Question 2

The electric field in the dielectric region between the cylinders ($r_1 = \sqrt{2}R$ and $r_2 = 2R$) is given by Gauss's Law: $E = \frac{Q}{2\pi \kappa \epsilon_0 \ell r}$

Consider the plane at a distance $R$ from the axis. A point on this plane at a distance $y$ from its center has a radial distance from the axis $r = \sqrt{R^2 + y^2}$.

The flux element through a strip of width $dy$ and length $\ell$ is $d\Phi = \vec{E} \cdot d\vec{A} = E \cos \theta dA$, where $\cos \theta = R/r$.

$d\Phi = \left( \frac{Q}{2\pi \kappa \epsilon_0 \ell r} \right) \left( \frac{R}{r} \right) (\ell dy) = \frac{QR}{2\pi \kappa \epsilon_0 (R^2 + y^2)} dy$

The plane intersects the dielectric region where $\sqrt{2}R \le \sqrt{R^2 + y^2} \le 2R$.

This gives the limits for $y$: Inner limit: $R^2 + y^2 = 2R^2 \Rightarrow y = \pm R$

Outer limit: $R^2 + y^2 = 4R^2 \Rightarrow y = \pm \sqrt{3}R$

The total flux is the integral over both segments (left and right sides of the chord):

$\Phi = 2 \int_{R}^{\sqrt{3}R} \frac{QR}{2\pi \kappa \epsilon_0 (R^2 + y^2)} dy = \frac{QR}{\pi \kappa \epsilon_0} \left[ \frac{1}{R} \tan^{-1} \left( \frac{y}{R} \right) \right]_{R}^{\sqrt{3}R}$

$\Phi = \frac{Q}{\pi \kappa \epsilon_0} (\tan^{-1} \sqrt{3} - \tan^{-1} 1) = \frac{Q}{\pi \kappa \epsilon_0} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \frac{Q}{12\kappa\epsilon_0}$

Substituting $\kappa = 5$: $\Phi = \frac{Q}{12(5)\epsilon_0} = \frac{Q}{60\epsilon_0}$ Thus, the correct option is (C).