A positive point charge is released from rest | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Vedclass · Answer

(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

Similar Questions

The figure shows the graph of electric field $E(r)$ versus distance $(r)$ from the center of an object. Therefore,...

$15$ charges,each of value $q$,are placed on the $X$-axis at an equal distance of $0.5R$. The electric flux associated with a spherical closed surface of radius $1.5R$,which has one of the charges at its center,is:

The given figure shows two parallel plates $A$ and $B$ with charge densities $+\sigma$ and $-\sigma$ respectively. The electric intensity will be zero in which region?

The electric field at a distance of $20 \ cm$ from the center of a charged spherical shell of radius $10 \ cm$ is $100 \ V/m$. What is the electric field at a distance of $3 \ cm$ from the center?

$A$ hollow metal sphere of radius $R$ is uniformly charged. The electric field due to the sphere at a distance $r$ from the centre is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module