# The Naked Scientists Forum

### Author Topic: Where does the charge lie  (Read 5824 times)

#### syhprum

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##### Where does the charge lie
« on: 20/09/2009 23:12:52 »
If I construct a capacitor with two metal sheets 1mm apart and of such an area to have a capacitance of 1nF and charge it to 1KV and then disconnect it from any external circuit (except that required to measure the voltage which draws no current)I am now storing 500 micro joule of energy.
If I then interpose a sheet of dielectric materiel between the plates of dielectric constant 10 what is the resultant energy stored ?
« Last Edit: 21/09/2009 15:43:15 by syhprum »

#### Geezer

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##### Re: Where does the charge lie
« Reply #1 on: 21/09/2009 04:49:04 »
The original dielectric is air, correct?

I'm going to say the energy is still 500 μJ. The charge (Q) remains the same, however, the capacitance has changed because the dielectric changed, so the voltage must also have changed.

It's not so different from a variable capacitor which changes the distance between the plates to alter the capacitance, except in this case we are modifying the dielectric to alter the capacitance.

Hope I'm right

#### JP

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##### Re: Where does the charge lie
« Reply #2 on: 21/09/2009 06:52:59 »
Do you leave this hooked up to the battery when you add the dielectric, or is it an isolated capacitor at that point?

If it's still hooked to the battery, the voltage remains constant, but the capacitance should increase x10, so the energy stored goes up by to (Energy=1/2 capacitance*voltage^2).

If you isolate the capacitor, the energy goes down since the charge on the plates is what is held constant.
Since Voltage=charge/capacitance, the voltage decreases 10-fold.  Therefore, according to
Energy=1/2*capacitance*voltage^2, the total energy decreases 10-fold.

In the isolated system, the dielectric would be pulled into the space between the plates by a force, whereas in the system attached to the batteries, you'd have to force the plate in between the plates against an opposing force.

#### Geezer

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##### Re: Where does the charge lie
« Reply #3 on: 21/09/2009 07:54:29 »
I think syhprum is leading us into a trap here.

Energy does not typically "evaporate", although if it does that could be really exciting because the converse may also be true.

Unless energy was somehow "carried away" when we changed the dielectric, it is still stored on the plates. We know C changed, and therefore V can have changed. Are we making an assumption about Q?

#### syhprum

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##### Re: Where does the charge lie
« Reply #4 on: 21/09/2009 11:08:59 »
When the dielectric is inserted assuming no energy is lost I would expect the voltage to fall to 1/10-.5 KV ie 316.2 V so that CV^2/2 remains the same.
Now we come to the more subtle case what happens if the plates are moved to a quasi infinite distance apart not only is the capacitance reduced to near zero but energy is put into the system pulling the attracting plates apart.

PS this experiment was tried about 150 years ago when it was found care had to be taken that the dielectric sheet had to be perfectly dry otherwise it carried away some of the charge.
As a boy I tried to test this with a variable air spaced capacitor but had no suitable means to measure the voltage
« Last Edit: 21/09/2009 11:16:43 by syhprum »

#### JP

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##### Re: Where does the charge lie
« Reply #5 on: 21/09/2009 15:11:54 »
When the dielectric is inserted assuming no energy is lost I would expect the voltage to fall to 1/10-.5 KV ie 316.2 V so that CV^2/2 remains the same.

This isn't the case in either scenario.  Energy is only conserved in an isolated system.
The capacitor setup isn't isolated because you are doing something to it (pushing the dielectric in).  The two possible constants are

Voltage: if you leave the battery hooked up, you know the voltage drop across the capacitor has to stay constant from Kirchoff's law.

Charge: if you remove the capacitor from the circuit after it's charged, you know the charge on the capacitor plates cannot go anywhere.

In either case, you can solve E=1/2CV2 and q=CV to figure out how the energy changes.

#### syhprum

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##### Where does the charge lie
« Reply #6 on: 21/09/2009 15:45:09 »
I have modified the question slightly to remove any ambiguity

#### Geezer

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##### Where does the charge lie
« Reply #7 on: 21/09/2009 16:08:38 »
Ah! So the change in energy is accounted for as work. (Doh!) I suppose the same would be true in the case of a variable air spaced capacitor.

