Question

25) An infinite plane sheet of charge density 10-
Cm-2 is held in air. In this situation how far
apart are two equipotential surfaces, whose
p.d is 5V?
(Ans: 8.85 mm)​

Answer 1

Step-by-step explanation:

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

−8

Cm

−2

According to Gauss Theoram,

E = \frac{\sigma}{2 \epsilon_0}

2ϵ

0

σ

(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}

△r

△V

(eq 2)

from above eq 1 and eq 2,

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

2ϵ

0

σ

=

△r

△V

on putting the values,

\epsilon_0 = 8.85*10^{-12}ϵ

0

=8.85∗10

−12

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

10

−8

2∗8.85∗10

−12

∗5

= 8.85*10^{-3}m8.85∗10

−3

m

= 8.85 mm8.85mm