question_answer

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

D)     \[+\frac{Q}{2}(1+2\sqrt{2})\]

Correct Answer: A

Solution :

For equilibrium Net force on each charge must be zero. For -Q due to other three -Q charge net force will be repulsive and its value will be             \[\frac{k{{Q}^{2}}}{{{l}^{2}}}\,\left( \sqrt{2}\,+\frac{1}{2} \right)\] Hence nature of q at centre should be negative and hence it attracts with             \[\frac{KQq}{{{\left( \frac{1\sqrt{2}}{2} \right)}^{2}}}\] So that net force on -Q will be zero. \[\frac{k{{Q}^{2}}}{{{l}^{2}}}\,\left( \sqrt{2}\,+\frac{1}{2} \right)\,=\frac{2kQq}{{{t}^{2}}}\] \[\Rightarrow \,\,q=\frac{Q}{4}\,(2\sqrt{2}+1)\] and is negative \[\Rightarrow \,\,q=-\frac{Q}{4}\,(2\sqrt{2}+1)\].