To find the net electric flux through the cylinder due to an infinite line charge at its axis, we can follow these steps:
### Step-by-Step Solution:
1. **Identify Given Values**:
- Length of the cylinder, \( L = 1 \, \text{m} \)
- Radius of the cylinder, \( r = 7 \, \text{cm} = 0.07 \, \text{m} \)
- Electric field at the curved surface, \( E = 250 \, \text{N/C} \)
2. **Understand Electric Flux**:
- The electric flux \( \Phi \) through a surface is given by the formula:
\[
\Phi = E \cdot A \cdot \cos(\theta)
\]
- Where \( E \) is the electric field, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the electric field and the normal to the surface.
3. **Calculate the Area of the Curved Surface**:
- The area \( A \) of the curved surface of the cylinder is given by:
\[
A = 2 \pi r L
\]
- Substituting the values:
\[
A = 2 \pi (0.07 \, \text{m}) (1 \, \text{m}) = 2 \pi (0.07) \approx 0.4398 \, \text{m}^2
\]
4. **Calculate the Electric Flux through the Curved Surface**:
- Since the electric field is radially outward and the area vector of the curved surface is also outward, the angle \( \theta = 0^\circ \) (which means \( \cos(0) = 1 \)).
- Therefore, the electric flux through the curved surface is:
\[
\Phi_{\text{curved}} = E \cdot A \cdot \cos(0) = E \cdot A
\]
- Substituting the values:
\[
\Phi_{\text{curved}} = 250 \, \text{N/C} \cdot 0.4398 \, \text{m}^2 \approx 109.95 \, \text{N m}^2/\text{C}
\]
5. **Final Result**:
- The net electric flux through the cylinder is approximately:
\[
\Phi \approx 1.1 \times 10^2 \, \text{N m}^2/\text{C}
\]
