Question

A particle of mass 'm' and charge 'q', initially at rest, is accelerated by a uniform electric field 'E' through a distance 'D' and is then allowed to approach a fixed static charge 'Q' of the same sign. The distance of the closest approach of the charge q is
[ \(ε_0\) = permittivity of free space ]

• A. \(Q/4πε_0D\)
• B. \(Q/4πε_0ED\)
• C. \(Q/2πε_0D^2\)
• D. \(Q/4πε_0E\)

Hint:

Work done by the field becomes kinetic energy, which is then converted to potential energy at the closest approach.

Answer 1

The Correct Option is B

Step 1: Work-energy theorem
The change in kinetic energy equals the net work done, first by the uniform field and then by the repulsion of $Q$.

Step 2: Two stages
Stage one: $\frac{1}{2}mv^2 = qED$. Stage two: the particle slows down until its speed is zero, giving $\frac{1}{2}mv^2 = \dfrac{qQ}{4\pi\varepsilon_0r}$.

Step 3: Equate
$qED = \dfrac{qQ}{4\pi\varepsilon_0r}$, so $r = \dfrac{Q}{4\pi\varepsilon_0ED}$.

Step 4: Dimension check
$ED$ is a potential difference (volts) and $\dfrac{Q}{4\pi\varepsilon_0}$ is a potential times a length. So $\dfrac{Q}{4\pi\varepsilon_0ED}$ has the dimension of length, as a distance of closest approach must.

Final Answer:
The closest approach is Q/(4 pi epsilon_0 E D). This is option (B). \[ \boxed{\text{(B) }\frac{Q}{4\pi\varepsilon_0ED}} \]