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Question

(C) The electric field $E$ at a distance $r$ from a line charge with linear charge density $\lambda$ is given by $E = \frac{\lambda}{2\pi\varepsilon_0

Vedclass · Accepted Answer

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

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(C) The electric field $E$ at a distance $r$ from a line charge with linear charge density $\lambda$ is given by $E = \frac{\lambda}{2\pi\varepsilon_0 r}$.The potential difference $\Delta V$ between distance $r_0$ and $r$ is $\Delta V = -\int_{r_0}^{r} E \, dr = -\int_{r_0}^{r} \frac{\lambda}{2\pi\varepsilon_0 r} \, dr = -\frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r}{r_0} \right) = \frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r_0}{r} \right)$.Using the work-energy theorem,the change in kinetic energy equals the work done by the electric field: $\frac{1}{2}mv^2 = q(V_i - V_f) = q \Delta V$.Substituting the potential difference: $\frac{1}{2}mv^2 = q \left( \frac{\lambda}{2\pi\varepsilon_0} \ln \left( \frac{r}{r_0} \right) \right)$.Since $m, q, \lambda, \pi, \varepsilon_0$ are constants,we have $v^2 \propto \ln \left( \frac{r}{r_0} \right)$,which implies $v \propto \sqrt{\ln \left( \frac{r}{r_0} \right)}$.

$A$ 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 the line charge,is proportional to

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