Question

An infinite nonconducting sheet has a surface charge density $\sigma =0.10\mu C/m^2$ on one side. How far apart are equipotential surfaces whose potentials differ by $50V$?
A. $8.8mm$
B. $8.8cm$
C. $8.8m$
D. $8.8\mu m$

Answer 1

Hint: Potential of a charge conducting field is the product of the electric field and the distance between the parallel plates of the charged conductors. And the electric field on an infinite sheet is the ratio of its charge density to the relative permittivity.

Formula Used: $V=Ed$
Where,
$V=$ Potential
$E=$ Electric field
$d$ is the separation between the plates of the charged conductor

$E={\displaystyle \frac{\sigma }{2{\epsilon }_0}}$
Where,
$E$ is electric field due to infinite sheet
$\sigma$ is the surface charge density
${\epsilon }_0$ is the relative permittivity of free space

Complete step by step answer:Observe the diagram

We know that, for an infinite charged sheet
$E={\displaystyle \frac{\sigma }{2{\epsilon }_0}}$
and $V=Ed$
where,
$E$ is electric field
${\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}$ is the permittivity of free space
$\sigma$ is surface charge density of the sheet
$V$ is potential
$d$ is separation between the sheets
Substituting the value of $E$ into the equation of $V$ we get
$V={\displaystyle \frac{\sigma d}{2{\epsilon }_0}}$
By rearranging it, we get
$d={\displaystyle \frac{2V{\epsilon }_0}{\sigma }}$
It is given to us that,
$\sigma =0.10\mu C/m^2$
$\Rightarrow \sigma =0.10\times {10}^{-6}C/m^2$
$\Rightarrow \sigma ={10}^{-7}C/m^2$
$V=50V$
Substituting these values in the above equation, we get
$d={\displaystyle \frac{2\times 50\times 8.85\times {10}^{-12}}{{10}^{-7}}}$
by simplifying it we get
$d=100\times 8.85\times {10}^{-5}$
$=8.85\times {10}^{-3}m$
$\Rightarrow d=0.88mm$
Therefore from the above explanation the correct option is (B) $8.8cm$

Note:Any surface over which the potential is constant is called an equipotential surface. In other words, the potential difference between any two points to an equipotential surface is zero. For this question, we considered the relativity of free space and it was not mentioned that there is any dielectric between the plates. But it is not always necessary. Sometimes, you might have a dielectric between the plates. Then the formula of electric field will be
$E={\displaystyle \frac{\sigma }{2\epsilon }}$
Where, $\epsilon$ is the permittivity of dielectric.