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

We know that the length of diagnal of the Distance between centre of the cube and each vertex `r=(bsqrt(3))/(2)`
`V=(1)/(4piepsi_0)(q)/(r)`
and 8 charges each of value q are present at the eight vertices of the cube therefore `therefore V=(1)/(4piepsi_0)(8q)/(bsqrt(3)//2) or V=(4q)/(sqrt(3)piepsi_0b)`
Further electric field intensity at the centre due to all the eight charge is zero because the fields due to individual charges cancel in pairs.