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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 A. `8 q_(2)`B. `8 q_(1)`C. `6 q_(2)`D. `6 q_(1)` |
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Answer» Correct Answer - A 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)`. |
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