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Physics · Electrostatics

JEE Advanced 2024 — Paper 2 — Question 21

A charge is kept at the central point PP of a cylindrical region. The two edges subtend a half-angle θ\theta at PP, 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 ϕn\frac{\phi}{\sqrt{n}}, where the value of nn is _____\_\_\_\_\_

Question figure

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

ϕ1=qε0−qε0(1−cos⁡30∘2)×2=qε0[1−1+32]=32qε0\quad \phi_{1}=\frac{q}{\varepsilon_{0}}-\frac{q}{\varepsilon_{0}}\left(\frac{1-\cos 30^{\circ}}{2}\right) \times 2=\frac{q}{\varepsilon_{0}}\left[1-1+\frac{\sqrt{3}}{2}\right]=\frac{\sqrt{3}}{2} \frac{q}{\varepsilon_{0}}

ϕ2=qε0−qε0(1−cos⁡60∘2)×2=12qε0\phi_{2}=\frac{\mathrm{q}}{\varepsilon_{0}}-\frac{\mathrm{q}}{\varepsilon_{0}}\left(\frac{1-\cos 60^{\circ}}{2}\right) \times 2=\frac{1}{2} \frac{\mathrm{q}}{\varepsilon_{0}}

ϕ2ϕ1=1/23/2=13\frac{\phi_{2}}{\phi_{1}}=\frac{1 / 2}{\sqrt{3} / 2}=\frac{1}{\sqrt{3}} ϕ2=ϕ13⇒N=3\phi_{2}=\frac{\phi_{1}}{\sqrt{3}} \Rightarrow N=3

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electric flux and Gauss's Law

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