Electric Flux through a Spherical Surface at the Origin

[Black Armadillo]

Homework Statement

A point charge q_1 = 3.45 nC is located on the x-axis at x = 1.90 m, and a second point charge q_2 = -6.95 nC is on the y-axis at y = 1.20 m.

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_2 = 1.65 m?

Homework Equations

\Phi=\oint E_\bot dA
A=4\pi r^2
E=\frac{kq}{r^2}

The Attempt at a Solution

I started with:
\Phi=\oint \frac{kq}{r^2} dA
\Phi=A \oint \frac{kq}{r^2} dy
\Phi=4\pi r^2 \oint \frac{kq}{r^2} dy

To get r I did:
x^2+y^2=r^2
x^2+y^2=1.65^2
x=\sqrt{1.65^2-y^2}

r=\sqrt{(\sqrt{1.65^2-y^2})^2+(y-1.20)^2}

So:
\Phi=4\pi r^2 \oint \frac{kq}{1.65^2-y^2+(y-1.20)^2} dy

Evaluating this integral from -1.65 to 1.65 gives -1992.28 Nm^2/C

I'm pretty sure I'm setting up this integral completely wrong. Any help on how to do it correctly would be greatly appreciated. Thanks in advanced for your help.

 

[Doc Al]

Why mess around with integrals? Use Gauss's law.

 

[CPL.Luke]

remember that the flux through a closed surface is equal to the charge enclosed divided by epsilon_not. aka gauss's law

 

[Black Armadillo]

By use Gauss's law do you mean \Phi=\frac{q}{\epsilon_0}? If so don't I need to know that permittivity of free space (\epsilon_0), which isn't given in the problem?

 

[Black Armadillo]

Alright I found that epsilon_0 = 8.854E-12. Thanks for your help.