# Stable equilibrium and non stable equilibrium question in electrostatics

• Jenny0000
In summary: Please correct that (one way or the other) then insert a step that demonstrates more convincingly what sign dU/dx takes. Specifically, combine the two fractions into one, cancelling as appropriate.In summary, the equilibrium point between two charges is stable if the electric field between the two charges is in the opposite direction of the force of repulsion.
Jenny0000
Homework Statement
How to determine whether it is stable equilibrium or non stable equilibrium by using potential energy equation derivative
Relevant Equations
F=-dU/dX
F=kqQ/x^2
U=kqQ/x
Hi all ,
please refer to the picture regarding my working.
please correct me if My working is wrong.

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Delta2
Your conclusion looks ok, but I do not understand your diagrams. They all seem to show a negative charge between two positives.
Also, a single diagram and expression for U will serve. The signs of the charge and the displacement need not be determined by that. You can break it into the various cases later.

Theorem(little bit informally)
If in a Hamiltonian system with a smooth Hamiltonian ##H(p,x)=T+V## the function ##V## has a local isolated minimum at a point ##x_0## then ##x_0## is a stable equilibrium.
If ##H## is an analytic function then the inverse is also true.

haruspex said:
Your conclusion looks ok, but I do not understand your diagrams. They all seem to show a negative charge between two positives.
Also, a single diagram and expression for U will serve. The signs of the charge and the displacement need not be determined by that. You can break it into the various cases later.
It is a charge which repulsion refer to positive charge at the middle while attraction refer to negative charge at the middle.

Can anyone shows me what is wrong is my working,feel something wrong

Delta2
Jenny0000 said:
It is a charge which repulsion refer to positive charge at the middle while attraction refer to negative charge at the middle.
Yes, I deduced that from your working, but it is not clear from the diagrams.

Also, I notice you are not assuming the two outside charges are the same, yet you do asume the equilibrium point is in the middle.
Please correct that (one way or the other) then insert a step that demonstrates more convincingly what sign dU/dx takes. Specifically, combine the two fractions into one, cancelling as appropriate.

can show the step？

Jenny0000 said:
can show the step？
You make it difficult by posting your working as an image and not numbering the steps. Per forum rules, images are for diagrams and textbook extracts.
It's the step from the first line containing dU/dx.

There is no stable equilibrium for electrostatics systems. If you found otherwise you made a mistake somewhere.

nasu said:
There is no stable equilibrium for electrostatics systems. If you found otherwise you made a mistake somewhere.
As a problem posted in the Introductory Physics Homework Help forum, let's keep in mind the likely level at which this problem is presented.

I'm pretty sure that this is a "one-dimensional" problem. So, assume that the middle charge is constrained to positions along the line determined by the locations of the other two charges, which have fixed position. So, stable equilibrium is possible for this problem.

I also presume that discussion of Hamiltonians is quite a bit beyond this level. (Post #3)

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Lnewqban
SammyS said:
I also presume that discussion of Hamiltonians is quite a bit beyond this level. (
I agree. It would be fine if you explained the concept of Lyapunov stability at this level. I give up to do that in advance

So you have two positive charges, q1 and q3 and you want to find whether a third negative charge q2 can be in one-dimensional equilibrium at a point between the two positive charges and whether that equilibrium is stable or unstable. Furthermore, the problem instructs you to do this "by using potential energy equation derivative".

Hint: What's another name for the "potential energy derivative" per unit charge?
The electric field.
How can you use this hint to determine what kind of equilibrium there is at that point? It seems to me that you can do that qualitatively and still use the "potential energy derivative" albeit with a different name.

Just derive the U (potential energy) to dU/dx and d2U/dx2 .But I am not sure my sign is correct not ,my equation is correct not

Jenny0000 said:
Just derive the U (potential energy) to dU/dx and d2U/dx2 .But I am not sure my sign is correct not ,my equation is correct not
I would start over because in your original attempt ##x## stands for half the distance between the two outer charges which is constant and cannot be varied.
1. Call the separation between the two outer charges ##L##.
2. Put charge ##q_3## in between at some arbitrary point. Call its distance from one of the charges ##x## in which case the distance to the other charge would be ##L-x##.
3. Write the potential energy function ##U(x)## and take derivatives with respect to ##x##.

The key questions to answer are (a) how do you find the location of the equilibrium point? (it's not necessarily in the middle); (b) once you have the location of the equilibrium point, what must be true for equilibrium to be stable or unstable at that point?

You can check your work because the negative first derivative of ##U# is the electric field which you can find using superposition at a point between the two outer charges. What is the value of the electric field at the equilibrium point?

