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:
A
If $h>2 R$ and $r>R$ then $\Phi=\frac{Q}{\epsilon_0}$
B
If $h<\frac{8 R}{5}$ and $r=\frac{3 R}{5}$ then $\Phi=0$
C
If $h>2 R$ and $r=\frac{4 R}{5}$ then $\Phi=\frac{Q}{5 \epsilon_0}$
D
If $h>2 R$ and $r=\frac{3 R}{5}$ then $\Phi=\frac{Q}{5 \epsilon_0}$
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
