In this way, the dielectric will increase the capacitance of the capacitor.

3.14 Combination Of Capacitors

In series combinations of capacitors, all capacitors could have identical charges. i.e., the resultant capacitance of series combination C=Q/V is the ratio of charge to total ability difference throughout the two capacitors connected in series.

3.14.1 Capacitors In Series

The overall voltage difference from end to end is apportioned to every capacitor in step with the inverse of its capacitance. The whole series acts as a capacitor smaller than any of its components. To know more, students may refer to NCERT Solutions Class 12 Physics Chapter 2.14.1.

3.14.2 Capacitors In Parallel

When capacitors are linked in parallel, the total capacitance is the sum of the individual capacitors’ capacitances. If or greater capacitors are connected in parallel, the general impact is that of a single equal capacitor having the sum general of the plate areas of the individual capacitors.

3.15 Energy Stored In A Capacitor

The electricity you see saved in a capacitor is electrostatic potential energy associated with the charge Q and voltage V among the plates of capacitor A charged capacitor stores energy withinside the electric field among its plates. As the capacitor gets charged, the electric field builds up.

3.16 Van De Graaff Generator

An electrostatic generator or Van de Graaff generator uses a transferring belt to build up electric charge on a hollow metallic globe at the top of an insulated column, developing excessive electric potentials. It produces excessive voltage direct current (DC) electricity at low current levels.

Students may refer to NCERT Solutions Class 12 Physics Chapter 2 provided by Extramarks to get detailed study material on Electrostatic Potential and Capacitance.

NCERT Solutions Class 12 Physics Chapter 2: Article link for Electrostatic Potential and Capacitance NCERT solutions

Click on the below links to view NCERT Solutions Class 12 Physics Chapter 2:

Class 12 Physics Chapter 2: Exercise 3.1
Class 12 Physics Chapter 2: Exercise 3.2
Class 12 Physics Chapter 2: Exercise 3.3
Class 12 Physics Chapter 2: Exercise 3.4
Class 12 Physics Chapter 2: Exercise 3.5
Class 12 Physics Chapter 2: Exercise 3.6
Class 12 Physics Chapter 2: Exercise 3.7
Class 12 Physics Chapter 2: Exercise 3.8
Class 12 Physics Chapter 2: Exercise 3.9
Class 12 Physics Chapter 2: Exercise 3.10
Class 12 Physics Chapter 2: Exercise 3.11
Class 12 Physics Chapter 2: Exercise 3.12
Class 12 Physics Chapter 2: Exercise 3.13
Class 12 Physics Chapter 2: Exercise 3.14
Class 12 Physics Chapter 2: Exercise 3.15
Class 12 Physics Chapter 2: Exercise 3.16

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Key Features of NCERT Solutions Class 12 Physics Chapter 2

NCERT Solutions Class 12 Physics Chapter 2 gives you detailed information about electrostatic potential and capacitance. Students will learn the basic concepts about capacitors, insulators, and how the potential energy works. Going through NCERT Solutions Class 12 Physics Chapter 2 will help students understand the topic in a better way.

The key features of NCERT Solutions Class 12 Physics Chapter 2 provided by Extramarks are

The Solutions are compiled by some of the highly qualified subject matter experts
Every minute detail of the chapter is included to ensure that students understand the chapter.
NCERT Solutions Class 12 Physics Chapter 2 by Extramarks can be used as revision notes during the examination.
They also contain important questions, sample questions, past years’ question papers for the students to refer to.

Q.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.

Ans-

$\begin{matrix} Here, q_{1} = 5 \times 10^{- 8} C q_{2} = 3 \times 10^{- 8} C Distance between the two charges, d = 16 cm = 0 . 16 m Letpotentialbezero at adistancescmfromcharge q_{1} = 5 \times 10^{- 8} C ∴ r_{1} = s \times 10^{- 2} m; r_{2} = (16 - s) \times 10^{- 2} m Let electric potentialatdistances = 0 Electric potential at a point is the sum of potentials caused by charges q_{1} and q_{2} respectively . ∴ V = \frac{q_{1}}{4 {πε}_{0} r} + \frac{q_{2}}{4 {πε}_{0} (d - r)} \to (i) Where, ε_{0} = Permittivity of free space For V = 0, equation (i) becomes 0 = \frac{q_{1}}{4 {πε}_{0} r} + \frac{q_{2}}{4 {πε}_{0} (d - r)} \frac{q_{1}}{4 {πε}_{0} r} = - \frac{q_{2}}{4 {πε}_{0} (d - r)} \frac{q_{1}}{r} = - \frac{q_{2}}{(d - r)} \frac{5 \times 10^{- 8}}{r} = - \frac{(- 3 \times 10^{- 8})}{(0 .16 - r)} \frac{0 .16}{r} - \frac{0 .16}{r} = \frac{8}{5} ∴ r = 0 .1 m = 10 cm ∴ The potential is zero at a distance of 1 0 cm from the {positive charge q}_{1} If the point lies outside the charges, {\frac{1}{4 {πε}_{°}}} [\frac{5 \times 10^{- 8}}{r} - \frac{3 \times 10^{- 8}}{(r - 0 .16)}] = 0 \Rightarrow r = 40 cm ∴ r = 40 cm from positive charge towards negative charge on extended line. \end{matrix}$ $\begin{array}{l} Here, q_{1} = 5 \times 10^{- 8} C \\ q_{2} = 3 \times 10^{- 8} C \\ Distance between the two charges, d = 16 cm \\ = 0 . 16 m \\ Let potential be zero at a distance s cm from charge q_{1} \\ = 5 \times 10^{- 8} C \\ ∴ r_{1} = s \times 10^{- 2} m; \\ r_{2} = (16 - s) \times 10^{- 2} m \\ Let electric potential at distance s = 0 \\ Electric potential at a point is the sum of potentials \\ caused by charges q_{1} and q_{2} respectively . \\ ∴ V = \frac{q_{1}}{4 {πε}_{0} r} + \frac{q_{2}}{4 {πε}_{0} (d - r)} \to (i) \\ Where, ε_{0} = Permittivity of free space \\ For V = 0, equation (i) becomes \\ 0 = \frac{q_{1}}{4 {πε}_{0} r} + \frac{q_{2}}{4 {πε}_{0} (d - r)} \\ \frac{q_{1}}{4 {πε}_{0} r} = - \frac{q_{2}}{4 {πε}_{0} (d - r)} \\ \frac{q_{1}}{r} = - \frac{q_{2}}{(d - r)} \\ \frac{5 \times 10^{- 8}}{r} = - \frac{(- 3 \times 10^{- 8})}{(0 .16 - r)} \\ \frac{0 .16}{r} - \frac{0 .16}{r} = \frac{8}{5} \\ ∴ r = 0 .1 m \\ = 10 cm \\ ∴ The potential is zero at a distance of 1 0 cm from the \\ {positive charge q}_{1} \\ If the point lies outside the charges, \\ {\frac{1}{4 {πε}_{°}}} [\frac{5 \times 10^{- 8}}{r} - \frac{3 \times 10^{- 8}}{(r - 0 .16)}] = 0 \\ \Rightarrow r = 40 cm \\ ∴ r = 40 cm from positive charge towards negative \\ charge on extended line. \end{array}$

Q.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.

