question_answer

Four charges of magnituded -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) \[-\frac{Q}{4}(1+2\sqrt{2})\]

D)  \[\frac{Q}{4}(1+\sqrt{2})\]

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 is on it. If we consider the charge at comer D. This charge will experience following forces                                 \[{{F}_{A}}=k\frac{{{Q}^{2}}}{{{a}^{2}}},\]                            \[{{F}_{C}}=\frac{k{{Q}^{2}}}{{{a}^{2}}}\]                 \[{{F}_{B}}=\frac{k{{Q}^{2}}}{{{(a\sqrt{2})}^{2}}}\]                 and        \[{{F}_{O}}=\frac{kQq}{{{(al\sqrt{2})}^{2}}}\]                 Force at B away from the centre \[={{F}_{A}}+{{F}_{B}}+{{F}_{C}}\]                 \[=\sqrt{F_{A}^{2}+F_{C}^{2}}+{{F}_{B}}\] \[=\sqrt{2}\frac{k{{Q}^{2}}}{{{a}^{2}}}+\frac{k{{Q}^{2}}}{2{{a}^{2}}}\]                 \[=\frac{k{{Q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)\]                 Force at D towards the centre \[{{F}_{O}}\frac{2k\,qQ}{{{a}^{2}}}\]                 For equilibrium of charge at B.                 \[{{F}_{C}}+{{F}_{A}}+{{F}_{B}}={{F}_{O}}\]                 \[\frac{k{{Q}^{2}}}{{{a}^{2}}}\left( \sqrt{2}+\frac{1}{2} \right)=\frac{2kQq}{{{a}^{2}}}\]                 \[q=\frac{Q}{4}(1+2\sqrt{2})\]