#### JP

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##### Where does the charge lie
« Reply #8 on: 21/09/2009 16:24:04 »
Ah.  For constant charge, then

Initially, without dielectric
q=C0*V0=1nF*1kV=10-6 Coulombs

After the dielectric is in place
C=10*C0=10nF
V=q/C=10-6 Coulombs /10nF=100 V

So the energy after the dielectric is in place is
E=1/2 C*V2=1/2*(10 nF)*(100 V)^2=50 micro Joules

Macroscopically, the electric field between the plates decreases because of the presence of the dielectric, so the energy stored there has to decrease.

Microscopically, the polarizations of the molecules making up the dielectric align with the field, and when a polarization rotates to align with a field, there is a drop of potential energy.  Therefore, the net energy decreases.

#### Geezer

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##### Where does the charge lie
« Reply #9 on: 21/09/2009 16:35:14 »
Presumably the energy would also change if we removed all the air molecules instead of inserting a different dielectric?

#### JP

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##### Where does the charge lie
« Reply #10 on: 21/09/2009 17:08:42 »
I was assuming that it was essentially in a vacuum, which air basically is.  You could model air by giving it a dielectric constant that is just a tiny bit larger than one, and removing it would then change the charge.  I think it would be more complicated due to the forces involved and since air isn't a rigid solid.

#### Geezer

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##### Where does the charge lie
« Reply #11 on: 21/09/2009 17:26:42 »
I was wondering if there is a simple method of independently quantifying the change in energy. Not sure why I care really, because I'm sure the result would be the same

Could we do something with a liquid dielectric, oil, for example?

#### syhprum

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##### Where does the charge lie
« Reply #12 on: 21/09/2009 18:12:56 »
Suppose we increase the capacitance to 10nF by allowing the plates to get closer, there will of course be some loss of energy to the restraints of the plates.
« Last Edit: 21/09/2009 19:32:47 by syhprum »

#### Geezer

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##### Where does the charge lie
« Reply #13 on: 21/09/2009 18:25:45 »
I should really go and read the books. However, it's also interesting to learn how much I have forgotten, if I ever knew it at all!

If memory serves, the charge on the plates produces a repulsive force between the plates. So, if we reduce the spacing, work is done and we have added energy. Does the work done account for the total change in energy?

#### Bored chemist

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##### Where does the charge lie
« Reply #14 on: 21/09/2009 18:50:52 »
"If memory serves, the charge on the plates produces a repulsive force between the plates. "
Wrong way round, but the right idea.

#### Geezer

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##### Where does the charge lie
« Reply #15 on: 21/09/2009 19:14:22 »
Dang! Well, I'm off to the library now

#### syhprum

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##### Where does the charge lie
« Reply #16 on: 21/09/2009 19:23:47 »
I am very puzzled by the opinion of JP that 90% of the energy in the system has disappeared by the voltage being reduced to 10% of what it was while the capacitance is increased 10 times.
Where has it gone ? it cannot surely be the work done moving the dielectric sheet in.
One could easily contrive an experiment where the plates are horizontal in a tank and a dielectric fluid such as water (dielectric constant 80)is poured in where little or no mechanical energy would be involved.
I know there is a hysteresis effect in dielectrics having often getting a shock from CRT's despite having discharged them!

Are you saying that 90% of the energy in the system has been dissipated in the dielectric sheet ?
« Last Edit: 21/09/2009 19:27:48 by syhprum »

#### lightarrow

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##### Where does the charge lie
« Reply #17 on: 21/09/2009 19:35:32 »
I am very puzzled by the opinion of JP that 90% of the energy in the system has disappeared by the voltage being reduced to 10% of what it was while the capacitance is increased 10 times.
Where has it gone ? it cannot surely be the work done moving the dielectric sheet in.
Why? A dielectric with ε>ε0 polarizes under the action of the field, so it's attracted between the plates, so it gains kinetic energy. Neglecting the energy conveyed into heat because of Joule effect inside the dielectric and into radiated electromagnetic waves because of time variable currents, this kinetic energy is what you loose from the capacitor.
« Last Edit: 21/09/2009 19:37:43 by lightarrow »

#### Geezer

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##### Where does the charge lie
« Reply #18 on: 21/09/2009 19:53:29 »
Why? A dielectric with ε>ε0 polarizes under the action of the field, so it's attracted between the plates, so it gains kinetic energy. Neglecting the energy conveyed into heat because of Joule effect inside the dielectric and into radiated electromagnetic waves because of time variable currents, this kinetic energy is what you loose from the capacitor.
I was just about to say that, but Lightarrow beat me to it  (Yeah! Sure you were Geezer!)

