An infinite line charge of uniform electric c | vedclass

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

Question

(C) Let the linear charge density be $\lambda_0$ at $t=0$ and $\lambda(t)$ at any time $t$.Using Gauss's law,the electric field $E$ at a distance $r$

Vedclass · Accepted Answer

Vedclass · Answer

(C) Let the linear charge density be $\lambda_0$ at $t=0$ and $\lambda(t)$ at any time $t$.Using Gauss's law,the electric field $E$ at a distance $r$ from the axis is given by $E = \frac{\lambda}{2 \pi \varepsilon r}$.According to Ohm's law,the current density $j$ is $j = \sigma E = \frac{\sigma \lambda}{2 \pi \varepsilon r}$.The rate of change of charge per unit length is equal to the current flowing out through a cylindrical surface of radius $r$ and length $l$:$\frac{dq}{dt} = -j(2 \pi r l) = -\left( \frac{\sigma \lambda}{2 \pi \varepsilon r} \right) (2 \pi r l) = -\frac{\sigma \lambda l}{\varepsilon}$.Since $q = \lambda l$,we have $\frac{d\lambda}{dt} = -\frac{\sigma \lambda}{\varepsilon}$.Solving this differential equation,we get $\lambda(t) = \lambda_0 e^{-\sigma t / \varepsilon}$.Substituting this into the expression for current density,we get $j(t) = \frac{\sigma \lambda_0}{2 \pi \varepsilon r} e^{-\sigma t / \varepsilon} = j_0 e^{-\sigma t / \varepsilon}$.This represents an exponentially decaying function,which corresponds to the graph shown in option $C$.

Similar Questions

The reading of an ideal voltmeter in the circuit shown is.....$V$.

In the given circuit, the current through the $2 \Omega$ resistor is: (in $\text{ A}$)

$A$ cell of emf $6\, V$ and internal resistance $0.5\, \Omega$ is short-circuited. The current in the cell is ............... $A$.

In the circuit shown in the figure,the current flowing in the $2\,\Omega$ resistance is ............... $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module