How to Calculate Electric Flux for a Rotated Square in a Uniform Electric Field?

[xcvxcvvc]

A uniform electric field of 25kN/C is directed parallel to the z-axis. A square of sides 5.0 cm is submersed in this field and initially located in the xy-plane. (a) Determine the electric flux for the square. (b) Repeat if the square is rotated 40.0 degrees about the x-axis. (c) Repeat if the square is moved again and this time its normal is parallel to the x-axis.




(a) 25*10^3 N/C * (5.0*10^-2 m)^2 = 62.5 = 63 Nm^2/C (this is right)

I don't understand how to set up (b) or (c). I'm trying to visualize rotating the square about the x-axis, but doesn't that make only four very slender slices of the square come in contact with the z-axis directed field?

answers:
1. a. 63 N m2/C
b. 48 N m2/C
c. 0

 

[blindside]

For part b), you can probably see there's a relationship involving the angle. As the square is rotated throughout 90 degrees, the flux that passes through the square gradually decreases to 0. Thus, the trigonometric function you're looking for is cos:

63\cos{40} = 48

For c), the only situation in which the normal of the square will be parallel to the x-axis is when it has been rotated 90 degrees about the y-axis. Thus it no longer lies in the xy plane and no flux can pass through it.

 

[xcvxcvvc]

LOL i thought the surface was a cube and i became really confused. Thanks, you helped. I completely understand it now.