# Coulombs Law: magnitude of the force between two point charges

1. Aug 28, 2012

### Pruddy

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

Determine the magnitude of the force between two point charges, Q1 = -2e located at (-3.60 cm, +4.10 cm) and Q2 = -8e located at (-0.60 cm, -8.60 cm). Give your answer in the form "a.bc x 10^(y) N".

2. Relevant equations

F = kq1q2/r2

3. The attempt at a solution
a.) I changed the unit vectors to meters, then I added, squared, and got the square roots of the vectors to get r = sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2))
b.) Then I multiplied the value of "K" by the value of "8e" and "2e" and divided all three values by r^2

2. Aug 29, 2012

### Simon Bridge

Re: Coulombs Law

Sounds reasonable - what was your question?
 see ehild below - it is not really all that clear what your reasoning is.

Converting everything to SI units first does help make sure you get SI units out. However, you don't have to.

Where qi=nie (n is an integer and e is the elementary charge) then |F1,2| = (ke2)n1n2/r2

Since: ke2 = 2.307x10-24N.cm2 you can do everything in cm.

Last edited: Aug 29, 2012
3. Aug 29, 2012

### ehild

Re: Coulombs Law

Why did you add the two vectors? The Coulomb force is inversely proportional to the square of the distance between two point charges. How do you get the distance between two points from their location?

ehild

4. Aug 29, 2012

### hms.tech

Re: Coulombs Law

Treat the two given points as two position vectors in 2D.

Find the magnitude of the displacement between these two position vectors !

That gives you "r"

5. Aug 29, 2012

### Pruddy

Re: Coulombs Law

Hi, hms.tech
Thanks for your reply. I solved the two positions as two vectors using pythogoras theory. For example sqrt(((-0.036 +(- 0.006))^2)+((0.041+(-0.08))^2)), I do not know if it gives this solution gives me the distance between two point charges.

6. Aug 29, 2012

### Simon Bridge

Re: Coulombs Law

If you don't know if the method gives the answer you want, then why did you use it?

In general, if you have points $P=(p_x,p_y)$ and $Q=(q_x,q_y)$ defined against an origin $O=(0,0)$, then the distance between them is the length of the vector from P to Q which is given by change in position

$$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = (q_x-p_x,q_y-p_y)$$... that length would be: $$|\overrightarrow{PQ}| = \sqrt{(q_x-p_x)^2 + (q_y-p_y)^2}$$... your working does not look like this.

Last edited: Aug 29, 2012
7. Aug 31, 2012

### Pruddy

Hi,
I have tried solving this problem several times but my answers are still wrong.I dont know if the error is from the radius or if i am supposed multiply each charge by 1.602 x 10^(-19), since e = 1.602 x 10^(-19)

Attemped effort:
8.99 x 10^(9) x 2(1.602 x 10^(-19)) x 8(1.602 x 10^(-19))/.13050^2

= 2.17 x 10^(-26)

I have solved this problem a billion times but its still wrong. Please someone help. I will really appreciate it....:happy:)

8. Aug 31, 2012

### ehild

The magnitude is wrong. Try again.

ehild

9. Aug 31, 2012

### Pruddy

Alright. But this what I got for the distance (r)

r^2 = (-0.006m-(-0.036m))^2 + (-0.086m-0.041m)^2
=.01703
square root of .01703 = .13050

10. Aug 31, 2012

### Muphrid

Just put it in your calculator again and make sure you don't mistype and that all your order of operations is correct. You're off by some power of 10, which usually means you just didn't enter the calculation correctly.

11. Aug 31, 2012

### Pruddy

I have resolved it and my answer is 2.17 x 10^(-25). But the answer is still wrong...

12. Aug 31, 2012

### ehild

If you calculate accurately, it is 2.16 x 10^(-25) N. Have you added the unit (N)?

ehild

13. Aug 31, 2012

### Simon Bridge

I get 2.1676x10-25N ... how many sig fig should you keep?
What makes you think this is the wrong answer?

14. Aug 31, 2012

### Pruddy

Yes your right, the answer is correct... I was having a confusion with the units... Thank you

15. Aug 31, 2012

### Simon Bridge

a.bc x 10^(y) N

ah riight. easy mistake ;)
One of the disciplines is to keep the units with the numbers when you substitute them in to the equation. You get to use the laws of algebra on units too.

It's all good though.