Consider a uniform electric field `E = 3xx10^(3) hat(i) N//C`. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ? (b) What is the flux through the same square if the normal to its plane makes a `60^(@)` angle with the x-axis ?

Options

A. `10 N C^(-1) m^(2)`
B. `20 N C^(-1) m^(2)`
C. `30 N C^(-1) m^(2)`
D. `40 N C^(-1) m^(2)`

Answer

To solve the problem, we will break it down into two parts as specified in the question.

### Given:
- Electric field, \( E = 3 \times 10^3 \, \hat{i} \, \text{N/C} \)
- Side of the square, \( s = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Part (a): Flux through a square whose plane is parallel to the YZ plane

1. **Calculate the area of the square**:
\[ A = s^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

2. **Determine the angle between the electric field and the normal to the square**:
Since the square is parallel to the YZ plane, the normal to the square is along the X-axis. The electric field is also along the X-axis. Therefore, the angle \( \theta \) between the electric field and the normal is:
\[ \theta = 0^\circ \]

3. **Calculate the electric flux**:
The electric flux \( \Phi \) is given by the formula:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]
Substituting the values:
\[ \Phi = (3 \times 10^3 \, \text{N/C}) \cdot (0.01 \, \text{m}^2) \cdot \cos(0^\circ) \]
Since \( \cos(0^\circ) = 1 \):
\[ \Phi = 3 \times 10^3 \cdot 0.01 \cdot 1 = 30 \, \text{N m}^2/\text{C} \]

### Part (b): Flux through the same square if the normal makes a \( 60^\circ \) angle with the x-axis

1. **Determine the angle**:
Here, the angle \( \theta \) is given as \( 60^\circ \).

2. **Calculate the electric flux**:
Using the same formula for electric flux:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]
Substituting the values:
\[ \Phi = (3 \times 10^3 \, \text{N/C}) \cdot (0.01 \, \text{m}^2) \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ \Phi = 3 \times 10^3 \cdot 0.01 \cdot \frac{1}{2} = 15 \, \text{N m}^2/\text{C} \]

### Final Answers:
- (a) The flux through the square parallel to the YZ plane is \( 30 \, \text{N m}^2/\text{C} \).
- (b) The flux through the square making a \( 60^\circ \) angle with the x-axis is \( 15 \, \text{N m}^2/\text{C} \).