$(A)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{\sqrt 3 R} \\ (B)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{3R} \\ (C)\; \frac{1}{4 \pi \in _0} \frac{q\;90}{ R} \\ (D)\;\frac{1}{4 \pi \in _0} \frac{q\;90}{2 R} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Potential due to ring at distance $\sqrt 3 R$

$\qquad= V_1=\large\frac{1}{4 \pi \in _0} \frac{q}{\sqrt {R^2+3R^2}}$

$\qquad= \large\frac{1}{4 \pi \in _0} \frac{q}{2R}$

At centre potential $V_2 =\large\frac{1}{4 \pi \in _0} \frac{q}{R}$

$K.E= q_0 (V_2-V_1)=\large\frac{1}{4 \pi \in _0} \frac{q 90}{2R}$

Hence D is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...