# Homework Help: Net electric force in particles in triangular formation

1. Mar 28, 2009

### nn3568

1. The problem statement, all variables and given/known data
Three positive point charges are arranged in a triangular pattern in a plane, as shown below. The Coulomb constant is 8.98755 × 109 N · m2/C2.
Find the magnitude of the net electric force on the 2 nC charge. Answer in units of N.

2. Relevant equations
Fe = (kq1q2) / (d2)

3. The attempt at a solution
I was looking at this thread with the exact same problem and I followed the directions. However, I don't know what I am doing wrong.

My diagram: (it's like the picture given in the thread I looked at but just with different numbers)

3nC
(+)
|
2m
||---2m---|
-------------(+) 2nC
|
2m
|
(+)
6nC

What I tried to do:
((8.9875e-9)(2e-9)(3e-9))/((sqroot(8))2) = 6.740625e-9
((8.9875e-9)(2e-9)(6e-9))/((sqroot(8))2) = 1.348125e-8
cos(45)6.740625e-9 + sin(45)6.740625e-9 = 9.53268329e-9
-cos(45)1.348125e-8 + sin(45)1.348125e-8 = 0

After that I am confused.

Last edited by a moderator: Apr 24, 2017
2. Mar 28, 2009

### LowlyPion

I'd write it out more simply to keep track of what's going on.

F1 = k*2*nC*(3/(2*4) i - 3/(2*4) j)
F2 = k*2*nC*(6/2*4) i + 6/(2*4) j)

Fnet = (k*nC/4)*(9 i + 3 j)

|Fnet| = (k*nC/4)*(81 + 9)1/2

3. Mar 28, 2009

### nn3568

I still don't understand what i and j are and where all the other numbers came from.

4. Mar 28, 2009

### LowlyPion

F is a vector right?

i,j are the x,y components of the vector.

The d or r radius is to the charges that are √2*2 away.
When you square that it's 2*4 right?

5. Mar 28, 2009

### nn3568

ok, then what about the "2*nC" part? nC is nano-Coulombs though...
Am I supposed to find the force between 3nC and 2nC and then the force between 2nC and 6nC? After that, do I use the Pythagorean theorem to get the resultant force?

6. Mar 28, 2009

### nn3568

OK, nevermind. I did what I just posted and I got it right!

F1 = ((8.9875e-9)(2e-9)(3e-9))/((sqroot(8))2) = 6.740625e-9
F2 = ((8.9875e-9)(2e-9)(6e-9))/((sqroot(8))2) = 1.348125e-8
sqroot ((F1)^2 + (F2)^2) = approx. 1.51e-8

YAY!

7. Mar 28, 2009

### nn3568

NOw, another question.

How do I find the direction of this force?

I did
inversetan(f1/f2) = 26.6 but it's wrong

8. Mar 28, 2009

### LowlyPion

That would be the method.

I didn't run the math. Apparently it works for you. So good.

9. Mar 28, 2009

### LowlyPion

Aren't the components in the ratio (3 j)/(9 i) = 1/3 ?

Arctan(1/3) is angle with the positive x axis?

10. Mar 28, 2009

### nn3568

oh i see. thanks so much for your help! i've learned more from you than from my teacher...