If the electric field is given by `vec(E) = 8 hat(i) + 4 hat(j) + 3 hat(k) NC^(-1)`, calculate the electric flux through a surface of area `100 m^(2)` lying in X-Y plane.

Explanation

To calculate the electric flux through a surface of area \(100 \, m^2\) lying in the X-Y plane when the electric field is given by \(\vec{E} = 8 \hat{i} + 4 \hat{j} + 3 \hat{k} \, N/C\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Area Vector**:
Since the surface lies in the X-Y plane, the area vector \(\vec{S}\) will be perpendicular to this plane. The normal vector to the X-Y plane is in the Z direction, represented by \(\hat{k}\). Therefore, the area vector can be expressed as:
\[ \vec{S} = 100 \hat{k} \, m^2 \]

2. **Calculate the Electric Flux**:
The electric flux \(\Phi\) through the surface is given by the dot product of the electric field vector \(\vec{E}\) and the area vector \(\vec{S}\):
\[ \Phi = \vec{E} \cdot \vec{S} \]
Substituting the values of \(\vec{E}\) and \(\vec{S}\):
\[ \Phi = (8 \hat{i} + 4 \hat{j} + 3 \hat{k}) \cdot (100 \hat{k}) \]

3. **Perform the Dot Product**:
The dot product can be calculated as follows:
\[ \Phi = 8 \hat{i} \cdot 100 \hat{k} + 4 \hat{j} \cdot 100 \hat{k} + 3 \hat{k} \cdot 100 \hat{k} \]
Since \(\hat{i} \cdot \hat{k} = 0\) and \(\hat{j} \cdot \hat{k} = 0\), we only consider the last term:
\[ \Phi = 0 + 0 + 3 \cdot 100 = 300 \, N \cdot m^2/C \]

4. **Final Result**:
Thus, the electric flux through the surface is:
\[ \Phi = 300 \, N \cdot m^2/C \]