Potential Energy of three charged particles

In summary, the author is trying to solve an equation that has no solutions, and is having difficulty understanding what may be wrong.
  • #1
feynman_fan
7
4
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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  • #2
feynman_fan said:
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
feynman_fan said:
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.
 
  • #6
haruspex said:
Yes, but a quadratic produces two answers.
Oh, right. The other answer would be 1.9 cm.
 
  • #7
Measured from where? What is the equation that has this solution?
 
  • #8
What's the general formula for potential energy again?
Next, what is it in terms of x?
 
  • #9
robphy said:
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?
 
  • #10
haruspex said:
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:
  • #11
robphy said:
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.
 
  • #12
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?
 

Related to Potential Energy of three charged particles

1. What is potential energy?

Potential energy is the stored energy an object possesses due to its position or configuration.

2. How is potential energy of three charged particles calculated?

The potential energy of three charged particles is calculated using the Coulomb's law formula, which takes into account the charges and distances between the particles.

3. What factors affect the potential energy of three charged particles?

The potential energy of three charged particles is affected by the magnitude of the charges, the distance between the particles, and the medium in which the particles are located.

4. How does the potential energy change when the distance between the particles is altered?

The potential energy of three charged particles is inversely proportional to the distance between them. As the distance increases, the potential energy decreases and vice versa.

5. Can potential energy of three charged particles be negative?

Yes, the potential energy of three charged particles can be negative if the particles have opposite charges and are attracted to each other. This indicates that work must be done to separate the particles, resulting in a decrease in potential energy.

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