Correct Answer - C
The initial change distribution is as shown in the figure.
Initial electrostatic potential energy is
`U_(i) = - (kq^(2))/(a) + (kq^(2))/(b)-(kq^(2))/(b)+(kq^(2))/(2a)+(kq^(2))/(2b)`
`= (kq^(2))/(2b)-(kq^(2))/(2a) = (q^(2))/(8pi epsilon_(0)) [(1)/(b)-(1)/(a)]`
Final potential energy `U_(f) = 0`
`:. DeltaW = U_(f) - U_(i) = (q^(2))/(8pi epsilon_(0)) [(1)/(a)-(1)/(b)]`