Question

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer 1

Diagonal of the cube face, $d=\sqrt{(}{b}^2+b^2)=b\sqrt{2}$
Diagonal of cube $l=\sqrt{(}{d}^2+b^2)=b\sqrt{3}$
Distance between center of cube and vertices, $r=l/2=b\sqrt{3}/2$
Potential, $v=8q/4\pi {ϵ}_{o}r=4q/\sqrt{3}\pi {ϵ}_{o}b$ (8 due to presence of 8 charges at the vertices)
by symmetry of charge electric field, E=0.