# Can an open circuit voltage field charge a capacitor?

pzlded
What happens when you sandwich an uncharged capacitor between two charged capacitors?

1. Just as a piece of Styrofoam can have a static electric charge, a capacitor's dielectric material can be charged, without the requirement of capacitor plates.

2. Given 3 equal pieces of dielectric material, with two of the pieces fully charged with static electricity. If the uncharged dielectric material is sandwiched between the two charged pieces, without creating a closed circuit. Will the uncharged dielectric material gain a static electric charge?

3. Does the charged versus uncharged state of the center piece of dielectric material depend on the polarity of the charge of its neighboring charged pieces of dielectric material?

4. If the center piece of dielectric material indeed becomes charged, will shorting this capacitor discharge all three pieces of dielectric material?

5. Does the charging of the center piece of dielectric material depend on its distance from neighboring charged capacitors? In other words, does charging the center capacitor within an open circuit require exposure of the electrostatic field of its neighboring charged dielectric material.

I claim that w/o a closed circuit, the center piece of dielectric material will not become charged. Can you prove me wrong?

Homework Helper
1. Just as a piece of Styrofoam can have a static electric charge, a capacitor's dielectric material can be charged, without the requirement of capacitor plates.
The dielectric material is strictly insulative. It would not be charged so long as dielectric breakdown does not occur. However the presence of an electric field would polarise the dielectric medium.

2. Given 3 equal pieces of dielectric material, with two of the pieces fully charged with static electricity. If the uncharged dielectric material is sandwiched between the two charged pieces, without creating a closed circuit. Will the uncharged dielectric material gain a static electric charge?
If they are all strictly insulative materials, then the centre item would not be charged. It would, however be "stuck" due to electrostatic attraction with the other 2 materials.

3. Does the charged versus uncharged state of the center piece of dielectric material depend on the polarity of the charge of its neighboring charged pieces of dielectric material?
I don't get this one. If it's uncharged it'll stay uncharged until some of the electrons from its neighbouring insulators slowly "leak" to/from the item. If it's charged, it wil stay as charged until again leakage occurs.

4. If the center piece of dielectric material indeed becomes charged, will shorting this capacitor discharge all three pieces of dielectric material?
Charges can't flow freely because they are all insulative materials. How do you propose to "short" them?

5. Does the charging of the center piece of dielectric material depend on its distance from neighboring charged capacitors? In other words, does charging the center capacitor within an open circuit require exposure of the electrostatic field of its neighboring charged dielectric material.
Again, how do you charge a dielectric? The dielectric for a capacitor is just to ensure the charges on each plate "stay" where they are, thereby creating a potential difference via electrostatic attraction. And in a capacitor, it is the conductive plates which are charged, not the dielectric.

Mentor
I claim that w/o a closed circuit, the center piece of dielectric material will not become charged. Can you prove me wrong?
Assuming the voltage applied is lower than the dielectric breakdown voltage then it won't become charged, but it will become polarized. Do you understand the difference between "charged" and "polarized" for non-conductive materials?

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pzlded
Originally Posted by Defennder
The dielectric material is strictly insulative. It would not be charged so long as dielectric breakdown does not occur. However the presence of an electric field would polarise the dielectric medium.

Perhaps the following URL provides an overview of methods to charge insulative materials.
http://www.experiencefestival.com/a/Electrostatic_generator_-_Description/id/1357749 [Broken]

Perhaps I should provide some detail on the necessity of plates in the Wimshurst generator example. The following is Wimshurst’s original publication:
The following is a quote from the Wimshurst’s publication:

“Another feature is that the neutralizing current may also be broken without reducing the excitement, but then the charges alternate from positive to negative with each half revolution of the disks.”

That means that without neutralizing bars, sparks occur each half revolution of the disks, regardless of the number of plates on each disk.

The following URL mentions some causes of triboelectric effects.
http://archive.evaluationengineering.com/archive/articles/1100cope.htm [Broken]

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Homework Helper
I don't see how that is relevant. Your original setup consisted of a dielectric material sandwich in between a capacitor. There was no mention of friction. I don't doubt that you can charge an insulator with static electricity; examples of this phenomenon are common in everyday lives as well as introductory physics courses. I answered your questions based on the setup of a dielectric material in a capacitor or sandwiched between 2 charged insulating materials.

