A plane area of `100cm^(2)` is placed in uniform electric field of `100N//C` such that the angle between area vector and electric field is `60^(@)`. The electric flux over the surface is

Options

A. `1Nm^(2)//C`
B. `2Nm^(2)//C^(2)`
C. `3Nm^(2)//C^(2)`
D. `0.5Nm^(2)//C`

Explanation

To find the electric flux (\( \Phi \)) through a plane area placed in a uniform electric field, we can use the formula:

\[ \Phi = E \cdot A \cdot \cos(\theta) \]

where:
- \( E \) is the magnitude of the electric field,
- \( A \) is the area through which the field lines pass,
- \( \theta \) is the angle between the electric field and the normal (perpendicular) to the surface.

### Step 1: Convert the area from cm² to m²
The area given is \( 100 \, \text{cm}^2 \). We need to convert this to square meters.

\[ A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \]

### Step 2: Identify the electric field strength
The electric field strength is given as \( E = 100 \, \text{N/C} \).

### Step 3: Identify the angle
The angle \( \theta \) between the electric field and the area vector is given as \( 60^\circ \).

### Step 4: Calculate the cosine of the angle
We need to find \( \cos(60^\circ) \):

\[ \cos(60^\circ) = \frac{1}{2} \]

### Step 5: Substitute values into the electric flux formula
Now we can substitute the values into the electric flux formula:

\[ \Phi = E \cdot A \cdot \cos(\theta) = 100 \, \text{N/C} \cdot 0.01 \, \text{m}^2 \cdot \frac{1}{2} \]

### Step 6: Perform the calculation
Calculating the above expression:

\[ \Phi = 100 \cdot 0.01 \cdot 0.5 = 1 \, \text{N m}^2/\text{C} \]

### Final Answer
The electric flux over the surface is \( \Phi = 1 \, \text{N m}^2/\text{C} \).

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