Post by: Eternal Student on 22/03/2023 05:02:21
The Tech section doesn't get used as much as the Physics section. So I've put this in here rather than in the usual Physics section.
Declaration: This has been set as an exercise by a UK University.
Background:
1. It's not my homework, just something I saw a month ago, became interested in and still can't explain to my satisfaction.
2. It was never assessed but just intended as something a student should do in their own time, for their own benefit.
3. The general spirit of the exercise is that a student should be able to explain the video to a friend.
Overall, I would think that a discussion here causes no harm and is perfectly consistent with the spirit of the exercise. Even if next year's students find this thread, it should help not hinder.
Here's the original phrasing of the task:
Quote
This is the video we watched at the start of the live session. You should be able to explain what happens to the current through the circuit and the voltage between the plates at the following times:
a) 0 -> 1m, the plates are connected to a power supply and moved apart from / towards each other;
b) 1m5s -> 1m45s, The capacitor is detached from the power supply, a fixed amount of charge is deposited, and the plates are moved apart from each other;
c) 1m45s -> 2m05s, with the capacitor plates still isolated, a sheet of plexiglass is placed between the plates.
Here's the video: "MIT Physics Demo -- Adjustable Capacitor with Dielectric", available on YouTube, duration ~2 minutes.
- - - - - - - - - - - - - - -
Brief response:
a) We have C = εA/d [1]
V = Q/C [2]
d increases, Eqn [1] => C decreases => Eqn [2] Q must decrease (because V is held constant) => we have a current flow.
OK... so the Ammeter should show a deflection in the same direction from the central position every time the experimenter turns the handle to move the plates apart.
BUT....if you look carefully when the plates move apart (at about time 0:27), the first deflection of the Ammeter during the first twist is to the right. All the other twists and you have a deflection to the left. WHY? i.d.k. The demonstrator knocked the handle? I am missing some physics?
I would blame it on a faulty Ammeter but the Ammeter is actually pretty good in that it manages to show something else that demanded precision from the instrument: Combining [Eqn 1] and [Eqn 2] we obtain Q ~ 1/d meaning that the biggest changes in Q (associated with the bigger currents and deflection) should be seen when the gap was smallest. (Assuming they turn the handle roughly the same number of rotations and roughly the same speed each time they twist it). You DO see that in the video clearly enough...... the deflections on the Ammeter are bigger when the plates are close.
Reverse the behaviour when the plates are moved back together. That bit seems to give the expected results. We get deflections to the right only and they become massive, almost to the limit of deflection, when the plates are closest.
----The post is already getting long, so I'll leave discussion of parts b and c for now ---
Main question from the first part is Why does the ammeter deflect right on the first twist of the handle when the plates were moved apart?
Best Wishes.
Post by: alancalverd on 22/03/2023 07:59:13
I've not seen a Braun electroscope before - neat piece of kit! It looks as though it is connected to the insulated capacitor plate, so bringing the experimenter's hand close to it will definitely affect the charge on the insulated plate.
Crap demo!
Post by: Eternal Student on 22/03/2023 13:13:59
Static charge on the experimenter might be as good as any other explanation and I hadn't thought of that.
In favour: They did walk fast. The lab coat looks shiny, could be polyester.
Concerns: I don't know where the ammeter is connected. Current must have passed through that when they touched the handle. This suggests the ammeter does allow a good path between the two plates - which will be a problem for later.
I was going to move on to part (b) of the exercise:
The capacitor is disconnected from the Power supply and a fixed charge Q is deposited on it.
Taken at face value, we have the same equations as before Q = VC [Eqn 2' ]
C =
[Eqn 1]When moving the plates apart, d increases => C decreases from [Eqn 1] => V must increase from [Eqn 2'] since Q was constant.
Overall, V ~ d but the electroscope has no scale and it's not obvious.
