Challenge Questions

Total amount of energy will remain unchanged, however movement of plates will increase potential difference and will decrease capacitance,

And still before and after movement, total amount of energy can be described by (CV²/2).

Change in voltage will be (√(D2/D1) times (√2.5 Times in this case)

Change in capacitance will be (D1/D2 times (0.4 times in this case)

Please note at the time of discharge energy can be converted in many form not just spark, what I am talking about here is total transfer of energy.

I think the form Energy = Q²/C/2 is more illuminating (pun intended). And the point of the question appears to be that the additional energy comes from work done pulling the plates apart, which you can regard as the electrical equivalent to lifting a mass (and the discharge energy as equivalent to dropping the mass).

Simple answer:

If you have to do work on the two plates to move them apart, the work must show up as potential energy stored in the plates. Conservation of energy. W = F*ds.

So yes, there will then be more energy released when they are discharged.

For theoretical explanation- whats going to be the force F on a frictionless horizontal surface?

Zero!!!!!!

Of course there'll be a force between the plates - they are charged, dammit!

F = ma, friction does not enter the equation.

At constant speed, force F is going to be zero on a frictionless horizontal surface.Next time you better sign in before raising a question, don't scared of being wrong.More or less that happen with everyone.

I doubt he was scared of being wrong - just in a hurry. Please remember that friction is not the only source of force. Try jumping and see if you keep on moving upwards - th reason you come down again is the force of gravity. The force between the plates (against which you are doing work) is electrostatic attraction.

Electrostatic force is known to every body.but can you elaborate how this work done is going to toped up on the electrical energy stored inside plates? I am really here to learn more from you, and again please sign in before you reply.no reason to hurry.

Not the guests - but everything you need can be found in the combination of BobD's reply to you in post #13 and mine in #24. The values and spacings of the questions make proper equations too complex (you need elliptic functions - yuck) to present here. Hopefully, someone will address my revised question with full arithmetic, and all will become clear...

As I see it, this would be similar to the following: You have a 1 Kg ball sitting on a stand 1 m above the ground. You then raise the ball to a height of 10 Km.

The force of gravity is stronger on the ball when it is closer to the Earth (since gravity goes as 1/r²), but the impact on the ground will be much greater if it is dropped from 10 Km, than if it were dropped from 1 m. That's because you did work on the ball to raise it to 10 Km, and that work is stored as potential energy in the ball. When it is released, that stored energy causes a bigger impact.

"10cm to 25cm" would be equivalent to "1m to 2.5m", not really sure how the analogy to moving a mass vertically, then letting it fall towards the pull of Gravity has any bearing on the question though.

Unless You routinely let Your Charged plates collide that is...

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I like some of the Answers I have seen, but Maybe I've forgotten a few things since studying electronics 20+ Years ago...

Energy is like Watts, right? Watts = Voltage x Current.

But in relation to Capacitance E=0.5 of CV with V being squared, and C represents "Capacitance", right?

My thought is if Voltage increases but Power (or Energy) remains constant (according to some posts I've read), then Current must be diminished, or is it just the Rate, or Speed of Discharge is slowed down because of the change in spacing, & the less pressure each plate exerts on the other through the Electrolyte?

Energy is like Joules, right? For this situation, any Joules are given by
Charge x Voltage change ("gradual change" being the reason for the factor 1/2)
Watts are power, = Joules/second (or Voltage x current)

Not if You are wearing the proper Gloves...

As the plates are attracted to one another seperating the plates is inputting some energy into the capacitor, so the energy will increase.

With that logic just put plates on long drive and store unlimited energy in plates,finally use as big a battery.

I don't think you understood my post, please read it again.

Number of electrons are fixed, each electron have fix eV. how the energy is adding inside of plate in terms of charge? if not charge then in what form?

You are so close, so it's worth taking a little more care and getting all the way.

