Two charged beads with a third bead in equilibrium

[putongren]

Homework Statement

Two small beads having positive charges 3q and q are fixed a the opposite ends of a horizontal, insulating rod, extending from the origin to the point x=d. A third small charged is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?

Relevant Equations

E = kq/r^2

Let a = distance between the charge on the left to the third bead.
Since the electric field is equilibrium, we equate the electric field emanating from the left charge to the right charge at the location of the third bead. I want to try to find the ratio of d with a.

E=E

Kq/(d-a)^2 = K*3q/a^2

1/(d-a)^2 = 3/a^2

3(d-a)^2 = a^2

3(d^2 - 2ad +a^2) = a^2

3d^2 - 6ad + 3a^2 = a^2

3d^2 - 6ad + 2a^2 = 0

This is as far as I got. I'm not sure how to proceed.

 

[TSny]

If you think of $a$ as the unknown, then your last equation is a quadratic equation that you can use to solve for $a$ in terms of $d$.

However, another approach is to back up to your earlier equation $1/(d-a)^2 = 3/a^2$. Take the square root of both sides and then try to solve for $a$.

 

[kuruman]

Of course, whichever way you solve this, you will obtain two values for $a$. Which one is the correct value and why?

 

[gneill]

Also, ask yourself what happens when the third bead is negatively charged rather than positively charged. Is there a difference?

 

[putongren]

If I take TSny's advice, I take the square root of both sides of the equation:

1/{d-a} = \sqrt 3 /a

a = \sqrt 3 * (d-a)

{1 + \sqrt 3} * a = \sqrt 3 * d

{1 + \sqrt 3} / \sqrt 3 = d

The correct value of a is positive, and there is no difference if the third bead is negatively charged rather than positively charged.

 

[kuruman]

{1 + \sqrt 3} / \sqrt 3 = d

You are looking for $a$ not $d$ which is given.

 

[putongren]

OK let me try again:

OK let me try again:

1/(d - a) = sqrt(3)/a
a = sqrt(3) (d - a)
(1 + sqrt(3)) * a = sqrt(3) d
a = sqrt(3)/(1+ sqrt(3)) d

So that's the ratio between a and d

 

[haruspex]

OK let me try again:

OK let me try again:

1/(d - a) = sqrt(3)/a
a = sqrt(3) (d - a)
(1 + sqrt(3)) * a = sqrt(3) d
a = sqrt(3)/(1+ sqrt(3)) d

So that's the ratio between a and d

When you take a square root, as @kuruman pointed out, there is a certain ambiguity. You have chosen the right solution, but for completeness you should consider both and deduce which to use.
What about the next part, can the equilibrium be stable?

 

[putongren]

What does it mean to be in an equilibrium? When the electric field emanating from both charges at the ends equal at the point of the location of the third bead? Since I calculated the point where the electric fields are equal, the equilibrium should be stable at that point.

 

[Frabjous]

What does it mean to be in an equilibrium? When the electric field emanating from both charges at the ends equal at the point of the location of the third bead? Since I calculated the point where the electric fields are equal, the equilibrium should be stable at that point.

Equilibriums can be stable or unstable. Think about a ball at the bottom of a valley (stable) or the top of the hill (unstable). The question is whether the ball stays near the equilibrium point if it is slightly perturbed. What happens if the third bead is moved slightly? Think about whether the bead is positively or negatively charged.

 

[putongren]

If it is slightly moved from it's equilibrium location, it would not be stable since the electric fields emanating from both charges at the end wouldn't equal and would result in a net force from an imbalance of the fields.

So I guess that would be it would not be stable.

 

[Frabjous]

If it is slightly moved from it's equilibrium location, it would not be stable since the electric fields emanating from both charges at the end wouldn't equal and would result in a net force from an imbalance of the fields.

So I guess that would be it would not be stable.

Let the third bead be positively charged. You move it slightly toward the second bead. Does the resultant force cause the third bead to move toward the second bead or back towards the equilibrium point?
What if you moved it toward the first bead?
What if it was negatively charged?

 

[putongren]

I think i understand what you mean by being balanced now. Is there a way mathematically to calculate and determine whether the bead is balanced in its equilibrium position? Just by intuition, i think it is balanced since if you move the bead, the bead would eventually migrate until it would settle into its equilibrium point.