I think the original exercise wanted E fields and electrostatics discussed directly instead of just electronics. So, we have some large plates. Near the centre of a plate draw a Gaussian surface, usual textbook stuff, we have E = σ / ε in the direction normal to the surface (with σ = charge density = Q / Area ). The E field is completely independent of distance from the plane (for a large plane and staying near the centre). σ never changes, so over most of the central region of the plates at least we obtain
E(at any location between the plates) = a constant independent of d (the distance between plates).
Then the potential difference, V, from one plate to the other is just E . d . So we have V ~ d exactly as before.
We see the electroscope lift up and down in the right places, so taking everything at face value, that's all fine.
Now the main question has to be: How did they tell the charge Q to stay on the plate?
There is an ammeter in the circuit and they claim it measures current flowing from one plate to the other. So there is at least one path from plate to plate through the ammeter. I would have expected the capacitor to discharge in the usual pattern, an exponential decay with time. I'm not seeing much evidence of that in the video. The ammeter has some huge resistance? - but then it's unlike the properties you would want or expect in an ammeter. It shows a flow of current from one plate to the other during many parts of the video, it clearly doesn't put a stop to all of them.
The second question would be: Is there any discernible pattern or explanation for the deflections shown on the ammeter during this part of the demonstration? It seems to move left or right a bit almost at random. Just noise or just there to deliberately make you ask why the system can't satisfy the restrictions from [Eqn 1] and [Eqn 2] by simply discharging some Q from the first plate to the other when the distance between plates is increased. i.d.k.
Let's take an argument based on E fields. The charge on the first plate is essentially on the surface of the plate only (because of the usual arguments about conductors and electrostatics). So those surplus charges are naturally tending to repel each other and would ordinarily flow through the conductor, ultimately reaching the other plate through some path like that which the ammeter should provide. However, there's a non-zero E field between the plates and it's precisely enough to match the force that would propel those surface charges away, they are held on the surface. If this was a stable equilibrium we'd all be happy - no current will appear. However it's not, it's an unstable equilibrium. If some small charge δQ is able to leave the first plate then it can reach the other plate and that will reduce the E field between the plates rather than increasing it. Basically, capacitors do discharge when there is a path from one plate to the other and, as far I can see, they would discharge even faster when you move the plates further apart.
Now, despite giving the capacitor every encouragement to discharge while moving the plates apart, at the end of the entire experiment it is still claimed that the charge Q is still there.
However, when the plexi is removed, the voltage rises back up again, showing that the charge is still there.
- taken from the info. box on the YouTube video.
I must be missing some physics here.... some clever way in which they have connected this whole circuit together, used the electroscope and/or galvanometer. i.d.k.
Best Wishes.
Post by: vhfpmr on 22/03/2023 14:06:44
the first deflection of the Ammeter during the first twist is to the right. All the other twists and you have a deflection to the leftNo, you can see that every time his finger makes contact with the knob, the meter gives a kick to the right.
The adjuster knob is on the earth side of the capacitor, but the live side of the electrometer, so any charge on the experimenter will leak to the power supply earth through the meter.
Post by: vhfpmr on 22/03/2023 14:35:48
I don't know where the ammeter is connected.Yes you do, follow the wiring:
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So there is at least one path from plate to plate through the ammeter.No there isn't, the power supply has been disconnected, so the only connnection to the live plate is the insulated support on the stand, and the electroscope.
Is there any discernible pattern or explanation for the deflections shown on the ammeter during this part of the demonstration?Yes, as above, the knob is on the live side of the meter.
Post by: alancalverd on 22/03/2023 15:04:29
How did they tell the charge Q to stay on the plate?If the plate is very large and the electrometer has a very high impedance, relatively little charge will flow when the plate is moved. Keithley electrometers are pretty good in that respect but you can do better in principle with a potentiometric system and electrostatic null detector. It's the way we used to measure charge in ionisation chambers used for radiation measurement, but it all gets a bit complicated for modern undergraduates who would probably find it offensive because it takes actual skill, can't be done with a mobile phone, and is therefore noninclusive of clumsy idiots.
