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

`sqrt(b^(2)+2b^(2))/(2)=b/2 sqrt3` is the centre
Distance of centre of the cube form one vertex is
`b/2 sqrt3`. Hence potential due to charge `=(1)/(4pi epsi_(0)) q/(b/2 sqrt3)`
Total potential `=(8 xx q)/(4pi epsi_(0) ""b/2 sqrt3)`
`=(2 xx 8 xx q)/(4pi epsi_(0)b sqrt3)=(4q)/(sqrt3 b pi epsi_(0))`
Field =zero