A cylinder of radius R and length L is placed in a uniform electric field E parallel to the axis. The total flux for the surface of the cylinder is given by

Options

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

Answer

To find the total electric flux through the 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 of radius \( R \) and length \( L \).
- The electric field \( E \) is uniform and parallel to the axis of the cylinder.

2. **Apply Gauss's Law**:
- Gauss's Law states that the total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{\text{enclosed}} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

3. **Determine the Charge Enclosed**:
- In this case, since the cylinder is placed in a uniform electric field and there are no charges inside the cylinder, we have:
\[ Q_{\text{enclosed}} = 0 \]

4. **Calculate the Total Flux**:
- Substituting the value of \( Q_{\text{enclosed}} \) into Gauss's Law gives:
\[ \Phi = \frac{0}{\epsilon_0} = 0 \]

5. **Conclusion**:
- Therefore, the total electric flux through the surface of the cylinder is:
\[ \Phi = 0 \]

### Final Answer:
The total flux for the surface of the cylinder is \( 0 \).

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