Question

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 the figure. If the ring is fixed and carrying a charge $+Q$, the kinetic energy of charge $-q$ when it is released and reaches the centre of the ring will be

• A. $\frac{qQ}{2\pi {ϵ}_{0}r}$
• B. $\frac{qQ}{6\pi {ϵ}_{0}r}$
• C. $\frac{qQ}{4\pi {ϵ}_{0}r}$
• D. $\frac{qQ}{12\pi {ϵ}_{0}r}$

Answer 1

The correct option is B $\frac{qQ}{6\pi {ϵ}_{0}r}$


From work-energy theorem,

Gain in KE $=$ Loss in PE

KE of $-q$ at $\text{O}=q(V_O-V_P)$

$=q[\frac{Q}{4\pi {ϵ}_{0}r}-\frac{Q}{4\pi {ϵ}_0\sqrt{r^2+(2\sqrt{2}r{)}^2}}]$

$=\frac{qQ}{4\pi {ϵ}_0}[\frac{1}{r}-\frac{1}{\sqrt{r^2+8r^2}}]$

$=\frac{qQ}{4\pi {ϵ}_0}[\frac{1}{r}-\frac{1}{3r}]=\frac{qQ}{6\pi {ϵ}_{0}r}$