91Ó°ÊÓ

The charge, $q_1=3.40pC=3.40×{10}^{-12}C$

The charge, $q_2=6.00pC=6.00×{10}^{-12}C$

The distance between two charges,$a=39.0cm=0.39m$

The potential due to a collection of point charges is,

$\mathit{V}\mathbf{=}\mathit{k}\mathbf{∑}\frac{{\mathbf{q}}_{\mathbf{i}}}{{\mathbf{r}}_{\mathbf{i}}}$

Here, V is the potential, k is the Coulombs constant having a value $\mathbf{9}\mathbf{×}{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{N}{\mathit{m}}^{\mathbf{2}}\mathbf{/}{\mathit{C}}^{\mathbf{2}}\mathbf{,}\mathbf{}{\mathit{q}}_{\mathbf{i}}$ is the charge, and r_i is the distance between two point charges.

Draw the free body diagram as drawn below.

Let assume $V→0$as $r→∞$.

All corner particles are equidistant from P. And their total charge is,

$\begin{array}{rcl}q& =& 2q_1-3q_1+2q_1-q_1\\ & =& 0\end{array}$

The potential due to a collection of point charges:

$V=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}∑\frac{q_i}{r_i}$

Therefore, V = 0 as q =0

The net potential is then due to the two 4q₂ charges at point B and point E at a distance a/2 from the center.

$\begin{array}{rcl}V& =& k\left(\frac{4q_2}{a/2}+\frac{4q_2}{a/2}\right)\\ & =& k\left(\frac{16q_2}{a}\right)\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}V& =& \frac{\left(9×{10}^{9}N·m^2/C^2\right)×16×\left(6×{10}^{-12}C\right)}{0.39m}\\ & =& 2.21V\end{array}$

Hence, the net electric potential at the rectangle’s center is 2.21 V.