Please help understanding relative permiivity

[RoGeorge]

So what happens if you charge up an air gap capacitor, and remove and short the plates ?

The main idea is that energy is in the fields, while the wires guide the location in space for those fields.  If you don't have guiding wires, then the field will spread all over and propagate just like the light or the radio waves propagates all over.

About the air (or vacuum) capacitor, if you dismantle/separate the plates, then the field will follow the new position of the plates.  The energy is not "anchored" to empty space.  The energy is in the fields, the fields are "anchored" to the charges, and the charges are "anchored" with the metal (as in, the charges prefer to stay with the metal plates of the capacitor).  Overall, the field appear "anchored" between the two plates of the capacitor.  So, in this case, the field will follow the plate movement, and if you touch the plates, the capacitor discharges.

Then, why did the energy in that video remained in the glass, and didn't remain "anchored" to the plates?  The field was trapped in the glass, because when the glass dielectric was removed, charges from the plates were transferred (by induction) to the glass, and the induced charges in the glass remained attached to the glass (glass is a very good insulator), until the capacitor was re-assembled.  Vacuum space does not have charges, so the dissectable capacitor trick doesn't work in vacuum.

Air do have charges, so the trick should work with air, just that the air is a gas, the molecules and the charges attached to the air will mix very fast.  Any induced charges during removing an air dielectric, will move with the air and neutralize each other very fast.  So it won't work with gas dielectric, and most probably won't work with a liquid dielectric either.

The glass was able to keep the induced charges separated because it is a very good insulator and a solid.  Any charge placed on glass remains "sticked" in its place.  Can not move.


However, there's another very funny experiment one can do while trying to dismantle an air capacitor:
- Imagine you have two horizontal metal plates, one above the other, so a capacitor
- you charge your capacitor from a 9V battery, then remove the battery
- the capacitor is now charged, if you connect a very high impedance voltmeter, it will measure 9V (the voltmeter has to be very high impedance, so it won't discharge the capacitor much during measurement)

- now, leave the voltmeter connected (still showing 9V), and try to disassemble the capacitor by lifting the top plate away
- keep an eye on the voltmeter while lifting the top plate

To everybody's surprise (well, at least to my surprise when I've seen this for the first time) when you lift the top plate, the voltage increases like crazy! 

And if you bring back the top plate in the initial position, the voltage is 9V again.  And if you lower the top plate even more (under the initial position, but without touching the bottom plate), the voltage goes less than 9V.  In fact, you can generate any voltage (of the same polarity) by simply changing the distance between the plates. 


So, the voltage changes with distance, in theory between zero and infinity.  But why always the same sign of voltage.  Why if you go to "negative distance" the other side of the plate, the sign of voltage doesn't change?  That's another can of worms question.

[watchmaker]

Fascinating.  Such a ubiquitous device that gets so little respect.

[wasedadoc]

Charge on a capacitor is capacitance times voltage. If the voltmeter takes negligible current the charge is not reduced. Changing the distance between the plates changes the capacitance. To keep the product of capacitance and voltage constant, the voltage must change.

Capacitance depends on magnitude, not sign, of separation. Thus voltage does not change polarity if top plate taken below the other one.

[mawyatt]

Conservation of charge!

Since the charge didn't change with ideal voltmeter, then voltage must change with distance between plates as wasedadoc indicated.

Think of this in discrete terms, as breaking up the large parallel plate capacitor into N smaller plates. Now charge up all N capacitor plates to Vi with distance between parallel plates X. Then place each smaller plate capacitor in series and you get N* Vi as output voltage with effective distance between original plates now becomes N*X.

Since charge is conserved as Q = C*Vi, or Vi = Q/C, then C was reduced by N, so now Vf = Q/(C/N), or N*(Q/C), or N*Vi as final voltage Vf.

You can do the reverse starting with a N series of C, charge to Vi, then placing all C in parallel and you should get Vf = Vi/N.

These are basic concepts with charge pumps by changing capacitance the output voltage is changed since charge is conserved.

All this can get confusing when one starts to consider conversation of energy and such, as mentioned moving the plates/charge entails energy use. In the charge pump case the available current (dq/dt) drains the voltage faster with a voltage step up rather than voltage step down since the more "effective source capacitance" for this current is higher in the voltage step down (paralleled C) rather than voltage step up (series C) cases.

Anyway, in the end there's no "free lunch", however fun stuff to think about tho!!

