Point charges on equilateral triangle no idea what's wrong.

1. Jul 21, 2012

1MileCrash

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

These are really frustrating because they seem so simple, I work them all out without a hitch only to find that I'm wrong half the time and left wondering why.

In fig 21-26a, particles 1 and 2 have charge 20 microcoulombs each and are held at separation distance d = 1.5 meters.

(1 is located above 2)

a.) What is the magnitude of the electrostatic force on 1 due to 2?
b.) Particle 3 is also of charge 20 microcoulombs and is placed to the right-center of 1 and 2, forming an equilateral triangle with 1 and 2. Find the net electrostatic force on 1 due to 2 and 3.

2. Relevant equations

3. The attempt at a solution

The first part, I have correct.

$F = kq_{1}q_{2}r^{-2} = (8.99x10^{9})(20x10^{6})(20x10^{6})(1.5)^{2} = 1.6 N$

The second part, I do not have correct.

Since the magnitude of the force from 2 is 1.6 N and it is located below 1, the vector of the force is just 1.6j.

Since the particles make an equilateral triangle, every angle is 60*. Since 3 is to the right, the positive angle to 1 from 3 is 120*.

Therefore the unit vector that points from 3 to 1 is cos(120)i + sin(120)j.

Since the magnitude of force from 3 to 1 is also 1.6, the vector of the force due to 3 is:

$1.6cos(120)i + 1.6sin(120)j = -.8i + 1.38j$

Adding the j components, the vector of all force on 1 is:

-.8i + 2.98j which has a magnitude of roughly 3.09 newtons.

The book answer is 2.77. I don't even understand how their result makes sense. 2 and 3 are purely additive, they are both below 1 and 2 has no x component while 3 does.

Help? TIA

2. Jul 21, 2012

TSny

Hi, 1MileCrash. 120o??

3. Jul 21, 2012

1MileCrash

Was my explanation poor?

4. Jul 21, 2012

TSny

I just wanted you to focus on the 120o. I don't think 120o is the angle you want for constructing the unit vector from 3 to 1. I'm sure you've drawn a sketch, but you might look at it again.

5. Jul 21, 2012

1MileCrash

Nevermind, I get you!

It's not 120* because I was subtracting the full 60 from 180 when I should only be subtracting half of that 60 from 180, to get 150. Right?

6. Jul 21, 2012

TSny

Try it and see

7. Jul 21, 2012

1MileCrash

Fair enough, thank you sir!