Given:
Sides of the cube $=b$
Charge at the vertices $=q$
Diagonal of one of the sides of the cube
$d^{2}=\sqrt{b^{2}+b^{2}}=\sqrt{2 b^{2}}$
$\mathrm{d}=\mathrm{b} \sqrt{2}$
Length of the diagonal of the cube
$l^{2}=\sqrt{d^{2}+b^{2}}=\sqrt{2 b^{2}+b^{2}}=\sqrt{3 b^{2}}$
$1=b \sqrt{3}$
The distance between one of the vertices and the centre of the cube is
$r=1 / 2=(b \sqrt{3 / 2})$
The electric potential $(V)$ in the cube’s centre is caused by the eight charges at the cube’s vertices, which add up to one volt.
$V=8 q / 4 \pi \varepsilon_{0}$ $V=\frac{8 q}{4 \pi \epsilon_{0}\left(b \frac{\sqrt{3}}{2}\right)}=\frac{4 q}{\sqrt{3} \pi \epsilon_{0} b}$
Therefore, potential at the centre of the cube is $\frac{4 q}{\sqrt{3} \pi \epsilon_{0} b}$
In the centre of the cube, where the eight charges are concentrated, the electric field strength is zero. The charges are dispersed in a symmetrical manner with respect to the cube’s centre of mass. As a result, they are no longer valid.