Two co-axial conducting cylinders of same length $\ell$ with radii $\sqrt{2} R$ and $2 R$ 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):

(C) $\frac{Q}{60 \epsilon_{0}}$

(A) $\frac{Q}{30 \epsilon_{0}}$

(B) $\frac{Q}{15 \epsilon_{0}}$

(D) $\frac{Q}{120 \epsilon_{0}}$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Electrostatics

Two co-axial conducting cylinders of same length

ℓ

with radii

2 R

and

2 R

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

κ = 5

. Consider an imaginary plane of the same length

ℓ

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 (

ϵ_{0}

is the permittivity of free space):

(A) $\frac{Q}{30 ϵ _{0}}$
(B) $\frac{Q}{15 ϵ _{0}}$
(C) $\frac{Q}{60 ϵ _{0}}$ ✓ Correct
(D) $\frac{Q}{120 ϵ _{0}}$

Answer: (C)

Solution

The electric field due to the inner cylinder (line charge density

λ = Q / ℓ

) at distance

r

E = \frac{λ}{2 π ϵ _{0} ϵ _{r} r}

. The plane at distance

R

intersects the field in two regions (AB and CD) where the field is non-zero.

For each segment, integrating the flux contribution over the angular range from

π /4

π /3

(determined by geometry):

ϕ_{A B} = 2 k λ ℓ [\frac{π}{3} - \frac{π}{4}] = 2 k λ ℓ \cdot \frac{π}{12}

With

k = \frac{1}{4 π ϵ _{0} ϵ _{r}}

and

ϵ_{r} = 5

ϕ_{A B} = \frac{Q}{120 ϵ _{0}}

Since

ϕ_{C D} = ϕ_{A B}

, the total flux is:

ϕ_{total} = 2 ϕ_{A B} = \frac{Q}{60 ϵ _{0}}

The answer is (C).

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Electrostatics

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