In the rectangle, shown below, the two corners have charges `q_(1) = - 5mu C` and `q_(2) = + 2.0 mu C`. The work done in moving a charge `+3.0 mu C` from `B` to `A` is (take `1//4 pi epsilon_(0) = 10^(10) N-m^(2)//C^(2))`

Options

A. `2.8 J`
B. `3.5 J`
C. `4.5 J`
D. `5.5 J`

Answer

A

Explanation

Work done `W = 3 xx 10^(-6) (V_(A) - V_(B))`, where
`V_(A) = 10^(10) [((5 xx 10^(-6)))/(15 xx 10^(-2)) + (2 xx 10^(-6))/(5xx10^(-2))] = (1)/(15) xx 10^(6)` volt
and `V_(B) = 10^(10) [((2xx 10^(-6)))/(15 xx 10^(-2)) + (5 xx 10^(-6))/(5xx10^(-2))] = - (13)/(15) xx 10^(6)` volt
`:.W = 3 xx 10^(-6) [(1)/(15) xx 10^(6) - (-(13)/(15) xx 10^(6))] = 2.8 J`