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(A) For a point outside the cylinder at a distance $r$ from the axis,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_{0}}$.Con

Vedclass · Accepted Answer

$\frac{\rho q R^{2}}{4 \varepsilon_{0}}$

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(A) For a point outside the cylinder at a distance $r$ from the axis,we use Gauss's Law: $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_{0}}$.Considering a Gaussian surface of radius $r$ and length $\ell$,the enclosed charge is $q_{enc} = \rho \cdot \pi R^{2} \ell$.Thus,$E(2 \pi r \ell) = \frac{\rho \pi R^{2} \ell}{\varepsilon_{0}}$,which gives the electric field $E = \frac{\rho R^{2}}{2 \varepsilon_{0} r}$.The electrostatic force provides the necessary centripetal force for circular motion: $qE = \frac{mv^{2}}{r}$.Substituting $E$: $q \left( \frac{\rho R^{2}}{2 \varepsilon_{0} r} \right) = \frac{mv^{2}}{r}$.Simplifying,we get $mv^{2} = \frac{q \rho R^{2}}{2 \varepsilon_{0}}$.The kinetic energy $K = \frac{1}{2} mv^{2} = \frac{q \rho R^{2}}{4 \varepsilon_{0}}$.

$A$ long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of the cylindrical volume is $R$. $A$ charged particle $(q)$ revolves around the cylinder in a circular path at a distance $r$ from the axis. The kinetic energy of the particle is:

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