# Electric Fields- point charges

1. Mar 7, 2008

### n77ler

[SOLVED] Electric Fields- point charges

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

Two point charges of charge 7.40 μC and -1.60 μC are placed along the x axis at x = 0.000 m and x = 0.360 m respectively. Where must a third charge, q, be placed along the x axis so that it does not experience any net force because of the other two charges?

2. Relevant equations

E= k (q_1q_2) / (r^2)

3. The attempt at a solution

E = (8.99x10^9 Kgm^2/C^2) x (7.40x10^-6)(-1.60x10^-6) / (0.360m)^2

I calculated this value but I'm not really sure if I actually need it or what needs to be done in the problem

2. Mar 7, 2008

### Hootenanny

Staff Emeritus
No net force means that if you add the force acting on the charge from the two other charges, the result will be zero. So, you a looking for the point x=a such that;

$$F_{7.40\mu C} + F_{-1.60\mu C} = 0$$

3. Mar 7, 2008

### n77ler

Ok so:

F=k (lq1l lq2l)/ r^2

=(8.99x10^9)x(7.40x106-6)x(1.60x10^-6) / (0.360)^2
= 0.821N
So, net force between these two point charges is 0.821N

F$$_{7.40uc}$$F$$_{-1.60uc}$$= 0

Can I just let force= 0.821 and using the above equation solve for r as the unknown? Thats all I can come up with but it's wrong. Any suggestions?

Edit: these superscripts should be subscripts I think they are called? Below the F

Last edited: Mar 7, 2008
4. Mar 8, 2008

### Hootenanny

Staff Emeritus
Here you have calculated the force between the 7.40 μC and the -1.60 μC charges. You need the net force (from these to charges) acting on the third charge, q, to be zero. Do you understand why?

I'll help you set up the problem. Let r1 and r2 be the positions of the positive and negative charges respectfully. Further, let r0 be the position of the third charge q such that the net force acting on q is zero. Then, by Coulomb's law (we can cancel the constants),

$$\frac{q\cdot 7.40}{\left(r_0-r_1\right)^2}\cdot10^{-6} + \frac{-q\cdot 1.60 }{\left(r_0-r_2\right)^2}\cdot 10^{-6} = 0$$

You now need to solve for r0. Do you follow? HINT: You may find it easier to solve if you does some cancellation first.

As an aside, since the electrostatic force is conservative and hence defined as $F=-\nabla\left(E\right)$, we could equally approach the problem by find the position at which the electric field due to the two charges is zero, r0. However, I feel that it make more intuitive sense (Newton's 3rd law) if we consider forces rather than fields.

Yes, they are called sub-scripts.

5. Mar 8, 2008

### n77ler

How will i solve for r$$_{}o$$ if I don't know q either?

6. Mar 8, 2008

### kbaumen

Another way to think is this: On the charge you place in, the intensity must be 0. E=k*q/r^2 where q is the charge that creates this electrical field. I think that's a bit easier way to solve, because there are less variables than in the Coulomb's law.

It doesn't matter whether you know the third charge or not. It can be 10^-20 as well as 10^20 C (though that's pretty much). The aim is to find a place where the other two charges don't interact with this third charge electrically.

Last edited: Mar 8, 2008
7. Mar 8, 2008

### n77ler

I still dont understand though. I don't know the q charge. I dont know the radius, or E ?

8. Mar 8, 2008

### kbaumen

The third charge doesn't matter. Let's say that the third charge is r meters away from q1. In that case, it's 0.36 - r meters away from q2. So what you have to do, is solve $$\frac{k*q1}{r^{2}}$$ = $$\frac{k*q2}{(0.36 - r)^{2}}$$ in terms of r. k is a constant you probably know - 9 * 10^9 N*m^2/C^2 and both charges are given.

Remember that it's further from the biggest charge, so don't forget to compare r and 0.36 - r.

I hope that helps.

9. Mar 8, 2008

### n77ler

I jsut cant get this, I worked out the equation you just wrote but it gets messy with all the big numbers. I come out with a quadratic and solve. r=2.59, r=-1.67 neither of them work and i compare them to 0.36-r and that doesnt work either :(

10. Mar 8, 2008

### kbaumen

Hmm... weird. It should work. Try first solving it algebraically in terms of r and then put in the numbers. If it still doesn't work, try the same idea with forces, but I can't see why it shouldn't work.

11. Mar 8, 2008

### n77ler

So when I solve for r, do I have to fill it into 0.36-r?

12. Mar 8, 2008

### kbaumen

Yes. r will be distance from one of the charges, and 0.36-r from the other. The biggest of those two numbers is the distance from the biggest charge.

13. Mar 8, 2008

### n77ler

well I've tried 8 times and can't get it, I dunno what else to do

14. Mar 8, 2008

### Hootenanny

Staff Emeritus
We start from the equation I posted previously,

$$\frac{q\cdot 7.40}{\left(r_0-r_1\right)^2}\cdot10^{-6} + \frac{-q\cdot 1.60}{\left(r_0-r_2\right)^2}\cdot 10^{-6} = 0$$

We divide through by q and multiply through by 106,

$$\frac{7.40}{\left(r_0-r_1\right)^2} - \frac{1.60}{\left(r_0-r_2\right)^2}= 0$$

We take the second quotient over to the RHS,

$$\frac{7.40}{\left(r_0-r_1\right)^2} = \frac{1.60}{\left(r_0-r_2\right)^2}$$

And now divide through by 7.40 & multiply though by $\left(r_0-r_2\right)^2$,

$$\frac{\left(r_0-r_2\right)^2}{\left(r_0-r_1\right)^2} = \frac{16}{74}$$

Can you take is from here?

15. Mar 8, 2008

### Hootenanny

Staff Emeritus
r0 is the distance from the origin to the equilibrium point.

16. Mar 8, 2008

### n77ler

Thank-you so much, now only 11 more problems to go :( lol

17. Mar 8, 2008

### kbaumen

Yeah, that's what I said, just rephrased better.