A uniform electric field of magnitude $600 N/C$ lies in the $x y$ ‑plane making an angle $3 0^{\circ}$ with the positive $x$ ‑axis. A square of side $0.10 m$ lies in the $x y$ ‑plane with its normal along the $+ z$ direction. What is the electric flux through the square?

medium MCQ 2 marks Class 12 · Physics · Electric Charges and Fields

Answer

The correct answer is 0 — Hence $Φ = E A cos θ = 600 \times (0.10)^{2} \times 0 = 0 N\cdotm^{2} / C$ .

Step-by-step solution

Flux through a surface is $Φ = E \cdot A = E A cos θ$ , where $θ$ is the angle between $E$ and the outward normal.
The normal of the square points along $+ z$ , while the electric field lies entirely in the $x y$ ‑plane, i.e., $E$ has no $z$ component.
Therefore the angle $θ = 9 0^{\circ}$ and $cos θ = 0$ .
Hence $Φ = E A cos θ = 600 \times (0.10)^{2} \times 0 = 0 N\cdotm^{2} / C$ .

Common mistake: Multiplying the field magnitude by the area without checking the angle between field and normal.

Concept tested: Electric flux through a surface