Electrons have fixed charge, and you are not adding or removing electrons when you pull the plates, the charge on the capacitor is fixed [Q=number_electrons x e^-]. What varies is the Voltage. However, that will not remain fixed during capacitor discharge, which is why you see the factor 1/2 in the equations for the energy stored on a capacitor. i.e. Energy=Q²/(2.C) [or Energy=CV²/2].

Because the plates attract [Force=Electric_field x Charge] you are doing work when you pull the plates apart. That work is exactly the same as the increase in the electrical energy stored on the plates. [If the author of the question had been a little more considerate and made the dimensions of the plates large compared with the separation,we could have presented some reasonable approximate equations...]

The charge remains the same, of course, so a bit different from a battery. But the energy will increase according to the work you do against the attraction between the plates. Once the separation exceeds the linear dimensions the attractive force is subject to the inverse square law (or worse if there are other conductors around). So the potential tends to a finite limit.

On the other hand, you might wish to look up Van der Graff generator and Wimshurst machine, both of which combine increasing charge by means of induction and increasing potential by mechanical means.

Hi Bob,

Thanks for such a detailed explanation.My GA but after verifying my ld physics book.

I am not sure about the discharging process, but the energy amount remains the same the voltage increase 1.5 times, so the spark will be bigger

Do you mean that the charge remains the same and the Voltage will increase? (That would of course represent more energy)

Energy is function of voltage, current and time,Not only voltage.

In this case there is trade off among these parameters. when we apart two plates there is going to be quicker discharge,

I had thought when reading the question that I'd be dealing with dielectrics, micro farads, Coulombs, all that old stuff.

But I'm not going into the detailed physics of capacitors to answer this one. I don't even care if a charge of 1 Coulomb passing through 10 cm. of air as a dielectric is a reasonable assumption or not. Since the question assumes that the 5 X 5 cm plates are capable of holding a charge that will jump 15 cm., I'm not even going to investigate that.

I just think the answer is that if it takes a certain charge in the plates to spark through 10 cm of dielectric, it takes a larger charge to spark-discharge through a 15 cm distance of the same dielectric.

Greater sparking distance, larger release of energy required to make it. Yes, more energy will have to be accumulated on the plates before they are able to discharge across the greater distance.

Mark

"Yes, more energy will have to be accumulated on the plates before they are able to discharge across the greater distance."

Mark! what I understand from original question is that the discharge is deliberate with physical means not the natural breakdown between plates.correct me if I am wrong.

Hi, rakesh semwal!

OK, if same sized plates, unchanged charge, at two different distances, are being discharged with a deliberate shorting device, like a conductor of uniform dimensions along its length, and the conductor's increased length matching the distance difference when shorting them, the spark will be infinitesmally weaker due to the added resistance of the lengthened conductor over the increased distance.

Is that what you mean?

Mark

The discharging circuit remains the same for each situation.

No it does not. If the Voltage is higher the discharging probes will be further apart when the discharge starts - the different discharge path is a change in the circuit.

Sorry I meant the conductors that lead to the discharge points are the same.

That way they don't interfere with the results by having different electrical properties.

Hi, rakesh semwal!

It looks like the discharge intended by the challenge was indeed due to the natural breakdown between the plates.

So the combined accumulated plate voltage had to be greater in order to overcome the greater dielectric resistance of distance after all.

Mark

No sign that I can see that the challenger meant natural breakdown between the plates, nor that he was aware that it would be impractical to charge such a small capacitor with one Coulomb.

From the equations, it would appear that the challenger supposed that capacitance varies inversely with plate separation even when the plates are this far apart. Also, that even with this simplification hshe didn't care to manipulate the equations into a straightforward format*.

Yet another "challenge answer" that my high-school class would have laughed out of court, I'm afraid.

