Help with an Irodov problem (Problem 3.3 electrodynamics )

[Brilli]

Homework Statement

Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length 1. The distance between the spheres x << 1. Find the rate dqldt with which the charge leaks off each sphere if their approach velocity v = a/ √x, where a is a constant.

Homework Equations

The Attempt at a Solution

The net acceleration required=dv/dt=-a^2/(2x^2)
This acceleration is given by electrostatic force and weight of the mass.
Is i take components and get the q and differentiate it the answer comes wrong.
Many solutions say that the body is constantly at equilibrium but i don't understand why.

 

[haruspex]

Many solutions say that the body is constantly at equilibrium but i don't understand why.

The question means that the system is at equilibrium initially, but as charge gradually leaks away the equilibrium position changes.

 

[Brilli]

The question means that the system is at equilibrium initially, but as charge gradually leaks away the equilibrium position changes.

But as per the solution we need to take net acceleration every instance 0. Why is it so? Why is it wrong to do that the rate of change of the given velocity is its acceleration and thus the forces along it should be giving it?

 

[haruspex]

But as per the solution we need to take net acceleration every instance 0. Why is it so? Why is it wrong to do that the rate of change of the given velocity is its acceleration and thus the forces along it should be giving it?

Consider a car jacked up. As you wind the jack down, the rate of descent is not controlled by forces on the car - it is controlled by the rate at which you choose to lower it.
It is the same here. The rate at which the spheres come together depends on the rate of leakage. At any instant it is very nearly in equilibrium.

 

[Brilli]

Consider a car jacked up. As you wind the jack down, the rate of descent is not controlled by forces on the car - it is controlled by the rate at which you choose to lower it.
It is the same here. The rate at which the spheres come together depends on the rate of leakage. At any instant it is very nearly in equilibrium.

Can you please clarify when should i use this logic and when should i use the force equation? Because i have been greatly confused as i first saw it as a standard pendulum problem with varying forces.

 

[haruspex]

Can you please clarify when should i use this logic and when should i use the force equation? Because i have been greatly confused as i first saw it as a standard pendulum problem with varying forces.

The question could have been better worded. Something like "the spheres hang in equilibrium, but charge gradually leaks from them, allowing them to come gradually closer together at speed ..."
If a small amount Δq of charge leaks at some moment then the system is not in equilibrium and the spheres will start to swing together. If no more charge is lost for a while, and no energy is lost, the spheres will oscillate about a new equilibrium forever. But in the real world energy is gradually lost to drag etc, so the spheres reach their new equilibrium, and some time later a further Δq is lost, etc.

I can't give you a simple rule here. You just have to understand the situation the question is describing and apply reason. In this case, that the spheres only lose charge gradually means that KE is being lost all the time. The only reason the spheres continue to get closer is that they continue to lose charge.

 

[Brilli]

The question could have been better worded. Something like "the spheres hang in equilibrium, but charge gradually leaks from them, allowing them to come gradually closer together at speed ..."
If a small amount Δq of charge leaks at some moment then the system is not in equilibrium and the spheres will start to swing together. If no more charge is lost for a while, and no energy is lost, the spheres will oscillate about a new equilibrium forever. But in the real world energy is gradually lost to drag etc, so the spheres reach their new equilibrium, and some time later a further Δq is lost, etc.

I can't give you a simple rule here. You just have to understand the situation the question is describing and apply reason. In this case, that the spheres only lose charge gradually means that KE is being lost all the time. The only reason the spheres continue to get closer is that they continue to lose charge.

so is it necessarily true that calculating it with force equations is 100% wrong?

 

[ehild]

Σ

Homework Statement

Two small equally charged spheres, each of mass m, are suspended from the same point by silk threads of length 1. The distance between the spheres x << 1. Find the rate dqldt with which the charge leaks off each sphere if their approach velocity v = a/ √x, where a is a constant.

Homework Equations

The Attempt at a Solution

The net acceleration required=dv/dt=-a^2/(2x^2)
This acceleration is given by electrostatic force and weight of the mass.
Is i take components and get the q and differentiate it the answer comes wrong.
Many solutions say that the body is constantly at equilibrium but i don't understand why.

When writing up the equation for the acceleration, you need to take also the tension in the strings into account.
Write the equation as mdv/dt=ΣF, use the relation given between v and x, and see what you get. Show your work.