Secondly you did not mention in your description of the setup that a conductor would be brought in contact briefly with the dielectric material sandwiched in between the 2 charged insulators. This step is necessary in order for the sandwiched dielectric to gain a net charge by electrostatic induction. I interpret the term "charged" to mean a "net charge", not mere polarisation itself, which is why I replied as above. I did mention that the dielectric material would become polarised in the capacitor setup, but not that it would be charged.

As DaleSpam said, you have to distinguish between "polarised" and "charged" for insulative materials.

pzlded
There is a difference between charged and polarized. Charged material contains voltage's energy. Polarized material contains polar molecules.

Some texts claim that dielectric material becomes polarized when exposed to an electromagnetic field.

Consider that when a vacuum capacitor becomes charged, the vacuum contains a voltage gradient, but the vacuum is not polarized. We know that the vacuum within vacuum tube diodes and triodes contained a voltage gradient, because the voltage gradient accelerated thermionic electrons. We know that the tube's vacuum was not polarized because the vacuum itself contained no atoms or molecules that could be polarized.

Anything (including conductors, resistors and capacitors) that contains a voltage gradient contains voltage’s energy. If polarization occurs w/o a voltage gradient, then that polarized material is w/o a voltage’s energy. The dielectric material within a charged capacitor indeed contains a voltage gradient.

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Homework Helper
Where did I say a vacuum can be polarised? I was referring to dielectric materials,not vacuum. Furthermore, where did I deny that the insulating dielectric cannot contain a potential gradient? The textbooks you cite are referring to dielectric insulative materials, and clearly not vacuum in particular. It is apparent that in the case of a vacuum, there is nothing to be polarised, so what point are you trying to make here?

I still don't see how this connects with your original questions.

pzlded
Where did I say a vacuum can be polarised? I was referring to dielectric materials,not vacuum. Furthermore, where did I deny that the insulating dielectric cannot contain a potential gradient? The textbooks you cite are referring to dielectric insulative materials, and clearly not vacuum in particular. It is apparent that in the case of a vacuum, there is nothing to be polarised, so what point are you trying to make here?

I still don't see how this connects with your original questions.

Perhaps Defennder should re-read his 1st response:
The dielectric material is strictly insulative. It would not be charged so long as dielectric breakdown does not occur. However the presence of an electric field would polarise the dielectric medium.

In my responses to Defennder’s 1st response, I (tried to) put forth the idea that:
1. Dielectric material can be charged w/o the requirement of dielectric breakdown, albeit one of the charging methods is the triboelectric effect.

2. The charge within dielectric material is not because of polarization of dielectric material, because a vacuum is dielectric material. A vacuum is dielectric material, because a vacuum is a poor conductor of electricity and a vacuum can store voltage’s energy.

Perhaps dielectric material simply contains voltage's energy, instead being a converter of voltage's energy to polarization energy.

Mentor
There is a difference between charged and polarized. Charged material contains voltage's energy. Polarized material contains polar molecules.
Almost, a charged material has a different number of positive and negative charges, there is a net charge. A polarized material has an uneven distribution of charges, but no net charge, e.g. a surface layer of polar molecules with all of the + ends pointed out and the - ends pointed in.

pzlded
Almost, a charged material has a different number of positive and negative charges, there is a net charge. A polarized material has an uneven distribution of charges, but no net charge, e.g. a surface layer of polar molecules with all of the + ends pointed out and the - ends pointed in.

Thank you DaleSpam for identifying which characteristics differ in a charged versus polar molecule. I really did not know that charged material has a different number of positive versus negative charges. I am having some trouble verifying your net charge hypothesis. Perhaps you could either provide some more detail, or provide references that support your hypothesis.

Your incite into the nature of polarized material is fantastic. Many people think that most polarized molecules are bipolar, thus result from displacement of a single charge. Do I understand you correctly; the surface layer of a polar molecule has multiple + ends pointed out and multiple – ends pointed in? I did even not know that the + ends of a polar molecule point out, or that the – ends point in.

