A) \[-4q\]
B) \[-2q\]
C) \[-\frac{q}{2}\]
D) \[-q\]
Correct Answer: D
Solution :
net\[F=0\] \[\therefore \] \[0=\frac{4q\times q}{{{l}^{2}}}+\frac{Q\times q}{{{(l/2)}^{2}}}\] \[=\frac{4{{q}^{2}}}{{{l}^{2}}}+\frac{4Qq}{{{l}^{2}}}=\frac{4{{q}^{2}}+4Qq}{{{l}^{2}}}\] \[\therefore \] \[Q=-q\]You need to login to perform this action.
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