 

[haruspex]

so is it necessarily true that calculating it with force equations is 100% wrong?

Yes, I think so. You never explained how you went from calculating a force to finding the leakage rate, so it is hard for me to say exactly why your method produced the wrong answer, but I do not see how it could get a correct one.

One can imagine that the charge leaks off at some other rate, maybe such that the spheres approach each other at constant speed. So you would deduce there is no force. How would you be able to find the leakage rate from that?

 

[haruspex]

Σ

When writing up the equation for the acceleration, you need to take also the tension in the strings into account.
Write the equation as mdv/dt=ΣF, use the relation given between v and x, and see what you get. Show your work.

Hi @ehild.
Are you seeing a way to solve this question using acceleration and forces? Have you reviewed the whole thread?

 

[ehild]

Hi @ehild.
Are you seeing a way to solve this question using acceleration and forces? Have you reviewed the whole thread?

As x<<L=1, only horizontal motion should be considered.
You need not solve the motion as it is given by the relation v=a/√x. From the forces, you get the acceleration in terms of x. From v=a/√x, you also know the acceleration in terms of x. Substituting into the first equation, you get the charge q in terms of x. You can get dq/dt in terms of x, or in terms of t solving the equation v=a/√x.
The relation v=a/√x could have arose if initially the charges were in equilibrium, and the leakage started at t=0.

 

[haruspex]

From the forces, you get the acceleration in terms of x.

From what forces? If you look at the system at any instant it appears to be in equilibrium.
To the extent that each increment in the leakage disturbs the equilibrium it allows the distance to reduce to a new equilibrium, but there is no F=ma relationship between the resulting acceleration and any force imbalance.
Consider my car jack example in post #4.

 

[ehild]

From what forces? If you look at the system at any instant it appears to be in equilibrium.
To the extent that each increment in the leakage disturbs the equilibrium it allows the distance to reduce to a new equilibrium, but there is no F=ma relationship between the resulting acceleration and any force imbalance.
Consider my car jack example in post #4.

What forces? T tension, weight W, Coulomb force Fe.
The net force = mass times acceleration, is not it?

I can not understand your car example as I can not drive.
Can you speak about equilibrium if a body accelerates?

 

[haruspex]

I can not understand your car example as I can not drive.

Then try this analogy.
A tank of water stands at rest on a spring balance. There is a slow leak in the tank. It is found that after time t the weight has reduced by at².
What has the magnitude of the upward acceleration of the tank to do with any imbalance of forces, in the sense of F=ma? It is surely determined purely by the rate of the leak.

 

[ehild]

Then try this analogy.
A tank of water stands at rest on a spring balance. There is a slow leak in the tank. It is found that after time t the weight has reduced by at².
What has the magnitude of the upward acceleration of the tank to do with any imbalance of forces, in the sense of F=ma? It is surely determined purely by the rate of the leak.

It is a problem with changing mass. You can solve using ΔP=FΔt.

 

[haruspex]

You can solve using ΔP=FΔt.

There is negligible momentum. And what are you proposing for F there?

Have you read my explanation in post #6? There is work being lost the whole time.

 

[ehild]

There is negligible momentum. And what are you proposing for F there?

Have you read my explanation in post #6? There is work being lost the whole time.

Forces are gravity and the spring force of the balance. The latter one can have a dissipative term.
What do you mean on work lost? Energy can be lost, not work.

Newtonian mechanics holds even for systems near equilibrium. Do you deny?
If a body accelerates there are unbalanced forces acting on it.
Motion does not happen by jumping from one equilibrium to the next one. It is continuous.
This example of yours is different from the original problem. You gave the time dependence of mass and asked the motion. The original problem gave the motion ( realistic or not, it does not matter) and asked the change of charge with time.

 

[haruspex]

Forces are gravity and the spring force of the balance.

Most of the time, in my analogy, those are in balance. The leak gradually forms a drip on the outside of the tank.
Every now and then a drip falls. There will then be an upward acceleration of the tank. In principle, this will overshoot the new equilibrium and oscillate, but quickly settle out to a new equilibrium. Thus, the motion can be described as "punctuated equilibrium" (pace the late S J Gould). Mechanical work is lost.
The rate of ascent of the tank is determined by the rate of loss of water mass and the spring constant, but it has nothing to do with F=ma.

