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At what point is the system in equilibrium?

  • Thread starter catch22
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  • #1
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Homework Statement


This is more of a math question but:

There are two particles fixed in place: a particle of charge +8q at the origin and a particle of charge -2q
at x=L. At what point (other than infinitely far away) can a proton be placed so that it is in equilibrium

Homework Equations




The Attempt at a Solution


The point should be at the right of the -2q charge, so it is farther away from the charge +8q.

upload_2015-10-26_7-33-56.png


upload_2015-10-26_7-34-40.png


upload_2015-10-26_7-35-2.png


diagram:
upload_2015-10-26_7-35-33.png


that was the answer, but would it be possible to use

8q / (L+x)2 = 2q / (x)2

where L is the distance between the charges 8q and -2q.

I get x = L using this set up instead.

what went wrong? Shouldn't I get the same answer?
 

Answers and Replies

  • #2
Doc Al
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that was the answer, but would it be possible to use

8q / (L+x)2 = 2q / (x)2

where L is the distance between the charges 8q and -2q.

I get x = L using this set up instead.

what went wrong? Shouldn't I get the same answer?
The answers are the same! Realize that in the second approach you are measuring x from the second charge, not the origin.
 
  • #3
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The answers are the same! Realize that in the second approach you are measuring x from the second charge, not the origin.
Ah, so the point is L away from the -2q and 2L away from the origin.
 
  • #4
Doc Al
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44,940
1,201
Ah, so the point is L away from the -2q and 2L away from the origin.
Exactly. What's confusing you is that in the second method, "x" stands for "unknown", not the x-coordinate of the final location. You'll have to translate to that to report your answer, if you want them to match.
 
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