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(A-D) According to Gauss's Law,the flux $\Phi$ through a closed surface is $\frac{q_{enclosed}}{\epsilon_0}$.$(1)$ If $h > 2R$ and $r > R$,the entire

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(A-D) According to Gauss's Law,the flux $\Phi$ through a closed surface is $\frac{q_{enclosed}}{\epsilon_0}$.$(1)$ If $h > 2R$ and $r > R$,the entire shell is enclosed by the cylinder. Thus,$q_{enclosed} = Q$ and $\Phi = \frac{Q}{\epsilon_0}$. This is correct.$(2)$ If $h $(3)$ If $h > 2R$ and $r = \frac{4R}{5}$,the cylinder intersects the shell. The flux is due to the charge on the spherical caps cut by the cylinder. The solid angle subtended by a cap is $\Omega = 2\pi(1 - \cos	heta)$,where $\sin	heta = \frac{r}{R} = \frac{4}{5}$,so $\cos	heta = \frac{3}{5}$. The flux through one cap is $\frac{Q}{4\pi\epsilon_0} 	imes \Omega = \frac{Q}{2\epsilon_0}(1 - \cos	heta)$. For two caps,$\Phi = \frac{Q}{\epsilon_0}(1 - \frac{3}{5}) = \frac{2Q}{5\epsilon_0}$. This is correct.$(4)$ If $h > 2R$ and $r = \frac{3R}{5}$,then $\sin	heta = \frac{3}{5}$,so $\cos	heta = \frac{4}{5}$. The flux is $\Phi = \frac{Q}{\epsilon_0}(1 - \frac{4}{5}) = \frac{Q}{5\epsilon_0}$. This is correct.

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