Question

Charges of $+\frac{10}{3} \times 10^{-9} C$ are placed at each of the four corners of a square of side $8\, cm$. 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 $O,$
$V=4 \times \frac{1}{4 \pi \varepsilon_{0}} .\frac{Q}{a / \sqrt{2}}$
where $Q=\frac{10}{3} \times 10^{-9} C$
and $a=8\, cm =8 \times 10^{-2} m$

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