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.