Question

An infinite line charge of uniform electric charge density $\lambda$ lies along the axis electrically conducting an infinite cylindrical shell of radius $r$. At time $t=0$, the space in the cylinder is filled with a material of permittivity $\epsilon$ and electrical conductivity $\sigma$. The electrical conduction in the material follows Ohm's law. Which one of the following best describes the subsequent variation of the magnitude of current density $j\left(t\right)$ at any point in the material?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Explanation for correct option:

Step 1: Given

Uniform electric charge density $\lambda$

Infinite cylindrical shell of radius $r$

Permittivity $\epsilon$

Electrical conductivity $\sigma$

Step 2: Formula

Let us start by calculating the current density in the cylinder. Since the electric field due to a line charge inside the cylinder will be

$E=\frac{2k\lambda }{r}$

We can determine the charge density as

$J=\sigma \left(\frac{2k\lambda }{r}\right)$ Where $\mathrm{J}\mathrm{is}\mathrm{current}\mathrm{density},\mathrm{K}\mathrm{is}\mathrm{constant}$

Step 3: Calculation

Now we know that the current density is the ratio of the current in the circuit $I$ to the area of the cylinder $A$ . So we can write

$$\begin{gathered} J=\frac{I}{A} \\ \Rightarrow I=JA \end{gathered}$$

Substituting the value of current density in the above equation, we get

$I=A\sigma \left(\frac{2k\lambda }{r}\right)$

Since $I=\frac{dq}{dt}$ we have

$\frac{dq}{dt}=\sigma A\left(\frac{2k\lambda }{r}\right)$

Line charge density and length can be used to compute the charge in a line charge.

$q=\lambda l$

Therefore

$\frac{d\left(\lambda l\right)}{dt}=\sigma 2\pi rl\left(\frac{2k\lambda }{2\pi {\epsilon }_{0}r}\right)\left[\because A=2\pi rl\mathrm{and}\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right]$

Since the line's length is still fixed, the variable in the equation above must be the line charge density, which changes over time.
By integrating the aforementioned equation with time, we obtain

$\lambda ={\lambda }_0{e}^{-\frac{\sigma t}{{\epsilon }_0}}$ where ${\lambda }_0$ is the term containing all the constants.

Multiplying both sides by $\frac{\sigma 2k}{r}$ we get

$J=J_0{e}^{-\frac{\sigma t}{\epsilon }}$

Hence the graph of current density with time is exponentially decaying with time which corresponds to option (A).