A uniform electric field of magnitude \(E\) is directed horizontally. A square sheet of area \(A\) is placed so that its surface makes an angle of \(30^\circ\) with the electric field. What is the electric flux through the sheet?
Correct Answer: \( \boxed{\mathrm{C}} \)
Electric flux is
\(\Phi=EA\cos\theta,\)
where \(\theta\) is the angle between the electric field and the area vector (normal to the surface).
The surface makes \(30^\circ\) with the field, so the normal makes \(60^\circ\).
\(\Phi=EA\cos60^\circ=\dfrac12EA.\)
A cube is placed in a uniform electric field directed vertically upward. Which face has the greatest positive electric flux?
Correct Answer: \( \boxed{\mathrm{A}} \)
The outward area vector of the top face points upward, parallel to the electric field.
\(\Phi=EA\cos0^\circ=EA,\)
which is the maximum positive flux.
Two identical square surfaces are placed in the same uniform electric field. Surface I has its area vector making an angle of \(30^\circ\) with the field, while Surface II has its area vector making an angle of \(60^\circ\). What is the ratio of the electric flux through Surface I to that through Surface II?
Correct Answer: \( \boxed{\mathrm{B}} \)
\(\Phi_1=EA\cos30^\circ=\dfrac{\sqrt3}{2}EA.\)
\(\Phi_2=EA\cos60^\circ=\dfrac12EA.\)
Therefore,
\(\dfrac{\Phi_1}{\Phi_2}=\sqrt3.\)
