Question

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 :

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

At any instant ${}^{\prime }{t}^{\prime }$
Total energy of charge distribution is constant
i.e. $\frac{1}{2}m{V}^2+\frac{K{Q}^2}{2R}=0+\frac{K{Q}^2}{2R_0}$
$\therefore \frac{1}{2}m{V}^2=\frac{K{Q}^2}{2R_0}-\frac{K{Q}^2}{2R}$
$\therefore V=\sqrt{\frac{2}{m}\frac{K{Q}^2}{2}.\left(\frac{1}{R_0}-\frac{1}{R}\right)}$
$\therefore V=\sqrt{\frac{K{Q}^2}{m}\left(\frac{1}{R_0}-\frac{1}{R}\right)}=C\sqrt{\frac{1}{R_0}-\frac{1}{R}}$
Also the slope of v-s curve will go on decreasing