There exists an electric field of `1 NC^(-1)` along Y direction. The flux passing through the square of 1 m placed in `xy-`plane inside the electric field is

Options

A. `1.0 Nm^(2)C^(-1)`
B. `10.0 Nm^(2)C^(-1)`
C. `2.0 Nm^(2)C^(-1)`
D. zero

Answer

To solve the problem of finding the electric flux passing through a square of area 1 m² placed in the xy-plane in an electric field of 1 N/C directed along the y-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Electric Flux**:
Electric flux (Φ) through a surface is given by the formula:
\[ Φ = \vec{E} \cdot \vec{A} \]
where \( \vec{E} \) is the electric field vector and \( \vec{A} \) is the area vector of the surface.

2. **Define the Area Vector**:
The area vector \( \vec{A} \) for a surface is defined as being perpendicular to the surface. For the square placed in the xy-plane, the area vector points in the z-direction. Thus, we can write:
\[ \vec{A} = A \hat{k} = 1 \, \text{m}^2 \hat{k} \]
where \( \hat{k} \) is the unit vector in the z-direction.

3. **Identify the Electric Field Vector**:
The electric field \( \vec{E} \) is given as 1 N/C in the y-direction, which can be represented as:
\[ \vec{E} = 1 \, \text{N/C} \hat{j} \]
where \( \hat{j} \) is the unit vector in the y-direction.

4. **Calculate the Dot Product**:
To find the electric flux, we need to calculate the dot product \( \vec{E} \cdot \vec{A} \):
\[ Φ = \vec{E} \cdot \vec{A} = (1 \, \text{N/C} \hat{j}) \cdot (1 \, \text{m}^2 \hat{k}) \]
Since the vectors \( \hat{j} \) and \( \hat{k} \) are perpendicular to each other, their dot product is zero:
\[ \hat{j} \cdot \hat{k} = 0 \]

5. **Conclusion**:
Therefore, the electric flux through the square is:
\[ Φ = 0 \, \text{N m}^2/\text{C} \]

### Final Answer:
The electric flux passing through the square is \( 0 \, \text{N m}^2/\text{C} \).

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