A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric field of magnitude 20 N/C . You may want to review (Page) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Electric flux through a disk.
A. Find the magnitude of the electric flux through the sheet.B.Does the answer to part A depend on the shape of the sheet?C.For what angle \phi between the normal to the sheet and the electric field is the magnitude of the flux through the sheet largest?D.For what angle \phi between the normal to the sheet and the electric field is the magnitude of the flux through the sheet smallest?

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0