In a region, the intensity of an electric field is given by `vecE = 2hati + 3hatj + hatk` in . `NC^(-1)` The electric flux through a surface `vecS =10hati m^2` in the region is

Options

A. `5 Nm^(2) C^(-1)`
B. `10 Nm^(2) C^(-1)`
C. `15 Nm^(2) C^(-1)`
D. `20 Nm^(2) C^(-1)`

Explanation

To find the electric flux through the surface given the electric field and the area vector, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electric Field and Area Vector**:
- The electric field is given as:
\[ \vec{E} = 2 \hat{i} + 3 \hat{j} + 1 \hat{k} \quad \text{(in N/C)} \]
- The area vector is given as:
\[ \vec{A} = 10 \hat{i} \quad \text{(in m}^2\text{)} \]

2. **Use the Formula for Electric Flux**:
- The electric flux (\(\Phi\)) through a surface is given by the dot product of the electric field (\(\vec{E}\)) and the area vector (\(\vec{A}\)):
\[ \Phi = \vec{E} \cdot \vec{A} \]

3. **Calculate the Dot Product**:
- The dot product is calculated as follows:
\[ \Phi = (2 \hat{i} + 3 \hat{j} + 1 \hat{k}) \cdot (10 \hat{i}) \]
- Expanding the dot product:
\[ \Phi = 2 \cdot 10 + 3 \cdot 0 + 1 \cdot 0 \]
- Simplifying this gives:
\[ \Phi = 20 + 0 + 0 = 20 \quad \text{(in N m}^2/\text{C)} \]

4. **Final Result**:
- Therefore, the electric flux through the surface is:
\[ \Phi = 20 \, \text{N m}^2/\text{C} \]