A positive point charge is released from rest | vedclass

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Question

$\frac{1}{2} m V^{2}=-q\left(V_{f}-V_{i}\right)$

$E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}$

$\Delta \mathrm{V}=\frac{\lambda}{2 \pi \vareps

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{1}{2} m V^{2}=-q\left(V_{f}-V_{i}\right)$

$E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}$

$\Delta \mathrm{V}=\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \left(\frac{\mathrm{r}_{0}}{\mathrm{r}}\right)$

$\frac{1}{2} m v^{2}=\frac{-q \lambda}{2 \pi \varepsilon_{0}} \ln \left(\frac{r_{0}}{r}\right)$

$v \propto \sqrt{\ln \left(\frac{r}{r_{0}}\right)}$

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

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