Question

A square of side $\sqrt{2}$m has charges of $+2\times {10}^{-9}C,$ $1\times {10}^{-9}C,$ $-2\times {10}^{-9}C,$ and $-3\times {10}^{-9}C,$ respectively at its corners. What is the potential at the centre of the square?

• A. $+8V$
• B. $-8V$
• C. $+18V$
• D. $-18V$

Answer 1

The correct option is A $+8V$

Side of the square $=\sqrt{2}m$

charge $=+2\times {10}^{-9}C$

$2\times {10}^{-9}C$

$-2\times {10}^{-9}C$

$-3\times {10}^{-9}C$ respectively

Now,

$V_1=\frac{K_q}{r}$

$=\frac{K\times 2\times {10}^{-9}}{\sqrt{2}}$

$V_2=\frac{K\times 1\times {10}^{-9}}{\sqrt{2}}$

$V_3=\frac{K\times (-2\times {10}^{-9})}{\sqrt{2}}$

$V_4=\frac{K\times 3\times {10}^{-9}}{\sqrt{2}}$

$V_1+V_2+V_3+V_4=\frac{9\times {10}^9\times {10}^{-9}}{\sqrt{2}}=(1-2+2+3)=8V$