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

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Question

(B) Inside a uniformly charged sphere of radius $R$ and total charge $Q$,the electric field at a distance $r$ from the center 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) Inside a uniformly charged sphere of radius $R$ and total charge $Q$,the electric field at a distance $r$ from the center is given by $E = \frac{Qr}{4\pi \epsilon_0 R^3}$.The force on a charge $q$ is $F = qE = \frac{qQr}{4\pi \epsilon_0 R^3}$.Since $q$ and $Q$ have opposite signs,the force is attractive towards the center,$F = -\frac{qQ}{4\pi \epsilon_0 R^3} r$.This is the equation of Simple Harmonic Motion $(SHM)$ where $F = -m\omega^2 r$.Comparing the two,we get $m\omega^2 = \frac{qQ}{4\pi \epsilon_0 R^3}$,so $\omega = \sqrt{\frac{qQ}{4\pi \epsilon_0 m R^3}}$.The particle starts from the center and reaches the boundary,which corresponds to one-quarter of the time period $T$ of the $SHM$.$t = \frac{T}{4} = \frac{1}{4} \left( \frac{2\pi}{\omega} \right) = \frac{\pi}{2\omega}$.Substituting the value of $\omega$,we get $t = \frac{\pi}{2} \sqrt{\frac{4\pi \epsilon_0 m R^3}{qQ}}$.

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