An infinite nonconducting sheet has a surface charge density is$\mathit{\sigma }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathit{\mu }\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$on one side. How far apart are equipotential surfaces whose potentials differ by 50 V?

The electric field produced by an infinite sheet of charge is normal to the sheet and is uniform.

The magnitude of the electric field produced by the infinite sheet of charge is

$E=\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$2{\epsilon }_0$}\right.$

Where, $\sigma$is the surface charge density and ${\epsilon }_0$is the permittivity of free space having a value $8.85\times {10}^{-12}{C}^2/N·m^2$.

Place the origin of a coordinate system at the sheet and take the x-axis to be parallel to the field and positive in the direction of the field. Then the electric potential is

$\begin{array}{rcl}V& =& V_s-\underset{0}{\overset{x}{\int }}Edx\\ & =& V_s-Ex\end{array}$

Where, V_s is the potential at the sheet.

The equipotential surfaces are surfaces of constant x; that is, they are planes that are parallel to the plane of charge.

If two surfaces are separated by $∆x$then their potentials difference in magnitude by,

$\begin{array}{rcl}∆V& =& E∆x\\ & =& \left(\raisebox{1ex}{$\sigma $}\!\left/ \!\raisebox{-1ex}{$2{\epsilon }_0$}\right.\right)∆x\end{array}$

Rearrange the above equation for the distance between two surfaces as below.

$∆x=\frac{2{\epsilon }_{\circ }∆V}{\sigma }$

Substitute known values in the above equation.

$\begin{array}{rcl}∆x& =& \frac{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)\left(50V\right)}{0.10\times {10}^{-6}C/m^2}\\ & =& 8.8\times {10}^{-3}m\end{array}$

Hence, the equipotential surface is $\begin{array}{rcl}& & 8.8\times {10}^{-3}m\end{array}$far apart.