question_answer

It is required to hold equal charges \[q\] in equilibrium at the comers of a square. What charge when placed at the center of the square will do this?

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

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

C) \[\frac{q}{4}(1+2\sqrt{2})\]       

D) \[-\frac{q}{4}(1+2\sqrt{2})\]

Correct Answer: D

Solution :

[d] \[AC=\sqrt{2l}=BD\] or \[BO=\frac{1}{\sqrt{2}}\] \[{{F}_{BO}}={{F}_{BD}}+({{F}_{BA}}+{{F}_{BC}})cos45{}^\circ \] Solving, we get \[Q=\frac{q}{4}(1+2\sqrt{2})\]and Q should be negative of q.