Question

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

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

Answer 1

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

Given $Q=\frac{10}{3}\times {10}^{-9}C$ and $a=8cm=8\times {10}^{-2}m$

Potential due to a point charge $V=\frac{1}{4\pi {ϵ}_0}.\frac{Q}{R}$




Potential at the centre $O$ due to four charges
$V=4\times \frac{1}{4\pi {ϵ}_0}.\frac{Q}{a/\sqrt{2}}$

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