Charges `+q` and `-q` are placed at points `A` and `B` respectively which are a distance `2L` apart, `C` is the midpoint between `A` and `B`. The work done in moving a charge `+Q` along the semicircle `CRD` is

Options

A. `(qQ)/(4pi epsilon_(0)L)`
B. `(qQ)/(2pi epsilon_(0)L)`
C. `(qQ)/(6pi epsilon_(0)L)`
D. `(qQ)/(6pi epsilon_(0)L)`

Answer

D

Explanation

Work done is equal to charge in potential energy.
In Ist case, when charge `+Q` is situated at `C`.
Electric potential energy of system
`U_(1) = (1)/(4pi epsilon_(0)) ((q)(-q))/(2L) + (1)/(4pi epsilon_(0)) ((-q)Q)/(L)+(1)/(4pi epsilon_(0)).(qQ)/(L)`
In IInd case, when charge `+Q` is moved from `C` to `D`.
Electric potential energy of system in that case
`U_(2) = (1)/(4pi epsilon_(0)). ((q)(-q))/(2L) + (1)/(4pi epsilon_(0)) ((q)Q)/(3L)+(1)/(4pi epsilon_(0)).((-q)(Q))/(L)`
work done `= DeltaU = U_(2) - U_(1)`
`= (-(1)/(4pi epsilon_(0))(q^(2))/(2L)+(1)/(4pi epsilon_(0))(qQ)/(3L)-(1)/(4pi epsilon_(0))(qQ)/(L))`
`= (-(1)/(4pi epsilon_(0))(q^(2))/(2L)-(1)/(4pi epsilon_(0)).(qQ)/(L)+(1)/(4pi epsilon_(0))(qQ)/(L))`
`= (qQ)/(4pi epsilon_(0)).[(1)/(3L)-(1)/(L)] = (qQ)/(4pi epsilon_(0)) ((1-3))/(3L)`
`= (-2qQ)/(12pi epsilon_(0)) = -(qQ)/(6pi epsilon_(0)L)`