Question

A positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform charge density. The speed $(v)$ of the point charge, as a function of instantaneous distance ' $r$ ' from line charge, is proportional to $\sqrt{\ln {\left(\frac{r}{r_0}\right)}^N}:$ The value of $\frac{1}{2N}$ is ________.

Answer 1

Answer: 0.5

At any point $P$, the force $\left(f_P\right)$ experienced by charge $=\frac{2k\lambda q}{x}$
Using Newton's second law of motion,
$mv\frac{dv}{dx}=\frac{2k\lambda q}{x}$
$\dots .\left(\because \frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}=v\frac{dv}{dx}\right)$
$\therefore \underset{0}{\overset{v}{\int }}mvdv=2k\lambda q\underset{r_0}{\overset{r}{\int }}\frac{dx}{x}$
${\therefore \frac{m{v}^2}{2}|}_0^v={2k\lambda q\ln x|}_{r_0}^r$
$\therefore \frac{1}{2}m{v}^2=2k\lambda q\ln \left(\frac{r}{r_0}\right)$
$\therefore v=\sqrt{\frac{4k\lambda q}{m}}\sqrt{\ln \left(\frac{r}{r_0}\right)}$
$\therefore v\propto \sqrt{\ln \left(\frac{r}{r_0}\right)}$
$\therefore N=1$
$\Rightarrow \frac{1}{2N}=0.5$