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

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Question

(A) Inside a uniformly charged sphere of radius $R$ and total charge $Q$,the electric field at a distance $x$ from the center is given by $\vec{E} = \

Vedclass · Accepted Answer

$t = \frac{\pi }{2}\sqrt {\frac{{4\pi { \in _0}m{R^3}}}{{qQ}}} $

Vedclass · Answer

(A) Inside a uniformly charged sphere of radius $R$ and total charge $Q$,the electric field at a distance $x$ from the center is given by $\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{Qx}{R^3} \hat{x}$.The force on the particle is $\vec{F} = q\vec{E} = \frac{qQ}{4\pi\epsilon_0 R^3} x \hat{x}$.Since $q$ and $Q$ have opposite signs,the force is directed towards the center,$\vec{F} = -(\frac{qQ}{4\pi\epsilon_0 R^3}) x \hat{x}$.This is the condition for Simple Harmonic Motion $(SHM)$ with $F = -kx_{eff}$,where $k_{eff} = \frac{qQ}{4\pi\epsilon_0 R^3}$.The angular frequency is $\omega = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{qQ}{4\pi\epsilon_0 m R^3}}$.The time period of oscillation is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi\epsilon_0 m R^3}{qQ}}$.The particle starts from the center (mean position) and reaches the boundary (extreme position). The time taken for this is $t = \frac{T}{4} = \frac{2\pi}{4} \sqrt{\frac{4\pi\epsilon_0 m R^3}{qQ}} = \frac{\pi}{2} \sqrt{\frac{4\pi\epsilon_0 m R^3}{qQ}}$.

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