Post by: alancalverd on 22/03/2023 15:11:21
Is there any discernible pattern or explanation for the deflections shown on the ammeter during this part of the demonstration?Badly constructed equipment. The whole kit should be enclosed in a grounded Faraday cage with an extension rod to the adjusting knob. Precision electrostatics is not for amateurs.
It's a pity that they have assembled some very nice kit but not spent a few dollars on a simple dog cage to make it work properly!
Post by: vhfpmr on 22/03/2023 17:33:05
The Keithley electrometer is the ammeter with a low impedance, and the electroscope is the voltmeter with a high impedance, but yes, the 'scope draws a small current, like all voltmeters. Measuring off the screen, the electroscope must be 2000-3000 times smaller than the capacitor.How did they tell the charge Q to stay on the plate?If the plate is very large and the electrometer has a very high impedance, relatively little charge will flow when the plate is moved. Keithley electrometers are pretty good in that respect but you can do better in principle with a potentiometric system and electrostatic null detector. It's the way we used to measure charge in ionisation chambers used for radiation measurement, but it all gets a bit complicated for modern undergraduates who would probably find it offensive because it takes actual skill, can't be done with a mobile phone, and is therefore noninclusive of clumsy idiots.
Post by: Eternal Student on 22/03/2023 20:11:02
Hi.
Thanks for all replies.
No, you can see that every time his finger makes contact with the knob, the meter gives a kick to the right.Allowing for noise, taking an average etc. , you could be right. It's not crystal clear to me but I'm old and can't really watch two places (hands and ammeter) at the same time. The static charge and capacitance of the experimenter together with a wiring diagram something like yours is beginning to look like a clear favourite to explain the right deflections at the beginning when the plates were moved apart.
Yes you do, follow the wiring: (with a wiring diagram shown)I'm not sure that is the wiring.
If the electroscope (the leaf style voltmeter) works the way I think it does then it doesn't even need to be connected more than once, it may not have a positive and negative lead (or an earth side and live side using your terminology). There can be one lead from the outer part (or top part) of the stem of the device to one place in the circuit where you want to determine the voltage. The device is really responding to charge that is driven into the deeper part of the stem and thus also the leaf which was connected to that stem at a pivot point. It's like a capacitor with just one plate. It is an indicator of surplus (or net) charge being present on the stem (and leaf) sections that are deep inside the device.
It is indirectly useable as a voltage indicator since (negative or positive) charge will tend to be driven into it if the place it's connected to in the circuit evolves over time to have a different (lower or higher, respectively) voltage.
However, I just don't know how the thing was wired up, the video never bothered to show any of that. We seem to be left guessing. Forced to accept that somehow every piece of kit does precisely and only what they claim, charge Q stays on the plate etc. I'm grateful for any attempt to show what the wiring may have been like, thank you.
ES said: So there is at least one path from plate to plate through the ammeter.If that's correct then the ammeter is virtually useless, unable to tell us anything from the moment the power supply was disconnected. At least that would solve one problem, the deflections were random during this part of the experiment. However, they did seem to make sense for the later part where the dielectric was inserted and removed.
vhfpmr replied: No there isn't, the power supply has been disconnected, so the only connection to the live plate is the insulated support on the stand, and the electroscope.
Best Wishes.
Post by: alancalverd on 22/03/2023 23:10:51
If the current meter had been wired "above" the capacitor in the diagram, the moveable plate would have been at ground potential, the insulating block on which the capacitor is mounted would have been unnecessary, and the ghostly hand could have been conveniently grounded with an antistatic strap. Problem is that the a maximum common mode voltage of the electronic ammeter might then have been exceeded, so (guessing a bit at the dimensions of the capacitor) you'd need a battery-powered microammeter floating at around 1000 V to show the effect - not a common piece of kit.
