If electric field is `3hati+4hatj` then find out ratio of flux when a rectangular plane is put on `x - z` plane and when same plane is put on `y-z` plane.

Options

A. `phi_1: phi_2 = 4:3`
B. `phi_1: phi_2 = 3:4`
C. `phi_1: phi_2 = 2:3`
D. `phi_1: phi_2 = 3:2`

Answer

To solve the problem of finding the ratio of electric flux through two rectangular planes placed on the x-z and y-z planes, we will follow these steps:

### Step 1: Understand the Electric Field and Area Vectors
The electric field is given as **E = 3i + 4j**. We need to determine the area vectors for the two surfaces:
- For the rectangular plane on the **x-z plane**, the area vector **A1** will be in the **y-direction** (perpendicular to the x-z plane).
- For the rectangular plane on the **y-z plane**, the area vector **A2** will be in the **x-direction** (perpendicular to the y-z plane).

### Step 2: Define the Area Vectors
- **A1 = A * j** (where A is the area of the rectangle)
- **A2 = A * i**

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

#### For Surface 1 (x-z plane):
- The area vector **A1** is in the **y-direction** (A1 = A * j).
- The electric field **E** has components in both the x and y directions.
- The angle θ between **E** and **A1** is 0° for the j component of E (4j) and 90° for the i component (3i).

Calculating the flux:
\[ \Phi_1 = \mathbf{E} \cdot \mathbf{A1} = (3i + 4j) \cdot (A * j) \]
\[ \Phi_1 = 3A \cdot 0 + 4A \cdot 1 = 4A \]

#### For Surface 2 (y-z plane):
- The area vector **A2** is in the **x-direction** (A2 = A * i).
- The angle θ between **E** and **A2** is 0° for the i component of E (3i) and 90° for the j component (4j).

Calculating the flux:
\[ \Phi_2 = \mathbf{E} \cdot \mathbf{A2} = (3i + 4j) \cdot (A * i) \]
\[ \Phi_2 = 3A \cdot 1 + 4A \cdot 0 = 3A \]

### Step 4: Find the Ratio of the Fluxes
Now we can find the ratio of the fluxes:
\[ \frac{\Phi_1}{\Phi_2} = \frac{4A}{3A} = \frac{4}{3} \]

### Final Answer
The ratio of the electric flux when the rectangular plane is placed on the x-z plane to when it is placed on the y-z plane is:
\[ \frac{\Phi_1}{\Phi_2} = \frac{4}{3} \]