An elementary particle of mass m and charge + e is projected with velocity v at a much more massive

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2. Electric Potential and Capacitance

medium

An elementary particle of mass $m$ and charge $ + e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$What is the closest possible approach of the incident particle

$\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$

$\frac{{Ze}}{{4\pi {\varepsilon _0}m{v^2}}}$

$\frac{{Z{e^2}}}{{8\pi {\varepsilon _0}m{v^2}}}$

$\frac{{Ze}}{{8\pi {\varepsilon _0}m{v^2}}}$

Solution

(a) Suppose distance of closest approach is $r$, and according to energy conservation applied for elementary charge.
Energy at the time of projection $=$ Energy at the distance of closest approach
$==>$ $\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{(Ze).e}}{r} \Rightarrow r = \frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$

Standard 12

Physics