# Electric Charges and Fields Chapter 1 Class 12 Physics NCERT Textbook PDF

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### Chapter 1: Electric Charges and Fields

#### 1.1 INTRODUCTION

All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater, particularly in dry weather.

This is almost inevitable with ladies’ garments like a polyester saree. Have you ever tried to find any explanation for this phenomenon?

Another common example of electric discharge is the lightning that we see in the sky during thunderstorms.

We also experience a sensation of an electric shock either while opening the door of a car or holding the iron bar of a bus after sliding from our seats.

The reason for these experiences is the discharge of electric charges through our body, which were accumulated due to the rubbing of insulating surfaces.

You might have also heard that this is due to the generation of static electricity.

This is precisely the topic we are going to discuss in this and the next chapter. Static means anything
that does not move or change with time.

Electrostatics deals with the study of forces, fields, and potentials arising from static charges.

#### 1.2 ELECTRIC CHARGE

Historically the credit for the discovery of the fact that amber rubbed with wool or silk cloth attracts light objects goes to Thales of Miletus, Greece, around 600 BC.

The name electricity is coined from the Greek word electron meaning amber. Many such pairs of materials were known that rubbing could attract light objects like straw, pith balls, and bits of paper.

You can perform the following activity at home to experience such an effect. Cut out long thin strips of white paper and lightly iron them.

Take them near a TV screen or computer monitor. You will see that the strips get attracted to the screen. In fact, they remain stuck to the screen for a while.

### 1.3 Conductors And Insulators

A metal rod held in hand and rubbed with wool will not show any sign of being charged. However, if a metal rod with a wooden or plastic handle is rubbed without touching its metal part, it shows signs of charging.

Suppose we connect one end of a copper wire to a neutral pith ball and the other end to a negatively charged plastic rod. We will find that the pith ball acquires a negative charge.

If a similar experiment is repeated with a nylon thread or a rubber band, no transfer of charge will take place from the plastic rod to the pith ball. Why does the transfer of charge not take place from the rod to the ball?

Some substances readily allow the passage of electricity through them, others do not. Those which allow electricity to pass through them easily are called conductors.

They have electric charges (electrons) that are comparatively free to move inside the material. Metals, human and animal bodies, and earth are conductors.

Most non-metals like glass, porcelain, plastic, nylon, and wood offer high resistance to the passage of electricity through them. They are called insulators.

Most substances fall into one of the two classes stated above. When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor.

In contrast, if some charge is put on an insulator, it stays in the same place. You will learn why this happens in the next chapter.

This property of the materials tells you why a nylon or plastic comb gets electrified on combing dry hair or on rubbing, but a metal article like a spoon does not.

The charges on metal leak through our body to the ground as both are conductors of electricity.

### NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields

Q 1.1) What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7 C placed 30 cm apart in the air?

Sol: Given,

The Charge on the 1st sphere and 2nd sphere is q1 = 2 x 10-7 C and q2 = 3 x 10-7 C

The distance between two charges is given by r = 30cm = 0.3m

The electrostatic force between the spheres is given by the relation :

F = \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}4πϵo​1​.r2q1​q2​​

Here,\epsilon _{o}ϵo​ = permittivity of free space and,\frac{1}{4\pi \epsilon _{o}}4πϵo​1​ = 9 x 109 Nm2C-2

Force, F = \frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}(0.3)29×109×2×10−7×3×10−7​ = 6 x 10‑3 N.

The force between the charges will be repulsive as they have the same nature.

Q 1.2) The electrostatic force on a small sphere of charge is 0.4 µC due to another small sphere of charge –0.8 µC in the air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Sol:

(a) Given,

The charge on the 1st sphere (q) and 2nd sphere (q2) is 0.4 µC or 0.4 × 10-6 C and -0.8 × 10-6C respectively.

The electrostatic force on the 1st sphere is given by F = 0.2N.

Electrostatic force between the spheres is given by the relation :

F = \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{r^{2}}4πϵo​1​.r2q1​q2​​

Here,\epsilon _{o}ϵo​ = permittivity of free space and,\frac{1}{4\pi \epsilon _{o}}4πϵo​1​ = 9 × 109 Nm2C-2

r2 = \frac{1}{4\pi \epsilon _{o}}.\frac{q_{1}q_{2}}{F}4πϵo​1​.Fq1​q2​​

= \frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^{9}}{0.2}0.20.4×10−6×8×10−6×9×109​ = 144 x 10-4

r = \sqrt{144 \times 10^{-4}}144×10−4​ = 12 x 10-2 = 0.12m

Therefore, the distance between the two spheres = 0.12 m

(b) Since the spheres have opposite charges, the force on the second sphere due to the first sphere will also be equal to 0.2N.

Q 1.3) Check that the ratio ke2/G memp is dimensionless. Lookup a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Sol.:

The ratio to be determined is given as follows :\frac{ke^{2}}{Gm_{e}m_{p}}Gmempke2​

where G is the gravitational constant in N m2 kg-2

me and mp is the masses of electron and proton in kg.

e is the electric charge (unit – C)

k = \frac{1}{4\pi \epsilon _{o}}4πϵo​1​  (unit – Nm2C-2)

Therefore, the unit of given ratio,\frac{ke^{2}}{Gm_{e}m_{p}}Gmempke2​ = \frac{[Nm^{2}C^{-2}][C^{-2}]}{[Nm^{2}kg^{-2}][kg][kg]}[Nm2kg−2][kg][kg][Nm2C−2][C−2]​ = M0L0T0

So, the given ratio is dimensionless.

Given,

e = 1.6 x 10-19 C

G = 6.67 x 10-11 N m2 kg-2

m= 9.1 x 10-31 kg

mp = 1.66 x 10-27 kg

Putting the above values in the given ratio, we get\frac{ke^{2}}{Gm_{e}m_{p}}Gmempke2​ = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}6.67×10−11×9.1×10−31×1.67×10−279×109×(1.6×10−19)2​ = 2.3 x 1039

So, the above ratio is the ratio of the electric force to the gravitational force between a proton and an electron when the distance between them is constant.

Q 1.4) (i) Explain the meaning of the statement ‘electric charge of a body is quantized’.
(ii) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large-scale charges?

Sol.:

(i) The ‘electric charge of a body is quantized’ means that only integral (1, 2, …n)  numbers of electrons can be transferred from one body to another.

Charges cannot get transferred to infractions. Hence, the total charge possessed by a body is only in integral multiples of electric charge.

(ii) In the case of large-scale or macroscopic charges, the charge which is used over there is comparatively too huge to the magnitude of the electric charge. Hence, on a macroscopic level, the quantization of charge is of no use Therefore, it is ignored and the electric charge is considered to be continuous.

Q 1.5) When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Sol.:

When two bodies are rubbed against each other,  a charge is developed in both bodies. These charges are equal but opposite in nature.

And this phenomenon of inducing a charge is known as charging by friction. The net charge on both of the bodies is 0 and the reason behind it is that an equal amount of charge repels it.

When we rub a glass rod with a silk cloth, a charge with the opposite magnitude is generated over there. This phenomenon is consistent with the law of conservation of energy.

A similar phenomenon is observed with many other pairs of bodies.

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