JEE Advanced 2025 · Paper 2 · Physics Q2

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Two co-axial conducting cylinders of same length $\ell$ with radii $\sqrt2R$ 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. 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):

Fig. 1: two coaxial cylinders of length ℓ, the inner one (radius √2R, charge Q) inside the outer one (radius 2R, grounded), with the annular gap shaded (dielectric); a transparent vertical plane of length ℓ cuts through the arrangement parallel to the axis. Fig. 2: cross-section showing the inner circle of radius √2R, the shaded annulus out to 2R, and the plane as a dashed vertical line at distance R from the centre, crossing the annulus.

\dfrac{Q}{30\epsilon_0}

\dfrac{Q}{15\epsilon_0}

\dfrac{Q}{60\epsilon_0}

\dfrac{Q}{120\epsilon_0}

Answer and worked solution

Official answer: C

Field in the dielectric. For $\sqrt2R<r<2R$ ,
$E=\frac{Q}{2\pi\epsilon_0\kappa\ell r}.$
The field is zero outside this annulus.

Geometry of the plane. At distance $s$ along the plane from its foot, a point is at $r=\sqrt{R^2+s^2}$ . The normal component of the field is $E\frac Rr$ . The plane is inside the dielectric for $R\le|s|\le\sqrt3R$ , on both sides of the foot.

Flux.
$\Phi=\frac{Q}{2\pi\epsilon_0\kappa}\cdot2\int_R^{\sqrt3R}\frac{R\,ds}{R^2+s^2}=\frac{Q}{\pi\epsilon_0\kappa}\left(\frac\pi3-\frac\pi4\right)=\frac{Q}{12\epsilon_0\kappa}.$

With $\kappa=5$ , $\Phi=\dfrac{Q}{60\epsilon_0}$ .

Topics:Electric flux and Gauss's law Dielectrics

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JEE Advanced 2025 · Paper 2, Physics question 2