A cylinder of length L and radius b has its axis coincident with the x-axis. The electric field in the region `vecE=200hati`. Find the flux through (i) the left end of cylinder (ii) the right end of cylinder (iii) the cylinder curved surface, (iv) the closed surface area of the cylinder .

Options

A. (i) `-100pib^(2)(ii) 100 pib^(2)(iii) 50pi b^(2) (iv) 50pib^(2)`
B. `(i)-200pib^(2)(ii)200pib^(2)(iii) 0(iv)0`
C. `(i)-100pib^(2)(ii) 100pib^(2)(iii) 100pib^(2) (iv) 100 pib^(2)`
D. `(i)-200pib^(2)(ii) 200pib^(2)(iii) 200pib^(2)(iv) 200pib^(2)`

Explanation

To solve the problem, we need to calculate the electric flux through different parts of a cylinder placed in a uniform electric field. The electric field is given as \( \vec{E} = 200 \hat{i} \).

### Step-by-Step Solution:

1. **Understanding the Cylinder's Orientation:**
- The cylinder has a length \( L \) and radius \( b \).
- Its axis is aligned with the x-axis.
- The left end of the cylinder is at \( x = 0 \) and the right end is at \( x = L \).

2. **Calculating the Flux through the Left End of the Cylinder:**
- The area vector \( \vec{A}_{\text{left}} \) for the left end points in the negative x-direction, which can be represented as \( \vec{A}_{\text{left}} = -\pi b^2 \hat{i} \).
- The electric flux \( \Phi_{\text{left}} \) through the left end is given by:
\[ \Phi_{\text{left}} = \vec{E} \cdot \vec{A}_{\text{left}} = (200 \hat{i}) \cdot (-\pi b^2 \hat{i}) = -200 \pi b^2 \]

3. **Calculating the Flux through the Right End of the Cylinder:**
- The area vector \( \vec{A}_{\text{right}} \) for the right end points in the positive x-direction, which is \( \vec{A}_{\text{right}} = \pi b^2 \hat{i} \).
- The electric flux \( \Phi_{\text{right}} \) through the right end is given by:
\[ \Phi_{\text{right}} = \vec{E} \cdot \vec{A}_{\text{right}} = (200 \hat{i}) \cdot (\pi b^2 \hat{i}) = 200 \pi b^2 \]

4. **Calculating the Flux through the Curved Surface of the Cylinder:**
- The area vector for the curved surface of the cylinder is perpendicular to the electric field lines at every point, which means the angle \( \theta \) between the electric field and the area vector is 90 degrees.
- Therefore, the flux through the curved surface \( \Phi_{\text{curved}} \) is:
\[ \Phi_{\text{curved}} = \vec{E} \cdot \vec{A}_{\text{curved}} = E \cdot A \cdot \cos(90^\circ) = 0 \]

5. **Calculating the Total Flux through the Closed Surface of the Cylinder:**
- The total flux through the closed surface \( \Phi_{\text{total}} \) is the sum of the fluxes through the left end, right end, and the curved surface:
\[ \Phi_{\text{total}} = \Phi_{\text{left}} + \Phi_{\text{right}} + \Phi_{\text{curved}} = (-200 \pi b^2) + (200 \pi b^2) + 0 = 0 \]

### Summary of Results:
- (i) Flux through the left end: \( \Phi_{\text{left}} = -200 \pi b^2 \)
- (ii) Flux through the right end: \( \Phi_{\text{right}} = 200 \pi b^2 \)
- (iii) Flux through the curved surface: \( \Phi_{\text{curved}} = 0 \)
- (iv) Total flux through the closed surface: \( \Phi_{\text{total}} = 0 \)