A fundamental particle of mass m and charge + | vedclass

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Question

(A) At the distance of closest approach $(d)$,the entire initial kinetic energy of the particle is converted into electrostatic potential energy.Initi

Vedclass · Accepted Answer

$\frac{Ze^2}{2\pi \varepsilon_0 mv^2}$

Vedclass · Answer

(A) At the distance of closest approach $(d)$,the entire initial kinetic energy of the particle is converted into electrostatic potential energy.Initial kinetic energy $K = \frac{1}{2}mv^2$.Electrostatic potential energy $U = \frac{1}{4\pi \varepsilon_0} \cdot \frac{(e)(Ze)}{d}$.Equating $K = U$:$\frac{1}{2}mv^2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Ze^2}{d}$.Solving for $d$:$d = \frac{Ze^2}{2\pi \varepsilon_0 mv^2}$.

$A$ fundamental particle of mass $m$ and charge $+e$ is projected towards a much heavier charged particle of charge $+Ze$,where $Z > 0$. The distance of closest approach of the incident particle is ........

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