Question

A conducting balloon of the radius $a$ is charged to potential $V_0$ and held at a large height above the earth ensures that charge distribution on the surface of the balloon remains unaffected by the presence of the earth. It is connected to the earth through a resistance $R$ and a valve in the balloon is opened. The gas inside the balloon escapes from valve and the size of the balloon decreases. The rate of decrease in radius of the balloon is controlled in such a manner that potential of the balloon remains constant. Assume the electric permittivity of the surrounding air equals to that of free space $\left({\epsilon }_0\right)$ and charge cannot leak to the surrounding air.



The rate at which radius $r$ of the balloon changes with time is best represented by the equation

• A. $\frac{dr}{dt}=\frac{1}{4\pi {\epsilon }_{0}R}$
• B. $\frac{dr}{dt}=-\frac{1}{4\pi {\epsilon }_{0}R}$
• C. $\frac{dr}{dt}=\frac{1}{4\pi {\epsilon }_{0}aR}$
• D. $\frac{dr}{dt}=-\frac{r}{4\pi {\epsilon }_{0}aR}$

Answer 1

Answer: (B)

For a sphere $V_0=\frac{q}{4\pi {\epsilon }_{0}r}$ , where $r$ is the radius
$\therefore q=V_{0}4\pi {\epsilon }_{0}r$
$-\frac{dq}{dt}=\frac{V_0}{R}$
$-V_{0}4\pi {\epsilon }_0\frac{dr}{dt}=\frac{V_0}{R}$
$\frac{dr}{dt}=-\frac{1}{4\pi {\epsilon }_{0}R}$ ​​​​