Electric flux through a plane described by an equation

[turquoisetea]

Homework Statement

The elctric field in a certain region of space has components Ex = 6.0 N/C, Ey = 7.0 N/C, and Ez = 0. Find the electric flux though the surface x = 6y, 0 < x < 6.0 m, 0 < z < 1.0 m. What is the angle between the electric field and the unit vector normal to the surface?

Homework Equations

Flux=E dot A

The Attempt at a Solution

Flux=(6N/C)(6m)(1m)= 36 Nm^2/C
This is the same as the answer in the back of the book but it seems too easy, and I think there should be a cosine in there.

 

[sandy.bridge]

You have \vec{E}(x, y, z)=(6.0, 7.0, 0) N/C.
\Phi=\vec{E}(x, y, z)\bullet\vec{S} where \vec{S} is a vector area. You can use the magnitudes of the vectors and the angle between them as EScos\phi, where \phi is the angle between the field lines and the normal to the the surface area S. Do you know how to find the normal to the surface you have?

 

[turquoisetea]

Yes, I understand where the normal is, I just don't understand how to find the angle between the E-field lines and the area vector (normal). I can find the angle between the e-field lines and the x and y-axis but I cannot find the angle of the plane.