JEE PYQ: Electrostatics Question 62
Question 62 - 2019 (12 Jan Shift 2)
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:
(1) Curve starting at $R_0$ increasing and saturating
(2) S-shaped curve
(3) Curve increasing and leveling off at $V_0$
(4) Curve starting steep then gradually leveling
Show Answer
Answer: (3)
Solution
Total energy is conserved. $\frac{1}{2}mV^2 + \frac{KQ^2}{2R} = \frac{KQ^2}{2R_0}$. $V = \sqrt{\frac{KQ^2}{m}\left(\frac{1}{R_0} - \frac{1}{R}\right)}$. As $R \to \infty$, $V \to V_{\max} = \sqrt{\frac{KQ^2}{mR_0}}$. The slope of $V$-$R$ curve decreases as $R$ increases, matching option (3).
BETA
