If identical charges `(-q)` are placed at each corner of a cube of side `b`, then electric potential energy of charge `(+q)` which is palced at centre of the cube will be

Options

A. `(8sqrt2q^2)/(4 pi epsilon_0b)`
B. `(-8sqrt2q^2)/( pi epsilon_0b)`
C. `(-4sqrt2q^2)/( pi epsilon_0b)`
D. `(-4q^2)/(sqrt3piepsilon_0b)`

Answer

D

Explanation

U = q (Potential at centre) `=qV_0`
`V_0=(-Kq)/(sqrt3b//2)=(-2Kq)/(sqrt3b){{:(d=sqrt(b^2+b^2+b^2)=sqrt3b),(r=d/2=(3sqrt3b)/2):}}`
`U_("system") = (-2Kq^2)/(sqrt3b)xx8`
`=(-8xx2q^2)/(4piepsilon_0sqrt3b)=(-4b^2)/(sqrt3piepsilon_0b)`