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

Answer

A

Explanation

The change in potential energy of the system is `U_(D) - U_(C)` as discussed under.
When charge `q_(3)` is at C, then its potential energy is
`U_(C) = (1)/(4 pi epsilon_(0))((q_(1)q_(3))/(0.4) +(q_(2) q_(3))/(0.1))`
When charge `q_(3)` is at D, then
` U_(C) = (1)/(4 pi epsilon_(0))((q_(1)q_(3))/(0.4) +(q_(2) q_(3))/(0.1))`
Hence, change in potential energy
`(1)/(4 pi epsilon_(0))((q_(2)q_(3))/(0.1)+(q_(2)q_(3))/(0.5))`
`:. (q_(3))/(4 pi epsilon_(0))=(1)/(4 pi epsilon_(0))((q_(2)q_(3))/(0.1)+(q_(2)q_(3))/(0.5))`
`rArr k=q_(2)(10 -2)=8q_(2)`.