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

The correct option is B $1500\sqrt{2}$ volt

Potential at the centre O, $V=4\times \frac{1}{4\pi {ϵ}_0}.\frac{Q}{\frac{a}{\sqrt{2}}}$
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