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) $\dfrac{4Q}{\sqrt{3}\,\pi\epsilon_0 a}$
Concept. Potentials add as scalars; each corner is $\dfrac{\sqrt3a}{2}$ from the centre.
Step 1. Sum. $8 \times \dfrac{Q}{4\pi\epsilon_0} \cdot \dfrac{2}{\sqrt3a} = \dfrac{4Q}{\sqrt3\pi\epsilon_0a}$.
Answer. $\dfrac{4Q}{\sqrt3\pi\epsilon_0a}$ (Option C).
Solution by Sreeraj P, M.Sc Physics