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

[1MileCrash]

Homework Statement

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.

Homework Equations

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

 

[TSny]

Hi, 1MileCrash. 120^o??

 

[1MileCrash]

Was my explanation poor?

 

[TSny]

I just wanted you to focus on the 120^o. I don't think 120^o 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.

 

[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?

 

[TSny]

Try it and see

 

[1MileCrash]

Fair enough, thank you sir!