The electric flux through a hemispherical surface of radius R placed in a uniform electric field E parallel to the axis of the circular plane is

Options

A. `(2piR)E`
B. `(piR^(2))E`
C. `(4/3piR^(3))E`
D. `(2/3piR^(3))E`

Explanation

To find the electric flux through a hemispherical surface of radius \( R \) placed in a uniform electric field \( E \) that is parallel to the axis of the circular plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux (\( \Phi \)) is defined as the integral of the electric field (\( \mathbf{E} \)) over a surface area (\( \mathbf{A} \)):
\[ \Phi = \int \mathbf{E} \cdot d\mathbf{A} \]

2. **Identifying the Geometry**:
We have a hemispherical surface with a flat circular base. The electric field \( \mathbf{E} \) is uniform and directed parallel to the axis of the circular base.

3. **Calculating the Flux through the Hemispherical Surface**:
The electric field is constant, so we can take it out of the integral. The area vector \( d\mathbf{A} \) for the hemispherical surface points outward, and the angle \( \theta \) between \( \mathbf{E} \) and \( d\mathbf{A} \) varies across the surface.

4. **Using Symmetry**:
The electric field contributes to the flux through the curved surface of the hemisphere. The contribution to the flux through the flat circular base is zero since the electric field is parallel to the base (angle \( \theta = 90^\circ \), \( \cos(90^\circ) = 0 \)).

5. **Calculating the Area of the Hemisphere**:
The surface area of the hemispherical part is given by:
\[ A_{\text{hemisphere}} = 2\pi R^2 \]

6. **Calculating the Flux through the Curved Surface**:
For the curved surface, we consider the average angle \( \theta \) between the electric field and the area vector. The average value of \( \cos(\theta) \) over the hemisphere can be shown to be \( \frac{1}{2} \).

Therefore, the electric flux through the hemispherical surface can be calculated as:
\[ \Phi = E \cdot A_{\text{hemisphere}} \cdot \text{average}(\cos(\theta)) = E \cdot (2\pi R^2) \cdot \frac{1}{2} \]
\[ \Phi = E \cdot \pi R^2 \]

7. **Final Result**:
Thus, the electric flux through the hemispherical surface is:
\[ \Phi = \pi R^2 E \]