Q 12-02-187JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
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:
Answer: (A) see figure
The self-energy of a charged shell is $\dfrac{Q^2}{8\pi\varepsilon_0R}$. Energy conservation from rest at $R_0$:
$$\frac12mV^2 = \frac{Q^2}{8\pi\varepsilon_0}\left(\frac1{R_0} - \frac1R\right) \Rightarrow V = V_0\sqrt{1 - \frac{R_0}{R}}$$
$V$ starts from zero at $R_0$, rises steeply at first and approaches a constant $V_0$ as $R \to \infty$, which is graph (1).
Solution by Sreeraj P, M.Sc Physics