If `vec E = 3/5 hati + 4/5 hatj` then find electric flux through an area of 0.4 m^2 parallel to y-z plane

Options

A. `0.12 Nm^2/C`
B. `0.24 Nm^2/C`
C. `0.36 Nm^2/C`
D. `0.48 Nm^2/C`

Answer

To solve the problem of finding the electric flux through an area of 0.4 m² parallel to the y-z plane given the electric field \(\vec{E} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}\), we can follow these steps:

### Step 1: Identify the Electric Field Components
The electric field is given as:
\[ \vec{E} = \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j} \]
This indicates that the electric field has components along the x-axis (\(\hat{i}\)) and y-axis (\(\hat{j}\)).

### Step 2: Understand the Area Orientation
The area is specified to be parallel to the y-z plane. The area vector \(\vec{A}\) for a surface parallel to the y-z plane will point along the x-axis, which can be represented as:
\[ \vec{A} = A \hat{i} = 0.4 \hat{i} \quad (\text{since the area is } 0.4 \, \text{m}^2) \]

### Step 3: Calculate the Electric Flux
The electric flux \(\Phi\) through a surface is given by the formula:
\[ \Phi = \vec{E} \cdot \vec{A} = |\vec{E}| |\vec{A}| \cos(\theta) \]
where \(\theta\) is the angle between the electric field vector and the area vector.

### Step 4: Determine the Angle \(\theta\)
In this case, the area vector \(\vec{A}\) is along the x-axis, and the electric field has a component along the x-axis as well. The angle \(\theta\) between the electric field vector and the area vector is 0 degrees because they are aligned in the x-direction.

### Step 5: Compute the Dot Product
Since \(\theta = 0\), we have:
\[ \cos(0) = 1 \]
Thus, the electric flux simplifies to:
\[ \Phi = |\vec{E}| |\vec{A}| \cdot 1 \]

### Step 6: Calculate the Magnitude of the Electric Field
The magnitude of the electric field \(\vec{E}\) can be calculated as:
\[ |\vec{E}| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = 1 \, \text{N/C} \]

### Step 7: Substitute Values into the Flux Equation
Now substituting the values:
\[ \Phi = |\vec{E}| \cdot |\vec{A}| = 1 \, \text{N/C} \cdot 0.4 \, \text{m}^2 = 0.4 \, \text{N m}^2/\text{C} \]

### Final Answer
Thus, the electric flux through the area is:
\[ \Phi = 0.4 \, \text{N m}^2/\text{C} \]