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## NCERT Class 12 Physics Textbook Chapter 2 With Answer PDF Free Download

### Chapter 2: Electrostatic Potential and Capacitance

#### 2.1 Introduction

In Chapters 6 and 8 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from one point to another against a force like spring force or gravitational force, that work gets stored as the potential energy of the body.

When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy.

The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces.

Coulomb force between two (stationary) charges is also a conservative force.

This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law.

Thus, like the potential energy of a mass in a gravitational field, we can define the electrostatic potential energy of a charge in an electrostatic field.

#### 2 Electrostaticla Potential

Consider any general static charge configuration. We define the potential energy of a test charge q in terms of the work done on the charge q.

This work is obviously proportional to q, since the force at any point is qE, where E is the electric field at that point due to the given charge configuration.

It is, therefore, convenient to divide the work by the amount of charge q, so that the resulting quantity is independent of q. In other words, work done per unit test charge is characteristic of the electric field associated with the charge configuration.

This leads to the idea of electrostatic potential V due to a given charge configuration. From Eq. (2.1), we get: Work done by external force in bringing a unit positive

charge from point R to P

= VP – V R

are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential the difference that is physically significant.

If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point.

Author | NCERT |

Language | English |

No. of Pages | 42 |

PDF Size | 1584 KB |

Category | English Book |

Source/Credits | ncert.nic.in |

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

**Q 2.5) A parallel plate capacitor with air between the plates has a capacitance of 8pF (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, C = 8pF.

In the first case, the parallel plates are at a distance ‘d’ and are filled with air.

Air has a dielectric constant, k = 1

Capacitance, C = \frac{k \times \epsilon _{o} \times A}{d}*d**k*×*ϵ**o*×*A* = \frac{\epsilon _{o} \times A}{d}*d**ϵ**o*×*A* … eq(1)

Here,

A = area of each plate\epsilon _{o}*ϵ**o* = permittivity of free space.

Now, if the distance between the parallel plates is reduced to half, then d_{1} = d/2

Given, the dielectric constant of the substance, k_{1} = 6

Hence, the capacitance of the capacitor,

C_{1} = \frac{k_{1} \times \epsilon _{o} \times A}{d_{1}}*d*1*k*1×*ϵ**o*×*A* = \frac{6 \epsilon _{o} \times A}{d/2}*d*/26*ϵ**o*×*A* = \frac{12 \epsilon _{o} A}{d}*d*12*ϵ**o**A* … (2)

Taking ratios of eqns. (1) and (2), we get,

C_{1} = 2 x 6 C = 12 C = 12 x 8 pF = 96pF.

Hence, the capacitance between the plates is 96pF.

**Q 2.6) Three capacitors connected in series have a capacitance of 9pF each.**

**(1) What is the total capacitance of the combination?**

**(2) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?**

**Solution:**

(1) Given,

The capacitance of the three capacitors, C = 9 pF

Equivalent capacitance (c_{eq}) is the capacitance of the combination of the capacitors given by\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} =\frac{1}{3}*C**e**q*1=*C*1+*C*1+*C*1=*C*3=93=31\frac{1}{C_{eq}} = \frac{1}{3}*C**e**q*1=31 = C_{eq} = 3 pF

Therefore, the total capacitance = 3pF.

(2) Given, supply voltage, V = 100V

The potential difference (V_{1}) across the capacitors will be equal to one–third of the supply voltage.

Therefore, V_{1} = \frac{V}{3}3*V* = \frac{120}{3}3120 = 40V.

Hence, the potential difference across each capacitor is 40V.

**Q 2.7) Three capacitors of capacitances 2 pF, 3 pF, and 4 pF are connected in parallel.**

**(1) What is the total capacitance of the combination?**

**(2) Determine the charge on each capacitor if the combination is connected to a 100 V supply.**

**Solution:**

(1) Given, C_{1 }= 2pF, C_{2} = 3pF and C_{3} = 4pF.

Equivalent capacitance for the parallel combination is given by C_{eq} .

Therefore, C_{eq} = C_{1} + C_{2} + C_{3} = 2 + 3 + 4 = 9pF

Hence, the total capacitance of the combination is 9pF.

(2) Supply voltage, V = 100V

The three capacitors are having the same voltage, V = 100v

q = VC

where,

q = charge

C = capacitance of the capacitor

V = potential difference

for capacitance, c = 2pF

q = 100 x 2 = 200pC = 2 x 10^{-10}C

for capacitance, c = 3pF

q = 100 x 3 = 300pC = 3 x 10^{-10}C

for capacitance, c = 4pF

q = 100 x 4 = 400pC = 4 x 10^{-10 }C

**Q 2.8) In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10 ^{-3} m ^{2,} 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?**

**Solution:** Given,

The area of plate of the capacitor, A = 6 x 10^{-3} m^{2}

Distances between the plates, d = 3mm = 3 x 10^{-3 }m

Voltage supplied, V = 100V

Capacitance of a parallel plate capacitor is given by, C = \frac{\epsilon \times A}{d}*d**ϵ*×*A*

Here,

ε_{o} = permittivity of free space = 8.854 × 10^{-12} N^{-1} m ^{-2} C^{-2}

C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}3×10−38.854×10−12×6×10−3 = 17.71 x 10^{-12} F = 17.71 pF.

Therefore, each plate of the capacitor is having a charge of

q = VC = 100 × 17.71 x 10^{-12} C = 1.771 x 10^{-9 }C

**Q 2.9: 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.*

*Answer 2.9:*

(a) Dielectric constant of the mica sheet, k = 6

If the voltage supply remained connected, the voltage between two plates will be

constant.

Supply voltage, V = 100 V

Initial capacitance, C = 1.771 × 10^{−11} F

New capacitance, C_{1 }= kC = 6 × 1.771 × 10^{−11} F = 106 pF

New charge, q_{1} = C_{1}V = 106 × 100 pC = 1.06 × 10^{–8} C

Potential across the plates remain 100 V.

(b) Dielectric constant, k = 6

Initial capacitance, C = 1.771 × 10^{−11} F

New capacitance, C_{1} = kC = 6 × 1.771 × 10^{−11} F = 106 pF

If the supply voltage is removed, then there will be a constant amount of charge

in the plates.

Charge = 1.771 × 10^{−9} C

Potential across the plates is given by,

V_{1} = q/C_{1} = \frac{1.771 \times 10^{-9}}{106 \times 10^{-12}}106×10−121.771×10−9

= 16.7 V

**Q 2.10) A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?**

**Solution:** Given,

Capacitance of the capacitor, C = 12pF = 12 x 10^{-12 }F

Potential difference, V = 50 V

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

E = \frac{1}{2}21 CV^{2} = \frac{1}{2}21 x 12 x 10^{-12} x (50)^{2} J = 1.5 x 10^{-8 }J

Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^{-8}

NCERT Class 12 Physics Textbook Chapter 2 With Answer PDF Free Download