Question

Two concentric rings placed in a gravity free region in yz-plane one of radius R carries a charge + Q and second of radius 4R and charge -8Q distributed uniformly over it. The minimum velocity with which a point charge of mass m and charge +q should be projected from a point at a distance 3R from the centre of rings on its axis so that it will reach to the centre of the rings is given by:

$v=\sqrt{\frac{Qq}{2\pi \epsilon mR}\left(\frac{3\sqrt{x}-5}{5\sqrt{x}}\right)}$

The value of $x$ is:

Answer 1

We know that potential on the axis of a ring having charge $Q$ and radius $r$ at distance $x$ is given by-

$V=\frac{Q}{4\pi {ϵ}_o\sqrt{r^2+x^2}}$ and hence potential energy of a particle of charge $q$ at that point is

$U=\frac{Qq}{4\pi {ϵ}_o\sqrt{r^2+x^2}}$

Hence, initial potential energy of the test charge due to both rings is

$U_1=\frac{Qq}{4\pi {ϵ}_o\sqrt{R^2+(3R{)}^2}}+\frac{-8Qq}{4\pi {ϵ}_o\sqrt{(4R{)}^2+(3R{)}^2}}$

$⟹U_1=\frac{Qq}{4\pi {ϵ}_{o}R}(\frac{1}{\sqrt{10}}-\frac{8}{5})$

Final potential energy when the test charge reaches the center of the rings-

$U_2=\frac{Qq}{4\pi {ϵ}_{o}R}+\frac{-8Qq}{4\pi {ϵ}_o\left(4R\right)}$

$⟹U_2=\frac{-Qq}{4\pi {ϵ}_{o}R}$

Now let initial velocity of projection be $u$ .

For, minimum speed, kinetic energy of test particle should be 0 when it reaches the center.

From Conservation of Energy:-

$U_1+K_1=U_2+K_2$

$⟹\frac{Qq}{4\pi {ϵ}_{o}R}(\frac{1}{\sqrt{10}}-\frac{8}{5})+\frac{1}{2}m{u}^2=\frac{-Qq}{4\pi {ϵ}_{o}R}$

$⟹\frac{1}{2}m{u}^2=\frac{Qq}{4\pi {ϵ}_{o}R}(\frac{3}{5}-\frac{1}{\sqrt{10}})$

$u=\sqrt{\frac{Qq}{2\pi {ϵ}_{o}mR}\left(\frac{3\sqrt{10}-5}{5\sqrt{10}}\right)}$

Hence, $x=10$.