Best,

[RoGeorge]

Fascinating.  Such a ubiquitous device

But wait, we were just starting.

Since the conservation of charge was brought to the topic, here's a funny experiment with charge conservation:
- take two identical capacitors, both discharged, say C1 and C2 of 100uF each
- charge one capacitor from a 9V battery, so C1 is charged now at 9V, then remove the battery
- now, connect the other capacitor in parallel with the charged one
- the charge is conserved, so it will distribute equally between the two capacitors
- by definition C=Q/V, total Q didn't change (because the law of conservation of charge), and we know two 100uF in parallel make 200uF total.  So now the voltage must be now 4.5V, and if you measure indeed it is.

The charged C1 discharged half into the empty C2, they shared all equally.  Makes sense.

- now, calculate the energy stored in capacitors before, and after putting them in parallel, formula is W=1/2C*V²
C1 at 9V and C2 at 0V means the stored energy (J) is:
     W1=1/2*100uF*9V*9V = 4.05mJ
     W2=1/2*100uF*0V*0V = 0mJ
so total W1+W2 is 4.05mJ

After they are connected together, the voltage is 4.5V on each, C1 at 4.5V and c2 at 4.5V means
    W1=1/2*100uF*4.5V*4.5V = 1.0125mJ
    W2=1/2*100uF*4.5V*4.5V = 1.0125mJ
so total W1+W2 is now only 2.025mJ    Half of the energy is missing!

Where energy?!? 

Funny thing is, similar happens every time you charge or discharge a capacitor.  Half of the energy is gone, no matter how you charge or discharge your capacitor. 

Unless...

[SteveThackery]

This is a fascinating riddle, @RoGeorge!

So what is the answer? Bearing in mind energy can't be created or destroyed, where has it gone?

[TimFox]

If you do the math for charging a capacitor from a constant voltage through a resistor, half the energy from the battery is dissipated in the resistor independent of the resistor value.
The smaller the resistance, the faster the approach to the asymptotic value.
If you want to get pedantic about the R=0 case, remember that although you could use superconducting wire, there will always be inductance.  The resonant circuit with the finite loop area will radiate energy to free space.

[SteveThackery]

If you do the math for charging a capacitor from a constant voltage through a resistor, half the energy from the battery is dissipated in the resistor independent of the resistor value.

So are you saying that the piece of wire you connect the two capacitors together with dissipates half the energy as heat?

I didn't think of that, but it does make sense. Your explanation about series resistors helped me visualise it. Thanks!

[Tation]

If you want to get pedantic about the R=0 case, remember that although you could use superconducting wire, there will always be inductance.  The resonant circuit with the finite loop area will radiate energy to free space.

I can (hardly) remember a discussion by Carver Mead about this, but more concerned in the theoretical minimum energy needed to store a single bit of information using an electronic device. Sorry to be such imprecise…

[MrAl]

We can talk about all this stuff until the proverbial cows come home but it does not mean any of it is true until we do actual experiments.  That's because intuition is often wrong.  Take a look at the Faraday Disk Paradox.  Defies intuition until a very, very careful analysis is done.

However, I think we can theorize, but then we have to indicate very specific setup descriptions which explains what we will ignore and what we will take as settled physics as well as any mechanical constructions and any movements of any kind.

[RoGeorge]

Sure you can do the experiment in practice.  Charge a capacitor then add another in parallel, and measure voltage.  Voltage will be half, therefore half of energy is now gone.  It happens to the book, and it's a very easy to reproduce experiment.

Where did half of the energy goes, it's not that easy to say.  Most probably it is dissipated as heat, and only a small part as radio waves.

Saying this only based on the fact that, to launch energy into space (as radio waves) is not easy.  Empty space has it's own impedance, and energy doesn't like impedance changes, it reflects back.  That is why a radio transmitter has to have an carefully designed antenna (matched to the impedance of space), and to the feeding cable/radio amplifier.  Everything has to be designed just right, and tuned to the working frequency.

The thought of managing to put half of energy into space (like with those 2 parallel capacitors), no matter the size of capacitors, no matter the size of wires, and no matter the resonant or the radiated frequency, seems like a very far fetch.  And to always put precisely half of the energy into radio?  I highly doubt.

Most probably that half of energy is lost as heat in the parasitic resistance of the wires/metal.  Then is when you lose precisely half of the energy:  when you charge (or discharge) a capacitor through a resistor.


