Magnitude of electric field at a point problem

[roam]

Homework Statement

If Q = 80 nC, a = 3.0 m, and b = 4.0 m in the figure, what is the magnitude of the electric field at point P?

[PLAIN]http://img19.imageshack.us/img19/1596/picat.jpg

Homework Equations

\vec{E} = k_e \frac{|q|}{r^2}

\left\{\begin{matrix}\vec{E}_x=k_e \frac{|q|}{a^2+b^2} \cos \ \phi<br /> \\ \vec{E}_y=k_e \frac{|q|}{a^2+b^2} \sin \ \phi\end{matrix}\right.

k_e = 8.9 x 10⁹ (Coulomb's constant)

The Attempt at a Solution

The charge -Q only has an x component, the electric field due to this charge is:

\vec{E}_{(-Q),x}= (8.9 \times 10^9) \times \frac{-80 \times 10^{-9}}{4^2}=-44.5

The two "2Q" charges cancel out each other's y-components, so we need to only consider their x components:

\vec{E}_{(2Q), x} = (8.9 \times 10^{9}) \times \frac{2 \times (80 \times 10^{-9})}{3^2+4^2} \cos 45 = 40.27

I got the 45 degree angle is from here:

[PLAIN]http://img200.imageshack.us/img200/5499/pic2ae.jpg

So the total electric field is:

\vec{E}_{(-Q), x}+2\vec{E}_{(2Q), x}=-44.5+ 2\times(40.27)=36.05 N/C

But this is wrong since the correct answer must be 47 N/C. What's wrong with my calculation? Any help is greatly appreciated.

 

[ehild]

why 45°?

ehild

 

[cupid.callin]

check angle again ...

tan 45 = 1

is a=b? no

do it again

 

[roam]

Thank you very much guys, I got it.