Electric flux `phi` through an element area `triangleS` when area is placed in region of uniform field E is

Options

A. `ExxtriangleS`
B. `EtriangleS`
C. `triangleSxxE`
D. `E(triangleS).sintheta`

Answer

To find the electric flux \( \phi \) through an elemental area \( \Delta S \) placed in a uniform electric field \( E \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
Electric flux \( \phi \) is defined as the number of electric field lines passing through a surface. It quantifies the effect of the electric field on a given area.

2. **Formula for Electric Flux**:
The electric flux through a surface is given by the formula:
\[ \phi = \mathbf{E} \cdot \Delta \mathbf{S} \]
where \( \phi \) is the electric flux, \( \mathbf{E} \) is the electric field vector, and \( \Delta \mathbf{S} \) is the area vector of the surface.

3. **Area Vector**:
The area vector \( \Delta \mathbf{S} \) is a vector whose magnitude is equal to the area \( \Delta S \) and whose direction is perpendicular to the surface.

4. **Dot Product**:
The dot product \( \mathbf{E} \cdot \Delta \mathbf{S} \) can be expressed as:
\[ \phi = E \Delta S \cos \theta \]
where \( \theta \) is the angle between the electric field vector \( \mathbf{E} \) and the area vector \( \Delta \mathbf{S} \).

5. **Special Cases**:
- If the electric field is perpendicular to the surface (\( \theta = 0^\circ \)), then \( \cos \theta = 1 \) and the flux is maximized:
\[ \phi = E \Delta S \]
- If the electric field is parallel to the surface (\( \theta = 90^\circ \)), then \( \cos \theta = 0 \) and the flux is zero:
\[ \phi = 0 \]

### Final Expression:
Thus, the electric flux \( \phi \) through the elemental area \( \Delta S \) in a uniform electric field \( E \) is given by:
\[ \phi = \mathbf{E} \cdot \Delta \mathbf{S} = E \Delta S \cos \theta \]