question_answer

Four charges equal to\[-Q\]are placed at the four comers of a square and a charge q is at its centre. If the system is in equilibrium, the value of q is

A) \[-\frac{Q}{4}(1+2\sqrt{2})\]

B)        \[\frac{Q}{4}(1+2\sqrt{2})\]       

C)        \[-\frac{Q}{2}(1+2\sqrt{2})\]

D)        \[\frac{Q}{2}(1+2\sqrt{2})\]

Correct Answer: B

Solution :

The system is in equilibrium means the force experienced by each charge is zero. It is clear that charge placed at centre would be in equilibrium for any value of q, so we are considering the equilibrium of charge placed at any corner. For equilibrium of charges, \[{{F}_{CD}}+{{F}_{CA}}\cos {{45}^{o}}+{{F}_{CO}}\cos {{45}^{o}}=0\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(-Q)(-Q)}{{{a}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-Q)(-Q)}{{{(\sqrt{2}a)}^{2}}}\]                 \[\times \frac{1}{\sqrt{2}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-Q)q}{{{(\sqrt{2}a/2)}^{2}}}\times \frac{1}{\sqrt{2}}=0\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}^{2}}}{{{a}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}^{2}}}{2{{a}^{2}}}.\frac{1}{\sqrt{2}}\]                                 \[-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2Qq}{{{a}^{2}}}\times \frac{1}{\sqrt{2}}=0\] \[\Rightarrow \] \[Q+\frac{Q}{2\sqrt{2}}-\sqrt{2}q=0\] \[\Rightarrow \] \[2\sqrt{2}Q+Q-4q=0\] \[\Rightarrow \]                               \[4q=(2\sqrt{2}+1)Q\] \[\Rightarrow \]                               \[q=(2\sqrt{2}+1)\frac{Q}{4}\]