A) \[-\frac{q}{2}\]
B) \[-q\]
C) \[\frac{+q}{2}\]
D) Zero
Correct Answer: A
Solution :
Potential energy of the system \[U=\frac{KQq}{l}+\frac{K{{q}^{2}}}{l}+\frac{KqQ}{l}=0\] \[\Rightarrow \] \[\frac{Kq}{l}(Q+q+Q)=0\] \[\Rightarrow \] \[Q=-\frac{q}{2}\]You need to login to perform this action.
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