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 from rest inside the cavity from point $P$ as shown in the figure. The centre of the sphere and centre of the cavity are separated by a distance $a$ . The time after which the electron again touches the sphere is

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

Answer 1

Answer: (A)

$E=\frac{\rho a}{3{ϵ}_0}F=eE=\frac{e\rho a}{3{ϵ}_0}$
acceleration $=\frac{\rho ea}{3m{ϵ}_0}$



$=\frac{1}{2}\times \frac{\rho ea}{3m{ϵ}_0}\times t^2=\sqrt{2}r$
$\Rightarrow t=\sqrt{\frac{6\sqrt{2}r{ϵ}_{0}m}{e\rho a}}$ The electron will move opposite to electronic field with constant acceleration along shown horizontal line of length $=\sqrt{2}r$