Question

A cavity of radius $r$ is made inside a solid sphere. The volume charge density of the remaining sphere is $\rho$. An electron (charge $e$ , mass $m$) is released inside the cavity from point $P$ as shown in figure. The center of sphere and center of cavity are seperated by a distance $a$. The time after which the electron again touches the sphere is

• A. $\sqrt{\frac{6\sqrt{2}{\epsilon }_{0}m}{e\rho }}$
• B. $\sqrt{\frac{\sqrt{2}{\epsilon }_{0}m}{e\rho }}$
• C. $\sqrt{\frac{6{\epsilon }_{0}m}{e\rho }}$
• D. $\sqrt{\frac{{\epsilon }_{0}m}{e\rho }}$

Answer 1

The correct option is A $\sqrt{\frac{6\sqrt{2}{\epsilon }_{0}m}{e\rho }}$

Given,

Radius of cavity =$r$



Electric field inside the cavity with respect to center of sphere , $$\overrightarrow{E}=\frac{\rho \overrightarrow{a}}{3{\epsilon }_0}....\left(1\right)$$
Since, acceleration of an electron $$\overrightarrow{A}=\frac{e\overrightarrow{E}}{m}....\left(2\right)$$
Using $\left(1\right)$ in $\left(2\right)$ we get,'

$\overrightarrow{A}=\frac{e\rho \overrightarrow{a}}{3{\epsilon }_{0}m}$ ( in backward direction )

Distance travelled by electron where it hits the wall of cavity is
$$S=2r\cos \theta$$
Substituting the value of $\left(\theta \right)$ we get,

$S=2r\cos {45}^{\circ }=\frac{r}{\sqrt{2}}$

Since, the acceleration is constant, we can use the kinematic relation
$$S=ut+\frac{1}{2}A{t}^2$$
From the data given in the question,

$S=\frac{1}{2}A{t}^2$

$\Rightarrow t=\sqrt{\frac{2S}{A}}$

Substituting the data obtained in the above equation we get,

$t=\sqrt{\frac{6\sqrt{2}r{\epsilon }_{0}m}{epa}}$

Hence, option (a) is the correct answer.