Flux from a charged shell

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JEE Advanced 2019 Paper 1, Question 8

A charged shell of radius R carries a total charge Q. Given \Phi as the flux of electric field through a closed cylindrical surface of height h, radius r
and with its center same as that of the shell. Here, the center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct?

[\epsilon_0 is the permittivity of free space]

  1. If h > 2R and r > R then \Phi = Q/\epsilon_0
  2. If h < 8R/5 and r = 3R/5 then \Phi = 0
  3. If h > 2R and r = 3R/5 then \Phi = Q/5\epsilon_0
  4. If h > 2R and r = 4R/5 then \Phi = Q/5\epsilon_0

Related problems:
Electric field from a sphere with a cavity
Electric field in a hollow region
Charge at one corner of a cube
Electric flux through a hemisphere

Solution:

The cylinder in option (A) completely encloses the sphere, which means \Phi = Q/\epsilon_0 indeed by Gauss’s law.

With a little geometry we can see that the cylinder in option (B) is contained within the sphere as shown below. Therefore the charge enclosed by the cyclinder is 0, which means \Phi = 0.

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The cylinder in option (C) contains a portion of the charged sphere as shown below (the angle \alpha = \tan^{-1}(3/4)). The charge enclosed is

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(1)   \begin{equation*} Q_{\rm enc} = \frac{\text{area of } AB + \text{area of } CD}{4 \pi R^2} \times Q \end{equation*}

We can compute the area of AB (which is also the same as that of CD) as shown in the figure below. Consider the shaded ring like segment of the region AB which subtends an angle d \theta at the center of the sphere. The radius of this ring is R \sin \theta and its width is R d \theta. Its area is therefore 2 \pi R \sin \theta \times R d \theta. The total area of AB is obtained by adding together the area of several such rings starting from \theta = 0 upto \theta = \alpha. That is,

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(2)   \begin{equation*} \text{area of } AB = \int_0^{\alpha} 2 \pi R^2 \sin \theta d \theta = 2 \pi R^2 (1 - \cos \alpha) . \end{equation*}

The charge enclosed by the cylinder is therefore

(3)   \begin{equation*} Q_{\rm enc} = \frac{2 \times 2 \pi R^2 (1 - \cos \alpha)}{4 \pi R^2} Q = \frac{Q}{5} \end{equation*}

where, in the last step we’ve substituted \cos \alpha = 4/5. Therefore, option (C) is correct.

Finally, the cylinder in option (D) encloses more area of the sphere and so Q_{\rm enc} > Q/5 which means (D) is incorrect.

The correct answers are (A), (B), (C).

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