Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Two equal point charges are fixed at x =- a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

IIT JEEIIT JEE 2002Electrostatic Potential and Capacitance

Solution:

$ U_i= \frac {2KQq}{a}+ \frac {K.q.q.}{2a} $ and $ U_f=KQ_q \bigg [\frac {1}{a+x}+ \frac {1}{a-x} \bigg ]+ \frac {K.q.q.}{2a} $ Here, $ K= \frac {1}{4 \pi \varepsilon_0} $ $ \Delta U=U_f-U_i $ or $ |\Delta U|= \frac {2KQqx^2}{a^3} $ $ For \, x < < a $ $\therefore \Delta Ux^2 $

Solution Image