# Potential Energy of three charged particles

• feynman_fan
feynman_fan
Homework Statement
Two identical charges q are placed on the x axis, one at the origin and the other at x = 5.0 cm.

A third charge -q is placed on the x axis so the potential energy of the three-charge system is the same as the potential energy at infinite separation.

What is the coordinate of the third charge?
Relevant Equations
##U=q_0\cdot V=k\frac{q\cdot q_0}{r}##
I set up an equation for the sum of all the potential energies and when cancelling out ##k## and ##q^2##, I got ##\frac{1}{0.05}-\frac{1}{x}-\frac{1}{0.05-x}=0##. However, this has no solutions, so I must've gone wrong somewhere. Could someone just give me a hint, not a solution, that would put me on the right track, because I want to really understand all the problems I do.

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Homework Statement:: Two identical charges q are placed on the x axis, one at the origin and the other at x = 5.0 cm.

A third charge -q is placed on the x-axis so the potential energy of the three-charge system is the same as the potential energy at infinite separation.

What is the coordinate of the third charge?
Relevant Equations:: ##U=q_0\cdot V=k\frac{q\cdot q_0}{r}##

I set up an equation for the sum of all the potential energies and when cancelling out ##k## and ##q^2##, I got ##\frac{1}{0.05}-\frac{1}{x}-\frac{1}{0.05-x}=0##. However, this has no solutions, so I must've gone wrong somewhere. Could someone just give me a hint, not a solution, that would put me on the right track, because I want to really understand all the problems I do.
You put the third charge between the first two. Is that the only place on the x-axis where you can put it?

feynman_fan
No, you could also put it on the outside. If you put it to the right, the equations would then be $$\frac{1}{0.05}-\frac{1}{x}-\frac{1}{x-0.05}=0,$$Which gives you an answer of 13 cm, is this correct? Thank you for your insight.

Edit, I meant .05, I accidentally wrote 5.05.

Homework Helper
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It is not correct. Where it did you get ##\dfrac{1}{5.05}##? Let L = 5.0 cm, draw a picture, figure out the equation in terms of ##L## and ##x##, solve it, then plug in the value for ##L## at the very end. That way you will see better what's going on.

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No, you could also put it on the outside. If you put it to the right, the equations would then be $$\frac{1}{0.05}-\frac{1}{x}-\frac{1}{x-0.05}=0,$$Which gives you an answer of 13 cm, is this correct? Thank you for your insight.

Edit, I meant .05, I accidentally wrote 5.05.

feynman_fan
Oh, right. The other answer would be 1.9 cm.

Homework Helper
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Measured from where? What is the equation that has this solution?

Homework Helper
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What's the general formula for potential energy again?
Next, what is it in terms of x?

Homework Helper
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What's the general formula for potential energy again?
Next, what is it in terms of x?
Do you disagree with the equation in post #3?

Homework Helper
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Do you disagree with the equation in post #3?
I haven't looked in detail... but something doesn't look correct to me.
I could be wrong.
I have a feeling that this equation is probably ok for some values of x on the x-axis.

Last edited:
Homework Helper
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I have a feeling that this equation is probably ok for some values of x on the x-axis.
A wrong equation could happen to give the right answer for certain values, but it would still be a wrong equation. The equation looks ok to me.