A charged particle $q$ is shot towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distances of approach would be
AIEEE 2004, Medium
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(d) Charge $q$ will momentarily come to rest at a distance r from charge $Q$ when all it's kinetic energy converted to potential energy i.e. $\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{qQ}}{r}$
Therefore the distance of closest approach is given by
$r = \frac{{qQ}}{{4\pi {\varepsilon _0}}}.\frac{2}{{m{v^2}}}$ ==> $r \propto \frac{1}{{{v^2}}}$
Hence if $v$ is doubled, $r$ becomes one fourth.
Therefore the distance of closest approach is given by
$r = \frac{{qQ}}{{4\pi {\varepsilon _0}}}.\frac{2}{{m{v^2}}}$ ==> $r \propto \frac{1}{{{v^2}}}$
Hence if $v$ is doubled, $r$ becomes one fourth.
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