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  4. 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 is shot with speed 2v, the closest distance of approach would be

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Q. 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 $q$ is shot with speed $2v$, the closest distance of approach would be

KEAMKEAM 2011Electrostatic Potential and Capacitance

Solution:

$ \frac{1}{2}m{{v}^{2}}=\frac{kQq}{{{r}_{1}}} $ ...(i)
$ \frac{1}{2}m.4{{v}^{2}}=\frac{kQq}{{{r}_{2}}} $ .. (ii)
From Eqs. (i) and (ii), we can say
$ \frac{{{r}_{1}}}{{{r}_{2}}}=4 $
$ {{r}_{2}}=\frac{r}{4} $