Set 54 Problem number 7

Problem

An electrostatic field has uniform magnitude 6.1 N/C.  A circular region whose radius is .8 meters lies within the field.  What is the electrostatic flux through the region if the field makes an angle of 86 degrees with a perpendicular to the plane of the region?

Solution

The area of the region is `pi ( .8 m) ^ 2 = 2.009 m ^ 2, so the flux of a perpendicular 6.1 N/C field would be

• flux of perpendicular field:  flux = ( 6.1 /C)( 2.009 m ^ 2) = 12.25 Nm ^ 2/C.

The 86 degree angle reduces the flux, changing it by factor cos( 86 deg) = .07073 so the actual flux is

• flux at given angle:  flux = ( 12.25 N m ^ 2/C)( .07073) = .8664 N m ^ 2/C.

Generalized Solution

The electrostatic flux of a field E (the field typically measured in Newtons / Coulomb) intercepted at a perpendicular by an area A is simply

• flux at perpendicular:  flux = E * A.

If the field makes angle `theta with the perpendicular, then flux is reduced according to the cosine of the angle:

• flux at angle `theta:   flux = (flux at perpendicular) * cos(`theta) = E * A * cos(`theta).