An infinite nonconducting sheet has a surface charge density$\mathit{\sigma }\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{}\mathit{p}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. (a) How much work is done by the electric field due to the sheet if a particle of charge $\mathit{q}\mathbf{=}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$is moved from the sheet to a point P at distance d = 3.5 cmfrom the sheet? (b) If the electric potential V is defined to be zero on the sheet, what is V at P ?

• Surface charge density of an infinite nonconducting sheet,
  $\sigma =+5.80pC/m^2=+5.80\times {10}^{-12}C/m^2$.
• The charge of a particle, $q=+1.60\times {10}^{-19}C$
• Distance between the sheet to a point P is $d=3.56cm=0.0356m$
• Initial electric potential, $V_i=0$
• Final electric potential,$V_f=V$

Using the equation of work done by the electric field and relation between work done & electric potential V , you find the work is done by the electric field due to the sheet and electric potential at a point respectively.

The work done by the electric field is,

$\begin{array}{rcl}W& =& \underset{i}{\overset{f}{\int }}qE·ds\\ & =& \frac{q\sigma }{2{\epsilon }_0}\underset{0}{\overset{d}{\int }}dz\\ & =& \frac{q\sigma d}{2{\epsilon }_0}\end{array}$

Here, ${\epsilon }_0$is the permittivity of free space having the value of $8.85\times {10}^{-12}{C}^2/N·m^2$.

Substitute known values in the above equation.

$\begin{array}{rcl}W& =& \frac{\left(1.60\times {10}^{-19}C\right)\left(5.80\times {10}^{-12}C/m^2\right)\left(0.0356m\right)}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}\\ & =& 1.87\times {10}^{-21}J\end{array}$

Hence, the work done by the electric field is $\begin{array}{rcl}& & 1.87\times {10}^{-21}J\end{array}$.

The potential difference between two points in an electric field is,

$\begin{array}{rcl}{V}_f-V_i& =& \frac{-W}{q}\\ & =& -\frac{q\sigma d}{2{\epsilon }_{0}q}\\ & =& -\frac{\sigma d}{2{\epsilon }_0}\end{array}$

Here, V_i is the initial electric potential, and V_f is the final potential, W is the work done and q is the charge.

Since V_i set to be zero on the sheet, the electric potential at P is,

$\begin{array}{rcl}V-0& =& -\frac{\left(5.80\times {10}^{-12}C/m^2\right)\left(0.0356m\right)}{2\left(8.85\times {10}^{-12}{C}^2/N·m^2\right)}\\ V& =& -1.17\times {10}^{-2}V\end{array}$

Hence, the electric potential at point P is $\begin{array}{rcl}& & -1.17\times {10}^{-2}V\end{array}$.