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(B) The electric field $E$ at a distance $r$ from an infinitely long line charge is given by $E = \frac{2 k \lambda}{r}$.The electrostatic force provi

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$T=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$

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(B) The electric field $E$ at a distance $r$ from an infinitely long line charge is given by $E = \frac{2 k \lambda}{r}$.The electrostatic force providing the centripetal force is $F = qE = \frac{2 k \lambda q}{r}$.Equating this to the centripetal force $m \omega^2 r$,we get $\frac{2 k \lambda q}{r} = m \omega^2 r$.Solving for angular velocity $\omega$,we have $\omega^2 = \frac{2 k \lambda q}{m r^2}$,which implies $\omega = \frac{1}{r} \sqrt{\frac{2 k \lambda q}{m}}$.Since the time period $T = \frac{2 \pi}{\omega}$,we substitute $\omega$ to get $T = 2 \pi r \sqrt{\frac{m}{2 k \lambda q}}$.

$A$ particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then the time period will be given as (Consider $k$ as Coulomb's constant).

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