Question

Cos2 sigma + Sin2 sigma

Answer 1

Step-by-step explanation:

Gaus's Theorem:-

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

i.e \bf\phi_{closed} = \dfrac{q_{net}}{\epsilon_{\circ}}.ϕclosed=ϵ∘qnet.

Where,

\phiϕ = Electric flux.

q = charge.

Electric Field Intensity derivation:-

(Diagram is in the ATTACHMENT)

By Gaus's theorem,

\begin{gathered}\\ \tt\mapsto \phi_{total} = \frac{q _{inside}}{ \epsilon_{ \circ}} ...(1)\end{gathered}↦ϕtotal=ϵ∘qinside...(1)

\begin{gathered}\\ \tt\mapsto \phi_{t} = \phi_1 + \phi_2 + \phi_3.\end{gathered}↦ϕt=ϕ1+ϕ2+ϕ3.

\begin{gathered}\\ \tt\mapsto \phi_t = \int EdA \cos \theta_1 + \int EdA \cos \theta_2 + \int EdA \cos \theta_3 .\end{gathered}↦ϕt=∫EdAcosθ1+∫EdAcosθ2+∫EdAcosθ3.

\begin{gathered}\\ \tt\mapsto \phi_t = \int EdA \cos0 {}^{ \circ} + \int EdA \cos0 \degree + \int EdA \cos90 \degree.\end{gathered}↦ϕt=∫EdAcos0∘+∫EdAcos0°+∫EdAcos90°.

\begin{gathered}\\ \tt\mapsto \phi_t = \int EdA + \int EdA + 0.\end{gathered}↦ϕt=∫EdA+∫EdA+0.

\begin{gathered}\\ \tt\mapsto \phi_t = 2EA.\end{gathered}↦ϕt=2EA.

\begin{gathered}\\ \tt\mapsto \frac{q_{inside}}{ \epsilon_{ \circ}} = 2EA. \\ \\ \rm \bigg(from \: (1) \bigg)\end{gathered}↦ϵ∘qinside=2EA.(from(1))

Also Superficial Charge density,

\begin{gathered}\\ \tt\mapsto \sigma = \frac{q}{A} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt\mapsto q = \sigma A \: \: . ..(2)\end{gathered}↦σ=Aq.↦q=σA...(2)

So from 2,

\begin{gathered}\\ \tt\mapsto \frac{ \sigma \cancel{A}}{ \epsilon_{ \circ}} = 2E \cancel{A}.\end{gathered}↦ϵ∘σA=2EA.

\begin{gathered}\\ \large\mapsto \boxed{ \bf E = \frac{ \sigma}{2 \epsilon_{ \circ.}}}\end{gathered}↦E=2ϵ∘.σ

Where E is electric field intensity.