Question

The electric flux due to an electric field \[ \vec{E} = (8\hat{i} + 13\hat{j}) \text{ NC}^{-1} \] through an area 3 m\(^2\) lying in the XZ plane is: 

• A. 39 Wb
• B. 24 Wb
• C. 63 Wb
• D. 15 Wb

Hint:

Electric flux is calculated using the dot product of the electric field and the area vector. Only the field component perpendicular to the surface contributes to the flux.

Answer 1

The Correct Option is A

Step 1: Understanding Electric Flux The electric flux (\(\Phi\)) is given by the dot product of the electric field vector (\(\vec{E}\)) and the area vector (\(\vec{A}\)): \[ \Phi = \vec{E} \cdot \vec{A} \] 

Step 2: Identifying the Perpendicular Component of \(\vec{E}\) Since the area lies in the XZ plane, its normal vector is along the \(\hat{j}\) direction (Y-axis). The electric field component along \(\hat{j}\) is 13 NC\(^{-1}\), which is the \(y\)-component of \(\vec{E}\). 

Step 3: Calculating Electric Flux The electric flux is: \[ \Phi = E_y \cdot A = 13 \times 3 = 39 \text{ Wb} \] Thus, the correct answer is: \[ \mathbf{39 \text{ Wb}} \]