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 centre same as that of the shell. Here center of 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? [${\in }_0$ is the permittivity of free space]

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

Answer 1

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

If $h>2R$ and $r>R$ then $\Phi =Q/{\in }_0$

For option $\left(A\right):(h>2Randr=\frac{3R}{5})$

$\sin \alpha =\frac{r}{R}=\frac{3}{5}$
$\cos \alpha =\frac{4}{5}$
Now, total charge inclosed by the system,
$Q_{\text{enclosed}}=\frac{Q}{5}$

Total flux of the system
$\therefore \varphi =\frac{Q}{5{\in }_0}$
Hence, option A is correct.

For option $\left(B\right)$:
$(h<\frac{8R}{5},r=\frac{3R}{5})$
$Q_{\text{enclosed}}=0$
and $\varphi =0$
Hence, option B is correct.

For option $\left(C\right):(h>2R\text{and}r>R)$

$Q_{\text{enclosed}}=Q$
so total flux of the system,
$\therefore \varphi =\frac{Q}{{ϵ}_0}$
Hence, option C is correct.


For option $\left(D\right):(h>2R,r=\frac{4R}{5})$


$Q_{\text{enclosed}}=Q[1-\cos \alpha ]$
$\sin \alpha =\frac{r}{R}=\frac{3}{5}$
and $\cos \alpha =\frac{5}{3}{Q}_{\text{enclosed}}=\frac{2Q}{5}$
$\therefore \varphi =\frac{2Q}{5{\in }_0}$
Hence, option D is wrong.