Question

A ring of radius $R$ carries a charge $+q$. A test charge $-q_0$ is released on its axis at a distance $\sqrt{3R}$ from its center. How much kinetic energy will be acquired by the test charge when it reaches at the center of the ring-

• A. $\frac{Kq{q}_0}{R}$
• B. $\frac{1}{2}\frac{Kq{q}_0}{R}$
• C. $\frac{Kq{q}_0}{\sqrt{3R}}$
• D. $\frac{2Kq{q}_0}{R}$

Answer 1

The correct option is B $\frac{1}{2}\frac{Kq{q}_0}{R}$

Only electrostatic force is acting, so we can use, $K_f+U_f=K_i+U_i$
We know, potential energy, $U=q\times V$ where $V$ is potential.

Now,$\frac{1}{2}m{v}^2+(-q_0)\frac{Kq}{R}=\frac{1}{2}m\times 0^2+(-q_0)\frac{Kq}{\sqrt{R^2+(\sqrt{3R}{)}^2}}$
Kinetic energy, $\frac{1}{2}m{v}^2=\frac{Kq{q}_0}{2R}$