Electric Flux Through Cylinder Intersecting Spherical Charged Shell

To determine the total electric flux $\Phi$ passing through the closed cylindrical surface, we apply Gauss's Law:

$\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$

where $q_{\text{enclosed}}$ is the net charge enclosed within the cylinder, and $\epsilon_0$ is the permittivity of free space.

The spherical shell has radius $R$ and total charge $Q$ uniformly distributed over its surface. Thus, its surface charge density is:

$\sigma = \frac{Q}{4\pi R^2}$

The center of the cylinder coincides with the center of the spherical shell at the origin $(0,0,0)$. The cylinder is defined by:

• Radius $r$ in the $xy$-plane ($\rho = \sqrt{x^2 + y^2} \le r$)
• Height $h$ along the $z$-axis ($-\frac{h}{2} \le z \le \frac{h}{2}$)

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Analysis of Options:

Option A: $h > 2R$ and $r > R$

• Since $h > 2R$, the height of the cylinder extends from $z = -h/2 < -R$ to $z = h/2 > R$. Thus, it completely covers the $z$-span of the spherical shell ($[-R, R]$).
• Since $r > R$, the radius of the cylinder is greater than the radius of the shell at all cross-sections.
• Consequently, the entire spherical shell is completely enclosed inside the cylinder:

$q_{\text{enclosed}} = Q \implies \Phi = \frac{Q}{\epsilon_0}$

Thus, Option A is correct.

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Option B: $h < \frac{8R}{5}$ and $r = \frac{3R}{5}$

• The spherical shell intersects the cylindrical boundary $\rho = \frac{3R}{5}$ at $z$-coordinates given by:

$z^2 = R^2 - r^2 = R^2 - \left(\frac{3R}{5}\right)^2 = \frac{16R^2}{25} \implies |z| = \frac{4R}{5}$

• The portions of the spherical shell that satisfy $\rho \le \frac{3R}{5}$ lie in the regions where $|z| \ge \frac{4R}{5}$.
• The cylinder extends along the $z$-axis only up to $|z| \le \frac{h}{2}$. Since $h < \frac{8R}{5}$, we have:

$\frac{h}{2} < \frac{4R}{5}$

• Hence, no part of the spherical shell falls within the cylindrical region ($|z| \le h/2$ and $\rho \le r$).
• Therefore, $q_{\text{enclosed}} = 0$, giving:

$\Phi = 0$

Thus, Option B is correct.

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Option C: $h > 2R$ and $r = \frac{3R}{5}$

• Since $h > 2R$, the cylinder covers the entire vertical extent of the sphere.
• The region of the spherical shell enclosed by the cylinder $\rho \le \frac{3R}{5}$ corresponds to two spherical caps located at:

$|z| \ge \frac{4R}{5}$

• The height $H$ of each spherical cap is:

$H = R - \frac{4R}{5} = \frac{R}{5}$

• The surface area of a single spherical cap of height $H$ is:

$A_{\text{cap}} = 2\pi R H = 2\pi R \left(\frac{R}{5}\right) = \frac{2\pi R^2}{5}$

• Since there are two such caps (top and bottom), the total enclosed surface area is:

$A_{\text{total}} = 2 \times A_{\text{cap}} = \frac{4\pi R^2}{5}$

• The enclosed charge is:

$q_{\text{enclosed}} = \sigma \cdot A_{\text{total}} = \left(\frac{Q}{4\pi R^2}\right) \left(\frac{4\pi R^2}{5}\right) = \frac{Q}{5}$

• Using Gauss's Law:

$\Phi = \frac{Q}{5\epsilon_0}$

Thus, Option C is correct.

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Option D: $h > 2R$ and $r = \frac{4R}{5}$

• The spherical shell intersects the cylinder boundary $\rho = \frac{4R}{5}$ at:

$|z| = \sqrt{R^2 - \left(\frac{4R}{5}\right)^2} = \frac{3R}{5}$

• The height of each spherical cap inside the cylinder is:

$H = R - \frac{3R}{5} = \frac{2R}{5}$

• The area of one cap is:

$A_{\text{cap}} = 2\pi R \left(\frac{2R}{5}\right) = \frac{4\pi R^2}{5}$

• The total enclosed area of both caps is:

$A_{\text{total}} = 2 \times \frac{4\pi R^2}{5} = \frac{8\pi R^2}{5}$

• The enclosed charge is:

$q_{\text{enclosed}} = \sigma \cdot A_{\text{total}} = \left(\frac{Q}{4\pi R^2}\right) \left(\frac{8\pi R^2}{5}\right) = \frac{2Q}{5}$

• Thus, the flux is:

$\Phi = \frac{2Q}{5\epsilon_0}$

Thus, Option D is incorrect.

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Correct Choice(s):

A, B, C