Please remember to show your work if you would like us to check it.

Lnewqban
Jenny0000 said:
Just derive the U (potential energy) to dU/dx and d2U/dx2 .But I am not sure my sign is correct not ,my equation is correct not
You have not supplied the original problem statement as given to you, and you have not responded to the points I raised in post #6. Are q1 and q3 equal? If not, the equilibrium point will not be equidistant from them.

Assuming they are supposed to be equal, you wrote expressions like ##U=kq_1q_2(\frac 1{(x+d)}-\frac 1{(x-d)})##.
I am guilty of not having looked closely enough at your working there. There is no justification for the minus sign on the second term. The potential due to one charge is ##kq_2\frac 1{(x+d)}## and due to the other charge is ##kq_2\frac 1{(x-d)}##. These should be added, to produce ##U=kq_1q_2(\frac 1{(x+d)}+\frac 1{(x-d)})##.

Then you differentiated wrt x to produce ##dU/dx=-kq_1q_2(\frac 1{(x+d)^2}+\frac 1{(x-d)^2})##, but in your assignment of variables, x is the fixed distance and d is the variable distance (the displacement from the equilibrium position). So any derivatives should be wrt d, not x.
Since that would involve writing things like dU/dd, it gets very confusing, so let's swap these over. In your diagram, swap x and d. So we have ##U(x)=kq_1q_2(\frac 1{(d+x)}+\frac 1{(d-x)})##.
Now differentiating wrt x produces the desired sign difference:
##dU/dx=kq_1q_2(-\frac 1{(d+x)^2}+\frac 1{(d-x)^2})##.
In short, you got a reasonable expression for dU/dx by sheer luck.

Edit: corrected below (thanks @SammyS ):

The next step I tried to get you to do is to collapse that into a single fraction:
##dU/dx=kq_1q_2\frac{ (d+x)^2-(d-x)^2}{(d^2-x^2)^2}##.
The numerator simplifies a bit, so that you can see easily whether this is an increasing or decreasing function for small |x|. It is not necessary to take a second derivative.

Better still, don't take any derivatives. Collapsing the expression for U we get:
##U(x)=kq_1q_2(\frac{ (d+x)+(d-x)}{(d^2-x^2)})=kq_1q_2(\frac{ 2d}{(d^2-x^2)})##. It is clear that if the charges have the same sign then this increases no matter which way we move from x=0, so it represents a minimum energy.

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SammyS and Lnewqban
Jenny0000 said:
... I am not sure my sign is correct or not ,..

Copied from
https://en.wikipedia.org/wiki/Coulomb's_law#Vector_form_of_the_law

"In the image, the vector F1 is the force experienced by q1, and the vector F2 is the force experienced by q2.
When q1q2 > 0 the forces are repulsive (as in the image) and when q1q2 < 0 the forces are attractive (opposite to the image).
The magnitude of the forces will always be equal."

## 1. What is stable equilibrium in electrostatics?

Stable equilibrium in electrostatics refers to a state in which a charged object is at rest and remains in its position when disturbed by a small force. This means that the object's potential energy is at a minimum and any slight displacement will result in a restoring force that brings the object back to its original position.

## 2. How is stable equilibrium different from non-stable equilibrium in electrostatics?

Non-stable equilibrium in electrostatics is a state in which a charged object is at rest but any slight disturbance will cause the object to move away from its original position. This means that the object's potential energy is at a maximum and there is no restoring force to bring the object back to its original position.

## 3. What factors affect the stability of equilibrium in electrostatics?

The stability of equilibrium in electrostatics is affected by the magnitude and distribution of charges, as well as the distance between the charged objects. Objects with larger charges and closer distances tend to have more stable equilibrium, while objects with smaller charges and farther distances have less stable equilibrium.

## 4. How can the stability of equilibrium in electrostatics be calculated?

The stability of equilibrium in electrostatics can be calculated by determining the potential energy of the charged object in its equilibrium position and then calculating the potential energy at a slightly displaced position. If the potential energy at the displaced position is higher, the equilibrium is stable. If the potential energy is lower, the equilibrium is non-stable.

## 5. What are some real-life examples of stable and non-stable equilibrium in electrostatics?

One example of stable equilibrium in electrostatics is a charged balloon hanging from a string. The balloon remains in its position even when slightly disturbed because the charges are distributed evenly and the distance between the balloon and the string is small. An example of non-stable equilibrium is a charged ball balanced on top of another charged ball. Any slight disturbance will cause the top ball to roll off because the charges are not evenly distributed and the distance between the balls is larger.

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