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]

• A. If $h>2R$ and $r>R$ then $\pi =Q/{\in }_0$
• B. If $h<8R/5$ and $r=3R/5$ then $\pi =0$
• C. If $h>2R$ and $r=3R/5$ then $\pi =Q/5{\in }_0$
• D. If $h>2R$ and $r=4R/5$ then $\pi =Q/5{\in }_0$

Answer 1

Answer: (A), (B), (C)

For option (A) : $\left(h>2Rr=\frac{3R}{5}\right)$
$sin\alpha =\frac{r}{R}=\frac{3}{5}$
$cos\alpha =\frac{4}{5}$
$Q_{enclosed}=\frac{Q}{5}$
$\therefore \varphi =\frac{Q}{5{\in }_0}$
For option (B) : $h<\frac{8R}{5},r=\frac{3R}{5}$
$Q_{enclosed}=0$
and $\varphi =0$
For option (C) : ($h>2R$ and $r>R$)
$Q_{enclosed}=Q$
$\therefore \varphi =\frac{Q}{{\in }_0}$
For option (D) : $h>2Rr=\frac{4R}{5}$
$Q_{enclosed}=Q\left[1-cos\alpha \right]$
$sin\alpha =\frac{r}{R}=\frac{3}{5}$
and $cos\alpha =\frac{3}{5}{Q}_{enclosed}=\frac{2Q}{5}$
$\therefore \varphi =\frac{2Q}{5{\in }_0}$