$\text{Two co-axial conducting cylinders of same length}\ell \text{with radii}\sqrt{2}R\text{and}2R\text{are kept, as shown in Fig. 1.}$

$\text{The charge on the inner cylinder is}Q\text{and the outer cylinder is grounded. The annular region between the}$

$\text{cylinders is filled with a material of dielectric constant}\kappa =5.\text{Consider imaginary plane of the same length}\mathrm{ \ell }\text{at a}$

$\text{distance}R\text{from the common axis of the cylinders. The plane is parallel to the axis of the cylinders.}$

$\text{The cross-sectional view of this arrangement is shown in Fig. 2. Ignoring edge effects, the flux of the electric}$

$\text{field through the plane is}({\epsilon }_0\text{is the permittivity of free space):}$

Correct Answer/Option:

The correct answer is:
$\frac{Q}{60{\epsilon }_0}$

Step-by-Step Explanation:

1. Understanding the Electric Field Distribution:
We are given two coaxial conducting cylinders of length ℓ. The inner cylinder has a radius of $\sqrt{2}R$ and carries a total charge Q. The outer cylinder has a radius of 2R and is grounded.
Since the inner cylinder is a conductor, all its charge Q resides on its outer surface (at $r=\sqrt{2}R$). Electrostatic theory dictates that:
• The electric field inside the inner conductor (for $r<\sqrt{2}R$) is zero: $E=0$.
• The electric field exists only in the annular region between the cylinders, i.e., for $\sqrt{2}R<r<2R$, which is filled with a dielectric of constant $\kappa =5$.

2. Finding the Electric Field in the Annular Region:
Applying Gauss's Law in the dielectric region for a coaxial cylinder of radius r and length ℓ:
$\oint D\cdot dA=Q_{\text{enclosed}}$
$D\left(2\pi r\ell \right)=Q$
Since $D=\kappa {\epsilon }_{0}E$, the electric field E(r) at a radial distance r is:
$E\left(r\right)=\frac{Q}{2\pi \kappa {\epsilon }_{0}r\ell }$

3. Determining the Intersection of the Plane with the Electric Field:
We consider an imaginary plane of length ℓ located parallel to the axis at a distance R from the center.
Let us choose a Cartesian coordinate system where the cylinder's axis is along the z-axis. The equation of the plane is then $x=R$.
The distance r of any point on the plane from the axis is:
$r=\sqrt{x^2+y^2}=\sqrt{R^2+y^2}$
As analyzed, the electric field is non-zero only in the region $\sqrt{2}R<r<2R$. Thus:
$\sqrt{2}R<\sqrt{R^2+y^2}<2R$
Squaring all terms:
$2R^2<R^2+y^2<4R^2$
$R^2<y^2<3R^2$
This yields two symmetric intervals along the y-direction where the field passes through the plane:
$y\in [-\sqrt{3}R,-R]\text{and}y\in [R,\sqrt{3}R]$

4. Calculating the Flux:
The normal vector to the plane at $x=R$ points in the x-direction. The component of the radial electric field perpendicular to this plane is:
$E_x=E\left(r\right)\cdot cos\theta =E\left(r\right)\cdot \frac{x}{r}=E\left(r\right)\cdot \frac{R}{r}$
Substituting E(r):
$E_x=\frac{Q}{2\pi \kappa {\epsilon }_{0}r\ell }\cdot \frac{R}{r}=\frac{QR}{2\pi \kappa {\epsilon }_0\ell r^2}=\frac{QR}{2\pi \kappa {\epsilon }_0\ell (R^2+y^2)}$
The total flux $\Phi$ through the plane is the sum of the fluxes through both symmetric segments:
$\Phi =2{\int }_0^{\ell }dz{\int }_R^{\sqrt{3}R}{E}_{x}dy$
$\Phi =2\cdot \ell \cdot \frac{QR}{2\pi \kappa {\epsilon }_0\ell }{\int }_R^{\sqrt{3}R}\frac{1}{R^2+y^2}dy$
Using the standard integration formula $\int \frac{1}{R^2+y^2}dy=\frac{1}{R}arctan\left(\frac{y}{R}\right)$:
$\Phi =\frac{Q}{\pi \kappa {\epsilon }_0}{\left[arctan\left(\frac{y}{R}\right)\right]}_R^{\sqrt{3}R}$
$\Phi =\frac{Q}{\pi \kappa {\epsilon }_0}\left(arctan\left(\sqrt{3}\right)-arctan\left(1\right)\right)$
Since $arctan\left(\sqrt{3}\right)=\frac{\pi }{3}$ and $arctan\left(1\right)=\frac{\pi }{4}$:
$\Phi =\frac{Q}{\pi \kappa {\epsilon }_0}\left(\frac{\pi }{3}-\frac{\pi }{4}\right)=\frac{Q}{\pi \kappa {\epsilon }_0}\left(\frac{\pi }{12}\right)=\frac{Q}{12\kappa {\epsilon }_0}$

5. Substituting the Given Value of Dielectric Constant:
Given $\kappa =5$:
$\Phi =\frac{Q}{12\cdot 5\cdot {\epsilon }_0}=\frac{Q}{60{\epsilon }_0}$