A) \[Q\]
B) \[\frac{Q}{2}\]
C) \[-\frac{Q}{2}\]
D) \[-Q\]
Correct Answer: D
Solution :
Total force acting on charge Q \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4Q.Q}{{{(l)}^{2}}}\] According to question \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4Q.Q}{{{(l)}^{2}}}=0\] \[\frac{4q}{{{l}^{2}}}+\frac{4Q}{{{l}^{2}}}=0\] \[\therefore \] \[q=-Q\]You need to login to perform this action.
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