91Ó°ÊÓ

• Surface charge density of an infinite nonconducting sheet,
  $\mathrm{σ}=+5.80pC/m^2=+5.80×{10}^{-12}C/m^2$.
• The charge of a particle, $q=+1.60×{10}^{-19}C$
• 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,

$\begin{array}{rcl}W& =& \underset{i}{\overset{f}{∫}}qE·ds\\ & =& \frac{q\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\underset{0}{\overset{d}{∫}}dz\\ & =& \frac{q\mathrm{σ}d}{2{\mathrm{Î\mu }}_0}\end{array}$

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

Substitute known values in the above equation.

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

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

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

$\begin{array}{rcl}{V}_f-V_i& =& \frac{-W}{q}\\ & =& -\frac{q\mathrm{σ}d}{2{\mathrm{Î\mu }}_{0}q}\\ & =& -\frac{\mathrm{σ}d}{2{\mathrm{Î\mu }}_0}\end{array}$

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,

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

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