Step-by-Step Solution

1. Understand Gauss's Law: According to Gauss's Law, the total electric flux ($\\\phi$) through a closed surface is equal to the net charge enclosed ($Q_{enc}$) by the surface, divided by the permittivity of free space ($\\\epsilon_0$). \\phi = \\frac{Q_{enc}}{\\\\[epsilon]_0}

2. Identify the Enclosed Charge ($Q_{enc}$): The closed surface is a cylinder of radius 'r' and height 'h'. The charge 'Q' is uniformly distributed on a spherical shell of radius 'R'. Both the shell and the cylinder are centered at the origin. The enclosed charge, $Q_{enc}$, is the charge on the portion of the spherical shell that lies inside the volume of the cylinder.

3. Determine the Geometry of the Enclosed Region: A point $(x, y, z)$ on the spherical shell is described by $x^2 + y^2 + z^2 = R^2$. A point is inside the cylinder if its cylindrical coordinates $(\\\\rho, z)$ (where $\\\\rho = \\\sqrt{x^2+y^2}$) satisfy $\\\\rho \\\le r$ and $|z| \\\le h/2$. For a point on the sphere, $\\\\rho^2 = R^2 - z^2$. The condition $\\\\rho \\\le r$ becomes $\\\\sqrt{R^2 - z^2} \\\le r$, which simplifies to $R^2 - z^2 \\\le r^2$, or $z^2 \\\ge R^2 - r^2$. This means $|z| \\\ge \\\sqrt{R^2 - r^2}$. So, the portion of the spherical shell inside the cylinder is where its z-coordinate satisfies both conditions: (i) $|z| \\\ge \\\sqrt{R^2 - r^2}$ (radially inside the cylinder) (ii) $|z| \\\le h/2$ (within the height of the cylinder)

4. Calculate the Enclosed Charge: The charge on a spherical zone is proportional to its height. The charge on the entire shell (height 2R) is Q. The charge on a zone of height H is $Q_{zone} = Q \\\cdot \\\frac{H}{2R}$. The enclosed region consists of two identical zones, one in the upper hemisphere and one in the lower. Let $z_{int} = \\\sqrt{R^2-r^2}$. The condition for the enclosed region is $z_{int} \\\le |z| \\\le h/2$. This region is non-empty only if $h/2 > z_{int}$.

   • If $h/2 \\\le z_{int}$, there is no part of the sphere inside the cylinder, so $Q_{enc} = 0$.
   • If $h/2 > z_{int}$, the height of each of the two enclosed zones is $H_{zone} = h/2 - z_{int}$. The total height of the enclosed surface area is $H_{total} = 2 \\\cdot H_{zone} = h - 2z_{int}$. The enclosed charge is $Q_{enc} = Q \\\frac{h - 2z_{int}}{2R}$.

5. Special Case for $h > 2R$ (Options A and B): If $h > 2R$, then $h/2 > R$. For any point on the sphere, $|z| \\\le R$. Thus, the condition $|z| \\\le h/2$ is always satisfied. The enclosed region is determined solely by $\\\\rho \\\le r$, which is $|z| \\\ge z_{int}$. This describes two full polar caps. The height of each cap is $H_{cap} = R - z_{int}$. The total height is $H_{total} = 2(R - z_{int})$. Q_{enc} = Q \\\frac{2(R - z_{int})}{2R} = Q \\\left(1 - \\\frac{z_{int}}{R}\\\\right) = Q \\\left(1 - \\\frac{\\\\[sqrt]{R^2-r^2}}{R}\\\\right).

Evaluate Each Option

A: If h > 2R and r = 4R / 5 Using the formula for $h > 2R$: $z_{int} = \\\sqrt{R^2 - (4R/5)^2} = \\\sqrt{R^2 - 16R^2/25} = \\\sqrt{9R^2/25} = 3R/5$. Q_{enc} = Q \\\left(1 - \\\frac{3R/5}{R}\\\\right) = Q(1 - 3/5) = 2Q/5. \\\\phi = \\\frac{Q_{enc}}{\\\\[epsilon]_0} = \\\frac{2Q}{5\\\\[epsilon]_0}. Option A states \\\\phi = Q / 5\\\\[epsilon]_0. Thus, Option A is incorrect.

B: If h > 2R and r = 3R / 5 Using the formula for $h > 2R$: $z_{int} = \\\sqrt{R^2 - (3R/5)^2} = \\\sqrt{R^2 - 9R^2/25} = \\\sqrt{16R^2/25} = 4R/5$. Q_{enc} = Q \\\left(1 - \\\frac{4R/5}{R}\\\\right) = Q(1 - 4/5) = Q/5. \\\\phi = \\\frac{Q_{enc}}{\\\\[epsilon]_0} = \\\frac{Q}{5\\\\[epsilon]_0}. Option B states \\\\phi = Q / 5\\\\[epsilon]_0. Thus, Option B is correct.

C: If h > 8R/5 and r = 3R / 5 Here, h is not necessarily greater than 2R. We use the general conditions from Step 4. $r = 3R/5 \\\implies z_{int} = \\\sqrt{R^2 - (3R/5)^2} = 4R/5$. The condition $h > 8R/5$ means $h/2 > 4R/5$. So, $h/2 > z_{int}$. This means there is a non-empty enclosed region on the sphere defined by $4R/5 \\\le |z| \\\le h/2$. Therefore, $Q_{enc}$ is not zero. The flux $\\\\phi$ is not zero. The specific enclosed charge would be $Q_{enc} = Q \\\frac{h - 2(4R/5)}{2R} = Q(\\\\frac{h}{2R} - \\\frac{4}{5})$, which is positive since $h/2R > 4/5$. Option C states $\\\\phi = 0$. Thus, Option C is incorrect as stated. (Note: It is highly likely there is a typo in this option. If the condition was $h \\\le 8R/5$, then $h/2 \\\le z_{int}$, which would make $Q_{enc}=0$ and the option correct. Given the provided answer key, this is a probable source of error in the question itself.)

D: If h > 2R and r = R In this case, $r=R$. The cylinder's radius is the same as the sphere's radius. The condition $\\\\rho \\\le r$ becomes $\\\\rho \\\le R$, which is always true for any point on the sphere. The other condition is $|z| \\\le h/2$. Since $h > 2R$, we have $h/2 > R$. For any point on the sphere, $|z| \\\le R$, so $|z| < h/2$ is always true. Both conditions are satisfied for all points on the sphere. Therefore, the entire spherical shell is enclosed within the cylinder. $Q_{enc} = Q$. \\\\phi = \\\frac{Q}{\\\\[epsilon]_0}. Option D states \\\\phi = Q/\\\\[epsilon]_0. Thus, Option D is correct.