Should it not be a gain in potential energy rather than kinetic energy?
Would we be able to quantify the change in energy if we can measure the force between the plates and the change in distance between them? (Neglecting the heating effects that Lightarrow points out.)
« Last Edit: 21/09/2009 20:53:05 by Geezer »

#### lightarrow

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##### Where does the charge lie
« Reply #19 on: 22/09/2009 15:13:39 »
I was just about to say that, but Lightarrow beat me to it  (Yeah! Sure you were Geezer!)

Should it not be a gain in potential energy rather than kinetic energy?
Of course it is potential energy at the beginning and kinetic energy with the dielectric exactly between the plates; in other positions it's both potential and kinetic, thanks for pointing it out .

#### Geezer

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##### Where does the charge lie
« Reply #20 on: 22/09/2009 16:14:17 »
I was just about to say that, but Lightarrow beat me to it  (Yeah! Sure you were Geezer!)

Should it not be a gain in potential energy rather than kinetic energy?
Of course it is potential energy at the beginning and kinetic energy with the dielectric exactly between the plates; in other positions it's both potential and kinetic, thanks for pointing it out .
Thanks! Syphrum suggested we alter the distance between the plates to alter the capacitance when the dielectric is air. There is an attractive force between the plates, so, as the distance between the plates decreases, the capacitor does mechanical work. I think this mechanical work will account for most of the difference in the energy between the two different spacings for a fixed amount of charge. If we could measure the attractive force between the plates we should be able to quantify the work done by purely mechanical means. Do you think that might be correct?

#### syhprum

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##### Where does the charge lie
« Reply #21 on: 22/09/2009 20:27:19 »
Yes I can well see that allowing the plates to move together under their attractive force would transfer out some of the energy stored in the electrostatic field as mechanical work to the supports but I am at a loss to see how allowing a dielectric fluid to flow in and out from between the plates could remove so much of the stored energy.
I would like to confirm this by experiment but do not have the facilities

Of course pulling the plates apart should increase the energy stored in the electrostatic field, is this not the principle behind the Wimshurst machine?

#### Geezer

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##### Where does the charge lie
« Reply #22 on: 22/09/2009 21:28:00 »
Ah! I liked your moving plate experiment better than the liquid dielectric experiment!

#### JP

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##### Where does the charge lie
« Reply #23 on: 22/09/2009 23:15:57 »
Think of it this way.  The field between the two plates points directly from one plate to the other, across the gap.  At the edges, however, the field bows out in arcs (see really bad diagram).  A dielectric fluid introduced near the plates will experience a field gradient because of these field arcs that are escaping from the plates, and will be pulled towards the gap between them until it reaches the equilibrium point in the center of the plates.  The liquid that got pulled in picked up kinetic energy in order to move, which it had to get from somewhere... and the only source of energy is the field in the
capacitor.  So in attracting the fluid into the gap, the capacitor gave up some energy.

« Last Edit: 23/09/2009 01:37:05 by jpetruccelli »

#### Geezer

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##### Where does the charge lie
« Reply #24 on: 23/09/2009 07:56:43 »
H'mm? I wonder if anyone ever ran an experiment where air was replaced with liquid and vice versa. I would have thought someone must have tried it buy now. It's not such a complicated experiment.

Here's a sketch of a possible experiment using Syhprum's original solid dielectric. This might allow us to measure the work done on the dielectric. The rectangle is the solid dielectric. It's suspended on a knife edge like a beam-balance. The two round things (are supposed to!) represent the plates of the cap. When we apply charge to the plates, the dielectric should be drawn into the space between the plates. If we know the mass of the dielectric, we should be able to determine the force from the angular deflection of the dielectric support rod.

Would something like this actually work? If it does, it's likely to be a well known experiment - just not well known to me!

« Last Edit: 23/09/2009 17:17:42 by Geezer »

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##### Where does the charge lie
« Reply #24 on: 23/09/2009 07:56:43 »