JEE Advanced 2025 Paper 2 · Q02 · Gauss's Law
Two co-axial conducting cylinders of same length $l$ with radii $\sqrt{2}\,R$ and $2R$ are kept (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, parallel to the axis (Fig. 2). Ignoring edge effects, the flux of the electric field through the plane is ($\epsilon_0$ is the permittivity of free space):
Reveal answer + step-by-step solution
Correct answer:C
Solution
Inside the dielectric ($\sqrt{2}\,R < r < 2R$) the radial electric field at distance $r$ from axis is $E(r) = Q/(2\pi\epsilon_0 \kappa l r)$. The plane lies parallel to the axis at distance $R$ from it. It intersects the dielectric in two strips on either side of axis; on each strip the angular extent of the strip subtended at the axis runs between $\theta_1 = \arccos(R/(\sqrt{2}\,R)) = \pi/4$ (where the plane meets the inner cylinder) and $\theta_2 = \arccos(R/(2R)) = \pi/3$ (where it meets the outer cylinder). Flux through one strip $= (Q/(2\pi\epsilon_0 \kappa))(\theta_2 - \theta_1) = (Q/(2\pi\epsilon_0 \cdot 5))(\pi/3 - \pi/4) = (Q/(10\pi\epsilon_0))(\pi/12) = Q/(120\,\epsilon_0)$. With two symmetric strips, total flux $= Q/(60\,\epsilon_0)$. Answer (C).
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