A cylinder of radius R and length l is placed in a uniform electric field E parallel to the axis of the cylinder. The total flux over the curved surface of the cylinder is

Options

A. zero
B. `piR^2E`
C. `2piR^2E`
D. `E//piR^2`

Explanation

To solve the problem of finding the total electric flux over the curved surface of a cylinder placed in a uniform electric field parallel to its axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a cylinder with radius \( R \) and length \( l \).
- The electric field \( \vec{E} \) is uniform and parallel to the axis of the cylinder.

2. **Identify the Surface Area**:
- The curved surface area of the cylinder can be calculated using the formula:
\[ A = 2\pi R l \]
- However, since we are interested in the flux through the curved surface, we will focus on the area vector direction.

3. **Determine the Area Vector Direction**:
- The area vector \( \vec{A} \) for the curved surface of the cylinder points radially outward, perpendicular to the surface.
- This means that the area vector is oriented perpendicular to the axis of the cylinder.

4. **Evaluate the Dot Product**:
- The electric flux \( \Phi \) through a surface is given by:
\[ \Phi = \vec{E} \cdot \vec{A} = \int \vec{E} \cdot d\vec{A} \]
- Since the electric field \( \vec{E} \) is parallel to the axis of the cylinder and the area vector \( d\vec{A} \) for the curved surface is perpendicular to the electric field, we have:
\[ \vec{E} \cdot d\vec{A} = |\vec{E}| |d\vec{A}| \cos(90^\circ) = 0 \]

5. **Conclusion**:
- Since the dot product is zero, the total electric flux \( \Phi \) through the curved surface of the cylinder is:
\[ \Phi = 0 \]

### Final Answer:
The total flux over the curved surface of the cylinder is **zero**.