*If we accept the oversimplification

We can clearly write
E = Q².d/(κ.ε₀.A) - or
E is proportional to d

P.S. If you tried to do this, what would actually happen is that the plates would discharge to an stable value at the 10-cm separation. The initial method of discharge would depend on how fast you tried to apply the 1-Coulomb, but the final state would be set by the point at which corona discharge ceases rather than by arcing. Then, as you pull the plates apart, the peak field would reduce, even as the total Voltage increased - so the direct discharge between the plates becomes ever more unlikely.
Conversely, if you charge the plates to the limit at 25-cm separation, reducing the separation to 10-cm might indeed be sufficient to cause sparking directly between the plates.

If You have a discharge between the plates, don't You have a failed Capacitor?

Light Green is a dielectric gel.

______________________________________________________________

I mistakenly said "Electrolyte" in an earlier post...

"Dielectric" is used in a Capacitor, "Electrolyte" is used in Batteries.

You said:

"Put two charged (one positive and the other negative)conducting plates facing each other at a distance of 10 cm. Assume the charge in each plate is 1 Coulomb and each plate has dimensions of 5 cm by 5 cm. If you discharge this capacitor certain amount of energy is released (the spark)."

Now you have a discharged cap.

Then you have us step back in time and add 15 cm to the 10 cm space between the plates of the 1 Coulomb charged cap making its separation 25 cm.

"If, before discharging the plates, you pull them apart to reach 15 more centimeters, the energy released when discharged will be smaller or bigger than in the first case?"

The capacitance has changed to a smaller value due to the added 15 cm of space between the plates and it would take more voltage to bring the charge up to 1 Coulomb as the charge is a function of the separation (no dielectric), area of the plates and potential applied.

The added 15 cm reduced the attractive force between the plates and would therefore reduce the energy of a discharge.

C = 0.0885 KS(N-1)/d.

C = capacitance in pF.

K = dielectric constant (air is 1).

S = area of one plate in square cm.

N = number of plates

d = thicknes of dielectric in cm (space beteen the plates for air)

It will be smaller with increased separation of the plates. Capacitance increases with an increase of charged plate area and decreases with increased separation of the charged plates. If you want the formula, look it up.

Less capacitance, same charge => more Volts => more Energy

In terms of formulae:
V=Q/C Energy=QV/2=Q²/C

This is not reasonable. 5-cm square plates charged with 1-Coulomb equates to an electric field greater than 2x10¹³ Volts/metre (that's dividing the charge by the sum of the areas on the two sides of the plate. (Another way of looking at it is that it represents about 1 free electron for each surface atom)
Field emission (discharge in vacuum) would start at about 10⁹ Volts/metre. (If it's in air, breakdown would occur around 3x10⁶V/m)

So I feel empowered to create a more reasonable version of the question - and one where we can present approximate (linear) equations so that we can show the equivalence of the mechanical work and the stored electrical energy.

So, I propose that we increase the dimensions of the plates to 1-metre square, and reduce the charge to 1-μCoulomb. Then I'd like to allow the equations to be linearised, so we should change the separations to 5-cm initially, and make the final separation 10-cm.
Any takers?

Hi, Physicist? !

Why resort to reality? I think the question just wants to know the effect of the greater distance on the discharge. No other change was made.

You're confusing things with the facts!

Mark

Agreed. But I was hoping you would appreciate that the non-physical bit was largely an excuse to justify reposing the question so that people could present linear (=simple) equations that show in detail how (that?) the energy balance works.

However, one of the facts was somewhat understated - I checked the numbers against graphite (chosen because graphite surface atoms are about closely-spaced as any), and it corresponds to about 40 electronic charges for each surface atom (at the facing centres of each of the plates - not the densest points).

The increase in distance between the plates will create more "spark". Just like when you arc weld, the more space you leave between your stick and the metal the more energy is needed to "complete the circuit" and keep an arc.

in other words, the spark will continue to increase as the plates are pulled apart until the amount of stored energy will not allow.

I think it's being assumed by some that the discharge would be between the plates. While this may be true with 1 Coulomb on the plates - I think it would be more helpful to think of this kind of set-up:

Or even the short circuit? I mean no spark?

You would need to make the contact spacing reduce from a separation large enough to prevent sparking to contact in less time than it takes for a spark to grow.

