Two charges `q_(1)` and `q_(2)` are placed `30 cm` apart, as shown in the figure. A third charge `q_(3)` is moved along the arc of a circle of radius `40 cm` from `C` to `D`. The change in the potential energy o fthe system is `(q_(3))/(4pi epsilon_(0))k`., where `k` is

Options

A. `8q_(1)`
B. `8q_(1)`
C. `4q_(1)`
D. `4q_(1)`

Answer

A

Explanation

Change in potential
Energy `=(q_(3))/(4pi epsi_(0))k`
Since CD is a part of circle, AD=AC=r=40cm
So, BD=10cm
Potential at `C= (1)/(4pi epsi_(0)) (q_(1))/((0.4)) + (1)/(4pi epsi_(0)) (q_(2))/((0.5))`
Potentiat at `D= (1)/(4pi epsi_(0)) (q_(1))/((0.4)) + (1)/(4pi epsi_(0)) (q_(2))/((0.1))`
`Delta V= (1)/(4pi epsi_(0)) (q_(1))/((0.1))- (1)/(4pi epsi_(0)) (q_(2))/((0.5)) rArr Delta U= q_(3) (Delta V)`
`rArr (q_(3))/(4pi epsi_(0)) k= (q_(3)q_(2))/(4pi epsi_(0)) ((1)/(0.1)- (1)/(0.5))`
`rArr k= q_(2) (10-2)`
`rArr k= 8q_(2)`