JEE Advanced_2019_S1 ( Physics 2019 Q-11 ) - Competishun – Live JEE & NEET Courses and Exam Prep

JEE Advanced 2019

Paper-1 2019

Multiple correct answers - Select all that apply

Question

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, 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]

Select ALL correct options:

If $h>2 R$ and $r>R$ then $\Phi=\frac{Q}{\epsilon_0}$

If $h<\frac{8 R}{5}$ and $r=\frac{3 R}{5}$ then $\Phi=0$

If $h>2 R$ and $r=\frac{4 R}{5}$ then $\Phi=\frac{Q}{5 \epsilon_0}$

If $h>2 R$ and $r=\frac{3 R}{5}$ then $\Phi=\frac{Q}{5 \epsilon_0}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

⚠ Partially correct. Some answers are missing.

Solution