Question

(i) Use Gauss's law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities ?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitances in the ratio 1: 2 so that the energy stored in the two cases becomes the same.

Answer 1

(i)

Consider a thin infinite uniformly charged plane sheet having the surface charge density of $\sigma$. The electric field is normally outward to the plane sheet and is same in magnitude but opposite in direction.

Now, draw a Gaussian surface in the form of cylinder around an axis. Let its cross-sectional area be A.

The cylinder is made from three surfaces A,$S_2$, and A' and the electric flux linked with $S_2$ is 0.So, the total electric flux linked through the Gaussian surface is

${\varphi }_E=\text{electric flux through A}+\text{electric flux through}{S}_2+\text{electric flux through A'}$

${\varphi }_E=EAcos{0}^{\circ }+EAcos{0}^{\circ }.$

$\varphi =2EA...\left(i\right)$

According to Gauss theorem,

$\varphi =\frac{q}{{\epsilon }_0}$

$\varphi =\frac{\sigma A}{2{\epsilon }_0}(\because q=\sigma A)...\left(ii\right)$
From equation (i) and (ii)
$2EA=\frac{\sigma A}{{ϵ}_0}$
$E=\frac{\sigma }{2{ϵ}_0}$
The direction of field for positive charge density is in outward direction and perpendicular to the plane infinite sheet whereas for the negative charge density the direction becomes inward and perpendicular to the sheet.

(ii) Given, $C_1:C_2=1:2$

$\Rightarrow C_2=2C_1$

For parallel combination of capacitor,
$C_p=C_1+C_2$

$=C_1+2C_1=3C_1$

The energy stored in capacitor
$E=\frac{1}{2}{C}_p{V}_p^2$

$=\frac{1}{2}3C_1{V}_p^2=\frac{3}{2}{C}_1{V}_p^2....\left(i\right)$

For series combination of capacitor,
$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}$

$C_s=\frac{2}{3}{C}_1$

The energy stored in capacitor
$E=\frac{1}{2}{C}_s{V}_s^2$

$E=\frac{C_1{V}_s^2}{3}...\left(ii\right)$

Equating equation (i) and (ii),we get

$\frac{3}{2}{C}_1{V}_p^2=\frac{C_1{V}_s^2}{3}$

$\frac{V_p}{V_s}=\frac{\sqrt{2}}{3}$