The electric field in a region is given by `E = 3/5 E_0hati +4/5E_0j` with `E_0 = 2.0 × 10^3 N//C`. Find the flux of this field through a rectangular surface of area `0.2 m^2` parallel to the y-z plane.

Explanation

To find the electric flux through a rectangular surface of area \(0.2 \, \text{m}^2\) parallel to the y-z plane, we can follow these steps:

### Step 1: Identify the Electric Field Vector
The electric field is given by:
\[ \mathbf{E} = \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \]
where \(E_0 = 2.0 \times 10^3 \, \text{N/C}\).

### Step 2: Calculate the Electric Field Components
Substituting the value of \(E_0\):
\[ \mathbf{E} = \frac{3}{5} (2.0 \times 10^3) \hat{i} + \frac{4}{5} (2.0 \times 10^3) \hat{j} \]
Calculating each component:
\[ \mathbf{E} = \frac{6}{5} \times 10^3 \hat{i} + \frac{8}{5} \times 10^3 \hat{j} = 1200 \hat{i} + 1600 \hat{j} \, \text{N/C} \]

### Step 3: Define the Area Vector
Since the surface is parallel to the y-z plane, the area vector \(\mathbf{A}\) will point in the x-direction:
\[ \mathbf{A} = 0.2 \, \text{m}^2 \hat{i} \]

### Step 4: Calculate the Electric Flux
The electric flux \(\Phi\) through the surface is given by the dot product of the electric field and the area vector:
\[ \Phi = \mathbf{E} \cdot \mathbf{A} \]
Calculating the dot product:
\[ \Phi = (1200 \hat{i} + 1600 \hat{j}) \cdot (0.2 \hat{i}) = 1200 \times 0.2 + 1600 \times 0 = 240 \, \text{N m}^2/\text{C} \]

### Final Answer
The electric flux through the rectangular surface is:
\[ \Phi = 240 \, \text{N m}^2/\text{C} \]