**JEE Advanced 2019 Paper 1, Question 8**

A charged shell of radius carries a total charge . Given as the flux of electric field through a closed cylindrical surface of height , radius

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?

[ is the permittivity of free space]

- If and then
- If and then
- If and then
- If and then

**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 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 .

The cylinder in option (C) contains a *portion* of the charged sphere as shown below (the angle ). The charge enclosed is

(1)

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

(2)

The charge enclosed by the cylinder is therefore

(3)

where, in the last step we’ve substituted . Therefore, option (C) is correct.

Finally, the cylinder in option (D) encloses more area of the sphere and so which means (D) is incorrect.

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