Question

A particle of mass $m$ and charge $-q$ moves along a diameter of a uniformly charged sphere of radius $R$ and carrying a total charge $+Q$. The frequency of S.H.M of the particle if the amplitude does not exceed $R$, is $\frac{1}{2\pi }\sqrt{\frac{qQ}{x\pi {\epsilon }_{0}m{R}^3}}$. Find $x$.

Answer 1

Using Gauss's law electric field inside solid sphere , $E.4\pi r^2=\frac{\rho .\left(4/3\right)\pi r^3}{{ϵ}_0}$ where $\rho =$ charge density and here $Q=\rho \left(4/3\right)\pi R^3$.
Thus, $E=\frac{Qr}{4\pi {ϵ}_0{R}^3}$
The force on charge $-q$ is $F=-qE=-\frac{qQr}{4\pi {ϵ}_0{R}^3}$
or $m\frac{d^{2}r}{d{t}^2}=-\frac{qQr}{4\pi {ϵ}_0{R}^3}\Rightarrow \frac{d^{2}r}{d{t}^2}=-\left(\frac{qQ}{4\pi {ϵ}_{0}m{R}^3}\right)r$
Thus, $\omega =2\pi f=\sqrt{\frac{qQ}{4\pi {ϵ}_{0}m{R}^3}}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{qQ}{4\pi {ϵ}_{0}m{R}^3}}$
thus $x=4$