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_(2)`
B. `8q_(1)`
C. `6q_(2)`
D. `6q_(1)`

Answer

A

Explanation

Change in potential energy `(DeltaU) = U_(f) -U_(i)`

`rArr DeltaU = (1)/(4pi epsilon_(0)) [((q_(1)q_(3))/(0.4)+(q_(2)q_(3))/(0.1))-((q_(1)q_(3))/(0.4)+(q_(2)q_(3))/(0.5))]`
`rArr DeltaU = (1)/(4pi epsilon_(0)) [8q_(2)q_(3)] = (q_(3))/(4pi epsilon_(0))(8q_(2))`
`rArr k = 4q_(2)`