Ans-

$\begin{matrix} Consider a regular hexagon ABCDEF, having an equal amount of charge q at each of its vertices . {Here, charge, q = 5 μC = 5\times10}^{-6} C Side of hexagon, l = 10 cm Distance of each vertex from the centre O, d = 10 cm = 0 .1 m As electric potential is scalar, ∴ Electric potential at O is given as V = \frac{6q}{4 π ε_{0} d} Here, ε_{0} = Permittivity of free space \frac{1}{4 π ε_{0}} {= 9\times10}^{9} {NC}^{-2} m^{-2} ∴ V = \frac{{6\times9\times10}^{9} {NC}^{-2} m^{-2} {\times5\times10}^{-6} C}{0 .1 m} = 2 {.7\times10}^{6} V ∴ The potential at the centre of the hexagon is 2 {.7\times10}^{6} V . \end{matrix}$ $\begin{array}{l} Consider a regular hexagon ABCDEF, having an equal amount of charge q \\ at each of its vertices . \\ {Here, charge, q = 5 μC = 5×10}^{-6} C \\ Side of hexagon, l = 10 cm \\ Distance of each vertex from the centre O, d = 10 cm = 0 .1 m \\ As electric potential is scalar, \\ ∴ Electric potential at O is given as \\ V = \frac{6q}{4 π ε_{0} d} \\ Here, ε_{0} = Permittivity of free space \\ \frac{1}{4 π ε_{0}} {= 9×10}^{9} {NC}^{-2} m^{-2} \\ ∴ V = \frac{{6×9×10}^{9} {NC}^{-2} m^{-2} {×5×10}^{-6} C}{0 .1 m} = 2 {.7×10}^{6} V \\ ∴ The potential at the centre of the hexagon is 2 {.7×10}^{6} V . \end{array}$

Q.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?

Ans-

(a) The plane normal to AB and passing through its middle point has zero potential everywhere and it represents an equipotential surface of the given system.
(b) The direction of the electric field at every point on this surface is along normal to the plane in the direction of AB.

Q.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C 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?

Ans-

$\begin{matrix} (a) Here,Radius of the spherical conductor, r = 12 cm = 0 . 12 m Charge over thespherical conductor, q = 1.6 \times 1 0^{- 7} C If there is field inside the conductor, then charges will move to neutralize it . ∴ Electric field inside the spherical conductor =0 (b) Electric field just outside the sphericalconductor is given as : E = \frac{q}{4 {πε}_{0} r^{2}} Here, ε_{0} = Permittivity of free space \frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2} ∴ E = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0.12)}^{2}} = 10^{5} {NC}^{- 1} ∴ The electric field just outside the sphere is 10^{5} {NC}^{- 1} . (c) LetElectric field at a point 18 m from the centre of the sphere = E_{1} Distance of thepoint from the centreofsphere, d = 18 cm = 0 . 18 m E_{1} = \frac{q}{4 {πε}_{0} r^{2}} = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0 .18)}^{2}} = 4 .4 \times 10^{4} {NC}^{- 1} ∴ The electric field at a point 18 cm from the centre of the sphere is 4 .4 \times 10^{4} {NC}^{- 1} . \end{matrix}$ $\begin{array}{l} (a) Here, Radius of the spherical conductor, r = 12 cm \\ = 0 . 12 m \\ Charge over the spherical conductor, q = 1.6 \times 1 0^{- 7} C \\ If there is field inside the conductor, then charges \\ will move to neutralize it . \\ ∴ Electric field inside the spherical conductor =0 \\ (b) Electric field just outside the spherical conductor is \\ given as : \\ E = \frac{q}{4 {πε}_{0} r^{2}} \\ Here, ε_{0} = Permittivity of free space \\ \frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2} \\ ∴ E = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0.12)}^{2}} \\ = 10^{5} {NC}^{- 1} \\ ∴ The electric field just outside the sphere is 10^{5} {NC}^{- 1} . \\ (c) Let Electric field at a point 18 m from the centre of \\ the sphere = E_{1} \\ Distance of the point from the centre of sphere, d \\ = 18 cm \\ = 0 . 18 m \\ E_{1} = \frac{q}{4 {πε}_{0} r^{2}} \\ = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0 .18)}^{2}} \\ = 4 .4 \times 10^{4} {NC}^{- 1} \\ ∴ The electric field at a point 18 cm from the centre of \\ the sphere is 4 .4 \times 10^{4} {NC}^{- 1} . \end{array}$

Q.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?

Ans-

$\begin{matrix} As initially air was filled betwen the plates of parallel plate capacitor, Dielectric constant of air, K = 1 Initial capacitance of the parallel plate capacitor = 8 pF Let area of each plate of capacitor = A Let initial distance between the plates of the parallel plate capacitor = d Capacitance of the parallel plate capacitor is given as : C_{1} = \frac{ε_{0} A}{d} = 8 pF \to (i) When distance between the plates is reduced to half, then new distance, d ‘ = \frac{d}{2} Dielectric constant of substance filled between the plates, K ‘ = 6 ∴ Capacitance of the parallel plate capacitor becomes, C_{2} = K ‘ \frac{ε_{0} A}{d ‘} = \frac{6 ε_{0} A}{\frac{d}{2}} = \frac{6 \times 2 \times ε_{0} A}{d} = 12 \times \frac{ε_{0} A}{d} = 12 \times 8 [Using equation (i)] = 96 pF ∴ The capacitance between the plates is 96 pF . \end{matrix}$ $\begin{array}{l} As initially air was filled betwen the plates of parallel \\ plate capacitor, \\ Dielectric constant of air, K = 1 \\ Initial capacitance of the parallel plate capacitor \\ = 8 pF \\ Let area of each plate of capacitor = A \\ Let initial distance between the plates of the parallel \\ plate capacitor = d \\ Capacitance of the parallel plate capacitor is given as : \\ C_{1} = \frac{ε_{0} A}{d} \\ = 8 pF \to (i) \\ When distance between the plates is reduced to half, \\ then new distance, d ‘ = \frac{d}{2} \\ Dielectric constant of substance filled between the \\ plates, K ‘ = 6 \\ ∴ Capacitance of the parallel plate capacitor becomes, \\ C_{2} = K ‘ \frac{ε_{0} A}{d ‘} \\ = \frac{6 ε_{0} A}{\frac{d}{2}} \\ = \frac{6 \times 2 \times ε_{0} A}{d} \\ = 12 \times \frac{ε_{0} A}{d} \\ = 12 \times 8 [Using equation (i)] \\ = 96 pF \\ ∴ The capacitance between the plates is 96 pF . \end{array}$

Q.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?

Ans-

$\begin{matrix} (a) Capacitance of each capacitor, C = 9 pF Since thethreecapacitorsareconnectedinseries, ∴ Equivalent capacitance of thecombination of capacitors is given by the relation, \frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} \frac{1}{C_{s}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} ∴ C_{s} = 3 pF ∴ Total capacitance of the combination is 3 pF . (b) Here, s upply voltage, V = 120 V Sincethe three capacitorsareconnectedinseries, ∴ The p otential difference (V ’) across each capacitor is equalto one - third of the supply voltage . ∴ V ’ = \frac{V}{3} = \frac{120}{3} = 40 V ∴ The potential difference across each capacitor is 40 V . \end{matrix}$ $\begin{array}{l} (a) Capacitance of each capacitor, C = 9 pF \\ Since the three capacitors are connected in series, \\ ∴ Equivalent capacitance of the combination of \\ capacitors is given by the relation, \\ \frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} \\ \frac{1}{C_{s}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \\ = \frac{1}{3} \\ ∴ C_{s} = 3 pF \\ ∴ Total capacitance of the combination is 3 pF . \\ (b) Here, s upply voltage, V = 120 V \\ Since the three capacitors are connected in series, \\ ∴ The p otential difference (V ’) across each capacitor is \\ equal to one - third of the supply voltage . \\ ∴ V ’ = \frac{V}{3} \\ = \frac{120}{3} \\ = 40 V \\ ∴ The potential difference across each capacitor is \\ 40 V . \end{array}$

