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):

Electrostatics PYQ — JEE Advanced 2025 · Shift 2 · Q34 with Solution

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Correct answer: C

Electric field between coaxial cylinders

Ignoring edge effects, the system is cylindrically symmetric. The dielectric fills the region between radii $a=\sqrt{2}R, \qquad b=2R$ with dielectric constant $\kappa=5.$

For a cylindrical Gaussian surface of radius $r$ and length $\ell$ such that $\sqrt{2}R<r<2R,$ Gauss' law for the electric displacement gives $\oint \vec D\cdot d\vec A = Q.$ Since $\vec D$ is radial and uniform over the cylindrical surface, $D(r)\,(2\pi r\ell)=Q$ so $D(r)=\frac{Q}{2\pi r\ell}.$

Now, $\vec E=\frac{\vec D}{\kappa \epsilon_0},$ therefore $E(r)=\frac{Q}{2\pi \kappa \epsilon_0\ell}\cdot \frac1r.$

With $\kappa=5$ , $E(r)=\frac{Q}{10\pi \epsilon_0\ell}\cdot \frac1r.$

Geometry of the imaginary plane

The plane is parallel to the axis and at perpendicular distance $R$ from the common axis. In cross-section, this corresponds to a vertical chord $x=R$ in the annular region.

At a point on this plane, let the transverse coordinate be $y$ . Then distance from axis is $r=\sqrt{R^2+y^2}.$

The plane intersects the dielectric annulus where $\sqrt{2}R \le r \le 2R.$

So for the inner boundary: $R^2+y^2=(\sqrt{2}R)^2=2R^2 \implies y=\pm R.$

For the outer boundary: $R^2+y^2=(2R)^2=4R^2 \implies y=\pm \sqrt{3}R.$

Thus the part of the plane lying inside the dielectric is made of two segments: $y\in[R,\sqrt{3}R] \quad \text{and} \quad y\in[-\sqrt{3}R,-R].$

Component of electric field normal to the plane

The plane is $x=R$ , so its area vector is along the $x$ -direction. The electric field is radial, so its $x$ -component is $E_x=E(r)\cos\theta = E(r)\frac{x}{r}=E(r)\frac{R}{r}.$

Using $E(r)=\frac{Q}{10\pi\epsilon_0\ell}\frac1r,$ we get

Since $r^2=R^2+y^2,$ $E_x=\frac{Q}{10\pi\epsilon_0\ell}\frac{R}{R^2+y^2}.$

Flux through the plane

An area element on the plane is $dA=\ell\,dy.$

So the flux is

=\ell \int E_x\,dy.$$ Because the two $y$-segments are symmetric, $$\Phi=2\ell \int_R^{\sqrt{3}R} \frac{Q}{10\pi\epsilon_0\ell}\frac{R}{R^2+y^2}\,dy.$$ Cancel $\ell$: $$\Phi=\frac{Q}{5\pi\epsilon_0}\int_R^{\sqrt{3}R}\frac{R}{R^2+y^2}\,dy.$$ Now, $$\int \frac{R}{R^2+y^2}\,dy = \tan^{-1}\left(\frac{y}{R}\right).$$ Hence $$\Phi=\frac{Q}{5\pi\epsilon_0}\left[\tan^{-1}\left(\frac{y}{R}\right)\right]_R^{\sqrt{3}R}.$$ Evaluate limits: $$\tan^{-1}(\sqrt{3})=\frac{\pi}{3}, \qquad \tan^{-1}(1)=\frac{\pi}{4}.$$ Therefore, $$\Phi=\frac{Q}{5\pi\epsilon_0}\left(\frac{\pi}{3}-\frac{\pi}{4}\right) =\frac{Q}{5\pi\epsilon_0}\cdot \frac{\pi}{12} =\frac{Q}{60\epsilon_0}.$$ --- 5. **Final answer** $$\boxed{\Phi=\frac{Q}{60\epsilon_0}}$$ So the correct option is **C**.

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Correct answer: C

Electric field between coaxial cylinders

Ignoring edge effects, the system is cylindrically symmetric. The dielectric fills the region between radii $a=\sqrt{2}R, \qquad b=2R$ with dielectric constant $\kappa=5.$

Now, $\vec E=\frac{\vec D}{\kappa \epsilon_0},$ therefore $E(r)=\frac{Q}{2\pi \kappa \epsilon_0\ell}\cdot \frac1r.$

With $\kappa=5$ , $E(r)=\frac{Q}{10\pi \epsilon_0\ell}\cdot \frac1r.$

Geometry of the imaginary plane

The plane is parallel to the axis and at perpendicular distance $R$ from the common axis. In cross-section, this corresponds to a vertical chord $x=R$ in the annular region.

At a point on this plane, let the transverse coordinate be $y$ . Then distance from axis is $r=\sqrt{R^2+y^2}.$

The plane intersects the dielectric annulus where $\sqrt{2}R \le r \le 2R.$

So for the inner boundary: $R^2+y^2=(\sqrt{2}R)^2=2R^2 \implies y=\pm R.$

For the outer boundary: $R^2+y^2=(2R)^2=4R^2 \implies y=\pm \sqrt{3}R.$

Thus the part of the plane lying inside the dielectric is made of two segments: $y\in[R,\sqrt{3}R] \quad \text{and} \quad y\in[-\sqrt{3}R,-R].$

Component of electric field normal to the plane

The plane is $x=R$ , so its area vector is along the $x$ -direction. The electric field is radial, so its $x$ -component is $E_x=E(r)\cos\theta = E(r)\frac{x}{r}=E(r)\frac{R}{r}.$

Using $E(r)=\frac{Q}{10\pi\epsilon_0\ell}\frac1r,$ we get

Since $r^2=R^2+y^2,$ $E_x=\frac{Q}{10\pi\epsilon_0\ell}\frac{R}{R^2+y^2}.$

Flux through the plane

An area element on the plane is $dA=\ell\,dy.$

So the flux is

=\ell \int E_x\,dy.$$ Because the two $y$-segments are symmetric, $$\Phi=2\ell \int_R^{\sqrt{3}R} \frac{Q}{10\pi\epsilon_0\ell}\frac{R}{R^2+y^2}\,dy.$$ Cancel $\ell$: $$\Phi=\frac{Q}{5\pi\epsilon_0}\int_R^{\sqrt{3}R}\frac{R}{R^2+y^2}\,dy.$$ Now, $$\int \frac{R}{R^2+y^2}\,dy = \tan^{-1}\left(\frac{y}{R}\right).$$ Hence $$\Phi=\frac{Q}{5\pi\epsilon_0}\left[\tan^{-1}\left(\frac{y}{R}\right)\right]_R^{\sqrt{3}R}.$$ Evaluate limits: $$\tan^{-1}(\sqrt{3})=\frac{\pi}{3}, \qquad \tan^{-1}(1)=\frac{\pi}{4}.$$ Therefore, $$\Phi=\frac{Q}{5\pi\epsilon_0}\left(\frac{\pi}{3}-\frac{\pi}{4}\right) =\frac{Q}{5\pi\epsilon_0}\cdot \frac{\pi}{12} =\frac{Q}{60\epsilon_0}.$$ --- 5. **Final answer** $$\boxed{\Phi=\frac{Q}{60\epsilon_0}}$$ So the correct option is **C**.

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