Help with an Irodov problem (Problem 3.3 electrodynamics )

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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.

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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 dont understand why.
 

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  • #2
haruspex
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Many solutions say that the body is constantly at equilibrium but i dont understand why.
The question means that the system is at equilibrium initially, but as charge gradually leaks away the equilibrium position changes.
 
  • #3
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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?
 
  • #4
haruspex
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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.
 
  • #5
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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.
 
  • #6
haruspex
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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.
 
  • #7
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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?
 
  • #8
ehild
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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 dont 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.
 
  • #9
haruspex
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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?
 
  • #10
haruspex
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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?
 
  • #11
ehild
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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.
 
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  • #12
haruspex
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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.
 
  • #13
ehild
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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?
upload_2018-5-6_7-5-20.png

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

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  • #14
haruspex
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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.
 
  • #15
ehild
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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.
 
  • #16
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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.
 
  • #17
ehild
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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.
 
  • #18
haruspex
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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.
 
  • #19
ehild
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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.
 
  • #20
haruspex
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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.
 
  • #21
ehild
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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.
 
  • #22
haruspex
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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.
 
  • #23
ehild
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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).
 
  • #24
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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.
 
  • #25
ehild
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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.
 
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