Q.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Ans-

$\begin{matrix} (a) Given, C_{1} = 2 pF C_{2} = 3 pF C_{3} = 4 pF For the parallel combination of the capacitors, the {totalcapacitanceofthecombination,C}_{p} = C_{1} + C_{2} + C_{3} = 2 + 3 + 4 = 9 pF ∴ Total capacitance of the combination is 9 pF . (b) Here, s upply voltage, V = 100 V The voltage througheachof the three capacitors is same, V = 100 V Charge on the capacitor of capacitance C and potential difference V is given by therelation, q = VC \dots (i) For C = 2 pF, Charge, q_{1} = VC =100 \times 2 =200pF =2 \times 10^{- 10} C For C = 3 pF, Charge, q_{2} = VC =100 \times 3 =300pF =3 \times 10^{- 10} C For C = 4 pF, Charge, q_{3} = VC =100 \times 2 =200pF =2 \times 10^{- 10} C \end{matrix}$ $\begin{array}{l} (a) Given, C_{1} = 2 pF \\ C_{2} = 3 pF \\ C_{3} = 4 pF \\ For the parallel combination of the capacitors, the \\ {total capacitance of the combination, C}_{p} = C_{1} + C_{2} + C_{3} \\ = 2 + 3 + 4 = 9 pF \\ ∴ Total capacitance of the combination is 9 pF . \\ (b) Here, s upply voltage, V = 100 V \\ The voltage through each of the three capacitors is \\ same, V = 100 V \\ Charge on the capacitor of capacitance C and potential \\ difference V is given by the relation, \\ q = VC \dots (i) \\ For C = 2 pF, \\ Charge, q_{1} = VC \\ =100 \times 2 \\ =200 pF \\ =2 \times 10^{- 10} C \\ For C = 3 pF, \\ Charge, q_{2} = VC \\ =100 \times 3 \\ =300 pF \\ =3 \times 10^{- 10} C \\ For C = 4 pF, \\ Charge, q_{3} = VC \\ =100 \times 2 \\ =200 pF \\ =2 \times 10^{- 10} C \end{array}$

Q.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?

Ans-

$\begin{matrix} Here, a rea of each plate of the parallel plate capacitor, A = 6 \times 1 0^{- 3} m^{2} Separation between the plates, d = 3 mm = 3 \times 1 0^{- 3} m Supply voltage, V = 100 V Capacitance of a parallel plate capacitor is given by therelation, C= \frac{ε_{0} A}{d} Here, ε_{0} = Permittivity of free space = 8.854 \times 1 0^{- 12} N^{- 1} m^{- 2} C^{- 2} ∴ C= \frac{8.854 \times 1 0^{- 12} \times 6 \times 1 0^{- 3}}{3 \times 1 0^{- 3}} = 17 .71 \times 1 0^{- 12} F = 17 .71 pF Potential is related to the charge q and capacitance C as : V = \frac{q}{C} ∴ q = VC = 100 \times 17 .71 \times 1 0^{- 12} = 1 .771 \times 1 0^{- 9} C ∴ Capacitance of the capacitor is 17 . 71 pF and charge on each plate is 1 . 771 \times 1 0^{- 9} C . \end{matrix}$ $\begin{array}{l} Here, a rea of each plate of the parallel plate capacitor, \\ A = 6 \times 1 0^{- 3} m^{2} \\ Separation between the plates, d = 3 mm \\ = 3 \times 1 0^{- 3} m \\ Supply voltage, V = 100 V \\ Capacitance of a parallel plate capacitor is given by \\ the relation, \\ C= \frac{ε_{0} A}{d} \\ Here, ε_{0} = Permittivity of free space \\ = 8.854 \times 1 0^{- 12} N^{- 1} m^{- 2} C^{- 2} \\ ∴ C= \frac{8.854 \times 1 0^{- 12} \times 6 \times 1 0^{- 3}}{3 \times 1 0^{- 3}} \\ = 17 .71 \times 1 0^{- 12} F \\ = 17 .71 pF \\ Potential is related to the charge q and capacitance C \\ as : V = \frac{q}{C} \\ ∴ q = VC \\ = 100 \times 17 .71 \times 1 0^{- 12} \\ = 1 .771 \times 1 0^{- 9} C \\ ∴ Capacitance of the capacitor is 17 . 71 pF and charge \\ on each plate is 1 . 771 \times 1 0^{- 9} C . \end{array}$

Q.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.

Ans-

$\begin{matrix} (a) While the voltage supply remained connected, potential across the plates remains 1 00 V . Here, d ielectric constant of the mica sheet, k = 6 Initial capacitance, C = 1.771 \times 1 0^{- 11} F New capacitance is given by, C’=kC C’=6 \times 1.771 \times 1 0^{- 11} = 106 p F Supply voltage, V = 100 V New charge is given b y, q’=C’V = 106 \times 1 0^{- 12} \times 100 = 1.06 \times 1 0^{- 8} (b) After the supply was disconnected, t h e a m o u n t o f charge remains constant . Here, d ielectric constant, k = 6 Initial capacitance, C = 1.771 \times 1 0^{- 11} F New capacitance is given as: C’=kC =6 \times 1.771 \times 1 0^{- 11} = 106 p F Charge, q = 1.771 \times 1 0^{- 9} C Potential across the plates is given by the relation: V’= \frac{q}{C ‘} = \frac{1.771 \times 1 0^{- 9}}{106 \times 1 0^{- 12}} = 16.7 V \end{matrix}$ $\begin{array}{l} (a) While the voltage supply remained connected, \\ potential across the plates remains 1 00 V . \\ Here, d ielectric constant of the mica sheet, k = 6 \\ Initial capacitance, C = 1.771 \times 1 0^{- 11} F \\ New capacitance is given by, \\ C’=kC \\ C’=6 \times 1.771 \times 1 0^{- 11} = 106 p F \\ Supply voltage, V = 100 V \\ New charge is given b y, \\ q’=C’V \\ = 106 \times 1 0^{- 12} \times 100 \\ = 1.06 \times 1 0^{- 8} \\ (b) After the supply was disconnected, t h e a m o u n t o f \\ charge remains constant . \\ Here, d ielectric constant, k = 6 \\ Initial capacitance, C = 1.771 \times 1 0^{- 11} F \\ New capacitance is given as: C’=kC \\ =6 \times 1.771 \times 1 0^{- 11} \\ = 106 p F \\ Charge, q = 1.771 \times 1 0^{- 9} C \\ Potential across the plates is given by the relation: \\ V’= \frac{q}{C ‘} \\ = \frac{1.771 \times 1 0^{- 9}}{106 \times 1 0^{- 12}} \\ = 16.7 V \end{array}$

Q.10 A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Ans-

$\begin{matrix} Here, c apacitor of the capacitance, C = 12 pF = 12 \times 1 0^{- 12} F Potential difference, V = 50 V Electrostatic energy stored in the capacitor is given as: E= \frac{1}{2} C V^{2} = \frac{1}{2} \times 12 \times 1 0^{- 12} \times {(50)}^{2} = 1.5 \times 10^{- 8} J ∴ The electrostatic energy stored in the capacitor is 1.5 \times 10^{- 8} J . \end{matrix}$ $\begin{array}{l} Here, c apacitor of the capacitance, C = 12 pF \\ = 12 \times 1 0^{- 12} F \\ Potential difference, V = 50 V \\ Electrostatic energy stored in the capacitor is given as: \\ E= \frac{1}{2} C V^{2} \\ = \frac{1}{2} \times 12 \times 1 0^{- 12} \times {(50)}^{2} \\ = 1.5 \times 10^{- 8} J \\ ∴ The electrostatic energy stored in the capacitor is \\ 1.5 \times 10^{- 8} J . \end{array}$

Q.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?

