question_answer

A charge \[-q\] is placed at the axis of a charged ring of radius r at a distance of \[2\sqrt{2}\,r\] as shown in figure. If ring is fixed and carrying a charge Q, the kinetic  energy of charge -q when it is released and reaches the centre of ring will be      

A)  \[\frac{qQ}{4\pi {{\varepsilon }_{0}}r}\]               

B)  \[\frac{qQ}{12\pi {{\varepsilon }_{0}}r}\]

C)  \[\frac{qQ}{6\pi {{\varepsilon }_{0}}}\]                 

D)  \[\frac{qQ}{2\pi {{\varepsilon }_{0}}r}\]

Correct Answer: C

Solution :

 KE of -q at \[O=q({{V}_{0}}-{{V}_{p}})\] (consider the potential at the axial point, p is \[{{v}_{p}}\])  \[=\frac{2}{3}\,\frac{qQ}{4\pi {{\varepsilon }_{0}}r}=\frac{qQ}{6\pi {{\varepsilon }_{0}}r}\]