Physics - Electrostatics Question with Solution | TestHub

The cube has side $\ell$, with +q at six corners and -q at the two ends of one body diagonal.

Adding the signed contributions of all pairs, the 12 edge pairs and 12 face-diagonal pairs cancel out completely.

Only the 4 body diagonals survive: three (+q,+q) pairs and one (-q,-q) pair, each contributing $+\frac{k{q}^2}{\sqrt{3}\ell }$.

So $U=\frac{4k{q}^2}{\sqrt{3}\ell }=\frac{4k{q}^2}{\sqrt{N}\ell }$, giving N = 3.