Question

The electric field \( \mathbf{E} \) in a region is given by \( \mathbf{E} = 3\hat{i} + 5\hat{j} \). The net electric flux through a square area of side 2 m parallel to y - z plane is

• A. 3 NC\(^{-1}\)m\(^2\)
• B. 6 NC\(^{-1}\)m\(^2\)
• C. 12 NC\(^{-1}\)m\(^2\)
• D. 24 NC\(^{-1}\)m\(^2\)

Hint:

For flux calculations, use the formula \( \Phi_E = E \times A \) when the electric field is perpendicular to the area vector.

Answer 1

The Correct Option is C

The electric flux \( \Phi_E \) is given by the formula: \[ \Phi_E = \mathbf{E} \cdot \mathbf{A} \] where \( \mathbf{E} \) is the electric field, and \( \mathbf{A} \) is the area vector. In this case, the area vector is perpendicular to the y-z plane, so it points along the x-axis. Since the electric field is along the \( x \)-axis, the angle between the electric field and the area vector is \( 0^\circ \). Thus, the flux is: \[ \Phi_E = E \times A \] where \( A = 2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{m}^2 \) is the area of the square, and \( E = 3 \, \text{NC}^{-1} \) is the electric field component along the \( x \)-axis. Thus, \[ \Phi_E = 3 \, \text{NC}^{-1} \times 4 \, \text{m}^2 = 12 \, \text{NC}^{-1} \, \text{m}^2 \] Hence, the correct answer is option (3) 12 NC\(^{-1}\)m\(^2\).

Answer 2

Step 1: Understand the given electric field.
The electric field is:
\( \mathbf{E} = 3\hat{i} + 5\hat{j} \)
This means the electric field has components in the x and y directions, but not in the z direction.

Step 2: Understand the orientation of the surface.
The surface is a square of side 2 m, and it lies parallel to the y–z plane.
So the normal vector to the surface is along the x-axis, i.e., \( \hat{n} = \hat{i} \).

Step 3: Use the electric flux formula.
Electric flux through a surface is given by:
\( \Phi = \mathbf{E} \cdot \mathbf{A} = |\mathbf{E}| \cdot |\mathbf{A}| \cdot \cos\theta \)
Here, the angle \( \theta = 0^\circ \) between \( \mathbf{E}_x = 3\hat{i} \) and area vector \( \hat{i} \), so:
\( \Phi = E_x \cdot A \)

Step 4: Calculate the area and flux.
Area of the square: \( A = 2 \times 2 = 4 \, \text{m}^2 \)
Only the x-component contributes to flux: \( E_x = 3 \, \text{N/C} \)

\( \Phi = 3 \times 4 = 12 \, \text{N} \cdot \text{m}^2/\text{C} \)

Final Answer: \( \boxed{12 \, \text{NC}^{-1} \text{m}^2} \)