1. Tardigrade
  2. Question
  3. Physics
  4. There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R (t) is :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R (t)$ is :

JEE MainJEE Main 2019Electrostatic Potential and Capacitance

Solution:

At any instant $'t'$
Total energy of charge distribution is constant
i.e. $\frac{1}{2} mV^{2} + \frac{KQ^{2}}{2R} = 0 + \frac{KQ^{2}}{2R_{0}} $
$ \therefore \frac{1}{2} mV^{2} = \frac{KQ^{2}}{2R_{0}} - \frac{KQ^{2}}{2R} $
$\therefore V = \sqrt{\frac{2}{m} \frac{KQ^{2}}{2}. \left(\frac{1}{R_{0}} - \frac{1}{R}\right)} $
$\therefore V = \sqrt{\frac{KQ^{2}}{m}\left(\frac{1}{R_{0}} -\frac{1}{R}\right)} =C \sqrt{\frac{1}{R_{0}} -\frac{1}{R}} $
Also the slope of v-s curve will go on decreasing