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

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$r/4$

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(D) By the law of conservation of energy,the initial kinetic energy of the particle $q$ is converted into electrostatic potential energy at the point of closest approach,where the particle momentarily comes to rest.Initial kinetic energy = $\frac{1}{2}mv^2$Electrostatic potential energy at distance $r$ = $\frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r}$Equating the two: $\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{qQ}{r}$From this,we can see that $r = \frac{qQ}{2\pi\varepsilon_0 mv^2}$,which implies $r \propto \frac{1}{v^2}$.If the speed is doubled $(v' = 2v)$,the new distance of closest approach $r'$ will be:$r' = r \cdot \left(\frac{v}{v'}\right)^2 = r \cdot \left(\frac{v}{2v}\right)^2 = r \cdot \frac{1}{4} = \frac{r}{4}$.Thus,the closest distance of approach becomes $r/4$.

$A$ charged particle $q$ is shot towards another charged particle $Q$ which is fixed,with a speed $v$. 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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