Q 12-02-020NEETEAMCET 2008 (Medical)Easy
A charge $Q$ is placed at each corner of a cube of side $a$. The potential at the centre of the cube is
Answer: (C) $\dfrac{4Q}{\sqrt{3}\,\pi\epsilon_0 a}$
Each corner is at $\dfrac{\sqrt{3}a}{2}$ from the centre. Potentials add as scalars:
$$V = 8 \times \frac{Q}{4\pi\epsilon_0}\cdot\frac{2}{\sqrt{3}a} = \frac{4Q}{\sqrt{3}\,\pi\epsilon_0 a}$$
Solution by Sreeraj P, M.Sc Physics