Question

A point charge $-q$ of mass $m$ is released with negligible speed from a distance $\sqrt{3}R$ on the axis of a fixed uniformly charged ring of charge $Q$ and radius $R$. Find out the velocity of point charge when it reaches the centre of ring.

• A. $\sqrt{\frac{KQq}{2mR}}$
• B. $\sqrt{\frac{KQq}{mR}}$
• C. $\sqrt{\frac{2KQq}{mR}}$
• D. $\sqrt{\frac{KQq}{3mR}}$

Answer 1

The correct option is B

$\sqrt{\frac{KQq}{mR}}$

As potential due to uniformly charged ring along its axis (at $x$ distance) is :

$V=\frac{kQ}{\sqrt{R^2+x^2}}[\because k=\frac{1}{4\pi {\epsilon }_0}]$



Let $\text{A}$ be the point where $-q$ charge is placed along the axis.

So, potential at point $\text{A}$ due to ring:

$V_A=\frac{kQ}{\sqrt{R^2+3R^2}}$

$\Rightarrow V_A=\frac{kQ}{2R}$

Potential at point $\text{B}$ due to ring:

$V_B=\frac{kQ}{R}$

So, potential energy of charge $-$q at point $\text{A}$,

${P.E}_A=\frac{-kQq}{2R}...\left(1\right)$

So, potential energy of charge $-q$ at point B:

${P.E}_B=\frac{-kQq}{R}...\left(2\right)$

Now by energy conservation:

${P.E}_A+{K.E}_A={P.E}_B+{K.E}_B...\left(3\right)$


Since, the particle is released with negligible speed 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 at the centre ( 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

$\frac{-kQq}{2R}+0=\frac{-kQq}{R}+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{kQq}{2R}$

$\therefore v=\sqrt{\frac{kQq}{mR}}$

Hence, option (b) is the correct answer.