Problem: What is the electrostatic flux of a field of magnitude 8.1 N/C through a circular region with radius .4000 meters, if the field makes an angle of 14 degrees with a perpendicular to the plane of the region?
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Solution: The area of the region is `pi ( .4000 m) ^ 2 = .5024 m ^ 2, so the flux of a perpendicular 8.1 N/C field would be ( 8.1 N/C)( .5024 m ^ 2) = 4.069 Nm ^ 2/C. The 14 degree angle entails an additional factor of cos( 14 deg) = .9703, so the flux is ( 4.069 N m ^ 2/C)( .9703) = 3.948 N m ^ 2/C.
Generalized Response: The flux of any vector field through an area A is equal to the product of the field strength and the area, as long as the vector field is directed perpendicular to the area. If the field is directed at angle `theta from perpendicular, the flux will be reduced by factor cos(`theta). In this case we have an electric field E directed at angle `theta from a perpendicular to an area A, so the flux is just flux = E * A * cos(`theta).
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