Question

A charge $‘Q^{\prime }$ is placed at each corner of a cube of side $‘a^{\prime }$. The potential at the centre of the cube is :

• A. $\frac{8Q}{\pi {\epsilon }_{0}a}$
• B. $\frac{4Q}{4\pi {\epsilon }_{0}a}$
• C. $\frac{4Q}{\sqrt{3}\pi {\epsilon }_{0}a}$
• D. $\frac{2Q}{\pi {\epsilon }_{0}a}$

Answer 1

Answer: (C)

$\frac{4Q}{\sqrt{3}\pi {\epsilon }_{0}a}$

Length of body diagonal $=\sqrt{3}a.$
$\therefore$ distance between charge Q and centre $=\frac{\sqrt{3}a}{2}$
We know $V=\frac{kQ}{r}$
As potential is scalar quantity $\Rightarrow$ potential due to all charges at the centre add up

$\Rightarrow V=8\times \frac{k\times Q}{\left(\frac{\sqrt{3}a}{2}\right)}$
$=\frac{4Q}{\sqrt{3}\pi {ϵ}_{o}a}$