Question

6. State Gauss' theorem and apply it to determine electric field
intensity due to non-conducting infinite plane sheet of charge.​

Answer 1

Gaus's Theorem:-

The net Electric flux through a closed surface (i.e. 3-D surface) is $\tt\dfrac{1}{\epsilon_{\circ}}$ times the net charge enclosed by the surface.

i.e $\bf\phi_{closed} = \dfrac{q_{net}}{\epsilon_{\circ}}.$

Where,

• $\phi$ = Electric flux.
• q = charge.

Electric Field Intensity derivation:-

(Diagram is in the ATTACHMENT)

By Gaus's theorem,

$\\ \tt\mapsto \phi_{total} = \frac{q _{inside}}{ \epsilon_{ \circ}} ...(1)$

$\\ \tt\mapsto \phi_{t} = \phi_1 + \phi_2 + \phi_3.$

$\\ \tt\mapsto \phi_t = \int EdA \cos \theta_1 + \int EdA \cos \theta_2 + \int EdA \cos \theta_3 .$

$\\ \tt\mapsto \phi_t = \int EdA \cos0 {}^{ \circ} + \int EdA \cos0 \degree + \int EdA \cos90 \degree.$

$\\ \tt\mapsto \phi_t = \int EdA + \int EdA + 0.$

$\\ \tt\mapsto \phi_t = 2EA.$

$\\ \tt\mapsto \frac{q_{inside}}{ \epsilon_{ \circ}} = 2EA. \\ \\ \rm \bigg(from \: (1) \bigg)$

Also Superficial Charge density,

$\\ \tt\mapsto \sigma = \frac{q}{A} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto q = \sigma A \: \: . ..(2)$

So from 2,

$\\ \tt\mapsto \frac{ \sigma \cancel{A}}{ \epsilon_{ \circ}} = 2E \cancel{A}.$

$\\ \large\mapsto \boxed{ \bf E = \frac{ \sigma}{2 \epsilon_{ \circ.}}}$

Where E is electric field intensity.