Question

A circular ring of radius $R$ with uniform positive charge density $\lambda$ is fixed in the $y-z$ plane with its centre at the origin $O$. A particle of mass $m$ and charge $+q$ is projected from point $P$ $(\sqrt{3}R,0,0)$ on the positive $x$-axis directly towards $O$ with initial velocity $v$ such that the particle does not come back to $P$. The minimum value of $v$ is

• A. $\sqrt{\frac{q\lambda }{2{ϵ}_{0}m}}$
  
• B. $\sqrt{\frac{2q\lambda }{{ϵ}_{0}m}}$
• C. $\sqrt{\frac{q\lambda }{{ϵ}_{0}m}}$
  
• D. $\sqrt{\frac{3q\lambda }{2{ϵ}_{0}m}}$

Answer 1

The correct option is A $\sqrt{\frac{q\lambda }{2{ϵ}_{0}m}}$


Given that,
Radius of the circular ring $=R$
Charge density of the ring $=\lambda$
Mass of the particle $=m$
Point of projection of particle $=P(\sqrt{3}R,0,0)$
Initial velocity of the particle $=v$

Let the total charge on the ring is $Q$,
$\Rightarrow Q=2\pi R\lambda$

The initial electric potential energy at point $P$ is, $U_i=q{V}_p$
$\Rightarrow U_i=q\times \frac{kQ}{\sqrt{R^2+{\left(\sqrt{3}R\right)}^2}}$

$\Rightarrow U_i=\frac{kQq}{2R}$

$\Rightarrow U_i=\frac{\left(2\pi R\lambda \right)q}{4\pi {ϵ}_o\times 2R}$

$\Rightarrow U_i=\frac{\lambda q}{4{ϵ}_0}$

Now, final potential energy when the particle is at the centre is,
$U_f=\frac{kQq}{R}$
$U_f=\frac{\left(2\pi R\lambda \right)q}{4\pi {ϵ}_{0}R}$
$\Rightarrow U_f=\frac{\lambda q}{2{ϵ}_0}$

Initial kinetic energy of the particle is,
$E_i=\frac{1}{2}m{v}^2$

For initial velocity to be minimum, the final velocity of the particle at the centre of the ring should be just zero.
$\Rightarrow E_f=0$

So, using law of conservation of mechanical energy,
$E_i+U_i=E_f+U_f$
$\Rightarrow \frac{1}{2}m{v}^2+\frac{\lambda q}{4{ϵ}_0}=0+\frac{\lambda q}{2{ϵ}_0}$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{\lambda q}{4{ϵ}_0}$
$\Rightarrow v=\sqrt{\frac{q\lambda }{2{ϵ}_{0}m}}$