Question

Show that the electric at the surface of a charged conductor is given by E = σ/ε0n cap , where where σ

is the surface charge density and n cap is a unit vector normal to the surface in the outward direction.​

Answer 1

Answer:

Explanation:

According to Gauss's theorem,

$\int\limits {E} \, ds$ = q/ε

Here, q is the charge of the conductor.

Now, σ=q/ds

         q=σds

Hence, E.dscosθ = σds/ε

As the unit vector is normal to to the surface in the upward   direction.

Thus, the angle between the electric field and area vector is 0.

E.ds = σds/ε

E=σ/ε

Along the vector-

E=(σ/ε)n cap

Answer 2

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