Question

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?

Short answer

The equipotential surface is $∆x=8.8\times {10}^{-3}m$ far apart.

Step-by-step solution

Given data:

• The surface charge density, $\sigma =0.10\mu C/m^2=0.10\times {10}^{-6}C/m^2$.
• The potential difference is V = 50 V.

Understanding the concept:

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

Calculate how far apart are equipotential surfaces whose potentials differ by 50 V:

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.