Ans-

$\begin{matrix} Here, c apacitance of the capacitor, C = 600 pF S u p p l y v o l t a g e, V = 200 V Electrostatic energy stored in the capacitor is given by the relation, E = \frac{1}{2} C V^{2} = \frac{(600 \times 10^{- 12}) \times {(200)}^{2}}{2} = 1.2 \times 10^{- 5} J When supply voltage is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C ’) of the combination is given a s : \frac{1}{C ‘} = \frac{1}{C} + \frac{1}{C} = \frac{1}{600} + \frac{1}{600} = \frac{1}{300} ∴ C ’ = 300 pF New electrostatic energy can be obtained as: E’= \frac{1}{2} \times C ’ \times V^{2} = \frac{1}{2} \times 300 \times {(200)}^{2} = 0.6 \times 10^{- 5} J L o s s o f e l ectrostatic energy = E - E ‘ = 1.2 \times 10^{- 5} - 0.6 \times 10^{- 5} = 0.6 \times 10^{- 5} = 6 \times 10^{- 6} J ∴ The electrostatic energy lost in the process is 6 \times 10^{- 6} J . \end{matrix}$ $\begin{array}{l} Here, c apacitance of the capacitor, C = 600 pF \\ S u p p l y v o l t a g e, V = 200 V \\ Electrostatic energy stored in the capacitor is given by \\ the relation, \\ E = \frac{1}{2} C V^{2} \\ = \frac{(600 \times 10^{- 12}) \times {(200)}^{2}}{2} \\ = 1.2 \times 10^{- 5} J \\ When supply voltage is disconnected from the \\ capacitor and another capacitor of capacitance C = \\ 600 pF is connected to it, then equivalent capacitance \\ (C ’) of the combination is given a s : \\ \frac{1}{C ‘} = \frac{1}{C} + \frac{1}{C} \\ = \frac{1}{600} + \frac{1}{600} \\ = \frac{1}{300} \\ ∴ C ’ = 300 pF \\ New electrostatic energy can be obtained as: \\ E’= \frac{1}{2} \times C ’ \times V^{2} \\ = \frac{1}{2} \times 300 \times {(200)}^{2} \\ = 0.6 \times 10^{- 5} J \\ L o s s o f e l ectrostatic energy = E - E ‘ \\ = 1.2 \times 10^{- 5} - 0.6 \times 10^{- 5} \\ = 0.6 \times 10^{- 5} \\ = 6 \times 10^{- 6} J \\ ∴ The electrostatic energy lost in the process is \\ 6 \times 10^{- 6} J . \end{array}$

Q.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).

Ans-

$\begin{matrix} Here, c harge located at origin, q = 8 mC = 8 \times 1 0^{- 3} C Magnitude of the charge, which is to be carried from point P to point Q, q_{1} = - 2 \times 1 0^{- 9} C All the points are shown in the given figure . Distance of p oint P, d_{1} = 3 cm, from the origin along z - axis . Distance of p oint Q, d_{2} = 4 cm, from the origin along y - axis . Potential at point P is given as: V_{1} = \frac{q}{4 π ε_{0} \times d_{1}} Potential at point Q is given as: V_{2} = \frac{q}{4 π ε_{0} \times d_{2}} Since w ork done (W) by the electrostatic force is independent of the path, ∴ W = q_{1} [V_{2} - V_{1}] = q_{1} [\frac{q}{4 π ε_{0} \times d_{2}} - \frac{q}{4 π ε_{0} \times d_{1}}] = \frac{q_{1} q}{4 π ε_{0}} [\frac{1}{d_{2}} - \frac{1}{d_{1}}] \to (i) H e r e, \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2} ∴ W = 9 \times 10^{9} \times 8 \times 1 0^{- 3} \times (- 2 \times 1 0^{- 9}) [\frac{1}{0.04} - \frac{1}{0.03}] = - 144 \times 1 0^{- 3} \times (\frac{- 25}{3}) = 1.27 J ∴ W ork done during the process is 1 . 27 J . \end{matrix}$ $\begin{array}{l} Here, c harge located at origin, q = 8 mC \\ = 8 \times 1 0^{- 3} C \\ Magnitude of the charge, which is to be carried from \\ point P to point Q, q_{1} = - 2 \times 1 0^{- 9} C \\ All the points are shown in the given figure . \\ Distance of p oint P, d_{1} = 3 cm, \\ from the origin along z - axis . \\ Distance of p oint Q, d_{2} = 4 cm, \\ from the origin along y - axis . \\ Potential at point P is given as: V_{1} = \frac{q}{4 π ε_{0} \times d_{1}} \\ Potential at point Q is given as: V_{2} = \frac{q}{4 π ε_{0} \times d_{2}} \\ Since w ork done (W) by the electrostatic force is \\ independent of the path, \\ ∴ W = q_{1} [V_{2} - V_{1}] \\ = q_{1} [\frac{q}{4 π ε_{0} \times d_{2}} - \frac{q}{4 π ε_{0} \times d_{1}}] \\ = \frac{q_{1} q}{4 π ε_{0}} [\frac{1}{d_{2}} - \frac{1}{d_{1}}] \to (i) \\ H e r e, \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2} \\ ∴ W = 9 \times 10^{9} \times 8 \times 1 0^{- 3} \times (- 2 \times 1 0^{- 9}) [\frac{1}{0.04} - \frac{1}{0.03}] \\ = - 144 \times 1 0^{- 3} \times (\frac{- 25}{3}) \\ = 1.27 J \\ ∴ W ork done during the process is 1 . 27 J . \end{array}$

Q.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.

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$\begin{matrix} Here, side of the cube = b Charge at each vertex of t h e cube = q A cube of side b is shown in the given figure . Here, d = Diagonal of t h e cube of e a c h side b d = \sqrt{b^{2} + b^{2}} = \sqrt{2 b^{2}} = \sqrt{2} b ∴ d = b \sqrt{2} l = Length of the diagonal of the cube l = \sqrt{d^{2} + b^{2}} = \sqrt{{(\sqrt{2} b)}^{2} + b^{2}} = \sqrt{2 b^{2} + b^{2}} = \sqrt{3 b^{2}} = \sqrt{3} b ∴ D i s tan c e b e t w e e n t h e c e n t r e o f c u b e a n d e a c h v e r t e x, r = \frac{l}{2} = \frac{\sqrt{3} b}{2} Electric potential (V) at the centre of the cube is because of the presence of eight charges at the vertices . ∴ V = \frac{8 q}{4 π ε_{0} r} = \frac{8 q}{4 π ε_{0} (\frac{\sqrt{3} b}{2})} = \frac{4 q}{\sqrt{3} π ε_{0} b} ∴ The potential at the centre of the cube is \frac{4 q}{\sqrt{3} π ε_{0} b} . The electric field at the centre of the cube, due to the eight charges, gets cancelled, because the charges are distributed symmetrically w .r .t . the centre of the cube . ∴ The electric field is zero at the centre . \end{matrix}$ $\begin{array}{l} Here, side of the cube = b \\ Charge at each vertex of t h e cube = q \\ A cube of side b is shown in the given figure . \\ Here, d = Diagonal of t h e cube of e a c h side b \\ d = \sqrt{b^{2} + b^{2}} \\ = \sqrt{2 b^{2}} \\ = \sqrt{2} b \\ ∴ d = b \sqrt{2} \\ l = Length of the diagonal of the cube \\ l = \sqrt{d^{2} + b^{2}} \\ = \sqrt{{(\sqrt{2} b)}^{2} + b^{2}} \\ = \sqrt{2 b^{2} + b^{2}} \\ = \sqrt{3 b^{2}} = \sqrt{3} b \\ ∴ D i s \tan c e b e t w e e n t h e c e n t r e o f c u b e a n d e a c h \\ v e r t e x, r = \frac{l}{2} \\ = \frac{\sqrt{3} b}{2} \\ Electric potential (V) at the centre of the cube is \\ because of the presence of eight charges at the \\ vertices . \\ ∴ V = \frac{8 q}{4 π ε_{0} r} \\ = \frac{8 q}{4 π ε_{0} (\frac{\sqrt{3} b}{2})} \\ = \frac{4 q}{\sqrt{3} π ε_{0} b} \\ ∴ The potential at the centre of the cube is \frac{4 q}{\sqrt{3} π ε_{0} b} . \\ The electric field at the centre of the cube, due to the \\ eight charges, gets cancelled, because the charges are \\ distributed symmetrically w .r .t . the centre of the cube . \\ ∴ The electric field is zero at the centre . \end{array}$

Q.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.

