A plane surface of area `200cm^(2)` is kept in a uniform electric field of intensity 200 N/C. if the angle between the normal to the surface and field is `60^(@)`, the electric flux through the surface is

Options

A. `0.5Nm^(2)//C`
B. `4.5Nm^(2)//C`
C. `2Nm^(2)//C`
D. `3Nm^(2)//C`

Explanation

To solve the problem of finding the electric flux through a plane surface in a uniform electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Area of the surface, \( A = 200 \, \text{cm}^2 \)
- Electric field intensity, \( E = 200 \, \text{N/C} \)
- Angle between the normal to the surface and the electric field, \( \theta = 60^\circ \)

2. **Convert Area to Standard Units:**
- Convert the area from cm² to m²:
\[ A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2 = 0.02 \, \text{m}^2 \]

3. **Use the Formula for Electric Flux:**
- The electric flux \( \Phi \) through the surface is given by the formula:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]
- Here, \( \cos(\theta) \) accounts for the angle between the electric field and the normal to the surface.

4. **Calculate \( \cos(60^\circ) \):**
- We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]

5. **Substitute the Values into the Flux Formula:**
- Now substitute the values into the flux formula:
\[ \Phi = 200 \, \text{N/C} \times 0.02 \, \text{m}^2 \times \frac{1}{2} \]

6. **Perform the Calculation:**
- Calculate the electric flux:
\[ \Phi = 200 \times 0.02 \times \frac{1}{2} = 200 \times 0.01 = 2 \, \text{N m}^2/\text{C} \]

7. **Final Result:**
- The electric flux through the surface is:
\[ \Phi = 2 \, \text{N m}^2/\text{C} \]