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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2024 · Paper 2 · Physics · Electrostatics

A charge is kept at the central point P of a cylindrical region....

A charge is kept at the central point P\mathrm{P} of a cylindrical region. The two edges subtend a half-angle θ\theta at P\mathrm{P}, as shown in the figure. When θ=30∘\theta=30^{\circ}, then the electric flux through the curved surface of the cylinder is Φ\Phi. If θ=60∘\theta=60^{\circ}, then the electric flux through the curved surface becomes Φ/n\Phi / \sqrt{n}, where the value of nn is _______ . A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ at P , as shown

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A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ at P , as shown
Solid angle made by plane surfaces Ω=2×2π(1−cos⁡θ)⇒Ω=4π−4πcos⁡θ\Omega=2 \times 2 \pi(1-\cos \theta) \Rightarrow \Omega=4 \pi-4 \pi \cos \theta So solid angle made by curved surface =4π−Ω=4 \pi-\Omega=4π−(4π−4πcos⁡θ)=4πcos⁡θϕ30∘=ϕ=4πcos⁡30∘4πQϵ0=cos⁡30∘Qϵ0ϕ60=4πcos⁡60∘4πQϵ0=cos⁡60∘Qϵ0ϕ30ϕ60=cos⁡30∘cos⁡60∘=3ϕϕ60=3ϕ60=ϕ3⇒n=3\begin{aligned} & =4 \pi-(4 \pi-4 \pi \cos \theta)=4 \pi \cos \theta \\ & \phi_{30^{\circ}}=\phi=\frac{4 \pi \cos 30^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 30^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \\ & \phi_{60}=\frac{4 \pi \cos 60^{\circ}}{4 \pi} \frac{\mathrm{Q}}{\epsilon_0}=\cos 60^{\circ} \frac{\mathrm{Q}}{\epsilon_0} \\ & \frac{\phi_{30}}{\phi_{60}}=\frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\sqrt{3} \\ & \frac{\phi}{\phi_{60}}=\sqrt{3} \\ & \phi_{60}=\frac{\phi}{\sqrt{3}} \Rightarrow \mathrm{n}=3 \end{aligned}$$
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