Ans-

$\begin{array}{l} Two charges placed at points A and B are represented \\ in the given figure . O is the midpointof the line joining \\ the two charges . \\ Amount of charge located at A, q_{1} = 1 .5 μ C \\ Amount of charge located at B, q_{2} = 2.5 μ C \\ Distance between the charges, d = 30 cm = 0. 3 m \\ (a) {Let V}_{1} {and E}_{1} be the electric potential and electric \\ field respectively at O . \\ V_{1} = Potential due to charge at A + Potential due to \\ charge at B \\ V_{1} = \frac{q_{1}}{4 π ε_{0} (\frac{d}{2})} + \frac{q_{2}}{4 π ε_{0} (\frac{d}{2})} \\ = \frac{1}{4 π ε_{0} (\frac{d}{2})} (q_{1} + q_{2}) \\ Here, ε_{0} = Permittivity of free space \\ \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N C^{2} m^{- 2} \\ ∴ V_{1} = \frac{9 \times 10^{9} \times 10^{- 6}}{(\frac{0.30}{2})} (2.5 + 1.5) \\ = 2.4 \times 10^{5} V \\ E_{1} = Electric field due to {charge q}_{2} - Electric field \\ due to charge q_{1} \\ = \frac{q_{2}}{4 π ε_{0} {(\frac{d}{2})}^{2}} - \frac{q_{1}}{4 π ε_{0} {(\frac{d}{2})}^{2}} \\ = \frac{9 \times 10^{9}}{{(\frac{0.30}{2})}^{2}} \times 10^{6} \times (2.5 - 1.5) \\ = 4 \times 10^{5} V m^{- 1} \\ ∴ A t the mid - point, t he potential is 2 . 4 \times 1 0^{5} V \\ and electric field is 4 \times 1 0^{5} {V m}^{- 1} . \\ The direction of the e l e c t r i c field is from the larger \\ charge to the smaller charge . \end{array}$

$\begin{array}{l} (b) Consider a point Z such that the normal distance \\ OZ = 10 cm = 0. 1 m, as shown in the g i v e n \\ figure . \\ V_{2} {and E}_{2} respectively are the electric potential and \\ electric field at Z . \\ From the figure, i t can be observed that distance, \\ B Z = A Z \\ = \sqrt{{(0.1)}^{2} + {(0.15)}^{2}} \\ = 0.18 m \\ V_{2} = Electric potential due to A + Electric Potential due \\ to B \\ = \frac{q_{1}}{4 π ε_{0} (A Z)} + \frac{q_{1}}{4 π ε_{0} (B Z)} \\ = \frac{9 \times 10^{9} \times 10^{- 6}}{0.18} (1.5 + 2.5) \\ = 2 \times 10^{5} V \\ Electric field due to q at Z is given by, \\ E_{A} = \frac{q_{1}}{4 π ε_{0} {(A Z)}^{2}} \\ = \frac{9 \times 10^{9} \times 1.5 \times 10^{- 6}}{{(0.18)}^{2}} \\ = 0.416 \times 10^{6} V m^{- 1} \\ E = \sqrt{E_{A}^{2} + E_{B}^{2} + 2 E_{A} E_{B} \cos 2 θ} \\ Here, 2 θ is the angle, ∠ A Z B \\ From the figure, we get \\ \cos θ = \frac{0.10}{0.18} \\ = \frac{5}{9} \\ = 0.5556 \\ ∴ θ = \cos^{- 1} 0.5556 \\ = 56.25 \\ ∴ 2 θ = {112.5}^{o} \\ \cos 2 θ = - 0.38 \\ E= \sqrt{{(0 {.416×10}^{6})}^{2} × {(0 {.69×10}^{6})}^{2} +2×0 .416×0 {.69×10}^{12} × (-0 .38)} \\ = 6.6 \times 10^{5} V m^{- 1} \\ ∴ The potential at a point 1 0 cm \\ (perpendicular to the mid - point) \\ is 2 .0 \times 10 5 V and electric field is 6 . 6 \times 1 0^{5} {V m}^{- 1} . \end{array}$

Q.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.

Ans-

(a) The magnitude of charge placed at the centre of the shell is +q. Therefore, a charge −q will be induced to the inner surface of the shell.

$\begin{array}{l} ∴ Total charge on the inner surface of the shell = - q . \\ Surface charge density at the inner surface of the shell \\ is given as: \\ σ_{1} = \frac{Total charge}{Inner surface area} \\ = \frac{- q}{4 {πr}_{1}^{2}} \to (i) \\ A charge of magnitude + q induces on the outer surface \\ of the shell . \\ Since a charge of magnitude Q is placed on the outer \\ surface of the shell . \\ ∴ Total charge on the outer surface of the shell= Q + q \\ Surface charge density at the outer surface of the \\ shell is given as : \\ σ_{2} = \frac{Total charge}{Outer surface area} \\ = \frac{Q + q}{4 {πr}_{2}^{2}} \to (i) \end{array}$

(b) Yes, the electric field intensity inside a cavity with no charge is zero, even if the shell is not spherical and has any irregular shape. When a closed loop is taken such that a part of it is inside the cavity along a field line and the rest is inside the conductor; the net work done by the electric field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field inside a cavity with no charge is zero, whatever is the shape.

$\begin{array}{l} (a) Show that the normal component of electrostatic field has a discontinuity \\ from one side of a charged surface to another given by \end{array}$

$\begin{array}{l} ({\vec{E}}_{2} - {\vec{E}}_{1}) \cdot \hat{n} = \frac{σ}{ε_{0}} \\ Where, (\hat{n}) is a unit vector normal to the surface at a point and σ is the surface \\ charge density at that point . (The direction of (\hat{n}) is from side 1 to side 2 .) \\ Hence show that just outside a conductor, the electric field is σ \frac{\hat{n}}{ε_{0}} . \end{array}$

$\begin{array}{l} (b) Show that the tangential component of electrostatic field is continuous from \\ one side of a charged surface to another . \\ [Hint: For (a), use Gauss’s law . For, (b) use the fact that work done by electrostatic \\ field on a closed loop is zero .] \end{array}$

$\begin{array}{l} (a) Let e {lectric field on one side of a charged body is E}_{1} \\ and electric field on the other side of the same body is \\ E_{2} . If the infinite plane charged body has a uniform \\ thickness, then \\ Electric field due to one surface of the charged body is \\ given as: \\ \vec{E_{1}} = - \frac{σ}{2 ε_{0}} \hat{n} \to (i) \\ Here, \hat{n} = Unit vector normal to the surface at a point \\ σ = Surface charge density at that point \\ Electric field due to the other surface of the charged \\ body is given as: \\ \vec{E_{2}} = \frac{σ}{2 ε_{0}} \hat{n} \to (i i) \\ Electric field at any point due to the two surfaces is \\ given as: \\ \vec{E_{2}} - \vec{E_{1}} = \frac{σ}{2 ε_{0}} \hat{n} + \frac{σ}{2 ε_{0}} \hat{n} \\ = \frac{σ}{ε_{0}} \hat{n} \\ (\vec{E_{2}} - \vec{E_{1}}) . \hat{n} = \frac{σ}{ε_{0}} \\ As, inside the closed conductor, \vec{E_{1}} = 0, \\ ∴ \vec{E} = \vec{E_{2}} \\ = - \frac{σ}{2 ε_{0}} \hat{n} \\ ∴ Electric field just outside the conductor is \frac{σ}{ε_{0}} \hat{n} . \\ (b) If a charged particle is moved from one point to the \\ other on a closed loop, \\ Work done by the electrostatic field=0 \\ ∴ The tangential component of electrostatic field is \\ continuous from one side of a charged surface to the \\ other . \end{array}$

Q.16 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?

