In a negion the intensity of an electric field is given by `E=Shati+8hatj+hatk` in `NC^(-1)`. The electric flux a surface
`S=10hatim^(2)` in the region is

Options

A. `10Nm^(2)C^(-3)`
B. `80Nm^(2)C^(-3)`
C. `Nm^(2)C^(-3)`
D. None

Explanation

To find the electric flux through the surface given the electric field and the area vector, we can use the formula for electric flux:

\[ \Phi_E = \vec{E} \cdot \vec{A} \]

where:
- \(\Phi_E\) is the electric flux,
- \(\vec{E}\) is the electric field vector,
- \(\vec{A}\) is the area vector.

### Step 1: Identify the electric field vector
The electric field is given as:
\[ \vec{E} = 8\hat{i} + 8\hat{j} + \hat{k} \, \text{N/C} \]

### Step 2: Identify the area vector
The area vector is given as:
\[ \vec{A} = 10\hat{i} \, \text{m}^2 \]

### Step 3: Calculate the dot product \(\vec{E} \cdot \vec{A}\)
The dot product of two vectors \(\vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k}\) and \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\) is calculated as:
\[ \vec{E} \cdot \vec{A} = E_x A_x + E_y A_y + E_z A_z \]

In our case:
- \(E_x = 8\), \(E_y = 8\), \(E_z = 1\)
- \(A_x = 10\), \(A_y = 0\), \(A_z = 0\)

Now substituting the values:
\[ \vec{E} \cdot \vec{A} = (8)(10) + (8)(0) + (1)(0) = 80 + 0 + 0 = 80 \]

### Step 4: Write the final answer for electric flux
Thus, the electric flux \(\Phi_E\) through the surface is:
\[ \Phi_E = 80 \, \text{N m}^2/\text{C} \]

### Summary of Steps
1. Identify the electric field vector.
2. Identify the area vector.
3. Calculate the dot product of the electric field and area vectors.
4. Write the final answer for the electric flux.