question_answer

A charge \[(-q)\] and another charge \[(+Q)\] are kept at two points A and B respectively. Keeping the charge \[(+Q)\] fixed at B, the charge \[(-q)\] at A is moved to another point C such that ABC forms an equilateral triangle of side l. The net work done in moving the charge \[(-q)\] is [MP PET 2001]

A)            \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{l}\]

B)                                      \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{l}^{2}}}\]

C)            \[\frac{1}{4\pi {{\varepsilon }_{0}}}Qql\]

D)                                      Zero

Correct Answer: D

Solution :

             According to figure, potential at A and C are equal. Hence work done in moving ? q charge from A to C is zero.