The length and radius of a cylinder are L and R respectively. The total flux for the surface of the cylinder, when it is placed in a uniform electric field E parallel to the axis of the cylinder is

Options

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

Answer

To find the total electric flux through the surface of a cylinder placed in a uniform electric field \( E \) parallel to the axis of the cylinder, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Surfaces of the Cylinder**:
The cylinder has three surfaces:
- Two circular ends (let's call them Surface 1 and Surface 2).
- One curved surface (let's call it Surface 3).

2. **Understand the Orientation of the Electric Field**:
The electric field \( E \) is parallel to the axis of the cylinder. This means that the electric field lines are running along the length of the cylinder.

3. **Calculate Electric Flux for Each Surface**:
- **Surface 1 (one circular end)**:
The area vector \( \vec{A_1} \) is perpendicular to the surface and points outward. The angle between the electric field \( \vec{E} \) and the area vector \( \vec{A_1} \) is \( 180^\circ \) (since they are in opposite directions).
\[ \Phi_1 = \vec{E} \cdot \vec{A_1} = E \cdot A_1 \cdot \cos(180^\circ) = -E \cdot \pi R^2 \]

- **Surface 2 (the other circular end)**:
The area vector \( \vec{A_2} \) for this surface also points outward, but the electric field is in the same direction as the area vector. The angle between \( \vec{E} \) and \( \vec{A_2} \) is \( 0^\circ \).
\[ \Phi_2 = \vec{E} \cdot \vec{A_2} = E \cdot A_2 \cdot \cos(0^\circ) = E \cdot \pi R^2 \]

- **Surface 3 (the curved surface)**:
For the curved surface, the area vector is perpendicular to the electric field at every point on the surface. The angle between \( \vec{E} \) and the area vector \( \vec{A_3} \) is \( 90^\circ \).
\[ \Phi_3 = \vec{E} \cdot \vec{A_3} = E \cdot A_3 \cdot \cos(90^\circ) = 0 \]

4. **Sum the Electric Fluxes**:
Now, we can find the total electric flux \( \Phi \) through the entire surface of the cylinder by summing the contributions from all three surfaces:
\[ \Phi = \Phi_1 + \Phi_2 + \Phi_3 \]
Substituting the values we calculated:
\[ \Phi = (-E \cdot \pi R^2) + (E \cdot \pi R^2) + 0 = 0 \]

### Final Result:
The total electric flux through the surface of the cylinder is:
\[ \Phi = 0 \]