 

[Frabjous]

Is the equilibrium the same if the particle is negatively charged?
Mathematically, if you had written out the general force, you could look at it’s slope.

 

[haruspex]

I think i understand what you mean by being balanced now. Is there a way mathematically to calculate and determine whether the bead is balanced in its equilibrium position? Just by intuition, i think it is balanced since if you move the bead, the bead would eventually migrate until it would settle into its equilibrium point.

First, let me correct your terminology. Balanced and in equilibrium are the same. Stability is an extra condition.
Again, think about a ball either sitting in a bowl or on top of a hill. In the bowl, if it is perturbed the forces will tend to return it to the bottom, so that is stable. At the top of the hill, the forces will take it further from that equilibrium position, so that is unstable. Yes, the forces will take it to an equilibrium point, a stable one this time, but that is a different equilibrium point.
To determine stability, you can either consider the forces when slightly perturbed and see which way the net force acts, or use a virtual work method. The work method compares the potential energy at the equilibrium position with that at a perturbed position. The system will tend to the lower PE.

 

[jbriggs444]

If you are familiar with calculus then you can determine whether an equilibrium is stable or unstable by looking at the second derivative of potential. You can think of potential like the height of a hill.

Force is the additive inverse of the first derivative of potential. It is like how steeply the hill is sloping. The "gradient" of the slope. [It is the additive inverse because a rising slope has a positive gradient and yields a retarding force]. You have already solved for the position where force is zero -- an equilibrium point.

The second derivative of potential is like how the profile of the hill is curved. Positive for upward curvature. Negative for downward curvature. It is equal to the additive inverse of the first derivative of force.

If the second derivative of potential is positive (equivalently if the first derivative of force is negative) then the situation is like the bottom of a bowl. A stable equilibrium.

If the second derivative of potential is negative (equivalently if the first derivative of force is positive) then the situation is like a hill. An unstable equilibrium.

If the second derivative of potential is zero then the situation is more interesting. You can look for higher order derivatives to resolve the situation. But you could possibly encounter an inflection point where the equilibrium is stable on one side and unstable on the other. It depends on whether the first non-zero derivative is of odd order or even order.

This will not apply in the situation at hand. The derivative will be non-zero.In more than one dimension, the situation is also interesting. Direction becomes important and the notion of "partial derivatives" (a derivative in a particular direction) is useful. One can have "saddle points" where the second partial derivative in the x direction is positive and the second partial derivative in the y direction is negative.

This will not apply in the situation at hand. We are constrained to one dimension. If that constraint did not exist then you may be able to see that the equilibrium point would be a saddle point.

 

[putongren]

I am somewhat acquainted with calculus, but it has been a while.

Let me try to see if the third bead is stable mathematically.Here is a recap of what I did, sort of:

E₁ = Electric field from left charge at the third bead
E₂ = Electric field from right charge at the third bead

E₁ - E₂ = Net Electric Field (Force if charge is present) on third bead.

Equilibrium is when Net Electric Field = 0

The distance from where the bead is in equilibrium is a = sqrt(3)/(1+ sqrt(3)) * d as calculated from before.

So we found the distance where it is a equilibrium. How do you find the equation of the potential so we can take the second derivative and see if the position is stable? Do we use some potential energy equation?

 

[haruspex]

I am somewhat acquainted with calculus, but it has been a while.

Let me try to see if the third bead is stable mathematically.Here is a recap of what I did, sort of:

E₁ = Electric field from left charge at the third bead
E₂ = Electric field from right charge at the third bead

E₁ - E₂ = Net Electric Field (Force if charge is present) on third bead.

Equilibrium is when Net Electric Field = 0

The distance from where the bead is in equilibrium is a = sqrt(3)/(1+ sqrt(3)) * d as calculated from before.

So we found the distance where it is a equilibrium. How do you find the equation of the potential so we can take the second derivative and see if the position is stable? Do we use some potential energy equation?

As I wrote in post #15, if you work in terms of forces (or, equivalently, fields) then you need to write an expression for the net force/field when slightly displaced from the equilibrium position. If you work in terms of energy (or, equivalently, potential) then find an expression for that and determine whether that is at a local minimum or a local maximum at the equilibrium position.
The methods are pretty much the same.

