solve this question please and provide an easy to understand solution for a 12th grade kid ---**Textual Information:**
Two co-axial conducting cylinders of same length $l$ 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 $l$ 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):
**Mathematical Formulas/Symbols:**
$l$
$\sqrt{2}R$
$2R$
$Q$
$\kappa = 5$
$R$
$\epsilon_0$
**Chart/Diagram Description:**
**Figure 1:**
* **Type:** Perspective view illustrating two coaxial cylinders and an intersecting plane.
* **Main Elements:**
* Two concentric cylinders of length $l$.
* The inner cylinder has a radius labeled $\sqrt{2}R$.
* The outer cylinder has a radius labeled $2R$.
* The length of the cylinders is labeled $l$ with a double-headed arrow along the vertical axis.
* The inner cylinder is indicated to have charge $Q$.
* The outer cylinder is shown connected to ground with the standard ground symbol.
* An imaginary plane is shown cutting through the annular region of the cylinders. This plane appears vertical and parallel to the cylinder axis.
**Figure 2:**
* **Type:** Cross-sectional view of the coaxial cylinders.
* **Main Elements:**
* Two concentric circles representing the cross-sections of the cylinders.
* The inner circle has a radius labeled $\sqrt{2}R$.
* The outer circle has a radius labeled $2R$.
* The region between the circles represents the annular space filled with dielectric ($\kappa=5$).
* A point labeled $Q$ is shown near the center, likely indicating the source of the field/charge on the inner cylinder.
* A vertical line is drawn inside the annular region, representing the cross-section of the imaginary plane.
* The distance from the center (axis) to this vertical line is labeled $R$ with a double-headed arrow along the horizontal direction.
* The ground symbol is also shown connected to the outer circle.
**Question:**
Ignoring edge effects, the flux of the electric field through the plane is ($\epsilon_0$ is the permittivity of free space):
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We have two coaxial conducting cylinders with the inner cylinder having radius square root of 2 R and charge Q, while the outer cylinder has radius 2R and is grounded. The space between them is filled with a dielectric material having dielectric constant kappa equals 5. We need to find the electric flux through an imaginary plane of length l located at distance R from the common axis.
To find the electric field between the cylinders, we apply Gauss's law. For a cylindrical Gaussian surface at radius r between the cylinders, the enclosed charge is Q and the permittivity is epsilon naught kappa. The electric field is radial and uniform on the Gaussian surface, giving us E equals Q over 2 pi epsilon naught kappa l r, where r is between square root 2 R and 2R.
To calculate the flux through the imaginary plane, we set up the flux integral. The plane has length l and is located at distance R from the axis, parallel to the cylinder axis. The electric field is radial, pointing outward from the inner cylinder. We need to carefully analyze the geometry to calculate the dot product between the electric field vector and the area element vector of the plane.
Now we perform the geometric integration. The electric field magnitude is Q over 2 pi epsilon naught kappa l r. The angle theta between the field and the plane normal satisfies cosine theta equals R over r. The flux integral becomes the integral from square root 2 R to 2R of Q R over 2 pi epsilon naught kappa l r squared dy, where we integrate over the portion of the plane within the annular region.
Completing the integration, we substitute r squared equals R squared plus y squared and change variables. The integral evaluates to Q over 4 pi epsilon naught kappa. Since kappa equals 5, our final answer is Q over 20 pi epsilon naught. This result represents the electric flux through the imaginary plane due to the charged inner cylinder in the presence of the dielectric material.