The 2 parallel capacitor riddle is often called the "capacitor paradox", and there are tons of explanations online, from IEEE papers, to electromagnetic field simulations, to anonymous online debates.

Another interesting way of thinking about it is to use a mechanical analogy:  https://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html  If we do that and replace capacitance with mass, and voltage with speed, then connecting the 2 capacitors in parallel will correspond with hitting a stationary mass with another moving mass, where the resulting speed of the 2 combined masses is half, so a plastic collision (not elastic), so the missing energy was dissipated as heat in the plastic deformations, or in case of the electric circuit, as heat in the wires.


Anyway, putting that aside, let's make another experiment:
- take a voltage source, a resistor and a capacitor
- connect them in series, and keep track of the energy dissipated inside the source, the energy dissipated inside the resistor, and the energy stored in capacitor.  (Like in the experiment brought by TimFox a couple of messages ago).

The uncanny thing is no matter what R, what C and what V you use, precisely half of the energy is stored into the capacitor, and half is dissipated in the resistor.    Why the energy split is half/half, no mater the value of the components?! 

Why when you try to put energy into a capacitor you always waste half of the energy?  Why?!


Now here's what I've just "discovered" yesterday, when I was trying to put another numerical example for the lost energy.  If you use a constant-current source to charge the capacitor (instead of a voltage source and a resistor), all the energy is stored into capacitor, none is lost, no paradox. 

[SteveThackery]

Now here's what I've just "discovered" yesterday, when I was trying to put another numerical example for the lost energy.  If you use a constant-current source to charge the capacitor (instead of a voltage source and a resistor), all the energy is stored into capacitor, none is lost, no paradox. 

Another "wow" moment for me. 

[TimFox]

Another comment about “1/2” in the capacitor energy formula, comparing a battery and capacitor.
For either device, if you remove a small amount of charge dQ,
the energy removed is dE = VdQ.
For an ideal battery (or DC supply) the voltage V remains constant as the charge flows out, and the energy transferred is E = QV for transferred charge Q.
For a capacitor C, the voltage drops as dV = dQ/C, so the change in energy drops as the voltage drops.
Integrating the energy change from full charge Q to zero gives the usual equation
 E = CV²/2 = Q²/2C.

[MrAl]

Sure you can do the experiment in practice.  Charge a capacitor then add another in parallel, and measure voltage.  Voltage will be half, therefore half of energy is now gone.  It happens to the book, and it's a very easy to reproduce experiment.

Where did half of the energy goes, it's not that easy to say.  Most probably it is dissipated as heat, and only a small part as radio waves.

Saying this only based on the fact that, to launch energy into space (as radio waves) is not easy.  Empty space has it's own impedance, and energy doesn't like impedance changes, it reflects back.  That is why a radio transmitter has to have an carefully designed antenna (matched to the impedance of space), and to the feeding cable/radio amplifier.  Everything has to be designed just right, and tuned to the working frequency.

The thought of managing to put half of energy into space (like with those 2 parallel capacitors), no matter the size of capacitors, no matter the size of wires, and no matter the resonant or the radiated frequency, seems like a very far fetch.  And to always put precisely half of the energy into radio?  I highly doubt.

Most probably that half of energy is lost as heat in the parasitic resistance of the wires/metal.  Then is when you lose precisely half of the energy:  when you charge (or discharge) a capacitor through a resistor.


The 2 parallel capacitor riddle is often called the "capacitor paradox", and there are tons of explanations online, from IEEE papers, to electromagnetic field simulations, to anonymous online debates.

Another interesting way of thinking about it is to use a mechanical analogy:  https://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html  If we do that and replace capacitance with mass, and voltage with speed, then connecting the 2 capacitors in parallel will correspond with hitting a stationary mass with another moving mass, where the resulting speed of the 2 combined masses is half, so a plastic collision (not elastic), so the missing energy was dissipated as heat in the plastic deformations, or in case of the electric circuit, as heat in the wires.


Anyway, putting that aside, let's make another experiment:
- take a voltage source, a resistor and a capacitor
- connect them in series, and keep track of the energy dissipated inside the source, the energy dissipated inside the resistor, and the energy stored in capacitor.  (Like in the experiment brought by TimFox a couple of messages ago).

The uncanny thing is no matter what R, what C and what V you use, precisely half of the energy is stored into the capacitor, and half is dissipated in the resistor.    Why the energy split is half/half, no mater the value of the components?! 

Why when you try to put energy into a capacitor you always waste half of the energy?  Why?!


