Question

A uniformly charged ring of radius $3a$ and total charge $q$ is placed in $xy-$plane centred at origin. A point charge $q$ is moving towards the ring along the $z-$axis and has speed $v$ at $z=4a$. The minimum value of $v$ such that it crosses the origin is

• A. $\sqrt{\frac{2}{m}}{\left(\frac{1}{15}\frac{q^2}{4\pi {\in }_{0}a}\right)}^{1/2}$
• B. $\sqrt{\frac{2}{m}}{\left(\frac{2}{15}\frac{q^2}{4\pi {\in }_{0}a}\right)}^{1/2}$
• C. $\sqrt{\frac{2}{m}}{\left(\frac{1}{5}\frac{q^2}{4\pi {\in }_{0}a}\right)}^{1/2}$
• D. $\sqrt{\frac{2}{m}}{\left(\frac{4}{15}\frac{q^2}{4\pi {\in }_{0}a}\right)}^{1/2}$

Answer 1

The correct option is B $\sqrt{\frac{2}{m}}{\left(\frac{2}{15}\frac{q^2}{4\pi {\in }_{0}a}\right)}^{1/2}$

Given:
Radius of ring $=3a$
Charge on ring $=q$
Speed of point charge at $(z=4a)$ along $z-$axis $=v$
Potential at any point of the charged ring
$V_p=\frac{Kq}{\sqrt{R^2+Z^2}}$
$R=3a$
$Z=4a$
$l=\sqrt{R^2+Z^2}=5a$
$V_p=\frac{Kq}{5a}$

The minimum velocity $\left(v_0\right)$ should just sufficient to reach the point charge at the center, therefore
$\frac{1}{2}m{v}_0^2=q[V_C-V_P]=q[\frac{Kq}{3a}-\frac{Kq}{5a}]$

$\Rightarrow v_0^2=\frac{4K{q}^2}{15ma}=\frac{4}{15}\frac{1}{4\pi {\in }_0}\frac{q^2}{ma}$

$\Rightarrow v_0=\sqrt{\frac{2}{m}}{\left(\frac{2q^2}{15\times 4\pi {\in }_{0}a}\right)}^{1/2}$

Hence option $\left(C\right)$ is correct.