Question

A positive point charge is released from rest at a distance \(r_0\) from a positive line charge with uniform density. The speed (\(v\)) of the point charge, as a function of instantaneous distance \(r\) from line charge, is proportional to :-

• A. \(v \propto \left(\frac{r}{r_0}\right)\)
• B. \(v \propto \ln\left(\frac{r}{r_0}\right)\)
• C. \(v \propto e^{+r/r_0}\)
• D. \(v \propto \sqrt{\ln\left(\frac{r}{r_0}\right)}\)

Hint:

The field of a long line charge is \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\); equate the work done by this field to the kinetic energy gained.

Answer 1

The Correct Option is D

Electric field of a long uniformly charged line at distance \(r\): \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\), directed away from the line.
The positive charge \(q\) is repelled, so it moves from \(r_0\) to \(r \gt r_0\).
Work done by the field = gain in kinetic energy:
\(\frac{1}{2}mv^2 = \int_{r_0}^{r} qE\,dr = \frac{q\lambda}{2\pi\varepsilon_0}\int_{r_0}^{r}\frac{dr}{r} = \frac{q\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r}{r_0}\right)\).
\(v = \sqrt{\frac{q\lambda}{\pi\varepsilon_0 m}\ln\left(\frac{r}{r_0}\right)}\), so \(v \propto \sqrt{\ln\left(\frac{r}{r_0}\right)}\).
Answer: (4) \(v \propto \sqrt{\ln(r/r_0)}\)