Question

An infinite plane sheet of charge density 10^-8 is held in air. In this situation how far apart are two equipotential surfaces, whose p.d. is 5 V?

Answer 1

Given,

Surface density (sigma) = $10^{-8}Cm^{-2}$

According to Gauss Theoram,

E = $\frac{\sigma}{2 \epsilon_0}$                  (eq 1)

here,

let Δr be the distance between two equipotential surfaces.

Also given, Potential Difference (ΔV) = 5 V

We know,

E = $\frac{\triangle V}{\triangle r}$                     (eq 2)

from above eq 1 and eq 2,

$\frac{\sigma}{2 \epsilon_0}=\frac{\triangle V}{\triangle r}$

on putting the values,

$\epsilon_0 = 8.85*10^{-12}$

Δr = $\frac{2*8.85*10^{-12}*5}{10^{-8}}$

      = $8.85*10^{-3}m$

        = $8.85 mm$

Answer 2

Explanation:

let's see

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