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. Ignoring edge effects, the flux of the electric field through the plane is $ (\varepsilon_0 \text{ is the permittivity of free space}) $:

• A. \( \frac{Q}{30\varepsilon_0} \)
• B. \( \frac{Q}{15\varepsilon_0} \)
• C. \( \frac{Q}{60\varepsilon_0} \)
• D. \( \frac{Q}{120\varepsilon_0} \)

Hint:

When dealing with cylindrical symmetry and dielectrics, apply Gauss's law considering the fraction of enclosed charge, and account for dielectric constant using \( \varepsilon = \kappa \varepsilon_0 \).

Answer 1

The Correct Option is B

Step 1: Electric Field in the Dielectric
Using Gauss's law for a cylindrical surface of radius \( r \) (\( \sqrt{2}R<r<2R \)) and length \( l \): \[ E (2 \pi r l) = \frac{Q}{\kappa \epsilon_0} = \frac{Q}{5 \epsilon_0} \] \[ E = \frac{Q}{10 \pi \epsilon_0 r l} \] The electric field is radial.
Step 2: Flux Through the Plane
Consider a small area element \( dA = l \, dy \) on the plane at a distance \( y \) from the center of the plane and at a distance \( x = R \) from the axis. The distance from the axis to this element is \( r = \sqrt{R^2 + y^2} \). The electric field at this point is \( E = \frac{Q}{10 \pi \epsilon_0 \sqrt{R^2 + y^2} l} \). The angle \( \theta \) between the electric field and the normal to the plane is such that \( \cos \theta = \frac{R}{\sqrt{R^2 + y^2}} \). The flux through this small area element is: \[ d\Phi_E = \mathbf{E} \cdot d\mathbf{A} = E \cos \theta \, dA = \frac{Q}{10 \pi \epsilon_0 \sqrt{R^2 + y^2} l} \cdot \frac{R}{\sqrt{R^2 + y^2}} (l \, dy) = \frac{Q R}{10 \pi \epsilon_0 (R^2 + y^2)} dy \] The plane intersects the dielectric region from \( y = -\sqrt{(2R)^2 - R^2} = -\sqrt{3}R \) to \( y = +\sqrt{(2R)^2 - R^2} = +\sqrt{3}R \). The total flux through the plane is: \[ \Phi_E = \int_{-\sqrt{3}R}^{\sqrt{3}R} \frac{Q R}{10 \pi \epsilon_0 (R^2 + y^2)} dy = \frac{Q R}{10 \pi \epsilon_0} \int_{-\sqrt{3}R}^{\sqrt{3}R} \frac{1}{R^2 + y^2} dy \] \[ \Phi_E = \frac{Q R}{10 \pi \epsilon_0} \left[ \frac{1}{R} \arctan\left(\frac{y}{R}\right) \right]_{-\sqrt{3}R}^{\sqrt{3}R} = \frac{Q}{10 \pi \epsilon_0} \left( \arctan(\sqrt{3}) - \arctan(-\sqrt{3}) \right) \] \[ \Phi_E = \frac{Q}{10 \pi \epsilon_0} \left( \frac{\pi}{3} - (-\frac{\pi}{3}) \right) = \frac{Q}{10 \pi \epsilon_0} \left( \frac{2 \pi}{3} \right) = \frac{2 \pi Q}{30 \pi \epsilon_0} = \frac{Q}{15 \epsilon_0} \]