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\(^2\).