# Homework Help: Coulomb's Law and charged particles

1. Sep 29, 2007

### UNG

1. The problem statement, all variables and given/known data
Hi, I would appreciate it if someone could help me with this question, im a n00b here
Okay heres the question.

The particles have charges Q1 = -Q2 = 100nC and Q3 = -Q4 =200nC
and distance a = 5.0cm.

What are the x and y components of the net electostatic force on particle 3 ?

So basically a diagram is given with the question showing the 4 particles equally seperated from each other at a distance of 5.0cm, making a square. And I have to find the force on particle 3.

2. Relevant equations

This is the equation I used.

F1,3 = (8.854x10^-12) x (Q1 x Q3)/0.0707xR^2 (0.0707 comes from Phyagrous therom)
H= (square root) 0.05^2 + 0.05^2
F1,4 = (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2

Fnet = F1,3 + F1,4

= F1,3 + (8.854x10^-12) x (Q1 x Q3)/R^2 Cos 45 (45 degrees middle of square)
+ (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2 Sin 45

3. The attempt at a solution

I have a feeling this is seriously wrong and have made it more complicated than it is.

I placed the values into the equation shown and came out with

= -79.46 x10^3 + 79.46 x10^3

Heres a site I found afterwards...
http://physics.bu.edu/~duffy/PY106/Charge.html

I think it might go a little something like that instead^^

If you could help that would be great thanks.

2. Sep 29, 2007

### learningphysics

I'm confused by the "pythagorean theorem" part...

What is the distance between charge 1 and charge 3?

3. Sep 29, 2007

### UNG

Doesnt say.
But I worked it out by spliting the square in two (a triangle)
and used pythagoras theorem to find the longest side (diagonal between Q 1 and 3)

4. Sep 29, 2007

### UNG

As in h^2 = a^2 + b^2.

So the distance between 1 and 3 should be 0.0707

Last edited: Sep 29, 2007
5. Sep 29, 2007

### learningphysics

Why do you have 0.0707*R^2, instead of just (0.0707)^2?

The distance between 1 and 3 is... sqrt(0.05^2 + 0.05^2) = 0.0707

so F1,3 =

F1,3 = (8.854x10^-12) x (Q1 x Q3)/(0.0707)^2

And also the distance between 1 and 4 is 0.05m right? so how is the 0.0707 coming into that part?

6. Sep 29, 2007

### learningphysics

yes. so why the 0.0707*R^2?

7. Sep 29, 2007

### UNG

Oh yes sorry, realised that a while ago but needed to edit it.
Also I only put the 0.0707 bit in F1,4 not F1,3

8. Sep 29, 2007

### learningphysics

let's start fresh... what is the magnitude of the force that particle 1 exerts on particle 3?

9. Sep 29, 2007

### UNG

I worked it out as -71.92x10-3

10. Sep 29, 2007

### learningphysics

wait... you're making a mistake with the 8.854x10^-12.

force = kQ1Q2/r^2,where k = 9*10^9.

instead of k you can also use: 1/(4*pi*(8.854*10^-12))

Last edited: Sep 29, 2007
11. Sep 29, 2007

### UNG

I see,
it really should be 8.99x10^9 when using the 1/ 4x pi x8.854*10^-12

12. Sep 29, 2007

### learningphysics

not -12... k = 8.99*10^9.

k=1/(4pi*8.854*10^-12) (the 8.854*10^-12 is in the denominator).

13. Sep 29, 2007

### UNG

My mistake, another typing error

14. Sep 29, 2007

### learningphysics

no prob. I'm getting the force between 1 and 3 as having a magnitude of 0.036N.

15. Sep 29, 2007

### UNG

I did,

F1,3 = (8.99^9) x (100x10^-9)(200x10^-9)/ 0.05^2

= -71.92x10-3 N

Wrong Q 1 and 3 values I put into the equation perhaps?

16. Sep 29, 2007

### learningphysics

you should have 0.0707^2 not 0.05^2.

17. Sep 29, 2007

### UNG

Oh no.
Its Q1 to Q3 are the charges on the same line.

Q1 to Q4 are diagonal, being 0.0707^2

18. Sep 29, 2007

### learningphysics

Ah... ok. so the magnitude of the force of Q1 on Q3 is 71.92x10-3N. Is this in the y-direction or x-direction?

Use $$\vec{i}$$ and $$\vec{j}$$ vectors... to write this force in vector form... careful about directions and signs.

19. Sep 29, 2007

### UNG

Well Q1 and Q3 are together on the y axis
and the question asks for the force on Q3

20. Sep 29, 2007

### learningphysics

Is the force upward or downward?

what are the positions of the 4 charges. Is Q1 the upper left, Q3 bottom left...