Ans-

$\begin{array}{l} Charge density of the long charged cylinder of length L \\ and radius a= λ . \\ Another cylinder of same length surrounds the \\ previous cylinder, its radius= b \\ Let electric field produced in the space between the \\ two cylinders=E \\ According to Gauss ’ s theorem, e lectric flux through the \\ cylinderical Gaussian surface is given by, \\ ϕ = (2 π r) L \\ Here, r = Distance of a point from the common axis \\ of the two cylinders \\ Let total charge on the cylinder=q \\ According to Gauss’s theorem \\ ϕ = E (2 π r L) = \frac{q}{ε_{0}} \\ Here, q = Charge on the inner sphere of the outer \\ cylinder \\ Since, q= λ L \\ ε_{0} = Permittivity of free space \\ ∴ E (2 π r L) = \frac{λ L}{ε_{0}} \\ ∴ E = \frac{λ}{2 π ε_{0} r} \\ ∴ The electric field in the space between the two \\ cylinders is \frac{λ}{2 π ε_{0} r} . \end{array}$ 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Q.17 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?

Ans-

$\begin{array}{l} Here, distance between electron and proton in a \\ hydrogen atom, d = 0.53 \overset{o}{A} \\ Charge of electron, q_{1} = - 1.6 \times 1 0^{- 19} C \\ Charge of proton, q_{2} = + 1.6 \times 1 0^{- 19} C \\ (a) The p otential at infinity is zero . \\ P .E . of the system=P .E . a t \inf i n i t y - P . E . a t d i s \tan c e d \\ = 0 - \frac{q_{1} q_{2}}{4 π ε_{0} d} \\ Here, ε_{0} =Permittivity of free space \\ \frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2} \\ ∴ P . E . = 0 - \frac{9 \times 10^{9} \times {(1.6 \times 1 0^{- 19})}^{2}}{0.53 \times 10^{- 10}} \\ = - 43.47 \times 1 0^{- 19} J \\ S i n c e 1.6 \times 1 0^{- 19} J = 1 e V \\ ∴ P . E . = - 43.47 \times 1 0^{- 19} J \\ = \frac{- 43.47 \times 1 0^{- 19}}{1.6 \times 1 0^{- 19}} e V \\ = - 27 . 2 eV \\ ∴ P .E . of the system is - 27 . 2 eV . \\ (b) As, K .E . in the orbit= \frac{1}{2} (P . E .) \\ ∴ K . E . = \frac{1}{2} \times (- 27.2) \\ = 13.6 e V \\ Total energy = 13.6 - 27.2 \\ = 13 . 6 eV \\ ∴ The minimum work required to free the electron is \\ 13 . 6 eV . \\ (c) When zero of P .E . is taken, \\ d_{1} = 1.06 \overset{o}{A} \\ ∴ P .E . of the system = P .E {. at d}_{1} - P .E . at d \\ = \frac{q_{1} q_{2}}{4 π ε_{0} d_{1}} - 27.2 eV \\ = \frac{9 \times 10^{9} \times {(1.6 \times 1 0^{- 19})}^{2}}{1.06 \times 10^{- 10}} - 27.2 eV \\ = 21.73 \times 1 0^{- 19} J - 27.2 eV \\ = 13.58 e V - 27.2 eV \\ = - 13.6 e V \end{array}$ $\begin{array}{l} \end{array}$

Q.18 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-

$\begin{array}{l} Let the radius of a sphere A = a \\ Let charge on the sphere A = Q_{A} \\ Let capacitance of the sphere A = C_{A} \\ Let radius of a sphere B = b, \\ Let charge on sphere B = Q_{B} \\ Let capacitance of the sphere B = C_{B} \\ A s the two spheres are connected with a wire, their \\ potentials (V) become equal . \\ {Let electric field of sphere A=E}_{A} \\ Let {electric field of sphere B=E}_{B} \\ ∴ R atio, o f E_{A} t o E_{B}, \\ \frac{E_{A}}{E_{B}} = \frac{Q_{A}}{4 π ε_{0} \times a^{2}} \times \frac{b^{2} \times 4 π ε_{0}}{Q_{B}} \\ \frac{E_{A}}{E_{B}} = \frac{Q_{A}}{Q_{B}} \times \frac{b^{2}}{a^{2}} \to (i) \\ B u t, \frac{Q_{A}}{Q_{B}} = \frac{C_{A} V}{C_{B} V} \\ a n d \frac{C_{A}}{C_{B}} = \frac{a}{b} \to (i i) \\ Substituting the value of (i i) in (i), we get \\ \frac{E_{A}}{E_{B}} = \frac{a}{b} \times \frac{b^{2}}{a^{2}} \\ = \frac{b}{a} \\ ∴ The ratio of electric fields at the surface is \frac{b}{a} . \end{array}$

Q.19 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?

Ans-

$\begin{array}{l} (a) Here, charge - q is at point (0, 0, - a) and \\ charge + q is at point (0, 0, a) . \\ ∴ They form an electric dipole . \\ Point (0, 0, z) is located on the axis of this dipole and \\ point (x, y, 0) is normal to the axis of the dipole . \\ ∴ Electrostatic potential at point (x, y, 0) = 0 \\ Electrostatic potential at point (0, 0, z) is given as: \\ V= \frac{1}{4 π ε_{0}} (\frac{q}{z - a}) + \frac{1}{4 π ε_{0}} (- \frac{q}{z + a}) \\ = \frac{q (z + a - z + a)}{4 π ε_{0} (z^{2} - a^{2})} = \frac{2 q a}{4 π ε_{0} (z^{2} - a^{2})} \\ = \frac{p}{4 π ε_{0} (z^{2} - a^{2})} \\ Here, ε_{0} = Permittivity of free space \\ p = Dipole moment of the system of two charges \\ = 2qa \\ (b) Since the d istance r is much greater than half of \\ the distance between the two charges . \\ ∴ The potential (V) at a distance r is inversely \\ proportional to square of the distance i . e ., \\ V \propto \frac{1}{r^{2}} \\ (c) Electrostatic potential at point (5, 0, 0) i s g i v e n a s : \\ V_{1} = \frac{- q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(- a)}^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{{(- 7)}^{2} + {(a)}^{2}}} \\ = \frac{- q}{4 π ε_{0} \sqrt{49 + a^{2}}} + \frac{q}{4 π ε_{0}} \frac{1}{\sqrt{49 + a^{2}}} \\ = 0 \\ W o r k d o n e i n m o v i n g a s m a l l t e s t c h \arg e f r o m p o int \\ (5, 0, 0) t o p o int (- 7, 0, 0) = C h \arg e (V_{2} - V_{1}) = 0 \\ ∴ No work is done in moving a small test charge \\ from point (5, 0, 0) to point (- 7, 0, 0) along the x - \\ axis . \\ As work done by electrostatic field in moving a test \\ charge between two points is independent of the path \\ connecting the two points, \\ ∴ W o r k d o n e w i l l c o n t i n u e t o b e z e r o a l o n g e v e r y p a t h . \end{array}$

Q.20 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).

