Question

Four charges $q_1=2\times {10}^{-8}C,q_2=-2\times {10}^{-8}C,$ $q_3=-3\times {10}^{-8}C$ and $q_4=6\times {10}^{-8}C$ are placed at four comers of a square of side $\sqrt{2}m$ . What is the potential at the centre of the square?

• A. 270V
• B. 300V
• C. Zero
• D. 100 V

Answer 1

Answer: (A)

The potential at a point r is given by $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
$\therefore$ $OA=OB=OC=OD=\frac{AC}{2}=\frac{2}{2}$ $=1$ Potential at the centre $V=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1}{OA}+\frac{q_2}{OB}+\frac{q_3}{OC}+\frac{q_4}{OD}\right)$ $V=9\times {10}^9$ $\times \left(\begin{array}{rl}& \frac{2\times {10}^8}{1}+\frac{-2\times {10}^{-8}}{1}\\ & +\frac{-3\times {10}^8}{1}+\frac{6\times {10}^{-8}}{1}\end{array}\right)$