An infinitely long positively charged straigh | vedclass

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(B) The electric field $E$ at a distance $r$ from an infinitely long charged wire is given by $E = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2 k \lam

Vedclass · Accepted Answer

Vedclass · Answer

(B) The electric field $E$ at a distance $r$ from an infinitely long charged wire is given by $E = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2 k \lambda}{r}$,where $k = \frac{1}{4 \pi \epsilon_0}$.The electrostatic force $F$ acting on the electron of charge $e$ is $F = eE = \frac{2 k \lambda e}{r}$.For the electron to revolve in a circular path of radius $r$,this electrostatic force provides the necessary centripetal force:$F_c = \frac{m v^2}{r} = \frac{2 k \lambda e}{r}$From this,we can find the square of the velocity $v^2$:$v^2 = \frac{2 k \lambda e}{m}$The kinetic energy $KE$ of the electron is given by:$KE = \frac{1}{2} m v^2 = \frac{1}{2} m \left( \frac{2 k \lambda e}{m} \right) = k \lambda e$Since $k$,$\lambda$,and $e$ are constants,the kinetic energy $KE$ is independent of the radius $r$. Therefore,the graph of $KE$ versus $r$ is a horizontal straight line. Thus,the graph shown in option $(B)$ is correct.

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