Question

What is the electronic potential at the centre of square of side 1 m? The charges $1\times {10}^{-8}C,-2\times {10}^{-8}C,3\times {10}^{-8}C\text{and}2\times {10}^{-8}C$ are placed at the corners of the square.

Answer 1

$q_A={10}^{-8}C$

$q_B=-2\times {10}^{-8}C$

$q_B=3\times {10}^{-8}C$

$q_D=2\times {10}^{-8}C$

$a=1m$

$r=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}m$

$V_A=\frac{1}{4\pi {\epsilon }_0}.\frac{q_A}{r}$ $=9\times {10}^9\times {10}^{-8}\times \sqrt{2}=90\sqrt{2}J$

$V_B=\frac{1}{4\pi {\epsilon }_0}.\frac{q_B}{r}$ $=9\times {10}^9\times {10}^{-8}\times -2\sqrt{2}=-180\sqrt{2}J$

$V_C=\frac{1}{4\pi {\epsilon }_0}.\frac{q_C}{r}$ $=9\times {10}^9\times {10}^{-8}\times 3\sqrt{2}=270\sqrt{2}J$

$V_D=\frac{1}{4\pi {\epsilon }_0}.\frac{q_D}{r}$ $=9\times {10}^9\times {10}^{-8}\times 2\sqrt{2}=180\sqrt{2}J$

$V_{net}=V_A+V_B+V_C+V_D=360\sqrt{2}J$