question_answer

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                                      [AIEEE 2004]

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 :

 If all charges are in equilibrium, system is also in equilibrium. Charge at centre: charge q is in equilibrium because no net force acting on it corner charge: If we consider the charge at corner B. This charge will experience following forces \[{{F}_{A}}=k\frac{{{Q}^{2}}}{{{a}^{2}}},\]\[{{F}_{C}}=\frac{k{{Q}^{2}}}{{{a}^{2}}}\],\[{{F}_{D}}=\frac{k{{Q}^{2}}}{{{(a\sqrt{2})}^{2}}}\text{and}\,{{F}_{O}}=\frac{KQq}{{{(a\sqrt{2})}^{2}}}\] Force at B away from the centre = \[{{F}_{AC}}+{{F}_{D}}\] \[=\sqrt{F_{A}^{2}+F_{C}^{2}}+{{F}_{D}}=\sqrt{2}\frac{k{{Q}^{2}}}{{{a}^{2}}}+\frac{k{{Q}^{2}}}{2{{a}^{2}}}=\frac{k{{Q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)\] Force at B towards the centre \[={{F}_{O}}=\frac{2kQq}{{{a}^{2}}}\] For equilibrium of charge at B, \[{{F}_{AC}}+{{F}_{D}}={{F}_{O}}\] Þ \[\frac{K{{Q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)=\frac{2KQq}{{{a}^{2}}}\] Þ \[q=\frac{Q}{4}\left( 1+2\sqrt{2} \right)\]