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.

Explanation

Suppose, side of cube is b. The length of the diagonal of the cube `= sqrt(3) b`

The distance between the centre of the cube and one of the eight vertices `r= (sqrt(3)b)/(2)`
The electric potential at the centre of the cube is,
`U=(8kq)/(r) =(8kq)/(sqrt(3)b//2)=(16q)/(4piin_(0)xxsqrt(3)b)`
`:. U =(4q)/(sqrt(3)piin_(0)b)`
And electric field at the centre of the cube gets canceUed due to pair of charges produced electric field in opposite direction with equal magnitude and hence resultant electric field is zero.