Worth noting that the tuning capacitor of oldfashioned radios (the ones with knobs and analog dials) is a variable-area device, and in almost all cases the moving vanes are grounded so that the frequency doesn't drift when you put your hand on the knob. But you wouldn't expect MIT physicists to lower themselves to a 100-year-old technical trick!
Post by: Bored chemist on 22/03/2023 23:26:43
you'd need a battery-powered microammeter floating at around 1000 V to show the effect - not a common piece of kit.Not common.
But there's one in that video.
The Keithley 602
https://download.tek.com/manual/29111C(Model602).pdf
Post by: alancalverd on 23/03/2023 09:13:50
On closer inspection it seems that they are using it at about 3μA full scale, which corresponds with my estimate of the currents involved. The Keithley has a triaxial input, allowing differential connection at up to 1500V above the case potential, but if they were worried about someone overcooking the HV supply it's easy enough to lash up a small analog meter that can float at the high voltage end: microamps is "O" level electronics!
Come on MIT, get your act together! What does that T stand for?
Post by: Eternal Student on 23/03/2023 15:02:26
Thanks for all the replies.
I've recently found an MIT website where they describe the way the Braun electroscope was connected:
https://tsgphysics.mit.edu/demo.html?E_4
It looks like @vhfpmr got the diagram right in post #4 and deserves a star or medal of honour rather than the suspicion I displayed in post #8. I will edit my post to show that.
Overall, I think we have the right general idea of how the circuit was connected and where some of the problems have arisen mainly as a consequence of the capacitance and static charge on the experimenter together with the placement of the ammeter in the circuit.
Thank you to everyone.
Post by: Eternal Student on 23/03/2023 20:16:23
Just realised we haven't actually discussed part (c) of the original task.
So, the basic capacitor equations are, as before: C = εA/d [Eqn 1] and C = Q/V [Eqn 2]
Where this time, the insertion of the dielectric is going to change ε.
We have: insert dielectric => ε increases => [Eqn 1] C increases => [Eqn 2] Q increases OR V decreases.
We see from the experiment that both happens, Q increases (we saw the ammeter deflect one way in the video) AND ALSO V decreases (the electroscope leaf fell down).
The basic electronics equations accepts either or both of these but doesn't explain why it would be both and not just one or the other.
Some simple electrostatics arguments can be used to explain why you get both effects. Basically, the E field can and will change instantly when the dielectric is pushed further into the capacitor, since the dipole sources in the dielectric are changing their location in space. The Voltage is just an integral of the E field, so some change in V can and will happen instantly. Local changes force a current to flow from plate to plate, so you will also see a current flow and that would be immediate. However, a current does not mean the new charges arrive on the plates instantly, the current will bring them but over some time. So there was never any chance a new charge could arrive on the plates fast enough to maintain a constant V and satisfy the constraint of [Eqn 2] where C was increasing. We will see the voltage V change slightly. However the current was there, so Q will also change over time. Overall both Q and V will be seen to change as the dielectric is inserted.
We reverse the reasoning and expectations when the dielectric is removed. That's all fine and the video seems to show that.
Now the main question that remains is: What happens to the energy?
The Energy stored in a capacitor is given by
We had Q went up and V went down when the dielectric was inserted. How did the overall stored energy change?
Were there forces resisting the experimenter inserting the dielectric or did the apparatus try to pull the thing out of his hand?
Best Wishes.
Post by: alancalverd on 24/03/2023 05:55:11
Post by: paul cotter on 24/03/2023 12:35:19
Post by: Eternal Student on 24/03/2023 16:20:43
Welcome back @paul cotter .
About stress mentioned by @alancalverd : I can see there would be some extension ("stretch") along the axis from plate to plate but are there any other forces? Shear forces, torque around some axis, a net force acting radially? For example, does the dielectric need to be glued in to prevent the capacitor spitting out bits of dielectric?
Best Wishes.
Post by: paul cotter on 24/03/2023 20:09:16