With these dimensions we should expect this to be limited by a combination of series inductance and ionisation effects, but it should not take more than about 10-ns once the spacing is half of the breakdown spacing or less. If we drop back from the impossible charges specified for the original challenge to a field in the capacitor that is (say) 1/10th of the breakdown field, you would be looking at a minimum of 0.5-mm/ns (that is 300 miles/second in old money).

P.S. I think JohnDG has captured the intention exactly.

Hi, JohnDG!

Whoah-Ho!!! Now there's a question worthy of slightly deeper thought. Thanks. Haven't thought about it yet, and may not, depending on available time but thanks for a great response; and you can be sure I'll review previous and following postings from this point of view. Maybe this is what the OP had in mind.

Mark

My guess is that these plates are being held apart by the indestructible box with the atomic bomb inside. I get the attractive force between the plates to be something like 22.5 tera Newtons or 5x10^12 pounds. On second thought, I'm not sure how the electrical fields will behave around the indestructible box so these calculations are probably in error. If energy can not escape the box, I'm guessing the electric fields won't pass throught the box either.

REgards

#2,

will Energy remain same ? as Q=CV

by removing plates apart C will decrease by the Square of distance between plates

ie if D is doubled the C will be C/4

#7 & #10

The analogy is not correct:

For lifting the ball needs more energy feed which will add potential energy to ball.

Here in capacitor you are not adding potential energy. It needs to supply more Electrical-charge to plates which is not done.

By moving plates apart is decreasing C by the square of dis

I don't agree to their concept.

SEe:

http://s.maxthon.com/?q=E%3DCV*V

No GA

"Here in capacitor you are not adding potential energy ..."

Haajee:

1. Do you agree that these two plates, carrying equal and opposite charges, have an attractive force between them?

2. Keeping all other conditions the same (i.e. retaining the charge on the plates, etc.), do you acknowledge that energy has to be expended to move them further apart?

3. If you've got this far, what has been done, if not adding potential energy to the system?

Mechanical energy used to separate capacitor plates does not add electrical potential.

The charge on the plates is changed by connecting a voltage source to them.

A charged capacitor only has potential energy except when it is being discharged.

Moving the plates is not like turning a generator.

I did not suggest that the electric potential difference (i.e. voltage), or the charge on the plates, was changed.

The potential energy of the system, however, is increased. If not, where has that energy gone?

Are you being too easily diverted? (If the charge is constant, and the capacitance reduces - surely the Voltage must increase?)

I wasn't saying that the voltage doesn't increase, just pointing out that I didn't say it does!

_{Maybe didn't express myself very well.}

If the plates are isolated, pulling them apart does indeed increase the electrical potential. You can come at this in at least two ways:

1) Electrostatics:
. First consider a special case (not this one), in which the plates are close together as compared with their dimensions: as the plates are not electrically connected, the charge remains constant. Because the plates are close together, nearly all the field due to the charge is uniformly spread between the two plates - so the field does not change. As the field is uniform, the potential between the plates can be equated to field x separation.
. That demonstrates that there is a situation where you can easily show numerically exactly how much the Voltage increases as you separate the plates. The numbers are not as simple when the plates are so far apart that the field is non-uniform, but you can in principle integrate along the field lines - and the total will obviously increase as you increase the separation.

2) Conservation of energy:
. When you pull the plates apart, you are doing work against the attractive field. Provided that you do not discharge the plates, that energy can be recovered mechanically by bringing the plates back together again. If you discharge the plates, the energy is no longer available - so you can clearly dissipate the mechanically derived energy in an electrical circuit. So we have a situation where the total charge does not change, but the electrically-available energy has changed. As electrical energy is the integral of charge and Voltage, the only way that can happen is that the Voltage during discharge has increased.

