In a uniformly charged dielectric sphere, a very thin tunnel has been made along the diameter as shown in figure. A charge particle -q having mass m is released from rest at one end of the tunnel. For the situation described, mark the correct statement(s). (Neglect gravity.)

At a center, the force experienced by the charge particle is zero, so it is a position of equilibrium. As we displace the charge from the equilibrium position, electric force starts acting on it toward equilibrium position and hence equilibrium is a stable one. At a distance x from the center of sphere (equilibrium position), force experienced by the charge particle is

$\mathrm{F}=\frac{\mathrm{qQx}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}[\text{ for }\mathrm{x}<\mathrm{R}]$

As $\mathrm{F}\propto \mathrm{x}$, it performs SHM about the center. Time period can be calculated by using $\mathrm{F}={\mathrm{m\omega }}^2\mathrm{x}$.

Also conserving mechanical energy ,$\frac{1}{2}m{V}^2 =\frac{3}{2}\frac{Qq}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}-\frac{1}{4}\frac{Qq}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}} \Rightarrow V=\sqrt{\frac{Qq}{4{\mathrm{\pi \epsilon }}_0\mathrm{Rm}}}$