Question

A point charge $q_0$ is placed at the centre of uniformly charged ring of total charge $Q$ and radius $R$. If the point charge is slightly displaced with negligible force along the axis of the ring then find out its speed when it reaches a large distance $(x>>R)$.

$[\text{Take}K=\frac{1}{4\pi {\epsilon }_0}]$

• A. $\sqrt{\frac{KQ{q}_0}{mr}}$
• B. $\sqrt{\frac{KQ{q}_0}{2mr}}$
• C. $\sqrt{\frac{3KQ{q}_0}{mr}}$
• D. $\sqrt{\frac{2KQ{q}_0}{mr}}$

Answer 1

The correct option is D $\sqrt{\frac{2KQ{q}_0}{mr}}$

By energy conservation principle:

${P.E}_A+{K.E}_A={P.E}_B+{K.E}_B...\left(1\right)$


Here,
potential energy of charge $q_0$ at point $\text{A}$,

${P.E}_A=\frac{KQ{q}_0}{R}...\left(2\right)$

So, potential energy of charge $q_0$ at point $\text{B}$ which is very far from centre of the ring $\text{A}$:

${P.E}_B=0...\left(3\right)$

Since, the force on the particle is negligible at point $\text{A}$, so its kinetic energy at this point will be zero i.e.,

$K.E_A=0...\left(4\right)$

Let the speed of the particle at point $\text{B}$ be $v$.

$K.E_B=\frac{1}{2}m{v}^2...\left(5\right)$

Using $\left(1\right),\left(2\right),\left(3\right),\left(4\right)$ and $\left(5\right)$, we get

$\frac{KQ{q}_0}{R}+0=0+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{KQ{q}_0}{R}$

$\therefore v=\sqrt{\frac{2KQ{q}_0}{mR}}$

Hence, option (d) is the correct answer.