The original problem is almost exactly the same. Charge leaks gradually away, and it is this rate that determines the rate of approach of the spheres, not F=ma.

If you believe the problem can be solved the way you say, please post a solution in a private conversation.

 

[ehild]

If you believe the problem can be solved the way you say, please post a solution in a private conversation.

You will not accept my solution if it not based to "punctuated equilibrium", which yields a different result.

 

[haruspex]

You will not accept my solution if it not based to "punctuated equilibrium", which yields a different result.

First, I am not clear on how you get a solution at all. Secondly, maybe if I see your solution I can either accept that it is also valid (given the uncertainty in the problem statement), or perhaps find a way to show that it cannot be right.

 

[ehild]

First, I am not clear on how you get a solution at all. Secondly, maybe if I see your solution I can either accept that it is also valid (given the uncertainty in the problem statement), or perhaps find a way to show that it cannot be right.

My solution is outlined in Posts #11 and #13. I can not imagine what is not clear to you. Can you write the forces at a given position of the charged balls?
It is me who do not understand your solution, assuming equilibrium, that is, zero net force, for accelerating particles. Would you so kind to send your solution in pm?
It is clear that the relation v=a/√x can not be true, but the problem includes it.

 

[haruspex]

My solution is outlined in Posts #11 and #13.

The forces you list in post #13 give you a relationship between x, L, m, charge and acceleration. Unfortunately we are not given the initial charge, so this is not enough to determine the acceleration as a function of x.

It is clear that the relation v=a/√x can not be true

No, it can be any function at all if the discharge rate can vary. The 1/√x arises because the discharge rate is constant.

 

[ehild]

The forces you list in post #13 give you a relationship between x, L, m, charge and acceleration. Unfortunately we are not given the initial charge, so this is not enough to determine the acceleration as a function of x.

No, it can be any function at all if the discharge rate can vary. The 1/√x arises because the discharge rate is constant.

From the relation v=a/√x, the acceleration of the particles can be derived. @Brilli has shown that the acceleration was dv/dt=-a^2/(2x^2).

 

[haruspex]

From the relation v=a/√x, the acceleration of the particles can be derived. @Brilli has shown that the acceleration was dv/dt=-a^2/(2x^2).

Ok, I accept that you can obtain an answer that way, but the reason it is wrong is that it neglects losses.
As I explained earlier, this should be taken to be a gradual leakage of charge. The only reason the particles continue to get closer is that charge continues to leak. E.g. consider one electron departing each second.

Try this analogy: a block is being lowered down a ramp by a cable passing over a winch at the top. The rate of descent is controlled by the rate of turning of the winch, not by F=ma. It is the same here. The rate of loss of charge determines the rate of approach.

Note that your approach gives dq/dt as varying, whereas mine gives a constant. I think that proves which is intended by the question setter.

 

[ehild]

Ok, I accept that you can obtain an answer that way, but the reason it is wrong is that it neglects losses.

You also neglected losses. Loss forces can be connected to velocity. According to v=1/√x , there is velocity every time. You took the same forces into account as me, only assumed acceleration of the balls when they were in equilibrium. It is wrong. When there is acceleration, there are unbalanced forces in an inertial frame of reference.
The rate of approach is determined by the varying Coulomb force, not by the change of charge itself. And the electric force is instantaneous in Classical Mechanics, there is no time to reach equilibrium when the charge changed.
I think the problem maker intended to solve the problem as you. Even then, the "solution" is wrong.
And I suggest to stop this debate. You do not accept that ma=ΣF, but I do and I won't change.

 

[haruspex]

You also neglected losses.

No. Reaching the new equilibrium dissipates the KE. Those are the losses.

there is no time to reach equilibrium when the charge changed.

At one electron lost per second (for example) there is ample time to reach equilibrium.

 

[kimbyd]

Σ

When writing up the equation for the acceleration, you need to take also the tension in the strings into account.
Write the equation as mdv/dt=ΣF, use the relation given between v and x, and see what you get. Show your work.

There's has been a lot of debate on this thread about this solution, and I think there have been a number of mistakes made, so I wanted to chime in and (hopefully) clarify things.

