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 center of the cube.

Explanation

We know that the length of diagonal of the cube side b is `sqrt(3b^(2)) = b sqrt(3)`
`:.` Distance between center of cube and each vertex, `r = (b sqrt(3))/(2)`
AS `V = (1)/(4pi in_(0)) (q)/(r )`
and 8 charges each of value q are present at the eight verticies of the cube, therefore,
`V = (1)/(4pi in_(0)) (8q)/(b sqrt(3)//2) or V = (4q)/(sqrt(3) pi in_(0) b)`
Further electric field intensity at the center due to all the eight charges is zero, because the fileds due to individual charges cancel in pairs.