# Electrostatic Charges

1. Nov 15, 2006

### kevykevy

Question - Three charges lie along the x acis as shown below. The negative charge q1 = 25 uC is at x = 2.0 m, the negative charge q2 = 6.0 uC is at the origin. Where must a positive charge q3 be placed on the x axis such that the resultant force on it is zero?

Answer - I was thinking the third charge (q3) would be in the middle of the other two charges, my only doubt is that that would be too easy of an answer seeing how this is supposed to be harder than the normal homework. Am I right?

2. Nov 15, 2006

### kevykevy

After re-reading the question, I realize that the two negative charges are different. So should I place the q3 6/15 away from q2?

3. Nov 15, 2006

### OlderDan

Your leaning somewhat in the right direction, but the forces are inversely proportional to the square of the distance betwen charges. Use Coulomb's law to write the forces calling one distance x and the other distance 2m - x and make the magnitudes of the two forces equal.

4. Nov 15, 2006

### kevykevy

k so I'll have:

(9x10^9)(6x10^-6)(q3)/x = (9x10^9)(1.5x10^-5)(q3)/2 - x

54000(q3)/x = 135000(q3)/2 - x

54000/x = 135000/2 - x

108000/x = 135000/-x

108000(-x) = 135000(x)

How do I solve from there? Did I make an algebraic error?

5. Nov 15, 2006

### kevykevy

I tried again and found my mistake, so my final answer is:

x = 0.571428571 m

Correct?

6. Nov 15, 2006

### OlderDan

I don't know if that is the correct result, but in your previous post the denominators should all have been squared. Did you fix that?