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(D) By the principle of conservation of energy,the initial kinetic energy of the particle $q$ is converted into electrostatic potential energy at the

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$r/4$

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(D) By the principle of conservation of energy,the initial kinetic energy of the particle $q$ is converted into electrostatic potential energy at the point of closest approach.For speed $v$,the initial kinetic energy is $\frac{1}{2}mv^2$. At the closest distance $r$,the potential energy is $\frac{kQq}{r}$.Equating them: $\frac{1}{2}mv^2 = \frac{kQq}{r} \implies r = \frac{2kQq}{mv^2}$.This shows that $r \propto \frac{1}{v^2}$.If the speed is doubled to $2v$,the new closest distance $r'$ will be:$r' = \frac{2kQq}{m(2v)^2} = \frac{2kQq}{4mv^2} = \frac{r}{4}$.

$A$ charged particle $q$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be:

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