Homework Help: Magnitude of electric field at a point problem

1. Jul 21, 2011

roam

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant equations

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

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

ke = 8.9 x 109 (Coulomb's constant)

3. 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 [Broken]

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.

Last edited by a moderator: May 5, 2017
2. Jul 21, 2011

ehild

why 45°?

ehild

3. Jul 21, 2011

cupid.callin

check angle again ...

tan 45 = 1

is a=b? no

do it again

4. Jul 21, 2011

roam

Thank you very much guys, I got it.