Discussion
A rectangle of cross-sectional area 'A' is placed in a uniform constant electric field \(\overline{E}\). The plane of the rectangle makes an angle of 30° with the direction of the electric field. The electric flux through the rectangle is :
Answer: B
The flux is AE/2.
Flux is \(\Phi = EA\cos\theta\), where \(\theta\) is the angle between the field and the normal to the surface. The plane makes 30° with the field, so the normal makes 90° - 30° = 60° with it:
\[ \Phi = EA\cos 60^{\circ} = \frac{AE}{2} \]
The option \(AE\sqrt{3}/2\) is the common slip of putting 30° into the cosine, treating it as the angle with the normal. AE would need the field perpendicular to the plane, and zero would need it parallel.
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