Now here's what I've just "discovered" yesterday, when I was trying to put another numerical example for the lost energy.  If you use a constant-current source to charge the capacitor (instead of a voltage source and a resistor), all the energy is stored into capacitor, none is lost, no paradox. 

Hi,

I was not talking about that very simple case that is quite well known and valid.  One of the reasons capacitance in CPU's is such a problem

The time equation for the voltage across C2 for two capacitors in parallel coupled by the inevitable resistance R is:
vC2(t)=((v2-v1)*C1*e^(-(t*(C2+C1))/(C1*C2*R)))/(C2+C1)+(v2*C2+v1*C1)/(C2+C1)

where v1 is the initial voltage across C1, v2 is the initial voltage across C2, R is the resistance, and t is the time.
The simplest view is when C2=C1 so we can set C2=C1.
Since C2 will have zero volts across it at the start when t=0, we can set v2=0, and since it does not matter what the resistance is or what the value of the two caps are, we can set C1=1 and R=1, and it does not matter what voltage we start with so we can set v1=1, and the result is:
vC2(t)=1/2-e^(-2*t)/2

Since we don't mind waiting all day to find out the results with any values, we can take the limit as t goes toward infinity and of course we get:
vC2(inf)=1/2

So the voltage was one-half of what we started with.

If we instead start with two voltages for C1 and C2 that are not equal, then in the end we get the average voltage:
vC2(inf)=(v1+v2)/2

and if we make C1 larger than C2 such that C1=n*C2 then in the end we get:
vC2(inf)=(v2+n*v1)/(n+1)

All this does have the lumped circuit element model assumptions.

I was talking about the examples that involved more complicated physics though in the previous post.


Also, in circuit analysis, if you do not assume any resistance you get zero energy loss.

 

[RoGeorge]

Indeed, the simple experiments are the most fun, though, there is an equivalence between energy and information.  Equivalence in a similar manner as the most popular, incomplete and not Einstein's formula E=mc² that tells the equivalence between mass and energy.  There is an equivalence between information and energy, too, just like there is between mass and energy.  This makes a legit question from "how much energy do we need to erase a bit".

After all, the mythical philosophical idea of all being one, might eventually be proven correct by physics.  Might be all energy, or all information, depending how we want to call it, but I digress.

And since you mentioned the inaccuracies that lumped components may introduce when compared with real electronic components, I'll add another experiment with a capacitor.


But first, let's add two short visualization from Khan Academy, one for capacitors, and the second for relative permittivity, so to pretend the question about relative permittivity was not abandoned. 

https://youtube.com/watch?v=u-jigaMJT10

In the next video, the relative permittivity is noted with letter k instead of the Greek epsilon (\$\varepsilon_r\$), but they are the same thing:



And now, back to "fun at parties" with capacitors.  Here's another experiment:
- take a capacitor made from 2 parallel metal discs
- charge the capacitor from a battery
What is the shape of the voltage observed at the terminals of the capacitor?

Does the voltage raises smooth, or in steps? 

More than 70% will say the voltage follows a smooth charging curve, according to this poll:  https://www.eevblog.com/forum/chat/does-a-capacitor-charges-smooth-or-in-stairs/ but in fact the voltage jumps in stairs. 

I like to believe that the voltages on the capacitor raises in jumps, like stair steps (though, I didn't try to measure that in practice myself).  It makes sense to say it should be voltage steps, if the capacitor is looked like a transmission line.

When the voltage is applied, a wavefront travels from the middle of the plates (where the capacitor's terminals are) to the margins of the disks, and when the wavefront reaches the edge of the capacitor's plates, it encounter open space.  When the plates end, there is no way to produce charge displacement (currents) because there is no more metal, and no charges to displace in an empty space, so the wavefront reflects back, and so on, back and forth it reflects between the terminals and the edges of the plates.

Overall, the voltage stairs resemble/average to the known and expected smooth voltage raise, just that it is not smooth, it is constructed from small voltage steps, something like this:

Picture by rhb in https://www.eevblog.com/forum/chat/does-a-capacitor-charges-smooth-or-in-stairs/msg3093173/#msg3093173

Learned about capacitors that might, in fact, charging/discharging in stairs from Prof. Gregory Durgin's classes about transmission lines (see at minute 29:50, some voltage stairs are drawn at 43:00):  https://youtu.be/gdiPwbZm0aM?t=2580

[watchmaker]

Still here and taking it all in.  Thanks!

Regards,

Dewey