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 the line charge, is proportional to,

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

Answer 1

The correct option is C $v\propto \sqrt{\ln \left(\frac{r}{r_0}\right)}$

Applying conservation of energy

$[K+U{]}_i=[K+U{]}_f$

$\Rightarrow 0+qV=\frac{1}{2}m{v}^2+V^{\prime }q$

$\Rightarrow \frac{1}{2}m{v}^2=(V-V^{\prime })q$

Here, $(V-V^{\prime })$ is the potential difference between the initial and final positions of the charge $q$.

So, $V^{\prime }-V=-q{\int }_{r_0}^{r}Edr$

$\Rightarrow \frac{1}{2}m{v}^2=q{\int }_{r_0}^{r}Edr$

$=q{\int }_{r_0}^r\frac{\lambda }{2\pi {\epsilon }_{0}r}dr$

$=\frac{\lambda q}{2\pi {\epsilon }_0}\left[\ln \left(\frac{r}{r_0}\right)\right]$

$\Rightarrow v\propto \sqrt{\ln \left(\frac{r}{r_0}\right)}$

Hence, $\left(B\right)$ is the correct answer.