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JEE Main 2004 (AIEEE) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2004 (AIEEE)

Q62JEE Main 2004 · AIEEEMedium

Electrostatic Potential Energy

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 of $2v$ the closest distance of approach would be

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Answer: (d)

At closest approach: $\frac{1}{2}mv^2 = \frac{kQq}{r}$ , so $r \propto \frac{1}{v^2}$ .
Speed $2v \Rightarrow r' = \frac{r}{4}$
So the answer is (d) $r/4$ .

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