A) \[-\frac{Q}{4}\]
B) \[-\frac{Q}{2}\]
C) \[+\frac{Q}{2}\]
D) \[+\frac{Q}{4}\]
Correct Answer: A
Solution :
The force on Q at A due to Q at B is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{a}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}^{2}}}{{{a}^{2}}}\] ?(1) the force on Q at C due to q is given by \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{\left( \frac{a}{2} \right)}^{2}}}\] ?(2) These forces should be equal and opposite for the equilibrium of charge Q at A. This is only posible when \[q\] is negative \[\frac{{{Q}^{2}}}{{{a}^{2}}}=-\frac{Qq}{{{(a/2)}^{2}}}\]or \[q=-\frac{Q}{4}\] Thus \[q=-\frac{Q}{4}\]You need to login to perform this action.
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