Four charges `+q, -q, +q` and `-q` are placed in order on the four consecutive corners of a square of side a. The work done in interchanging the positions of any two neighboring charges of the opposite sign is

Answer

`(q^2)/(4 pi epsilon_0 a)(4 - 2 sqrt(2))`

Explanation

Let `U_1` be the potential energy of system I and `U_(11)` br the potential energy of system II.
Work done = change in `PE = U_(11) - U_1`
,
`U_1 = - (4q^2)/(4 pi epsilon_0 a) + (2 q^2)/(4 pi epsilon_0 sqrt(2) a)`
`U_(11) = (-2 q^2)/(4 pi epsilon_0 sqrt(2) a)`
`:. W = (-2q^2)/(4 pi epsilon_0 sqrt(2) a) + (4 q^2)/(4 pi epsilon_0 a) -(2 q^2)/(4 pi epsilon_0 sqrt(2) a)`
=`(q^2)/(4 pi epsilon_0 a)(4 - 2 sqrt(2))`.