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

`W=q(V_A-V_B)`
`=3xx10^(-6)[1/15+13/15]xx10^6 =42/15=2.83J`
As `V_A=(Kxx(-5muC))/(0.15)+(Kxx2muC)/(0.05)=1/15xx10^6 `Volt
And
`V_B=(Kxx(-5)xx10^6)/(0.05)+(Kxx2 xx10^(-6))/(0.15) =(-13)/15 xx10^(6)` Volt .