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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2025 · Paper 2 · Physics · Electrostatics

Two co-axial conducting cylinders of same length with radii 2 R and...

Two co-axial conducting cylinders of same length ℓ\ell with radii 2R\sqrt{2} R and 2R2 R are kept, as shown in Fig. 1. The charge on the inner cylinder is QQ and the outer cylinder is grounded. The annular region between the cylinders is filled with a material of dielectric constant κ=5\kappa=5. Consider an imaginary plane of the same length ℓ\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 ( ϵ0\epsilon_0 is the permittivity of free space): Two co-axial conducting cylinders of same length with radii 2 R and 2 R are kept, as shown in Fig. 1. The charge on the

Options

  1. AQ30ϵ0\frac{\mathrm{Q}}{30 \epsilon_0}
  2. BQ15ϵ0\frac{\mathrm{Q}}{15 \epsilon_0}
  3. CQ60ϵ0\frac{\mathrm{Q}}{60 \epsilon_0}
  4. DQ120ϵ0\frac{\mathrm{Q}}{120 \epsilon_0}

Correct answer

C. Q60ϵ0\frac{\mathrm{Q}}{60 \epsilon_0}

Step-by-step solution

Here we are assuming that " ℓ\ell " is very large just for the sake of symmetry. Outside cylinder will have zero electric field inside, so the flux generated on the plate will be due to inner cylinder only in sections AB and CD , as section be will be at that place where electric field is zero.flux through element will be dϕ=E→⋅dS→dϕ=2kλdy⋅ℓ⋅cos⁡θ.....(1)\begin{aligned} \mathrm{d} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}} \\ \mathrm{d} \phi & =\frac{2 \mathrm{k} \lambda}{} \mathrm{dy} \cdot \ell \cdot \cos \theta ..... \mathrm{(1)}\end{aligned} from figure we can say that cos⁡θ=Rr⇒r=Rsec⁡θtan⁡θ=yR⇒y=Rtan⁡θ⇒dy=Rsec⁡2θdθdϕ=2kλRsec⁡θR⋅sec⁡2θ⋅ℓ⋅cos⁡θdθdϕ=2kλℓdq∫0ϕABdϕ=2kλℓ∫π/4π/3dθϕAB=2kλℓ[π3−π4]\begin{aligned} & \cos \theta=\frac{\mathrm{R}}{\mathrm{r}} \Rightarrow \mathrm{r}=\mathrm{R} \sec \theta \\ & \tan \theta=\frac{\mathrm{y}}{\mathrm{R}} \Rightarrow \mathrm{y}=\mathrm{R} \tan \theta \\ & \Rightarrow \mathrm{dy}=\mathrm{R} \sec ^2 \theta \mathrm{ d} \theta \\ & \mathrm{ d} \phi=\frac{2 \mathrm{k} \lambda}{\mathrm{R} \sec \theta} \mathrm{R} \cdot \sec ^2 \theta \cdot \ell \cdot \cos \theta \mathrm{ d} \theta \\ & \mathrm{ d} \phi=2 \mathrm{k} \lambda \ell \mathrm{dq} \\ & \int_0^{\phi_{\mathrm{AB}}} \mathrm{d} \phi=2 \mathrm{k} \lambda \ell \int_{\pi / 4}^{\pi / 3} \mathrm{ d} \theta \\ & \phi_{\mathrm{AB}}=2 \mathrm{k} \lambda \ell\left[\frac{\pi}{3}-\frac{\pi}{4}\right]\end{aligned}ϕAB=2kQ[π12]ϕAB=2×14πϵ0ϵrQ[π12]ϕAB=Q120ϵ0ϕplate =ϕAB+ϕBC+ϕCD(ϕCD=ϕAB)ϕplate =Q60ϵ0\begin{aligned} & \phi_{\mathrm{AB}}=2 \mathrm{kQ}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=2 \times \frac{1}{4 \pi \epsilon_0 \epsilon_{\mathrm{r}}} \mathrm{Q}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=\frac{\mathrm{Q}}{120 \epsilon_0} \\ & \phi_{\text{plate }}=\phi_{\mathrm{AB}}+\phi_{\mathrm{BC}}+\phi_{\mathrm{CD}} \quad\left(\phi_{\mathrm{CD}}=\phi_{\mathrm{AB}}\right) \\ & \phi_{\text{plate }}=\frac{\mathrm{Q}}{60 \epsilon_0}\end{aligned}$
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