Electric flux through a surface of area ` 100m^2` lying in the xy plane is (in V-m) if `vec E= hat i+ sqrt 2hat j + sqrt 3 hat k-`.

Options

A. ` 100`
B. ` 141.4`
C. ` 17. 2`
D. ` 200`

Explanation

To find the electric flux through a surface of area \( 100 \, m^2 \) lying in the xy-plane due to the electric field \( \vec{E} = \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k} \), we can follow these steps:

### Step 1: Identify the Area Vector
Since the surface is lying in the xy-plane, the area vector \( \vec{A} \) will be perpendicular to the surface. Therefore, the area vector can be expressed as:
\[ \vec{A} = 100 \hat{k} \, m^2 \]

### Step 2: Write the Electric Field Vector
The electric field vector is given as:
\[ \vec{E} = \hat{i} + \sqrt{2} \hat{j} + \sqrt{3} \hat{k} \]

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

### Step 4: Compute the Dot Product
To compute the dot product, we can use the property that the dot product of two vectors is the sum of the products of their corresponding components:
\[ \Phi = 100 \cdot \left( \hat{i} \cdot \hat{k} + \sqrt{2} \hat{j} \cdot \hat{k} + \sqrt{3} \hat{k} \cdot \hat{k} \right) \]
Since \( \hat{i} \cdot \hat{k} = 0 \) and \( \hat{j} \cdot \hat{k} = 0 \), we have:
\[ \Phi = 100 \cdot 0 + 100 \cdot 0 + 100 \cdot \sqrt{3} = 100 \sqrt{3} \]

### Step 5: Calculate the Numerical Value
Now, we can calculate the numerical value of \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
Thus,
\[ \Phi = 100 \cdot 1.732 = 173.2 \, V \cdot m \]

### Final Result
The electric flux through the surface is:
\[ \Phi = 173.2 \, V \cdot m \]