First, a couple of points:
1) The tension on the rope is always perpendicular to the motion of the spheres, and thus plays no direct role on their motions. It is possible to write your equations in such a way that the tension does not appear, which makes all of the math simpler.
2) The question asks us to assume that the distance between the spheres is far smaller than their connecting point. This is another way of saying that we can safely ignore the fact that the spheres move in a circle: we can consider it to just be two spheres moving horizontally towards one another. This means that gravity plays no role.
3) The problem is asking us to match up an equation of motion ($v = a/\sqrt{x}$) with a force magnitude (related to the dissipation of charge). The concept of equilibrium plays no role, because the relationship between force and acceleration acts both in and out of equilibrium.

The way I would approach this problem is as follows:
1) Write down the equation for the relative distance between the spheres. This is a function for v(t) given x(t), i.e. a differential equation. It should be solvable to provide an equation x(t).
2) Write down the equation for the electrostatic force between the spheres. This is a function of F(t) given v(t) and q(t). You could use $\vec{F} = m\vec{a}$ to solve for q(t). Note that the problem structure has one-dimensional motion, so we don't need to worry about any vector sums, except for getting the signs right on the equations.

I haven't solved the problem, but one general comment I have is:
Try keeping everything in terms of differential equations as long as possible, to see if you can simplify things. Getting a picture of the whole problem may provide hints with ways to make your life easier. It may not be necessary to solve the differential equation for v(t) given x(t) at all. If that doesn't seem simpler, just solve the differential equation for x(t), which will make the equation for q(t) a matter of algebra.

 

[haruspex]

There's has been a lot of debate on this thread about this solution, and I think there have been a number of mistakes made, so I wanted to chime in and (hopefully) clarify things.

First, a couple of points:
1) The tension on the rope is always perpendicular to the motion of the spheres, and thus plays no direct role on their motions. It is possible to write your equations in such a way that the tension does not appear, which makes all of the math simpler.
2) The question asks us to assume that the distance between the spheres is far smaller than their connecting point. This is another way of saying that we can safely ignore the fact that the spheres move in a circle: we can consider it to just be two spheres moving horizontally towards one another. This means that gravity plays no role.
3) The problem is asking us to match up an equation of motion ($v = a/\sqrt{x}$) with a force magnitude (related to the dissipation of charge). The concept of equilibrium plays no role, because the relationship between force and acceleration acts both in and out of equilibrium.

The way I would approach this problem is as follows:
1) Write down the equation for the relative distance between the spheres. This is a function for v(t) given x(t), i.e. a differential equation. It should be solvable to provide an equation x(t).
2) Write down the equation for the electrostatic force between the spheres. This is a function of F(t) given v(t) and q(t). You could use $\vec{F} = m\vec{a}$ to solve for q(t). Note that the problem structure has one-dimensional motion, so we don't need to worry about any vector sums, except for getting the signs right on the equations.

I haven't solved the problem, but one general comment I have is:
Try keeping everything in terms of differential equations as long as possible, to see if you can simplify things. Getting a picture of the whole problem may provide hints with ways to make your life easier. It may not be necessary to solve the differential equation for v(t) given x(t) at all. If that doesn't seem simpler, just solve the differential equation for x(t), which will make the equation for q(t) a matter of algebra.

I have finally thought of the real reason the DE approach does not work.
The leakage is in itself dissipative. Each time an electron is lost it carries away electric potential energy. The particles get closer again, turning GPE into EPE, but then some more leaks away. It is like a slowly deflating balloon.

 

[Charles Link]

I think the form $v=\frac{a}{\sqrt{x}}$ makes this problem unphysical as $x \rightarrow 0$. The solution assumes the electrostatic and gravitational forces are always balanced, and thereby it is solved under the approximation that any acceleration is small. This is contrary to the statement of the problem that says $v=\frac{a}{\sqrt{x}} \rightarrow +\infty$ as $x \rightarrow 0$.

 

[kimbyd]

We've discussed this a bit in more detail, and I think the answer is that there is no correct solution to the question as stated.

The fundamental problem is that the question seems to be describing a system where two spheres are falling towards one another, but there is an electrostatic repulsion pushing them apart, causing them to approach slowly.

However, if you differentiate the provided $\dot{x} = a/\sqrt{x}$ value, you end up with two spheres accelerating towards one another very rapidly. That just can't physically happen with this system.

So, I think the answer in the end is: please check to see if you stated the problem incorrectly. If it was copied correctly, please contact the professor for clarification. As stated, this question cannot be answered.

