Question

A cube of side $x$ has a charge $q$ at each of its vertices. The potential due to this charge array at the centre of the cube is

• A. $\frac{4q}{3\pi {\epsilon }_{0}x}$
• B. $\frac{4q}{\sqrt{3}\pi {\epsilon }_{0}x}$
• C. $\frac{3q}{4\pi {\epsilon }_{0}x}$
• D. $\frac{2q}{\sqrt{3}\pi {\epsilon }_{0}x}$

Answer 1

Answer: (B)

The length of diagonal of the cube of each side $x$ is
$\sqrt{3x^2}=x\sqrt{3}$
$\therefore$ Distance between centre of cube and each vertex,
$r=\frac{x\sqrt{3}}{2}$
Now, potential, $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
Since cube has $8$ vertices and $8$ charges each of value $q$ are present there
$\therefore V=\frac{1}{4\pi {\epsilon }_0}\frac{8q}{\frac{x\sqrt{3}}{2}}<br>=\frac{4q}{\sqrt{3}\pi {\epsilon }_{0}x}$