Two point charges `100 mu C` and `5 mu C` are placed at points `A and B` respectively with `AB = 40 cm`. The work done by external froce in displacing the charge `5 mu C` from B to C, where `BC = 30 cm`, angle `ABC = (pi)/(2)` and `(1)/(4pi epsilon_(0)) = 9 xx 10^(9)Nm^(2)//C^(2)`

Options

A. `9 J`
B. `(81)/(20)J`
C. `(9)/(25) J`
D. `-(9)/(4) J`

Answer

D

Explanation

Work done in displacement charges of `5 mu C` from B to C is
`W = 5xx 10^(-6) (V_(C) - V_(B))` where

`V_(B) = 9 xx 10^(9) xx (100 xx 10^(-6))/(0.4) = (9)/(4) xx 10^(6) V`
and `V_(C) = 9 xx 10^(9) xx (100 xx 10^(-6))/(0.5) = (9)/(5) xx 10^(6) V`
So `W = 5 xx 10^(-6) xx ((9)/(5) xx 10^(6) - (9)/(4) xx 10^(6)) = - (9)/(4)J`