Question

There is a spherical cavity of radius R/2 in uniformly charged spherical region having charge density r and radius R. A small charged particle having q0 charge is released at point 0. It will collide with the wall of cavity with kinetic energy

Answer 1

$\begin{array}{c}\overrightarrow{E}=\frac{\rho \frac{R}{2}}{3t_0}=\frac{\rho R}{6t_0}\\ \overrightarrow{F}=\frac{\rho R{q}_0}{6E_0}\\ \Rightarrow a=\frac{\rho R{q}_0}{6t_{0}m}\\ \therefore v^2-0=2\times \frac{\rho R^2{q}_0}{6t_{0}m}\times \frac{R}{2}\\ \Rightarrow v^2=\frac{\rho R^2{q}_0}{6t_{0}m}\\ \frac{m{v}^2}{2}=\frac{\rho R^2{q}_0}{12t_0}\\ \therefore KE=\frac{\rho R^2{q}_0}{12t_0}\end{array}$