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Step-by-Step Detailed Solution
Step 1: Identify the Physical Principles
When a charged particle $q$ is projected towards a fixed charge $Q$, its initial kinetic energy is converted into electrostatic potential energy at the point of closest approach. The electrostatic potential energy of two point charges $q$ and $Q$ separated by a distance $r$ is given by
$$
\frac{kqQ}{r},
$$
where $k$ is a constant (often written as $k = \frac{1}{4\pi\epsilon_0}$ in SI units).
Step 2: Write the Energy Conservation for the First Case (Speed = $v$)
At the closest approach, the entire initial kinetic energy of the moving charge $q$ is converted into electrostatic potential energy. Mathematically:
$$
\frac{1}{2} m v^2 = \frac{k q Q}{r}.
$$
Here,
$m$ is the mass of the moving charge $q$.
$v$ is its initial speed.
$r$ is the given closest distance of approach.
Step 3: Write the Energy Conservation for the Second Case (Speed = $2v$)
If the same charge $q$ is now projected with a speed $2v$, at the new closest approach $r'$, energy conservation gives:
$$
\frac{1}{2} m (2v)^2 = \frac{k q Q}{r'}.
$$
Simplifying,
$$
\frac{1}{2} m \cdot 4v^2 = \frac{k q Q}{r'}.
$$
Step 4: Relate the Two Equations to Find the New Closest Distance
From the first case, we have:
$$
\frac{1}{2} m v^2 = \frac{k q Q}{r}.
$$
From the second case:
$$
2 m v^2 = \frac{k q Q}{r'}.
$$
Divide the second relation by the first:
$$
\frac{2 m v^2}{\frac{1}{2} m v^2} = \frac{\frac{k q Q}{r'}}{\frac{k q Q}{r}} \quad \Longrightarrow \quad 4 = \frac{r}{r'}.
$$
So,
$$
r' = \frac{r}{4}.
$$
Step 5: Conclusion
When the initial speed of the charged particle is doubled from $v$ to $2v$, the closest distance of approach becomes one-fourth of the original distance, i.e., $r/4$.
