The electric field in a region of space is given by, `vec(E ) = E_(0) hat(i) + 2 E_(0) hat(j)` where `E_(0)= 100N//C`. The flux of this field through a circular surface of radius 0.02m parallel to the Y-Z plane is nearly.

Options

A. `3.14 Nm^(2)//C`
B. `0.02 Nm^(2)//C`
C. `0.005Nm^(2)//C`
D. `0.125 Nm^(2)//C`

Answer

To find the electric flux through a circular surface of radius 0.02 m that is parallel to the Y-Z plane, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field is given by:
\[ \vec{E} = E_0 \hat{i} + 2E_0 \hat{j} \]
where \(E_0 = 100 \, \text{N/C}\). This means:
\[ \vec{E} = 100 \hat{i} + 200 \hat{j} \, \text{N/C} \]

### Step 2: Identify the Orientation of the Circular Surface
The circular surface is parallel to the Y-Z plane. This means that the normal vector to the surface will be along the X-axis (in the direction of \(\hat{i}\)).

### Step 3: Calculate the Area of the Circular Surface
The area \(A\) of a circle is given by:
\[ A = \pi r^2 \]
where \(r = 0.02 \, \text{m}\). Thus:
\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 0.00125664 \, \text{m}^2 \]

### Step 4: Determine the Angle Between the Electric Field and the Normal
Since the electric field has a component in the \(\hat{i}\) direction and the normal to the surface is also in the \(\hat{i}\) direction, the angle \(\theta\) between the electric field vector and the normal vector is \(0^\circ\). Therefore, \(\cos(0^\circ) = 1\).

### Step 5: Calculate the Electric Flux
The electric flux \(\Phi_E\) through the surface is given by:
\[ \Phi_E = \vec{E} \cdot \vec{A} = E \cdot A \cdot \cos(\theta) \]
Substituting the values:
\[ \Phi_E = (100 \, \text{N/C}) \cdot (0.00125664 \, \text{m}^2) \cdot 1 \]
\[ \Phi_E \approx 0.125664 \, \text{N m}^2/\text{C} \]

### Step 6: Round Off to the Required Precision
Rounding off the result, we get:
\[ \Phi_E \approx 0.126 \, \text{N m}^2/\text{C} \]

Thus, the flux of this field through the circular surface is approximately \(0.126 \, \text{N m}^2/\text{C}\).

### Final Answer:
The flux of the electric field through the circular surface is approximately \(0.126 \, \text{N m}^2/\text{C}\).

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