A hemispherical bowl of radius r is placed in uniform electric field E . The electric flux passing through the bowl is

Options

A. `pir^(2)E`
B. `2pirE^(2)`
C. `4pir^(2)E`
D. `4pirE^(2)`

Explanation

To find the electric flux passing through a hemispherical bowl of radius \( r \) placed in a uniform electric field \( E \), we can follow these steps:

### Step 1: Understand the Concept of Electric Flux
Electric flux (\( \Phi \)) through a surface is defined as the product of the electric field (\( E \)) and the area (\( A \)) through which the field lines pass, taking into account the angle (\( \theta \)) between the electric field and the normal (area vector) to the surface. Mathematically, it is given by:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]

### Step 2: Analyze the Orientation of the Electric Field
The orientation of the electric field relative to the hemispherical bowl is crucial. There are two main cases to consider:
1. **Electric Field Parallel to the Flat Surface of the Hemisphere**: In this case, the electric field lines run parallel to the flat circular base of the hemisphere.
2. **Electric Field Perpendicular to the Flat Surface of the Hemisphere**: Here, the electric field lines are directed straight through the center of the hemisphere.

### Step 3: Case 1 - Electric Field Parallel to the Flat Surface
If the electric field is parallel to the flat surface of the hemisphere:
- The angle \( \theta \) between the electric field and the area vector of the flat surface is \( 90^\circ \).
- Therefore, \( \cos(90^\circ) = 0 \).
- The electric flux through the hemispherical bowl is:
\[ \Phi = E \cdot A \cdot \cos(90^\circ) = E \cdot A \cdot 0 = 0 \]
Thus, the electric flux through the hemispherical bowl is zero in this orientation.

### Step 4: Case 2 - Electric Field Perpendicular to the Flat Surface
If the electric field is perpendicular to the flat surface of the hemisphere:
- The angle \( \theta \) between the electric field and the area vector of the flat surface is \( 0^\circ \).
- Therefore, \( \cos(0^\circ) = 1 \).
- The area \( A \) of the circular base of the hemisphere is \( \pi r^2 \).
- The electric flux through the hemispherical bowl is:
\[ \Phi = E \cdot A \cdot \cos(0^\circ) = E \cdot (\pi r^2) \cdot 1 = E \cdot \pi r^2 \]

### Conclusion
The electric flux passing through the hemispherical bowl depends on the orientation of the electric field:
- If the electric field is parallel to the flat surface, the flux is \( 0 \).
- If the electric field is perpendicular to the flat surface, the flux is \( E \cdot \pi r^2 \).

### Final Answer
The electric flux passing through the hemispherical bowl is:
\[ \Phi = E \cdot \pi r^2 \quad \text{(if perpendicular)} \]
\[ \Phi = 0 \quad \text{(if parallel)} \]