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(D) Initially,the particle $q$ is at an infinite distance from $Q$,so the initial potential energy $P.E._i = 0$.Case $1$: Initial velocity is $v$ and

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$r/4$

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(D) Initially,the particle $q$ is at an infinite distance from $Q$,so the initial potential energy $P.E._i = 0$.Case $1$: Initial velocity is $v$ and the closest distance is $r$.Applying the law of conservation of energy: $P.E._i + K.E._i = P.E._f + K.E._f$$0 + \frac{1}{2}mv^2 = \frac{kQq}{r} + 0$$\frac{1}{2}mv^2 = \frac{kQq}{r} \quad \dots (1)$Case $2$: Initial velocity is $2v$ and let the new closest distance be $r'$.Applying the law of conservation of energy: $P.E._i + K.E._i = P.E._f + K.E._f$$0 + \frac{1}{2}m(2v)^2 = \frac{kQq}{r'} + 0$$\frac{1}{2}m(4v^2) = \frac{kQq}{r'} \quad \dots (2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\frac{1}{2}mv^2}{\frac{1}{2}m(4v^2)} = \frac{\frac{kQq}{r}}{\frac{kQq}{r'}}$$\frac{1}{4} = \frac{r'}{r}$$r' = \frac{r}{4}$

$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be:

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