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JEE Advanced 2024 · Paper 2, Physics question 10

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 nn is _____.

A vertical cylinder with a dashed central axis; the point charge position P is at the centre of the axis. Dashed lines from P to the rims of the top and bottom faces make a half-angle θ with the axis, i.e. each circular face subtends a cone of half-angle θ at P.
Answer and worked solution

Official answer: 3

Each end face subtends a cone of half-angle θ\theta at P, i.e. a solid angle 2π(1−cos⁡θ)2\pi(1-\cos\theta).

The two faces together take a fraction 4π(1−cos⁡θ)4π=1−cos⁡θ\frac{4\pi(1-\cos\theta)}{4\pi}=1-\cos\theta of the total flux. The curved surface takes the rest:
Φcurved=qε0cos⁡θ.\Phi_{curved}=\frac{q}{\varepsilon_0}\cos\theta.

  • θ=30∘\theta=30^\circ: Φ=qε0⋅32\Phi=\frac{q}{\varepsilon_0}\cdot\frac{\sqrt3}{2}.
  • θ=60∘\theta=60^\circ: qε0⋅12=Φ3\frac{q}{\varepsilon_0}\cdot\frac12=\frac{\Phi}{\sqrt3}.

So n=3n=3.

Topics:Electric flux and Gauss's law
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