Potential Energy of three charged particles

[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.

 

[kuruman]

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.

 

[kuruman]

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.

 

[haruspex]

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.

Yes, but a quadratic produces two answers.

 

[feynman_fan]

Yes, but a quadratic produces two answers.

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

 

[kuruman]

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

 

[robphy]

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

 

[haruspex]

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?

 

[robphy]

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.

 

[haruspex]

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.

 

[robphy]

My concern was the missing absolute-values for the distances.
But I now see, comparing post 1 and post 3, that
the OP knows that there are three regions (to the left, between, and to the right)
and that the equation has a different form in terms of x (without using the absolute value).

Is there a reference for the question?