Question

Four charges ${10}^{-8},-2\times {10}^{-8},+3\times {10}^{-8}$ and $2\times {10}^{-8}C$ are placed at the four corners of a square of 1 meter $1/4\pi {\epsilon }_0$ $=9\times {10}^{9}N-m^2/C^2$ its potential at the centre Of the square is

• A. 206 V
• B. 308 V
• C. $360\sqrt{2}V$
• D. 90 V

Answer 1

The correct option is C $360\sqrt{2}V$

Potential at centre $O=k\Sigma \frac{q}{r}$

$\begin{array}{c}=\frac{1}{4\pi E_0}\left(\frac{{10}^{-8}}{4\sqrt{2}}-\frac{2\times {10}^{-8}}{4\sqrt{2}}+\frac{3\times {10}^{-8}}{4\sqrt{2}}+\frac{20\times {10}^{-8}}{4\sqrt{2}}\right)\\ =9\times {10}^9\times \sqrt{2}\times w^{-8}\left(1-2+3+2\right)\\ =36\sqrt{2}\times 10=360\sqrt{2}V\end{array}$