Two co-axial conducting cylinders of same length l with radii √2R and 2R are kept, as shown in Fig.

Q: Why is the electric field inside the inner cylinder zero?

The inner cylinder is a solid conductor. In electrostatics, charges reside entirely on the outer surface of a conductor to minimize repulsion. Consequently, by Gauss's Law, any Gaussian surface drawn entirely inside the conductor encloses zero net charge, resulting in a zero electric field inside.

Q: Why do we use the formula for an infinite line charge $E = \frac{\lambda}{2\pi\epsilon_0 \kappa r}$?

The problem states to "ignore edge effects," which implies we can treat the cylinders as infinitely long for the sake of the electric field calculation. For a long cylinder, the electric field outside it behaves exactly like that of a line charge located at its central axis.

Q: How did we get $dA = \ell \, dy$?

The imaginary plane is a rectangle of length $\ell$ (parallel to the cylinders' axis) and some width. If we take a tiny horizontal strip on this plane at a height $y$ with a thickness $dy$, its area is simply a rectangle's area: length $\times$ width, which is $\ell \times dy$.

Q: Why does the $\cos\theta$ term appear in the flux equation $d\phi = E \cos\theta \, dA$?

Electric flux is the dot product of the electric field vector $\vec{E}$ and the area vector $d\vec{A}$. The electric field is radial (pointing straight out from the center), while the area vector is perpendicular to the plane. The angle between them is exactly $\theta$, so the dot product yields $E \, dA \cos\theta$.

Q: Why do we multiply the final integral result by 2?

The plane extends both above and below the central axis. Our integration from $45^\circ$ to $60^\circ$ only calculated the flux for the upper segment (AB). By symmetry, the lower segment (CD) has the exact same geometry and electric field passing through it, so it contributes an equal amount of flux.

Question

Two co-axial conducting cylinders of same length $\ell$ with radii $\sqrt{2}R$ and $2R$ are kept, as shown in Fig. 1. The charge on the inner cylinder is $Q$ and the outer cylinder is grounded. The annular region between the cylinders is filled with a material of dielectric constant $\kappa = 5$. Consider an imaginary plane of the same length $\ell$ at a distance R from the common axis of the cylinders. This plane is parallel to the axis of the cylinders. The cross-sectional view of this arrangement is shown in Fig. 2. Ignoring edge effects, the flux of the electric field through the plane is ($\epsilon_0$ is the permittivity of free space):

Options:

(A) $\frac{Q}{30\epsilon_0}$

(B) $\frac{Q}{15\epsilon_0}$

(C) $\frac{Q}{60\epsilon_0}$

(D) $\frac{Q}{120\epsilon_0}$

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Coaxial Cylinders

Imagine a brilliant electrostatics setup right in front of you. We have two coaxial conducting cylinders, meaning they share the exact same central axis. We've given the inner cylinder a positive charge $Q$, and we've grounded the outer cylinder, forcing its potential to zero. In the empty space between them, we've filled a dielectric material with a constant $\kappa = 5$.

The problem introduces an imaginary plane placed at a distance $R$ from the axis. Our mission is to calculate the net electric flux passing through this entire plane. Don't get intimidated; we will break this down step by step.

### The Core Principle: Gauss's Law and the Conducting Core

First and foremost, we need to understand exactly where the electric field exists. Recall Gauss's Law. The inner cylinder is a solid conductor, and we know that the net electric field inside a conductor is always strictly zero.

This means, as long as our radial distance $r$ is less than $\sqrt{2}R$, we won't find any field lines. The entire electric field exists exclusively in the annular gap between the inner and outer cylinders. These field lines are radiating outward from the inner cylinder towards the outer one. Grasping this concept is absolutely crucial before we move forward.

### The Geometry of the Intersection

Now look closely at that imaginary plane. This plane is located at a distance of just $R$ from the central axis. But wait, the radius of our inner cylinder is $\sqrt{2}R$, which is mathematically greater than $R$!

This simply means that our plane is actually slicing right through the inner cylinder. The portion of the plane that lies inside the cylinder experiences an electric field of absolute zero. And if the field is zero, the flux passing through that specific section must also be zero. So, we don't need to worry about this middle section at all.

