# Magnitude and Direction of Charged Particles

1. Sep 20, 2009

### dmullin4

1. The problem statement, all variables and given/known data
Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m.
http://i565.photobucket.com/albums/ss99/dmullin4/HomeworkCH21P13.jpg

The charges are +7.0 microcoulombs, -8.0 microcoulombs, and -6.0 microcoulombs. Calculate the magnitude and direction of the net force on each due to the other two.

2. Relevant equations
F=k(Q1Q2)/r^2

3. The attempt at a solution
I've worked out the forces between the particles but I am having trouble with the directions.

I have:
F12 = 0.35N
F23 = 0.30N
F31 = 0.26N

I know that Q2 and Q3 will attract Q1 down towards them, a little more towards Q2.
I know that Q2 will repel Q3 but will be attracted by Q1.
I know that Q3 will be repelled by Q2 but attracted by Q1.

I remember vector components and such from spring semesters Physics I class, but I don't remember exactly how it worked. For example, I know that the force in the y direction that Q2 will act on Q1 will be 0.35 sin60, but in one of the solutions I was looking at on Cramster, they had it as negative.

I understand the *concepts* of most of the stuff I am doing, but when it comes down to putting the math on the paper, I am at a loss. Any help is appreciated.

Danny

2. Sep 20, 2009

### rl.bhat

If α is the angle between F1 and F12, then it can shown that
tan α = F2*sinθ/(F1 + F2*cosθ)

Last edited: Sep 20, 2009
3. Sep 20, 2009

### dmullin4

What do you mean the angle between F1 and F12?

My calculations for F12 is F12=k(Q1Q2)/r^2 = 9.0 N*m^2/C^2 * (7.0x10^-6C * 8.0x10^-6C)/(1.20m)^2 = 0.35 N

F23 and F31 are similiar but with the appropriate charges.

4. Sep 21, 2009

### rl.bhat

F12 = 0.35 N towards Q2
F13 = 0.26 N towards Q3
Angle between them is 60 degrees.
Net force on Q1 = F1 = Sqrt[ F12^2 + F13*2 + 2*F12*F13*cosθ]
To find the angle between F1 and F12 see my post #2