You say that "moving the plates is not like turning a generator". That is true only in the sense that this particular configuration cannot generate continuously increasing charge - that is because this isolated situation is the electrostatic equivalent of pulling a straight wire once across a magnetic field (the posts you were responding to were not claiming continuous generation). However, there are two well-known generators where the effect of plate separation is used to generate electrical energy continuously. The Wimshurst machine is an electrostatic generator that is in all respects analogous to a DC electromagnetic dynamo - it generates charge through electrostatic attraction, and then increases the potential by increasing the separation between the charges, which in turn provides the potential to drive the electrostatic attraction. The other is the Van der Graff generator, in which the belt acts in the same way as separating the capacitors does here.

Confusion about increasing potential by increasing the distance between capacitor plates:

Van de Graaff and Wimshurst machines.

The plates of the capacitor in the example have not been in contact and then drawn away and there is no dielectric and so the Van de Graaff and Wimshurst machines are not relevent to this discussion. Because:

A battery supplied the initial charge and there has been no subsequent triboelectric exchange since then and therefore increasing the distance of the plates does not add more electrons to the negative plate. The source of energy could have been generated by one of these machines though.

And: We can assume the plates are both of the same metalic property and having no dielectric material between them to produce a triboeletric effect when brought into contact and then separated.

Van de Graaff and Wimshurst machines do not generate a charge so much by electrostatic attraction as by stealing electrons via contact and separation (removed from contact rather than increased separation) between disimilar materials, one of them is a metalic brush in the case of the Van de Graaff and Wimshurst machines.

The relative position of two substances in the triboelectric series tells you how they will act when brought into contact. Glass rubbed by silk causes a charge separation because they are several positions apart in the table. The same applies for amber and wool. When two non-conducting materials come into contact with each other, a chemical bond, known as adhesion, is formed between the two materials. Depending on the triboelectric properties of the materials, one material may "capture" some of the electrons from the other material. If the two materials are now separated from each other, a charge imbalance will occur. The material that captured the electron is now negatively charged and the material that lost an electron is now positively charged. This charge imbalance is where "static electricity" comes from.

These machines are used to apply a charge to the Layden Jar capacitors until an arc is produced or your hair stands on end.

".....there is no dielectric between the plates....."

Of course there is..... its the space between them, whether its air or a vacuum it is the dielectric!!! It wouldn't be a capacitor capable of holding a charge if there was no dielectric!!

Thanks,

I should have said no significant dielectric since air is a 1 in the factor.

But whether the number is 1 or 1000 makes no difference to the argument

What you mean is, it's the bleedin' air !

If you want to make a case that the Voltage does not increase, you need to address both the issues of conservation of energy and the field x distance (in air dielectric).

In the meantime, I'll address your misinterpretation of the Wimshurst machine:
Once running, the brushes in a Wimshurst machine serve to transfer induced charge. This does not rely on a triboelectric effect - except at start-up (which is equivalent to a dynamo having a small magnetic remanence to get things started). Once running, the function of the brush is the electrostatic equivalent of the brushes in a dynamo.
Small Van der Graph generators do indeed use tribo-electricity to charge the belt - rather like some DC generators use permanent magnets to induce the Voltage. However, some large Van der Graph generators have used a combination of subsidiary induction machines and point discharge to charge the belt - though these days it is generally more convenient to use mains, rectifiers, and corotrons.

The energy has to be the same, only the form can change. The voltage amplitude will be greater in the latter case.

With a greater space between the plates you are able to charge them at a higher voltage. But in this case the voltage was not changed so there is no increase.

The electrostatic flux density was decreased when the space between the plates was increased.

The closer the plates the more energetic the influence they have during charge and discharge. The more distant the less influence they exert.

If you want a higher voltage rating at the same capacitance you have to increase the area of the plates to compensate for the effect of separation.

The plate area and the spacing determine the capacity.

Keeping the applied voltage and distance the same and changing the area of the capacitor allows it to charge and discharge at a rate corresponding to the flux density between the plates.

The same affect is achieved by varying the distance between the plates.

Both methods are used in selecting a station frequency on a radio.

Capacitors are more easily understood in ac applications.

