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

Answer

D

Explanation

Key Idea : Work done is equal to change in potential energy.
In Ist case, when charge +Q is situated at C. Electric potential energy of system in that case

`U_(1)=(1)/(4pi epsilon_(0))((q)(-q))/(2L)+(1)/(4 pi epsilon_(0))((-q)Q)/(L)+(1)/(4 pi 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)/(4piepsilon_(0))((q)(-q))/(2L)+(1)/(4pi epsilon_(0))(qQ)/(3L)+(1)/(4 pi epsilon_(0))((-q)(Q))/(L)`
`:.` Work done `=Delta U = U_(2) - U_(1)`
`=(-(1)/(4pi epsilon_(0))(q^(2))/(2L)+(1)/(4pi epsilon_(0))(qQ)/(3L)-(1)/(4 pi epsilon_(0))(qQ)/(L))`
`-(-(1)/(4 pi epsilon_(0))(q^(2))/(2L)-(1)/(4 pi epsilon_(0)).(qQ)/(L)+(1)/(4 pi epsilon_(0)).(qQ)/(L))`
`=(qQ)/(4 pi epsilon_(0))[(1)/(3L)-(1)/(L)]`
`=c(qQ)/(4 pi epsilon_(0))(((1-3))/(3L))=(-2qQ)/(12 pi epsilon_(0)L)=-(qQ)/(6 pi epsilon_(0)L)`.