An infinite line charge of uniform electric charge density lambda lies along the axis of an electric

Q: Why do we use $\varepsilon$ instead of $\varepsilon_0$ in Gauss's Law?

Because the space is filled with a dielectric material of permittivity $\varepsilon$. The electric field is modified by the polarization of the medium, so we must use the permittivity of the material itself.

Q: Why does the radius $r$ cancel out when calculating the total current?

The current density $j$ decreases as $1/r$, but the surface area of the cylinder increases as $r$. These two effects perfectly cancel each other out, meaning the total current crossing any cylindrical surface is the same, as long as no charge is accumulating in the bulk of the material.

Q: What does the negative sign in $I = -\frac{dq}{dt}$ mean?

The current $I$ represents the flow of charge *outwards* from the enclosed volume. Since charge is leaving, the total charge $q$ inside the volume is decreasing. The negative sign ensures that a positive outward current corresponds to a decreasing enclosed charge.

Q: Why doesn't the current density drop to a non-zero constant like in graph (a)?

The current is driven entirely by the electric field, which is produced by the line charge. As the line charge neutralizes, the electric field weakens, which in turn reduces the current. This process continues until the line charge is completely neutralized, at which point the electric field and current density both become exactly zero.

Q: What is the physical significance of the term $\frac{\varepsilon}{\sigma}$?

This term is known as the relaxation time of the material. It represents the characteristic time scale over which the material neutralizes any internal charge imbalances. A perfect conductor ($\sigma \to \infty$) has a relaxation time of zero, meaning it neutralizes instantly!

Question

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

Sigma Percentile · Accepted Answer

The problem of a discharging line charge inside a conducting medium is a beautiful intersection of electrostatics and current electricity. It forces us to think dynamically about a setup that usually remains static in our textbooks.

### Analyzing the Setup

Imagine an infinite line charge with a linear charge density $\lambda$ placed exactly at the axis of a cylindrical shell.

Normally, in a vacuum, this line charge would just sit there, creating a static electric field around it.

But here, the space is filled with a material that has both a permittivity $\varepsilon$ and a conductivity $\sigma$.

This means the material can conduct electricity! The electric field created by the line charge won't just sit there; it will push the free charge carriers in the material, creating a current.

### The Master Equation

Let's find the electric field at a distance $r$ from the axis. Using Gauss's Law for a cylindrical surface of length $l$ and radius $r$, the enclosed charge is $\lambda l$.

The electric field $E$ is purely radial, so the flux is $E \cdot 2\pi r l$.

Equating this to the enclosed charge divided by the permittivity of the medium $\varepsilon$, we get:

$$E = \frac{\lambda}{2\pi\varepsilon r}$$

Because the material is conducting, this electric field drives a current. According to the microscopic form of Ohm's Law, the current density $j$ is proportional to the electric field:

$$j = \sigma E = \frac{\sigma \lambda}{2\pi\varepsilon r}$$

### Conservation of Charge

Now, let's look at the total current $I$ flowing outwards through our Gaussian cylinder.

Current is simply the current density multiplied by the surface area:

$$I = j \cdot (2\pi r l) = \left( \frac{\sigma \lambda}{2\pi\varepsilon r} ight) (2\pi r l) = \frac{\sigma \lambda l}{\varepsilon}$$

Notice something magical? The radius $r$ completely cancels out! The total outward current depends only on the charge enclosed.

But where is this current coming from? It represents the flow of positive charge away from the axis (or equivalently, electrons flowing towards the axis and neutralizing the line charge).

By the principle of conservation of charge, the outward current must equal the rate at which the enclosed charge is decreasing:

$$I = -\frac{dq}{dt} = -l \frac{d\lambda}{dt}$$

### The Differential Equation

Equating our two expressions for the current $I$, we get a beautiful differential equation:

$$-l \frac{d\lambda}{dt} = \frac{\sigma \lambda l}{\varepsilon}$$

The length $l$ cancels out, leaving us with:

$$\frac{d\lambda}{dt} = -\frac{\sigma}{\varepsilon} \lambda$$

This is a classic first-order linear differential equation. It tells us that the rate of decay of the charge is directly proportional to the amount of charge present.

### Final Calculation

Let's integrate this equation from time $t=0$ (where $\lambda = \lambda_0$) to some time $t$:

$$\int_{\lambda_0}^{\lambda} \frac{d\lambda}{\lambda} = -\frac{\sigma}{\varepsilon} \int_0^t dt$$

$$\ln\left(\frac{\lambda}{\lambda_0}ight) = -\frac{\sigma}{\varepsilon} t$$

$$\lambda(t) = \lambda_0 e^{-\frac{\sigma}{\varepsilon} t}$$

The charge density decays exponentially!

Since the current density $j(t)$ is directly proportional to $\lambda(t)$, it must also follow the exact same exponential decay:

$$j(t) = \frac{\sigma}{2\pi\varepsilon r} \lambda_0 e^{-\frac{\sigma}{\varepsilon} t} = j_0 e^{-\frac{\sigma}{\varepsilon} t}$$

Looking at the given options, we need a graph that starts at a positive initial value and decays exponentially to zero.

Graph (d) perfectly matches this mathematical reality.

**Bonus Insight:** The quantity $\frac{\varepsilon}{\sigma}$ has the dimensions of time and is called the **relaxation time** ($	au$) of the material. It is a fundamental property that dictates how fast a material can neutralize any internal charge imbalance!

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Visualized Solution

$Physical Setup$

$Electric Field$

$Ohm’s Law$

$Total Current$

$Conservation of Charge$

$Differential Equation$

$Integration$

$Current Density Variation$

$Conclusion$

$Relaxation Time$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Analyzing the Setup

The Master Equation

Conservation of Charge

The Differential Equation

Final Calculation

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Comprehension Passage

$Δ V$ measured between $B$ and $C$ is

Two bars of radius $r$ and $2 r$ are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of following is correct ?

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ohm’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differential Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current Density Variation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relaxation Time