A) \[-\frac{Q}{2}\]
B) \[+\frac{Q}{2}\]
C) \[-\frac{Q}{4}\]
D) \[+\frac{Q}{4}\]
Correct Answer: C
Solution :
For system to be in equilibrium. The force between charges Q and q plus the force between charge Q and Q = 0. \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q\theta }{{{l}^{2}}}=0\] or \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{l}_{2}}}[4q+Q]=0\] or \[4q+Q=0\] \[q=-\frac{Q}{4}\]You need to login to perform this action.
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