Electrostatic Charges

  • Thread starter kevykevy
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  • #1
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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?
 

Answers and Replies

  • #2
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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
OlderDan
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kevykevy said:
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?
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
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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
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I tried again and found my mistake, so my final answer is:

x = 0.571428571 m

Correct?
 
  • #6
OlderDan
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kevykevy said:
I tried again and found my mistake, so my final answer is:

x = 0.571428571 m

Correct?
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?
 

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