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Two charges `+4e` and `+e` are at a distance `x` apart. At what distance,a charge `q` must be placed from charge `+e` so that is in equilibriumA. `x//2`B. `2x//3`C. `x//3`D. `x//4` |
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Answer» Correct Answer - D For equilibrium of `q` `|F_(1)|=|F_(2)|` `(BMS_OBJ_XII_C01_E01_066_S01.png" width="80%"> Which gives `x_(2)= (x)/(sqrt(Q_(1)/(Q_(2)))+1)=(x)/(sqrt(9e)/(e)+1)=(x)/(4)` |
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