Question

Charges of $+\frac{10}{3}\times 10^{-9}C$ are placed at each of the four corners of a square of side $8cm$. The potential at the intersection of the diagonals is

• A. $150\sqrt{2}$ volt
• B. $1500\sqrt{2}$ volt
• C. $900\sqrt{2}$ volt
• D. 900 volt

Answer 1

Answer: (B)

Potential at the centre $O,$
$V=4\times \frac{1}{4\pi {\epsilon }_0}.\frac{Q}{a/\sqrt{2}}$
where $Q=\frac{10}{3}\times 10^{-9}C$
and $a=8cm=8\times 10^{-2}m$

So $V=4\times 9\times 10^9\times \frac{\frac{10}{3}\times 10^{-9}}{\frac{8\times 10^{-2}}{\sqrt{2}}}$
$=1500\sqrt{2}$ volt