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

From the figure the distance of centre from each vertex in the cube can be calculated as:
`HC^(2)=HF^(2)+CF^(2)`
`rArr HC^(2)=(sqrt(2)b)^(2)+b^(2)=3b^(2)`
`rArr HC=sqrt(3)b`
`therefore HO=(sqrt(3))/(2)b`

Potential at the centre O can be calculated as:
`V=8xx(1)/(4pi epsilon_(0)) (q)/(r )=(4q)/(sqrt(3)pi epsilon_(0)b)`
Net electric field at the centre of the cube is zero due to symmetry.