A charged particle with charge q and mass m s | vedclass

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Question

(B) The electric field inside a uniformly charged sphere of radius $R$ and total charge $Q$ at a distance $r$ from the centre is given by $E = \frac{Q

Vedclass · Accepted Answer

$\frac{\pi}{2} \sqrt{\frac{4\pi \varepsilon_0 m R^3}{qQ}}$

Vedclass · Answer

(B) The electric field inside a uniformly charged sphere of radius $R$ and total charge $Q$ at a distance $r$ from the centre is given by $E = \frac{Qr}{4\pi \varepsilon_0 R^3}$.The force on the particle of charge $q$ is $F = qE = \frac{qQr}{4\pi \varepsilon_0 R^3}$.Since $q$ and $Q$ have opposite signs,the force is directed towards the centre,i.e.,$F = -\frac{qQ}{4\pi \varepsilon_0 R^3} r$.This is the equation of Simple Harmonic Motion $(SHM)$ of the form $F = -m\omega^2 r$,where $\omega = \sqrt{\frac{qQ}{4\pi \varepsilon_0 m R^3}}$.The particle starts from the centre $(r=0)$ and moves to the boundary $(r=R)$. This corresponds to one-quarter of the time period $(T/4)$.$t = \frac{T}{4} = \frac{1}{4} \left( \frac{2\pi}{\omega} \right) = \frac{\pi}{2\omega}$.Substituting the value of $\omega$:$t = \frac{\pi}{2} \sqrt{\frac{4\pi \varepsilon_0 m R^3}{qQ}}$.

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