Finding electric field from two point charges at point P

[RonakPat]

Homework Statement

What is the electric field at Point P shown below?
http://img685.imageshack.us/img685/2268/58580322.png

Homework Equations

$$\begin{flalign*}<br /> E & = & & k \ \frac{q}{r^2} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r^2}\\<br /> V & = & & k \ \frac{q}{r} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r}\\<br /> \end{flalign*}$$

The Attempt at a Solution

I figured that to find the electric field acting on point P, I would have to add the two separate fields from the charges.

Using the equation above, i got these answers
for the 5.6 uC I got 5,040,000 N/C
and for the -3.6 uC I got -9,000,000 N/C

and then I added them to get -3,960,000 N/C.

Maybe my thinking was wrong but, the answers my teacher gave me say that the answer to this problem is 7.20 x 10⁶ and the angle is 56degrees N of E.

I also have no idea how to find the angle. My only thought is taking the inverse tangent?

 

[Mindscrape]

Break the problem up completely into vectors and tell me each of the vectors you got. The only way you're going to get the answer is by adding vector components, and not the vector resultants.

 

[RonakPat]

I don't see how you can make vectors out of fields =\ my teacher says the solution doesn't have anything to do with vectors.

 

[Mindscrape]

Electric fields are inherently vectors. They are easy vectors to work with, and maybe that is why he says the solution has nothing to do with vectors. The simple fact of the matter is that you have one field making a contribution in one direction, and another making a contribution in a different direction. You have to know how to correctly add them. In geometrical terms, you simply can't add/subtract the hypotenuse and leg of triangles together.

 

[ideasrule]

Break each electric field into its x and y components. Add the two x components and the two y components. You'll get the resultant x component and the resultant y component.

This may seem to have nothing to do with vectors, but a vector IS a pair of components. The usefulness of breaking a quantity down into its components and adding components was why vectors were invented in the first place.

 

[RonakPat]

Thanks everyone! I got the answer! Ill just try to break every problem like this into components.