At what point is the system in equilibrium?

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catch22
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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.

Equation: 1/(4 pi epsilon_0) * 8qq_p / x^2 = 1/(4 pi epsilon_0) * 2qq_p / (x - L)^2


Equation: ((x - L)/x)^2 = 1/4, written as a squared ratio set equal to one quarter.


Handwritten equation: (x - L)/x = 1/2, followed by the result x = 2L.


diagram:
Diagram of a point charge q1 at the origin and charge q2 to its right on the x-axis, with a proton at the right of q2 and force arrows F1 and F2 pointing along the x-axis.


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?
 
catch22 said:
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.
 
Doc Al said:
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.
 
catch22 said:
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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