Electric Flux Through Parallel Plane Between Coaxial Cylinders with Dielectric

To find the flux of the electric field through the imaginary plane, we proceed step-by-step using Gauss's Law and surface integration.

Step 1: Electric Field in the Dielectric Region

The system consists of two coaxial conducting cylinders of length $\ell$. The inner cylinder has a radius $r_1 = \sqrt{2}R$ and carries a charge $Q$. The outer cylinder has a radius $r_2 = 2R$ and is grounded. The region between $r_1$ and $r_2$ is filled with a dielectric material of relative permittivity $\kappa = 5$.

By Gauss's law for the electric displacement field $\vec{D}$ in cylindrical coordinates: $\oint \vec{D} \cdot d\vec{A} = Q_{\text{free}}$ $D(2\pi r \ell) = Q \implies D(r) = \frac{Q}{2\pi r \ell}$

Since $\vec{D} = \kappa \epsilon_0 \vec{E}$, the electric field in the annular dielectric region ($\sqrt{2}R \le r \le 2R$) is: $\vec{E}(r) = \frac{Q}{2\pi \kappa \epsilon_0 \ell r} \hat{r}$

For $r < \sqrt{2}R$ (inside the inner conductor) and $r > 2R$ (outside the grounded outer cylinder), the electric field $\vec{E} = 0$.

Step 2: Parametrizing the Plane

Set up a Cartesian coordinate system with the origin at the center of the cross-section along the common axis:

• The common axis is along the $z$-axis ($0 \le z \le \ell$).
• The plane is parallel to the axis at a distance $R$ from it, so its equation is $x = R$.

The distance $r$ from any point $(R, y, z)$ on the plane to the axis is: $r = \sqrt{x^2 + y^2} = \sqrt{R^2 + y^2}$

The normal to the plane is in the direction of $\hat{i}$, so the area element is $d\vec{A} = dy \, dz \, \hat{i}$.

The component of the electric field normal to the plane is: $E_x = \vec{E} \cdot \hat{i} = E(r) \cos\theta = E(r) \frac{x}{r} = \frac{Q}{2\pi \kappa \epsilon_0 \ell r} \frac{R}{r} = \frac{Q R}{2\pi \kappa \epsilon_0 \ell (R^2 + y^2)}$

Step 3: Determining the Limits of Integration

The electric field is non-zero only within the dielectric region where $\sqrt{2}R \le r \le 2R$: $(\sqrt{2}R)^2 \le R^2 + y^2 \le (2R)^2$ $2R^2 \le R^2 + y^2 \le 4R^2$ $R^2 \le y^2 \le 3R^2$

This yields two symmetric segments for $y$: $y \in [-\sqrt{3}R, -R] \quad \text{and} \quad y \in [R, \sqrt{3}R]$

Step 4: Calculating the Total Flux

The total flux $\Phi$ through the plane is given by the integral of $E_x$ over the area of the plane where $\vec{E} \neq 0$: $\Phi = \int_{z=0}^{\ell} dz \int_{y \in \text{region}} E_x \, dy$

$\Phi = \ell \cdot \frac{Q R}{2\pi \kappa \epsilon_0 \ell} \left[ 2 \int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} \right]$

Evaluating the definite integral: $\int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} = \left[ \frac{1}{R} \arctan\left(\frac{y}{R}\right) \right]_{R}^{\sqrt{3}R} = \frac{1}{R} \left( \arctan(\sqrt{3}) - \arctan(1) \right)$

$\int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} = \frac{1}{R} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \frac{\pi}{12R}$

Substituting this back into the flux equation: $\Phi = \frac{Q R}{2\pi \kappa \epsilon_0} \cdot 2 \cdot \left( \frac{\pi}{12R} \right) = \frac{Q}{12 \kappa \epsilon_0}$

Given $\kappa = 5$: $\Phi = \frac{Q}{12 \times 5 \epsilon_0} = \frac{Q}{60\epsilon_0}$

Correct Answer: C ($\frac{Q}{60\epsilon_0}$)