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 & with its centre same as that of the shell. Here centre of cylinder is a point on the axis of the cylinder which is equidistant from its top & bottom surfaces. which of the followintg are correct.

Options

A. If ` h gt 2R ` and `r = 4R //5` then `phi = Q // 5 in_(0)`
B. If `h gt 2R and r = 3R// 5` then `phi = Q // 5 in_(0)`
C. If `h gt 2R and r gt R` then `phi = // in_(0)`
D. If `h lt 8 R// 5` and `r = 3R//5` then `phi = 0`

Explanation

To solve the problem step by step, we will analyze the situation based on the provided information about the charged shell and the cylindrical surface.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a charged shell of radius \( R \) carrying a total charge \( Q \).
- We need to find the electric flux \( \phi \) through a closed cylindrical surface of height \( h \) and radius \( r \) centered on the axis of the shell.

2. **Using Gauss's Law:**
- According to Gauss's Law, the electric flux \( \phi \) through a closed surface is given by:
\[ \phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
- Here, \( Q_{\text{enclosed}} \) is the charge enclosed by the Gaussian surface, and \( \epsilon_0 \) is the permittivity of free space.

3. **Case Analysis:**
- We will analyze different cases based on the height \( h \) of the cylinder and its radius \( r \).

### Case A: \( h > 2R \) and \( r = \frac{4R}{5} \)

- **Charge Enclosed:**
- The cylinder extends beyond the shell. The charge enclosed can be calculated using the angle subtended by the cylinder at the center of the shell.
- Using geometry, we find the angle \( \theta \) corresponding to \( r = \frac{4R}{5} \) and calculate the enclosed charge:
\[ Q_{\text{enclosed}} = 2 \cdot \left(1 - \cos(\theta)\right) \cdot \frac{Q}{4\pi} \]
- For \( r = \frac{4R}{5} \), \( \theta \) can be calculated as \( \tan^{-1}\left(\frac{4R/5}{R}\right) \).

- **Flux Calculation:**
- The flux through the cylindrical surface can be expressed as:
\[ \phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

### Case B: \( h > 2R \) and \( r = \frac{3R}{5} \)

- **Charge Enclosed:**
- Similar analysis as in Case A, but now with \( r = \frac{3R}{5} \).
- Calculate the angle \( \theta \) and the enclosed charge:
\[ Q_{\text{enclosed}} = 2 \cdot \left(1 - \cos(\theta)\right) \cdot \frac{Q}{4\pi} \]

- **Flux Calculation:**
- Again, use Gauss's Law to find the flux:
\[ \phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

### Case C: \( h > 2R \) and \( r > R \)

- **Charge Enclosed:**
- The entire charge \( Q \) is enclosed since the cylinder is outside the shell.

- **Flux Calculation:**
- The flux is:
\[ \phi = \frac{Q}{\epsilon_0} \]

### Case D: \( h < 4R \) and \( r = \frac{3R}{5} \)

- **Charge Enclosed:**
- If the cylinder is entirely within the shell, the enclosed charge is zero.

- **Flux Calculation:**
- The flux is:
\[ \phi = 0 \]

### Summary of Results:
- For Case A, the flux is \( \frac{2Q}{5\epsilon_0} \) (not correct).
- For Case B, the flux is \( \frac{Q}{5\epsilon_0} \) (correct).
- For Case C, the flux is \( \frac{Q}{\epsilon_0} \) (correct).
- For Case D, the flux is \( 0 \) (correct).

### Correct Statements:
- The correct options are B, C, and D.