# Charge Placement where Force is Zero

1. May 18, 2006

### jg95ae

I think that this problem should be easy but for some reason I don't think I really understand it.

A charge of +2q is placed at the origin and a second charge of -q is placed a x = 3cm. Where can a third charge +Q be placed so that it experiences a zero force?

I figure that F3 = F1 + F2 = k2q/x^2 - kq/(x+0.03)^2 = 0
Cancelling out and cross multiplying gives 2(x+0.03) = x^2
Therefore the eq'n is x^2 + 0.12x + 0.0018 = 0

Using the quadratic eq'n to solve I found that x = -1.8 cm or x = -10.3 cm.

Here's where I'm not sure, if they are both negative does that mean that they are to the left of the origin, and if so which is the correct one then??

2. May 18, 2006

### LeonhardEuler

Are you sure you solved that equation correctly? Check again. (How do you go from 2(x+0.03) = x^2 to x^2 + 0.12x + 0.0018 = 0?)

3. May 18, 2006

### jg95ae

Sorry I left out a ^2

The eq'n should be 2(x+0.03)^2 = x^2
Therefore 2(x^2 + 0.06x + 0.0009) - x^2 =0
Which then becomes x^2 + 0.12x + 0.0018 = 0

I can't see anything wrong with this so the quadratic is still giving me negative answers.

4. May 18, 2006

### LeonhardEuler

OK, I just noticed what your real mistake was. Draw a picture. What is the distance between the particle at position "x" and the charge at .03? It isn't x + .03.

5. May 18, 2006

### jg95ae

I think I see what you're saying that it should really be x - 0.03

Which would give me positive answers, one lays in between the charges, which is not correct and one be at x = 10.2 cm. Does this mean that the charge would be 10.2 cm to the right of f2, and 13.2 cm to the right of the origin?

6. May 18, 2006

### LeonhardEuler

Since x is measured from the origin, the distance from the charge at the origin when x=10.2cm is just 10.2cm. You've already given the formula that relates x to the distance from the second charge.

7. May 18, 2006

### jg95ae

Ok I think I'm confused because I'm not sure that I did the first part right. Should I not have multiple q's for each force. If so I think I may have done things wrong.

I'm so confused!

8. May 18, 2006

### LeonhardEuler

I don't understand what you mean by "multiple q's for each force". What you did looks fine, except that bit about the 13.2cm from the origin.

9. May 18, 2006

### jg95ae

I just meant that since its normally F = kQ1Q2/r^2

So I thought maybe I'm supposed to have F13 = k(+2q)(+q)/x^2 and so on?

10. May 18, 2006

### LeonhardEuler

OK, just to be sure, we're calling the charge at the origin Q1, the charge 3cm to the right Q2 and the test charge Q3. Then the force experienced by Q3 as a result of Q1 is:
$$F_{1,3}=\frac{Q_1Q_3}{x^2}$$
While the force on Q3 resulting from Q2 is:
$$F_{2,3}=\frac{Q_2Q_3}{(x-.03)^2}$$
So the total force is just the sum, keeping in mind different signs of Q1 and Q2. This is supposed to be zero. Does that clear up the confusion?

11. May 18, 2006

### jg95ae

Yep I got it now, thanks for the help.