Q160 / 160 · WBJEE 2011 · Physics
Four charges equal to \(-Q\) are placed at the four corners of a square and a charge \(q\) is at its centre. If the system is in equilibrium, the value of \(q\) is
- A\(\frac{-Q}{4}(1+2 \sqrt{2})\)
- Correct answerB\(\frac{Q}{4}(1+2 \sqrt{2})\)
- C\(\frac{-\mathrm{Q}}{2}(1+2 \sqrt{2})\)
- D\(\frac{\mathrm{Q}}{2}(1+2 \sqrt{2})\)
Solution
Hints: 
\(\begin{aligned} & \mathrm{F}^{\prime}=\sqrt{2} \mathrm{~F}+\frac{\mathrm{F}}{2}=\mathrm{F}\left(\sqrt{2}+\frac{1}{2}\right) \\ & \frac{\mathrm{q} Q}{\left(\frac{\sqrt{2} \mathrm{a}}{2}\right)^2}=\frac{\mathrm{Q}^2}{\mathrm{a}^2}\left(\sqrt{2}+\frac{1}{2}\right), \mathrm{q}=\frac{\mathrm{Q}}{2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & q=\frac{Q}{4}(2 \sqrt{2}+1)\end{aligned}\)
\(\begin{aligned} & \mathrm{F}^{\prime}=\sqrt{2} \mathrm{~F}+\frac{\mathrm{F}}{2}=\mathrm{F}\left(\sqrt{2}+\frac{1}{2}\right) \\ & \frac{\mathrm{q} Q}{\left(\frac{\sqrt{2} \mathrm{a}}{2}\right)^2}=\frac{\mathrm{Q}^2}{\mathrm{a}^2}\left(\sqrt{2}+\frac{1}{2}\right), \mathrm{q}=\frac{\mathrm{Q}}{2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & q=\frac{Q}{4}(2 \sqrt{2}+1)\end{aligned}\)
Hence, the correct answer is option: 'B'.