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,

Question

Aicharya · Accepted Answer

K.E. of $-q$ at $O=q\left(V_{O}-V_{P}ight)$ $=q\left[\frac{Q}{4 \pi \varepsilon_{0} r}-\frac{Q}{4 \pi \varepsilon_{0} \sqrt{r^{2}+(2 \sqrt{2} r)^{2}}}ight]$ $=\frac{2}{3} \frac{q Q}{4 \pi \varepsilon_{0} r}=\frac{q Q}{6 \pi \varepsilon_{0} r}$ $	herefore \quad$ (c)

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…

Solution