High voltage tuning capacitor.

The screw adjusts the spacing of these Arco trimer caps.

We can apply similar ideas in the construction and use of a coil by applying and removing a current and changing the number of turns per inch, the gage of the wire, the spacing and the diameter of the coil.

Hi K9,

You stated that "With a greater space between the plates you are able to charge them at a higher voltage. But in this case the voltage was not changed so there is no increase". I think that you are assuming that the voltage source remains connected to the capacitors as they are separated. The question implies (I think) that the voltage source is disconnected before the plates are separated in which case there will be a voltage increase. If the voltage source was not disconnected charge would flow out of the plates and into the voltage source as they were separated so that the voltage would remain the same, however, the total charge would be less. In this case you would also be putting energy into the voltage source (charging it if it was a battery) and this is why the energy used in separating the plates would not result in a higher energy discharge of the capacitor.

in this case the value of Capacity is e *S/ d where e is a value that depends of materials between the plates (air in this case), s is the common area and d is the distance between the plates. if d is increased the c value decrease so, the energy discharged is smaller than in the first case - the case when the plates were closer.

Right!

You are assuming the plates are connected across an applied Voltage throughout - in which case the Voltage is constant and the charge on the capacitor reduces.
However, the challenge clearly specifies a value for the charge - implying that the plates are electrically disconnected while you separate them.
Of course, if you have a source of potential connected permanently as you imply, the charge on the capacitor could become almost irrelevant - you will see whatever current the fixed Voltage can provide.

Hi, ion mihaiu!

OK, except that to discharge across the same dielectric at a greater distance, a larger charge is required. So the energy discharged would have to initiate from a greater potential or it couldn't discharge at all.

According to what you've just posted, does that leave us with an equal discharge (greater potential but weakened as it's dissipated over a larger distance) or a greater one (greater potential, larger spark, more energy dissipated)?

Just for kicks, I'm going with suggestion equal:

^-v₁/d₁ = ^-V₂/D₂

Mark

You don't need more charge to cross a larger dielectric - you only need increased potential.

If you did not intend what I thought I read, I apologise. (But I'd still be grateful if you could humour me and other pedantic idiots by using the technical terms in their conventional sense in the future?)

Having criticised the expression, I come to the mathematics - capable of being far clearer.
If the separation was small compared with the dimensions of the plates, I would agree your equations (constant charge => constant field => V1/D1 = field = V2/D2). Unfortunately, the poser of this challenge chose a nasty awkward intermediate separation that allows neither the constant field approximation of closely-spaced plates nor the inverse-square law of widely separated ones. & GRRR....

Discharging is done by putting a conductor (resistor and oscilloscope) accross the plates.

How else would we be able to determine the difference?

How about sticking with at least some of the intent of the challenge and measuring the distance between the discharge terminals? (Or listen to the loudness of the spark, or...)

Spark gap measuring is also good.

Where does the challenge specify the method for determining the energy of the discharge?

it is an equal discharge, but the spark is weaker.

The discharge would be less energetic because of the decreased capacitive reactance due to the increased distance and capacitance.

Even though most people agree that work needs to be done in separating the plates many still do not see that this extra energy translates to a higher discharge energy. I will try to use this example to illiustrate that the extra energy used in separating the plates results in more energy in the discharge. Assume that the two charged plates are in a zero gravity environment and are held apart at 10cm when they are charged at which time the voltage source is disconnected so that the two plates are effectively isolated. If the restraints holding the plates apart are removed the two plates will be attracted to each other and will develop some kinetic energy before they collide. Now if they are first discharged and then released they will not aquire any kinetic energy (because they no longer attract each other) and so the potential kinetic energy must have been converted into the discharge energy. Now take the same two charged plates and move them a further 15cm apart. If they are released they will move towards each other. When they are separated by 10cm they wil already have gained some kinetic energy and so their total kinetic energy when they collide will be greater than when they were released at 10cm separation. Once again if they are discharged first they do not move towards each other so this greater potential kinetic energy is once again converted into the discharge energy. Therefore the discharge energy is higher when the plates are further apart then when they are closer.

