1.

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»

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


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