The electric field in a region of space is given by `E=5hati+2hatjN//C`. The flux of E due ot this field through an area `1m^2`lying in the y-z plane, in SI units is

Options

A. 5
B. 10
C. 2
D. `5sqrt(29)`

Explanation

To find the electric flux through a given area in the presence of an electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electric Field and Area Vector**:
The electric field is given as:
\[ \mathbf{E} = 5 \hat{i} + 2 \hat{j} \, \text{N/C} \]
The area is specified to be \(1 \, \text{m}^2\) lying in the y-z plane. The area vector \(\mathbf{A}\) for a surface in the y-z plane points in the x-direction (i.e., along \(\hat{i}\)):
\[ \mathbf{A} = 1 \hat{i} \, \text{m}^2 \]

2. **Calculate the Electric Flux**:
The electric flux \(\Phi\) through the area is given by the dot product of the electric field \(\mathbf{E}\) and the area vector \(\mathbf{A}\):
\[ \Phi = \mathbf{E} \cdot \mathbf{A} \]
Substituting the values:
\[ \Phi = (5 \hat{i} + 2 \hat{j}) \cdot (1 \hat{i}) \]

3. **Perform the Dot Product**:
The dot product can be computed as follows:
\[ \Phi = (5 \hat{i} \cdot 1 \hat{i}) + (2 \hat{j} \cdot 1 \hat{i}) \]
Since \(\hat{j} \cdot \hat{i} = 0\) (the unit vectors are orthogonal), we have:
\[ \Phi = 5 \cdot 1 + 0 = 5 \]

4. **Final Result**:
Therefore, the electric flux through the area is:
\[ \Phi = 5 \, \text{N m}^2/\text{C} \, \text{(or V m)} \]