A) \[\frac{-Q}{2}\]
B) \[\frac{-Q}{4}\]
C) \[-4Q\]
D) \[\frac{+Q}{2}\]
Correct Answer: B
Solution :
Let the distance between the two charges Q and Q be r. Force on Q due to charge q at a distance (r/2) \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{qQ}{{{(r/2)}^{2}}}\] Force on Q due to other charge Q at a distance r \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{Q+Q}{{{(r)}^{2}}}\] \[\therefore \]\[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{qQ}{{{(r/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{Q\times Q}{{{r}^{2}}}=0\] \[4q=-Q\]or \[q=-Q/4\] Hence, the correction option is [b].
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