 

[putongren]

I'm not sure if this is right and I'm not sure if I'm understanding everything that has been said by others thus far.

I think we should sample what happens when we let a be greater than the equilibrium point and when a be less than the equilibrium point. If there is a positive net force when a is greater than the equilibrium point and when a is less than the equilibrium point, than the net force is at a minimum and the equilibrium point is stable.

If there is a net negative force when a is greater than the equilibrium point and when a is less than the equilibrium point, than the net force is at a maximum and the equilibrium is unstable.

Take the equation E1 - E2 = Net Force.

Let A > sqrt(3)/(1+ sqrt(3)) * d and then let A < sqrt(3)/(1+ sqrt(3)) * d. If Net Force > 0, then it is stable since Net Force is a minimum at the equilibrium point. The equilibrium point is unstable if the Net Force < 0 when we let A > sqrt(3)/(1+ sqrt(3)) * d and then let A < sqrt(3)/(1+ sqrt(3)) * d.

Someone suggested we use calculus to solve the problem. I know some calculus, but seems a little beyond my mathematical skill set to do that.

 

[haruspex]

Take the equation E1 - E2 = Net Force.

Let A > sqrt(3)/(1+ sqrt(3)) * d and then let A < sqrt(3)/(1+ sqrt(3)) * d. If Net Force > 0, then it is stable since Net Force is a minimum at the equilibrium point. The equilibrium point is unstable if the Net Force < 0 when we let A > sqrt(3)/(1+ sqrt(3)) * d and then let A < sqrt(3)/(1+ sqrt(3)) * d.

Yes, that's the idea (but E1-E2 is a net field, not a net force), so long as you get all signs right.
It will be simplest to work in terms of a displacement $x$ from the equilibrium position $a$. What is the net field at $a+x$?

 

[kuruman]

A complementary and instructive approach might be to plot the potential energy of the third bead as a function of the reduced distance $x/d$ in the region $0<x/d<1$. Make two plots, one assuming that its charge is $q_3=q$ and one assuming that $q_3=-q.$ See how the two plots differ with regards to the equilibrium position. I recommend using a spreadsheet, not sketches drawn by hand.

 

[putongren]

let
d = distance between charges
x = displacement from equilibrium
\begin{equation}
\notag a =point of equilibrium = \frac{\sqrt{3}*d}{(1+ \sqrt{3})}
\end{equation}
then
\begin{equation}
\notag \frac{k*3q}{(x+a)^2} - \frac{k*q}{(d-(x+a)^2} = Net Field
\end{equation}

Excuse my poor use of LaTeX. I'm still learning and trying to get used to using it. I can substitute d with a and simplify the algebraic equation. Am I on the right track? If I get Net Field > 0 for both sides of a, then the equilibrium is stable. The reverse is true if I get Net Field < 0 for both sides of a.

 

[haruspex]

If I get Net Field > 0 for both sides of a, then the equilibrium is stable.

No. It is stable if the field opposes the movement, in whichever direction. If a>0 then the net field is negative and v.v.

 

[putongren]

I want to use calculus to solve the stability issue. It seems like a more elegant way to do solve the problem. I want to find the first and second derivative of potential as mentioned in post #16. Also, when you mention potential, do you mean kq/r^2? How do we go about finding the first and second derivative? Any hint?

 

[haruspex]

when you mention potential, do you mean kq/r^2?

No, potential is $k \frac qr$. The first derivative, $-k\frac q{r^2}$, is the field.

 

[kuruman]

I want to use calculus to solve the stability issue. It seems like a more elegant way to do solve the problem. I want to find the first and second derivative of potential as mentioned in post #16. Also, when you mention potential, do you mean kq/r^2? How do we go about finding the first and second derivative? Any hint?

If you're wondering how one finds the derivative of a function that obeys a simple power law, you are probably better off setting elegance aside and interpreting a plot of the potential energy as a function of position of the third bead.

Shown below is a plot of two curves. The blue line shows the potential energy when the third charge is positive and the red line when it is negative. The abscissa (x-axis) is the distance of the third bead from charge $3q$ divided by the overall distance $d$. When $x/d$ is equal to zero, the third bead is at the charge $+3q$; when it is equal to 1 the third bead is at charge $+q$. The dashed line marks the position where the net force on the bead is zero.

