A charged particle $q$ is shot towards another charged particle $\mathcal{Q}$ which is fixed, with a speed$\ v$. It approaches$\mathcal{Q}$ upto a closest distance $r$ and then returns. If $q$ is shot with speed2$v$, the closest distance of approach would be

Question

Dr. Abubacker · Accepted Answer

(a) $\frac{1}{2}\ mv^{2} = \frac{k\mathcal{Q}q}{r_{1}}$
$\frac{1}{2}m\left( 4v^{2} \right) = \frac{k\mathcal{Q}q}{r_{2}}$
From Eqs.(i) and ($ii$), we get
$\frac{r_{1}}{r_{2}}$ =4
$r_{2} = \frac{r}{4}$

Q. A charged particle $q$ is shot towards another charged particle $\mathcal{Q}$ which is fixed, with a speed$\ v$. It approaches$\mathcal{Q}$ upto a closest distance $r$ and then returns. If $q$ is shot with speed2$v$, the closest distance of approach would be

Explanation

Reviewed by Dr. Abubacker

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