 

[Jahnavi]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

 

[Charles Link]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

That does simplify the problem, and that scenario does look like it would eliminate the inconsistencies. :)

 

[Jahnavi]

That does simplify the problem, and that scenario does look like it would eliminate the inconsistencies. :)

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

Feel free to say NO

 

[kimbyd]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

Yes. That would work.

 

[Charles Link]

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

Feel free to say NO

I actually think the author tried to be a little fancy, and didn't realize what he was proposing was unphysical. In the author's case, $\frac{dq}{dt}$ is a constant, while for your case $\frac{dq}{dt}$ is dependent on position.

 

[Jahnavi]

I actually think the author tried to be a little fancy, and didn't realize what he was proposing was unphysical. In the author's case, $\frac{dq}{dt}$ is a constant, while for your case $\frac{dq}{dt}$ is dependent on position.

You didn't answer my question

 

[Charles Link]

You didn't answer my question

The only thing missing is the word "No". :)

 

[Jahnavi]

The only thing missing is the word "No". :)

Very bad .

OK , one more chance . I might have something up my sleeve .

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

 

[Charles Link]

Very bad .

OK , one more chance . I might have something up my sleeve .

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

It certainly is a possibility. One other thing the author could have said is to avoid $x \approx 0$, and to only consider the region where $\frac{dv}{dt}$ is small, so that the approximation of equilibrium between the electrostatic force and the gravitational force is reasonably good. (So that any additional unbalanced horizontal force can be small, and the acceleration remains small). $\\$ In general, I think most of us are in agreement that the author's "empirical" formula $v=\frac{dx}{dt}=-\frac{a}{\sqrt{x}}$ simply can not work all the way to $x=0$.

 

[Jahnavi]

@haruspex , @Charles Link , @ehild , @kimbyd this is the actual question along with the original problem in Russian .

 

[Charles Link]

@Jahnavi So who is the translator who introduced a $v=-\frac{a}{\sqrt{x}}$? Was it a student who assisted with the publication? Is the problem submitted by the OP an English and later version of the same book? This is getting interesting. :)

 

[Jahnavi]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

In recent editions of the book , this has been corrected .

I also have a copy of this book . It has same question as given in the OP

 

[Charles Link]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

In recent editions of the book , this has been corrected .

I also have a copy of this book . It has same question as given in the OP

I give you +2 for this. A very good finding. :)

 

[Jahnavi]

I give you +2 for this

 

[ehild]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

Congratulation Jahnavi!
Somebody has changed the problem, and then followed the original solution that the acceleration was zero. In the original question it was because of the constant velocity. The English version explained it with equilibrium.
Moral: do not change Russian problems. Usually they are rather good, you can only mess up them.
I wonder how the problem was corrected in the recent version? To v=const?

 

[Jahnavi]

In the original question it was because of the constant velocity. The English version explained it with equilibrium.
Moral: do not change Russian problems.

No

In fact in the initial editions of the Russian version , it was given that v = a/√x . But , in the recent editions it has been corrected .

 

[ehild]

No

In fact in the initial editions of the Russian version , it was given that v = a/√x . But , in the recent editions it has been corrected .

My bad. I am totally confused. Was there an original Russian version with v=a/√x, and it was changed to v=const, and then an English version with v=a/√x, and then a new English version with v=?
When I was young, we used quite a few Russian textbooks and problem books ( usually translated to Hungarian, or in original, as we studied Russian from age 10) and I found them rather good.

 

[Jahnavi]

Firstly I would say that this is perhaps a rare problem you would find in this book with some issue . It is an excellent , superb book .Very accurate ! And mind boggling too .

I guess that the initial Russia version actually had v = a/√x which was later corrected to v = constant .

But the problem is that the English editions keep printing the initial Russian version . Even now the new English editions at my place has the same wording as given in the OP .

The copy of the book I am having also has the same incorrect version

 

[ehild]

It is amazing to see the English text of the problem with the solution which originally belonged to the corrected Russian version
http://exir.ru/3/resh/3_3.htm

 

[rude man]

But as per the solution we need to take net acceleration every instance 0. Why is it so?

Who says it is so? With x = x(t), neither a(x) nor a(t) is necessarily constant. But I'm referring to the original problem as posted in the OP's post 1 and I gather that has eo ipso been questioned. But the math looks menacing even so.