Ans-

$\begin{array}{l} The system of three charges forms an electric \\ quadrupole . \\ The given figure shows a quadrupole . \\ Here, c harge + q is located at point X, c harge - 2q is \\ locat ed at point Y and Charge + q located at point Z \\ Let XY = YZ \\ = a \\ Let YP = r \\ Let PX = r + a \\ Let PZ = r - a \\ Electrostatic potential caused by the system of three \\ charges at point P is given as: \\ V= \frac{1}{4 {πε}_{0}} [\frac{q}{XP} - \frac{2 q}{YP} + \frac{q}{ZP}] \\ = \frac{1}{4 {πε}_{0}} [\frac{q}{r + a} - \frac{2 q}{r} + \frac{q}{r-a}] \\ = \frac{q}{4 {πε}_{0}} [\frac{r (r-a) - 2 (r + a) (r-a) + r (r + a)}{r (r + a) (r-a)}] \\ = \frac{q}{4 {πε}_{0}} [\frac{r^{2} - ra - 2 r^{2} + 2 a^{2} + r^{2} + ra}{r (r^{2} - a^{2})}] \\ = \frac{q}{4 {πε}_{0}} [\frac{2 a^{2}}{r (r^{2} - a^{2})}] \\ = \frac{2 {qa}^{2}}{4 {πε}_{0} r^{3} (1 - \frac{a^{2}}{r^{2}})} \\ As \frac{r}{a} ≫ 1, \\ ∴ \frac{a}{r} ≪ 1 \\ \frac{a^{2}}{r^{2}} is negligibly small . \\ It is clear that, potential, V \propto \frac{1}{r^{3}} \\ However, for a dipole V \propto \frac{1}{r^{2}}, and for a monopole \\ V \propto \frac{1}{r} \end{array}$

Q.21 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.

Ans-

$\begin{array}{l} Here, t otal required capacitance, C = 2 μ F \\ Potential difference, V = 1 kV = 1000 V \\ Capacitance of each capacitor, C_{1} = 1 μ F \\ Maximum potential difference across e ach capacitor, \\ V_{1} = 400 V \\ Suppose a number of capacitors are connected in \\ series and these series circuits are connected in \\ parallel (row) to each other . \\ Since, potential difference across each row=1 000 V \\ ∴ Potential difference across each capacitor= 4 00 V . \\ ∴ Number of capacitors in each row= \frac{1000}{400} \\ = 2.5 \\ ∴ There are three capacitors in each row . \\ Capacitance of each row of 3 capacitors o f 1 μ F e a c h = \\ \frac{1}{1 + 1 + 1} = \frac{1}{3} μ F \\ Let there are n rows, each having three capacitors, \\ connected in parallel . \\ ∴ Equivalent capacitance of circuit= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + … n t e r m s \\ = \frac{n}{3} \\ B u t, capacitance of circuit= 2 μ F \\ ∴ \frac{n}{3} = 2 \\ ∴ n = 6 \\ ∴ Number of rows of three capacitors in the circuit=6 \\ ∴ Number of capacitors required for the given \\ arrangement= 6 \times 3 = 18 \end{array}$

Q.22 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 5F or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Ans-

$\begin{array}{l} Here, c apacitance of a parallel plate capacitor, V = 2 F \\ Separation between the two plates, d = 0 . 5 cm \\ = 0.5 \times 1 0^{- 2} m \\ Capacitance of the parallel plate capacitor is given as: \\ C= \frac{ε_{0} A}{d} \\ ∴ A = \frac{C d}{ε_{0}} \\ Here, ε_{0} = Permittivity of free space \\ = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2} \\ ∴ A = \frac{2 \times 0.5 \times 1 0^{- 2}}{8.85 \times 1 0^{- 12}} \\ = 1130 {km}^{2} \\ ∴ The area of the plates is too large . \\ In order to avoid this situation, the capacitance is \\ taken in the range of 5F . \end{array}$

$\begin{array}{l} Q.23 Show that the force on each plate of a parallel plate capacitor has a magnitude equal \\ to \frac{1}{2} QE, where Q is the charge on the capacitor, and E is the magnitude of electric field \\ between the plates . Explain the origin of the factor . \end{array}$ $\begin{array}{l} \end{array}$

Ans-

$\begin{array}{l} \end{array}$ $\begin{array}{l} Let F be the force applied to separate the plates of a parallel plate capacitor by a \\ distance of Δx . \\ ∴ Work done by the force = FΔx \\ As a result, the increase in potential energy of the capacitor= uAΔx \\ Here, \\ u = Energy density \\ A = Area of each plate \\ d = Separation between the plates \\ V = Voltage across the plates \\ Work done by force = Increase in the potential energy \\ ∴ FΔx = uAΔx \\ F = uA = (\frac{1}{2} ε_{0} E^{2}) A \\ Electric field intensity is given as: \\ E = \frac{V}{d} \\ E = \frac{1}{2} ε_{0} (\frac{V}{d}) EA = \frac{1}{2} (ε_{0} A \frac{V}{d}) E \\ Capacitance, C= \frac{ε_{0} A}{d} \\ ∴ F = \frac{1}{2} (CV) E \\ Charge on the capacitor is given as: \\ Q = CV \\ ∴ F = \frac{1}{2} QE \\ The physical origin of the factor, \frac{1}{2} in the force can be explained by the fact that just \\ outside the conductor, field is E and inside it is zero . \\ ∴ It is the average value, \frac{1}{2} of the field that contributes to the force . \end{array}$

Q.24 A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show that the capacitance of a spherical capacitor is given by

$\begin{array}{l} Ans-C = \frac{4 {πε}_{0} r_{1} r_{2}}{r_{1} - r_{2}} \\ where r_{1} and r_{2} are the radii of outer and inner \\ spheres, respectively . \end{array}$

$\begin{array}{l} Here, r adius of outer shell = r_{1} \\ Radius of inner shell = r_{2} \\ Charge on the inner surface of the outer shell= + Q . \\ Charge induced on the outer surface of the inner shell \\ = - Q . \\ Potential difference between the two shells is given as: \\ V= \frac{Q}{4 {πε}_{0} r_{2}} - \frac{Q}{4 {πε}_{0} r_{1}} \\ Here, ε_{0} = Permittivity of free space \\ ∴ V= \frac{Q}{4 {πε}_{0}} [\frac{1}{r_{2}} - \frac{1}{r_{1}}] = \frac{Q (r_{1} - r_{2})}{4 {πε}_{0} r_{1} r_{2}} \\ Capacitance of the given system is given as : \\ C = \frac{Charge (Q)}{Potential difference (V)} = \frac{4 {πε}_{0} r_{1} r_{2}}{r_{1} - r_{2}} \\ ∴ Capacitance of the given spherical capacitor is \\ \frac{4 {πε}_{0} r_{1} r_{2}}{r_{1} - r_{2}} . \end{array}$

Q.25 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C 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?