So where exactly is the flux passing through? The flux will only pass through the upper segment (let's call it $AB$) and the lower segment ($CD$). Let's use some basic geometry to find the angles. Where the plane exits the inner cylinder, it forms a right-angled triangle. The base is $R$ and the hypotenuse is $\sqrt{2}R$. So, $\cos	heta_1$ will be $\frac{R}{\sqrt{2}R}$, which is $\frac{1}{\sqrt{2}}$. This means $	heta_1$ is exactly $45^\circ$.

Similarly, where the plane touches the outer cylinder, $\cos	heta_2$ will be $\frac{R}{2R}$, which is $\frac{1}{2}$. So $	heta_2$ comes out to be $60^\circ$.

### The Master Equation: Setting up the Differential Flux

Now we are going to bring in some calculus. Focus on segment $AB$ and consider a very tiny element $dy$ there. The area of this tiny element, $dA$, will be the length of the plane $\ell$ multiplied by $dy$.

The tiny amount of flux passing through this small area, $d\phi$, will be the dot product of $\vec{E}$ and $d\vec{S}$. Expanding the dot product gives us:

$$d\phi = E \cos	heta \, dA$$

Here, $	heta$ is the angle that the electric field vector makes with the area vector. This is our master setup for the integration.

### The Magic of Calculus: Simplifying the Integral

Let's substitute the actual values into this equation. The electric field $E$ due to a long cylinder is:

$$E = \frac{\lambda}{2\pi\epsilon_0 \kappa r}$$

Here $\kappa$ is the dielectric constant. Now look at the geometry, the vertical distance $y$ can be written as $R 	an	heta$. If we differentiate this, $dy$ becomes $R \sec^2	heta \, d	heta$. And from the triangle, the radial distance $r$ can be written as $R \sec	heta$.

Now watch the magic happen! When we plug all these values of $E$, $dy$, and $r$ into our $d\phi$ expression, a lot of terms start canceling out.

$$d\phi = \frac{\lambda}{2\pi\epsilon_0 \kappa (R \sec	heta)} \cos	heta (\ell R \sec^2	heta \, d	heta)$$

The $R$ in the denominator and the $R$ in the numerator cancel each other. The $\sec	heta$ terms also simplify beautifully. After all the cancellations, we are left with an incredibly elegant and simple expression:

$$d\phi = \frac{\lambda \ell}{2\pi\epsilon_0 \kappa} d	heta$$

Such complex geometry simplified so beautifully!

### The Final Calculation: Bringing it all Together

We are very close to the destination now. To find the total flux, we need to integrate this expression. Our limits will be from $45^\circ$ to $60^\circ$, which is $\frac{\pi}{4}$ to $\frac{\pi}{3}$ in radians.

$\lambda$, which is the linear charge density, multiplied by the length $\ell$ will give us the total charge $Q$. When we integrate $d	heta$, $\frac{\pi}{3} - \frac{\pi}{4}$ gives us $\frac{\pi}{12}$. Solving this, the flux for segment $AB$ comes out to be:

$$\phi_{AB} = \frac{Q}{2\pi\epsilon_0 \kappa} \left( \frac{\pi}{12} ight) = \frac{Q}{24\epsilon_0 \kappa}$$

Finally, we need the flux for the entire plane. Because of symmetry, the exact same amount of flux passing through the upper segment $AB$ will also pass through the lower segment $CD$. So for the total flux, we simply multiply our answer by $2$.

$$\phi_{total} = 2 	imes \frac{Q}{24\epsilon_0 \kappa} = \frac{Q}{12\epsilon_0 \kappa}$$

The problem gives us the value of the dielectric constant $\kappa$ as $5$. So when we substitute $\kappa = 5$, our final answer is:

$$\phi_{total} = \frac{Q}{60\epsilon_0}$$

This was a truly magnificent problem that beautifully combined Gauss's Law, geometry, and calculus!

Select Answer:

Visualized Solution

\text{Analyzing the Geometry}

\text{Electric Field Region}

\text{Active Flux Segments}

\text{Intersection Angles}

\text{Differential Flux Setup}

\text{Variable Substitution}

\text{Simplifying the Integral}

\text{Integrating over Segment AB}

\text{Total Flux Calculation}

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

The Setup

The Core Principle

The Geometry of the Intersection

The Master Equation

The Magic of Calculus

The Final Calculation

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