JEE Advanced 2024 · Paper 2 · numerical answer

JEE Advanced 2024 Physics question on Electrostatics

A charge is kept at the central point P of a cylindrical region. The two edges subtend a half-angle θ\theta at 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 n is______.
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Answer: 3

[JEE Advanced 2024 · Paper 2]

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