A cube of side 'a' has a charge Q at each of its vertices, what is the potential due to this charge array at the centre of the cube?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4Q}{\sqrt{3}\pi\varepsilon_0 a} \)
Potential is a scalar, so the potentials of the eight charges at the centre add as numbers, with no directions to resolve. Every vertex is at the same distance from the centre, half the body diagonal.
\( \displaystyle r = \frac{\sqrt{3}\,a}{2} \)
\( \displaystyle V = 8 \times \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r} = \frac{8Q}{4\pi\varepsilon_0}\times\frac{2}{\sqrt{3}\,a} \)
\( \displaystyle V = \frac{16Q}{4\sqrt{3}\pi\varepsilon_0 a} = \frac{4Q}{\sqrt{3}\pi\varepsilon_0 a} \)