Look at the two plots. You know that the net force on the third bead is the negative slope (or gradient) of the potential energy. If you displace the bead from the dashed line in either direction, which curve predicts a net force that will return the third bead towards the equilibrium position and which curve predicts a net force that will push the third bead away from the equilibrium position?

 

[putongren]

Thanks kuruman for the graph. I think the blue curve is the curve that will return the third bead towards the equilibrium. The potential energy in both ends of the curve is positive. This means the third bead will have a tendency to move from positive potential energy to zero, the place of equilibrium. Honestly, I'm not sure what a negative potential energy actually implies about the behavior of the bead to the system.

 

[haruspex]

not sure what a negative potential energy actually implies

Potentials are relative. That applies to electric potential energy, electric potential, gravitational potential energy, gravitational potential… In principle, you can set the zero wherever you like. A lower potential than that is therefore negative, but it has no special significance.
In electrostatics, a common convention is to take the potential to be zero at infinity, but that is not always convenient. In cosmology, gravitational potential is taken to be zero at infinity, so anywhere else the potential is negative.

 

[jbriggs444]

Accordingly, rather than calling a potential "positive" or "negative", it is more meaningful and more intuitively helpful to speak about a potential "peak" from which an object may roll or slide away or a potential "valley" into which an object may settle.

Sometimes we may use other evocative terms such as a potential "barrier" across which an object does not have sufficient energy to pass. Or a potential "well" from which an object does not possess enough energy to escape.

 

[kuruman]

Honestly, I'm not sure what a negative potential energy actually implies about the behavior of the bead to the system.

As @haruspex and @jbriggs444 already indicated, do not focus on the negative sign of the potential. Here is a thought experiment to give you intuition. Take a hemispherical bowl and put it on a table with the rim facing you. Place a small bead in it and move it away from the equilibrium position. It it will roll (oscillate) back and forth. Now invert the bowl and put its rim flat on the surface of the table. Place the bead (if you can) at rest at the highest point of the bowl and displace it from the equilibrium position. It will roll off the bowl and fall on the table.

The potential energy of the bead-Earth system with the bead constrained to move on the bowl has a valley-like shape similar to the blue curve for the first case and a peak-like shape similar to the red curve for the second.

Because mechanical energy is conserved, when the bead is moved away from the equilibrium position in either case, it will move in the direction that will increase the kinetic energy at the expense of potential energy, i.e. in the direction of decreasing potential energy also known as the direction of the "downhill" force. It's the direction of "downhill" that counts not the sign of potential energy.

 

[putongren]

I understood the situation intuitively and conceptually concerning the equilibrium issue, but couldn't wrap my head around how the mathematics dictate whether the equilibrium is stable or not.

 

[kuruman]

I understood the situation intuitively and conceptually concerning the equilibrium issue, but couldn't wrap my head around how the mathematics dictate whether the equilibrium is stable or not.

Mathematically, one evaluates the second derivative of the potential energy at the equilibrium position.

• If the result is positive, the equilibrium is stable.
• If the result is negative, the equilibrium is unstable.

 

[haruspex]

Mathematically, one evaluates the second derivative of the potential energy at the equilibrium position.

• If the result is positive, the equilibrium is stable.
• If the result is negative, the equilibrium is unstable.

I'm not sure whether @putongren's difficulty is with the mathematical formulation or why it is correct.
At the equilibrium point, the first derivative is zero; i.e. the gradient is horizontal. If the second derivative is positive then the gradient is increasing as x increases. Since the gradient is zero at the equilibrium point, that means it is negative to the left of the point and positive to the right. Hence it is in the bottom of a dip.

There is also the possibility that the second derivative is also zero. In this case, you just have to keep taking more derivatives until you reach the first that is nonzero there.
Having reached it, it matters whether you had to differentiate an odd or even number of times. If an even number, positive indicates a minimum and negative indicates a maximum, as in the simple case discussed above.
But if an odd number of differentiations then it is neither a minimum nor a maximum. Instead, it is a "saddle", a kind of inflexion point.

E.g. consider $y=x^3$. The first nonzero derivative at x=0 is the third, making zero an inflexion. But for $y=x^4$ the first nonzero derivative at x=0 is the fourth, where it is 24, making it a minimum.