Class 12 Electric Charges and Fields NCERT Solutions: Free PDF, Important Topics and Formulas

Ans.1.1 Charge of the first sphere,  $q_1$ = 2 ×  ${10}^{-7}$ C

Charge of the second sphere,  $q_2$ = 3 ×  ${10}^{-7}$ C

Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

F =  $\frac{1}{4\pi {ϵ}_0}$  $\times \frac{q_1{q}_2}{r^2}$ , where  ${ϵ}_0$ = Permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

Hence, F =  $\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}$  $\times \frac{2\mathrm{}\times {10}^{-7}\times 3\mathrm{}\times {10}^{-7}}{{\left(0.3\right)}^2}$ = 5.99  $\times {10}^{-3}$ N

Therefore the force between the two small charged spheres is 5.99  $\times {10}^{-3}$ N. The charges are of same nature. Hence, force between them will be repulsive.

Q.1.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in 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?

Ans.1.2

(a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere,  $q_1$ = 0.4 μC = 0.4  $\times {10}^{-6}$ C

Charge on this sphere,  $q_2$ = - 0.8 μC = - 0.8  $\times {10}^{-6}$ C

Electronic force between the spheres is given by the relation

F =  $\frac{1}{4\pi {ϵ}_0}$  $\times \frac{q_1{q}_2}{r^2}$ , where  ${ϵ}_0$ = Permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

Or  $r^2$ =  $\frac{1}{4\pi {ϵ}_0}$  $\times \frac{q_1{q}_2}{F}$ =  $\frac{1}{4\pi \times 8.854\mathrm{}\times {10}^{-12}}$  $\times \frac{0.4\mathrm{}\times {10}^{-6}\times 0.8\mathrm{}\times {10}^{-6}}{0.2}$ = 0.01438 m2

r = 0.1199 m = 0.12 m

The distance between two spheres is 0.12 m

(b)

Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is also 0.2 N

Ans.1.3 The units of the given equation $\frac{{ke}^2}{G{m}_e{m}_p}$ is

e = Electric charge in C

k =  $\frac{1}{4\pi {ϵ}_0}$ in N  $m^2{C}^{-2},$ where  ${ϵ}_0$ = Permittivity of free space = 8.854  $\times {10}^{-12}$  $C^2{N}^{-1}$  $m^{-2}$

G = Gravitational constant = N  $$

$m^2{kg}^{-2}$

${\mathrm{m}}_{\mathrm{e}}=\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g},\mathrm{}\mathrm{}{\mathrm{m}}_{\mathrm{p}}=\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{t}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{k}\mathrm{g}$

Therefore

$\frac{{ke}^2}{G{m}_e{m}_p}$ =  $\frac{\left[\mathrm{N}{m}^2{C}^{-2}\right]\times \left[C^2\right]}{\left[\mathrm{N}{m}^2{kg}^{-2}\right]\left[kg\right]\left[kg\right]}$= $M^0{L}^0{T}^0$

So the given equation is dimensionless

We have the following values

e = 1.6 $\times {10}^{-19}$ C

G = 6.67   $\times {10}^{-11}$ ${\mathrm{Nm}}^2{kg}^{-2}$ kg

${\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31}$ kg

${\mathrm{m}}_{\mathrm{p}}=1.66\times {10}^{-27}$

The numerical value of this ratio is given by

$\frac{{ke}^2}{G{m}_e{m}_p}$ =  $\frac{\left[\mathrm{N}{m}^2{C}^{-2}\right]\times \left[C^2\right]}{\left[\mathrm{N}{m}^2{kg}^{-2}\right]\left[kg\right]\left[kg\right]}$ =  $\frac{1}{4\pi {ϵ}_0}$ =  $\times \frac{e^2}{G{m}_e{m}_p}$= 
$\frac{1}{4\times \pi \times 8.854\mathrm{}\times {10}^{-12}}$ $\times \frac{{(1.6\mathrm{}\times {10}^{-19})}^2}{6.67\times {10}^{-11}\times {9.1\times {10}^{-31}\times 1.66\times {10}^{-27}}_{}}$

 = 2.284 $\times {10}^{39}$

Q.1.4 (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.

          (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e. large scale charges?

Ans.1.4

(a) Electric charge of a body is quantized, this means that only integers (1,2,3,….n) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integers.

(b) In macroscopic i.e. large scale charges, the charges used are huge as compared to the electric charge of electrons or protons. Therefore, it is ignored and it is considered that electric charge is continuous.

Gauss’s Law

Applications of Gauss’s Law