A uniform electric field `E=2xx10^(3)NC^(-1)` is acting along the positive x-axis. The flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane is

Options

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

Explanation

To find the electric flux through a square of side 10 cm in a uniform electric field \( E = 2 \times 10^3 \, \text{N/C} \) acting along the positive x-axis, we can follow these steps:

### Step 1: Convert the side length of the square to meters
The side length of the square is given as 10 cm. We need to convert this to meters for consistency in SI units.

\[ \text{Side length} = 10 \, \text{cm} = 0.1 \, \text{m} \]

### Step 2: Calculate the area of the square
The area \( A \) of the square can be calculated using the formula:

\[ A = \text{side}^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Determine the angle between the electric field and the area vector
Since the electric field is directed along the positive x-axis and the square lies in the yz-plane, the area vector of the square will also be directed along the x-axis. Therefore, the angle \( \phi \) between the electric field vector and the area vector is 0 degrees.

\[ \phi = 0^\circ \]

### Step 4: Use the formula for electric flux
The electric flux \( \Phi_E \) through a surface is given by the formula:

\[ \Phi_E = E \cdot A \cdot \cos(\phi) \]

Substituting the values we have:

\[ \Phi_E = (2 \times 10^3 \, \text{N/C}) \cdot (0.01 \, \text{m}^2) \cdot \cos(0^\circ) \]

Since \( \cos(0^\circ) = 1 \):

\[ \Phi_E = (2 \times 10^3) \cdot (0.01) \cdot 1 \]

### Step 5: Calculate the electric flux
Now we can calculate the electric flux:

\[ \Phi_E = 2 \times 10^3 \times 0.01 = 20 \, \text{N m}^2/\text{C} \quad \text{(or equivalently, 20 V m)} \]

Thus, the electric flux through the square is:

\[ \Phi_E = 20 \, \text{N m}^2/\text{C} \]

### Final Answer
The flux of the electric field through the square is \( 20 \, \text{N m}^2/\text{C} \).

---