$\begin{array}{l} Ans - \end{array}$ $\begin{array}{l} (a) Here, Radius of the spherical conductor, r = 12 cm \\ = 0 . 12 m \\ Charge over the spherical conductor, q = 1.6 \times 1 0^{- 7} C \\ If there is field inside the conductor, then charges \\ will move to neutralize it . \\ ∴ Electric field inside the spherical conductor =0 \\ (b) Electric field just outside the spherical conductor is \\ given as : \\ E = \frac{q}{4 {πε}_{0} r^{2}} \\ Here, ε_{0} = Permittivity of free space \\ \frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} C^{- 2} \\ ∴ E = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0.12)}^{2}} \\ = 10^{5} {NC}^{- 1} \\ ∴ The electric field just outside the sphere is 10^{5} {NC}^{- 1} . \\ (c) Let Electric field at a point 18 m from the centre of \\ the sphere = E_{1} \\ Distance of the point from the centre of sphere, d \\ = 18 cm \\ = 0 . 18 m \\ E_{1} = \frac{q}{4 {πε}_{0} r^{2}} \\ = 9 \times 10^{9} \times \frac{1.6 \times 1 0^{- 7}}{{(0 .18)}^{2}} \\ = 4 .4 \times 10^{4} {NC}^{- 1} \\ ∴ The electric field at a point 18 cm from the centre of \\ the sphere is 4 .4 \times 10^{4} {NC}^{- 1} . \end{array}$

Q.26 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 Q₁Q₂/4πr 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).

Ans-

(a) When two charged spheres are brought close to each other, the charge distribution on them does not remain uniform. Therefore, the force between two conducting spheres is not exactly given by the given expression.
(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r³ dependence, instead of 1/r²dependence.
(c) Yes, if a small test charge is released at rest at a point in an electrostatic field configuration, then it will move along the line of force passing through that point, only if the field lines are straight. If the field lines are not straight, the charge will not go along the line. This is because the field lines give the direction of acceleration.
(d)The direction of force due to field is towards the nucleus, and the electron does not move along the direction of this force. Therefore, whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
(e) No, electric potential is continuous across the surface of a charged conductor.
(f) The capacitance of a single conductor implies a parallel plate capacitor with one of its two plates at infinity.
(g) A water molecule has an unsymmetrical shape as compared to that of mica. Therefore, it has a permanent dipole moment. That is why; it has a greater dielectric constant than mica.

Q.27 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?

$\begin{array}{l} Here, v oltage rating of a parallel plate capacitor, V \\ = 1 kV = 1000 V \\ Dielectric constant of t h e material, ε_{r} = 3 \\ Dielectric strength = 1 0^{7} {Vm}^{-1} \\ For safety, the electric field intensity should never \\ exceed 10 % of the dielectric strength . \\ ∴ E l e c t r i c f i e l d, E = 10 % of 1 0^{7} \\ = 10^{6} V m^{- 1} \\ Capacitance of parallel plate capacitor, C = 50 pF \\ = 50 \times 1 0^{- 12} F \\ Distance between the plates is given b y t h e r e l a t i o n, \\ d= \frac{V}{E} \\ = \frac{1000}{10^{6}} \\ = 10^{- 3} m \\ C a p a c i \tan c e i s g i v e n a s : \\ C = \frac{ε_{0} ε_{r} A}{d} \\ Here, A = Area of each plate of t h e capacitor \\ ε_{0} = Permittivity of free space \\ = 8.85 \times 1 0^{- 12} N^{- 1} C^{2} m^{- 2} \\ ∴ A = \frac{C}{ε_{0} ε_{r}} \\ = \frac{50 \times 1 0^{- 12}}{8.85 \times 1 0^{- 12} \times 3} \approx 19 c m^{2} \\ ∴ {The area of each plate is approximately 19 cm}^{2} . \end{array}$

Ans-

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Q.28 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.

Ans-

(a) The equipotential surfaces are the equidistant planes parallel to the x-y plane.
(b) The equipotential surfaces are the planes parallel to the x-y plane. When the field increases uniformly, planes get closer (differing by fixed potential).
(c) The equipotential surfaces are the concentric spheres centered at the origin.
(d) The equipotential surfaces have the periodically varying shape. The shape of equipotential surfaces becomes parallel to the grid at a far off distance.

As the charge lies always on the outer surface of the shell, therefore, when the sphere and the shell are connected by a wire, charge will flow from the sphere to the shell, whatever be the magnitude and sign of charge q₂.

Q.29 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 Cm⁻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.)

Ans-

(a) As we step out of our house, we do not get an electric shock. This is because the original equipotential surfaces of open air change keeps our body and the ground at the same potential.
(b) Yes, the man will get an electric shock because the steady discharging current in the atmosphere charges up the aluminium sheet and raises its voltage gradually. The rise in its voltage depends on the capacitance of the capacitor formed by the aluminium sheet and the ground.
(c) The atmosphere is charged continuously by occurrence of thunderstorms and lightning. Therefore, even with the presence of discharging current of 1800 A in the atmosphere, the atmosphere is not discharged completely. The two opposing currents are, on an average, in equilibrium and the atmosphere remains electrically charged.
(d) During a lightning, the electrical energy of the atmosphere is dissipated in the form of light energy, heat energy, and sound energy.

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Chapter 1 - Electric Charges and Fields

Chapter 3 - Current Electricity

Chapter 4 - Moving Charges and Magnetism

Chapter 5 - Magnetism And Matter

Chapter 6 - Electromagnetic Induction

Chapter 7 - Alternating Current

Chapter 8 - Electromagnetic Waves

Chapter 9 - Ray Optics and Optical Instruments

Chapter 10 - Wave Optics

Chapter 11 - Dual Nature of Radiation and Matter

Chapter 12 - Atoms

Chapter 13 - Nuclei

Chapter 14 - Semiconductor Electronic: Material, Devices And Simple Circuits

Chapter 15 - Communication Systems

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NCERT Solutions Class 12 Physics Chapter – 2

NCERT Solutions Class 12 Physics Chapter – 2 Electrostatic Potential and Capacitance

Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 2

3.1 Introduction

3.2 Electrostatic Potential

3.3 Potential Due To A Point Charge

3.4 Potential Due To An Electric Dipole

3.5 Potential Due To A System Of Charges

3.6 Equipotential Surfaces

3.6.1 Relation Between Field And Potential

3.7 Potential Energy Of A System Of Charges

3.8 Potential Energy In An External Field

3.8.1 Potential Energy Of A Single Charge

3.8.2 Potential Energy Of A System Of Two Charges In An External Field

3.8.3 Potential Energy Of A Dipole In An External Field

3.9 Electrostatics Of Conductors

3.10 Dielectrics And Polarization

3.11 Capacitors And Capacitance

3.12 The Parallel Plate Capacitor

3.13 Effect Of Dielectric On Capacitance

3.14 Combination Of Capacitors

3.14.1 Capacitors In Series

3.14.2 Capacitors In Parallel

3.15 Energy Stored In A Capacitor

3.16 Van De Graaff Generator

NCERT Solutions Class 12 Physics Chapter 2: Article link for Electrostatic Potential and Capacitance NCERT solutions

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Related Chapters

Chapter 1 - Electric Charges and Fields

Chapter 3 - Current Electricity

Chapter 4 - Moving Charges and Magnetism

Chapter 5 - Magnetism And Matter

Chapter 6 - Electromagnetic Induction

Chapter 7 - Alternating Current

Chapter 8 - Electromagnetic Waves

Chapter 9 - Ray Optics and Optical Instruments

Chapter 10 - Wave Optics

Chapter 11 - Dual Nature of Radiation and Matter

Chapter 12 - Atoms

Chapter 13 - Nuclei

Chapter 14 - Semiconductor Electronic: Material, Devices And Simple Circuits

Chapter 15 - Communication Systems

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1. How many topics are there in Chapter 2 Physics Class 12 NCERT Solutions?

2. Is Class 12th Physics easy?

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