# Equilbrium charge between 2 charges

1. Jan 16, 2005

### thursdaytbs

So what I did was:

Kq1/(1+r) + Kq2/r = 0, making r =the distance away from q2 outside of the system to be the equilibrium point. so q1 is 1m + r distance away from the point, and q2 = r distance away from the point.

K's cancel out, you get:
q1/(1+r) + q2/r = 0
Some algebra, turns out r = -0.2m.

Does that mean the point is 0.2m inwards from q2->q1 instead of outside of the system?
aka: (q1) -----0.8m----(qe)--0.2--(q2), where qe = equilbirium point?

Also, arent there two points which are in equliibrium of the system? We did a problem in class similar to this and had 2 pointsat equlibrium. Any help is greatly appreciated.

Last edited: Jan 16, 2005
2. Jan 16, 2005

### Curious3141

Are you sure you're supposed to be using the first power of r in your equations ? How does the inverse square law pertain here ? When should you instead use the first power as you did here (hint : potential).

I get a "nice" answer between the two charges.

3. Jan 16, 2005

### thursdaytbs

isn't the potential of a charge: kq/r?

well q(Vb - Va) = kq(q-initial) / ra - kq(q-initial) / rb
but thats for moving a charge from one place to another, right?

4. Jan 16, 2005

### thursdaytbs

OHhhhh is it:

Kq1 / r-from-q1 + kq2 / r-from-q2 = 0

so the K's cancel out leaving:

q1 / r1 + q2 / r2 = 0

q1 and q2 are knowns but r1 and r2 are unknowns. how would that be solved?

Would it be, kq1qe / r1^2 = kq2qe / r2^2
where q1 = q1
q2 = q2, and qe = equilbrium charge
r1 = distance from q1 - qe,
r2 = distance from q2 - qe

from that you get kq1qe / r1^2 = kq2qe / r2^2
becomes: q1 / r1^2 = q2 / r2^2
where q1, and q2 are known.

Then using system of equations, solve for r1 and r2?

Last edited: Jan 16, 2005
5. Jan 16, 2005

### thursdaytbs

This makes no sense to me...

equation 1) q1/r1 +q2/r2 = 0
equation 2) r1 [distance between equilibrium charge -> q1] = 2(R2)

plugging it in just gives me zero. help please, anyone?

6. Jan 16, 2005

### Curious3141

Potential yes, electric field strength, no.

What is the definition of electric field strength ? What is the electric field strength at distance r from point charge q ? Show some equations to prove understanding.

7. Jan 16, 2005

### thursdaytbs

Alright, well.

The definition of electric field strength? The amount of force that is pulling or pushing on charges surrounding a charge?

E=F/q, so the Electric Field strength would be the Force [kq1q2/r^2] divided by the charge, q.

That means, E = kq1 / r^2.

Oh, also - Eleectric Potential Energy = qV, where V = kq/r. So, EPotential Energy = kq^2/r.

Last edited: Jan 16, 2005
8. Jan 16, 2005

### Curious3141

This is correct. So you realise you were wrong to use the first power in your first attempt, correct ? Because there you were trying to find a zero potential point, which is NOT what the question was asking for.

So rewrite your first attempt using the second power of r and get a result and I'll check to see if it's right.

9. Jan 16, 2005

### thursdaytbs

OHHH wow. Thanks a lot.

Ok, so for the net electric field to be 0, kq1 / (1+r)^2 + kq2/(r^2) = 0

Alright so:
1) k's drop out, leaving q1 / (1+r)^2 + q2/(r^2) = 0
2) (6.8x10^-6) / (1+2r+r^2) + (-1.7x10^-6) / (r^2) = 0
- multiply through by (1+2r+r^2)(r^2)
3) (6.8x10^-6)(r^2) + (-1.7x10^-6)(1+2r+r^2) = 0
4) R becomes a quadratic formula giving +.66667 and -.66667??

I feel so lost. Does that seem remotely correct?

*For q2, should I be putting in the negative (-) in front of it since it's a negative charge??

Last edited: Jan 16, 2005
10. Jan 16, 2005

### Curious3141

I don't see why you're having so much trouble.

Let q1 (the $+6.8 \mu C$ charge) lie at position r = 0 and the other q2 ($-1.7 \mu C$)lie at position r = R. Here R = 1 m.

Then at a position r, the net electric field strength of the system is :

$$+ \frac{kq_1}{r^2} - \frac{kq_2}{(R - r)^2}$$

isn't it ?

Equate that expression to zero, and solve for r in terms of R. What do you get ?

You should get $$r = R(1 + (\frac{q_2}{q_1})^{\frac{1}{2}})^{-1}$$

BTW, q1 and q2 refer to the absolute numerical values of the charges, since I've taken care of the signs in my formulation.

Last edited: Jan 16, 2005
11. Jan 16, 2005

### thursdaytbs

Yep $$r = R(1 + (\frac{q_2}{q_1})^{\frac{1}{2}})^{-1}$$ equates to 0.66666. That's what I got too using what I said above. I think it's doing the same thing.

Anyway, thanks a lot for the help.

12. Jan 17, 2005

### Curious3141

Then you have basically the right idea, but you don't have to solve a quadratic to get the answer. Just do this :

$$+ \frac{kq_1}{r^2} - \frac{kq_2}{(R - r)^2}$$

$$\frac{q_2}{q_1} = (\frac{R}{r} - 1)^2$$

$$\frac{R}{r} = 1 + (\frac{q_2}{q_1})^{\frac{1}{2}}$$

and you can manipulate that easily.

But it's OK to do it via a quadratic, just more tedious. You do understand why you should ignore the negative root though, correct ?

Last edited: Jan 17, 2005