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

Explanation

Consider a cube ABCDEFGH of side b each and let a charge q be placed at all the 8 vertices of the cube.
We know that distance of centre point o of the cube from any vertex `r = (b sqrt3)/2`.
`:.` Total potential at point O
`V = 8 xx 1/(4pi epsi_0). q/r = 8/(4pi epsi_0) .q/((bsqrt3)/2) = (4q)/(sqrt3pi epsi_0b)`
The resultant electric field at point O is zero because electric field is a vector and the fields due to diagonally opposite charges of cube, being equal but opposite, nullify each other.