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.

Homework Helper
Gold Member
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
Gold Member
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.

Homework Helper
Gold Member
2022 Award
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
Gold Member
Measured from where? What is the equation that has this solution?

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

Homework Helper
Gold Member
2022 Award
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
Gold Member
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
Gold Member
2022 Award
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.