How are you able to correlate a bowling ball on an elevator or stretching a rubberband with charged metal plates changing proximity?

Kinetic energy does not translate to electrostatic charge or the uncharged plates could be charged by moving them toward each other.

You would see sparks between vehicles passing each other on the street or parking next to each other if this were true.

I like the zero gravity and electrostatic attraction idea. Like an electrostatic solenoid.

Hi K9,

The correlation is that in all of the cases energy is being added to the system. Clearly my example did not provide enough clarity. Try this reasoning. If the voltage source is removed after the plates are charged at 10cm separation then the charge on the plates cannot change until they are discharged. As you correctly pointed out when their separation is increased the capacitance is reduced which means that the only way for the plates to hold their charge is if the voltage between them increases. If the voltage did not increase then some of the charge would have to leave the plates which it can't until they are discharged. Higher voltage with the same charge means more energy in the discharge.

The same imbalance is still in the plates and if you move them all around the Mulberry bush without any electrons escaping and bring them back to thier original proximity there will still be the same charge?

Charge remains the same but voltage increases as the spacing between the plates increases so total energy in discharge is also greater when they are further apart.

Sooo... if moving them apart causes more voltage, then moving them closer would produce less voltage?

So if the there was 1 Coulomb of charge on the plates and before discharging the plates, you move them from 10 cm to 1 centimeter, the energy released when discharged will be smaller or bigger than in the first case?

Explain.

My take (hopefully, BobD will express it more clearly when he returns)

Electrically: same charge, closer spacing => larger capacitance => lower Voltage = less energy.

Mechanically: the plates are attracted together. If they are restrained as they approach each other they are doing work on the external system. Because the capacitor is electrically isolated while this is happening, this mechanical energy had to come from the capacitor - and the only source of potential energy available was the field/potential in/of the capacitor.

Hi, Physicist?

"Electrically: same charge, closer spacing => larger capacitance => lower Voltage = less energy...the only source of potential energy available was the field/potential in/of the capacitor."

Since the size of the plates didn't change, how is that conclusion different from evaluating the discharge using:

^-v₁/d₁ = ^-V₂/D₂ ?

Mark

The directions of the trends are the same. But (as BobD has clearly demonstrated) the combination of expression in different ways and repetition in context can sometimes be helpful.

And, of course, your equations assume constant field - and that is only accurate when the separation of the plates is small compared with their dimensions. Unfortunately, the setter of the challenge chose a separation which varies from twice to five-times the linear dimensions of the plates. The proportional change in potential will be closer to a factor of 1.6 than the factor of 2.5 of that the linearised equations would suggest.

Hi, Physicist?

OK. I'm not sure I follow...

"The proportional change in potential will be closer to a factor of 1.6 than the factor of 2.5 of that the linearised equations would suggest."

The equation was an attempt to bypass the details of the charge or any other form of energy accumulation on the plates to deal only with the strength of the discharge. The

^-V (negative V)

value is intended to be a measurement of the discharge voltage spark. It's compared only to the distance between the plates, and was intended to be independent of loading since the challenge specified a charge of 1 Coulomb at both distances.

My thinking was as follows.

The challenge asks: "energy released when discharged will be smaller or bigger than in the first case"

Using the equation only depends upon that same amount of energy expended regardless of the greater distance, provided it is not too great to permit a spark. The actual strength of the spark "energy released" would be inversely proportional to the distance, but the energy required to propel it that far through the dielectric had not been changed, resulting in a smaller, weaker spark ^-V₂ for farther distances and a larger stronger spark ^-V₂ for shorter distances compared to the datum spark ^-V₁ at the datum distance.

Mark

Hi Mark

For any of this to be valid, the plates would have to be isolated until you deliberately discharge them. So the length of the spark is appreciably less than the spacing between the plates. This is well illustrated by JohnDG in post #38, where the spark-balls are closer together than the plates.

So, the length of the spark depends on the Voltage between the plates, not the other way around. Expressing that in a more directly relevant way: in calculating the potential on the plates prior to discharge you can ignore breakdown effects.

Clearly, the Voltage between the plates depends on the average field between them. When the plates are close together, the displacement field confined between the plates - and so is equal to Charge/Area. When they are further apart the field can spread out. If the plates are a great distance apart, the field from each plate will spread out in all directions, which means it comes from both sides of the plate - simplistically, it would appear that the effective area is doubled; the upshot is that the component of the field perpendicular to the plate will be reduced by more than a factor of two (as compared to close-spaced plates). Now let us trace the field starting at one of the plates: as you go further from the plates, its field has more space to spread - so the field midway between the plates will be less than the field at either plate. That means that, as the plates are pulled further apart, the average field between them reduces. In fact, if you perform an experiment, it will become apparent (once the separation exceeds a few plate diameters) that the field at a distance reduces so rapidly that the Voltage between the plates is tending towards a finite limit; in fact, once you are in this region**, doubling the separation will simply halve the difference between the measured Voltage and that limit.
**That assumes of course that the fields are not disturbed by the presence of other charged or conductive objects

Regards

Fyz

The mechanical energy to move the plates is me moving one plate toward the other. Doing so does not change anything but the distance.

So if I pull on a rope and you allow me to move in the direction I am pulling, I am not doing any work? What then if I pull a cart up a hill instead of your allowing me to move away from you?

Kinetic energy and electrostatic energy are not related.

If it were you would see sparks as people and objects move around with out touching.

Applying a charge to plates and moving them is not the same as tribolelectric generation of sparks either.

So far as I can judge, the first two paragraphs are simply incorrect; the final one has a meaningful interpretation, but not one that is relevant to this thread.

One of the fundamentals of modern* physics (that underpins all its technological consequences) is equivalence of one sort or another. In that respect, all forms of energy are related; most (not all) can be readily converted from one to another in a manner that is in principle lossless.
These theoretically inter-convertible forms include gravitational potential, kinetic energy, and electromagnetic energy. Conversion between the first pair is commonplace - dropping a ball converts gravitational energy to kinetic energy. Electric generators and motors convert between mechanical and electromagnetic energy - and electromagnetic energy can be stored as electrostatic energy on a capacitor, in a chemical form in a battery, or as the magnetic field in a coil.

Obviously, the methods and characteristics of triboelectric generation differ from other methods. But you can use tribo-electricity to charge a capacitor, and the charge is then indistinguishable from charge generated in any other way.

*by which I mean the physics that has developed into currently accepted physics, and that dates at least as far back as Newton

Does that mean moving the charged plates back and forth from their original proximity changes something besides the capacitance and capacitive reactance?

I am sure the whole idea of the question has to do with the original charge and what happens when you change a characteristic of the construction of the device.

And are you implying that I can charge my capacitor by dropping a bowling ball on it?

The squirrel running in the wheel connected to an alternator and inverter connected to my PC works just fine. So yes, a running squirrel can charge my capacitors. Nothing complicated about that.

I see a lot of talk about moving the plates causing voltage changes. If I connect those plates to an infinite impedence voltmeter that charges the cap and shuffle the plates and the meter deflects or not due to a change in voltage that would be a good test?

"And are you implying that I can charge my capacitor by dropping a bowling ball on it?"

You could probably charge it up quite nicely by dropping the ball onto a piezo-electric generator connected to the plates.

Nope, there are those in this thread who seem to think so.

Piezoelectric stuff is pretty neat!

I was in one of our branches that did undersea sensors and transducers.

I was looking at a tall stack and was warned about the extreme voltages generated by us walking by and trucks rolling by out on the street.

Lots of eyes and ears stuff for civil and military applications.

You put a few of those on the sea bottom and use them to provide stable positioning of floating oil platforms and ships. Or even make sparks to light your barbeque.