Homework Helper
Perhaps Defennder should re-read his 1st response:
Perhaps pzlded should re-read his original post:

pzlded said:
What happens when you sandwich an uncharged capacitor between two charged capacitors?
1. Just as a piece of Styrofoam can have a static electric charge, a capacitor's dielectric material can be charged, without the requirement of capacitor plates.
My response which you convenient quoted completely out of context was written with the bolded part of your original post in mind. You provided the setup, neglected to mention friction or any way in which it could be used to charge the insulator, nor provided means by which charging by electrostatic induction can be used and then you baselessly accused me of claiming that insulators cannot be charged at all.

pzlded said:
In my responses to Defennder’s 1st response, I (tried to) put forth the idea that:
1. Dielectric material can be charged w/o the requirement of dielectric breakdown, albeit one of the charging methods is the triboelectric effect.
This is trivial. Nowhere have I denied that insulators may be charged. Again I reiterate that your setup says nothing about friction or partially discharging the polarised material, which can be done easily if it is conductor, but not if they are dielectrics. Quoting out of context is disingenous. I have already said this was very common and self-evident.

pzlded said:
2. The charge within dielectric material is not because of polarization of dielectric material, because a vacuum is dielectric material. A vacuum is dielectric material, because a vacuum is a poor conductor of electricity and a vacuum can store voltage’s energy.

Perhaps dielectric material simply contains voltage's energy, instead being a converter of voltage's energy to polarization energy.
Dear pzlded, can you distinguish the difference between "polarised" and "charged" ? Please look at DaleSpam's last post more carefully for the difference. There is no "charge" within a dielectric material in a capacitor setup. If you want to charge it, it is by means of as you said tribo-electric effect and electrostatic induction. As for your point of vacuum being dielectric material, you are splitting semantical hairs. You want to fault textbooks for leaving out the trival possibility that vacuum, which is literally empty space cannot be polarised, when it is readily apparent.

And it turns out, if I'm not mistaken, that there is actually something known "vacuum polarization" in quantum physics, thought I don't think that is relevant here.

Mentor
Perhaps you could either provide some more detail, or provide references that support your hypothesis.
It isn't really a hypothesis, it is just the definition of polarized and charged. For reference, what I am describing is straight out of my 2nd semester freshman physics textbook: Serway, "Physics for Scientists and Engineers" 3rd edition. Look at Chapter 26 with a special emphasis on section 26.7 and the figures on page 727.

pzlded
Hi DaleSpam,

The force between the charges is said to be the source of voltage’s energy. Near the + plate, there is a net excess of + charges; near the – plate, there is an excess of – charges. Kirchoff’s current law specifies that “current in equals current out”, therefore + charges in equal + charges out. In other words, if the + plate gets more charge than the – plate, the charges got there in violation of Kirchoff’s law.

In the diagram of a polarized molecule, the little +’s and – ‘s represent the field from the + charge at one end of the molecule and the – charge at the other end.

Mentor
In other words, if the + plate gets more charge than the – plate, the charges got there in violation of Kirchoff’s law.
That is OK, Kirchoff's law is not a fundamental law and gets broken all the time. For example, Kirchoff's law doesn't work in antennas either.

You use Kirchoff's law for analyzing small circuits and you usually only apply it at nodes of your circuit diagram.

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pzlded
Hi Defennder,

Perhaps I did misinterpret your response. Let us start over. I would like you (or others) to provide whatever assistance you can in clarifying a problem. I would like to know more about the nature of the field beyond a capacitor’s plates.

A vacuum separates two charged insulators. The charged insulators attract each other.
Does the portion of electrostatic field within the vacuum contain a voltage gradient?

If you claimed that electrostatic field between the insulators has a voltage gradient:
Why am I not been able to detect the voltage gradient with a high impedance voltmeter?

If you claimed that, the electrostatic field within the vacuum has no voltage gradient:
How do point charges produce both a field within an insulator with a voltage gradient, and the attractive field within a vacuum that does not have a voltage gradient?

Mentor
If you claimed that electrostatic field between the insulators has a voltage gradient:
Why am I not been able to detect the voltage gradient with a high impedance voltmeter?
Are you sure about that? I would think you would be able to detect it. Perhaps not with a standard voltmeter, but certainly with a device designed for the purpose.

Homework Helper
If you claimed that electrostatic field between the insulators has a voltage gradient:
Why am I not been able to detect the voltage gradient with a high impedance voltmeter?
You probably have to use a sensitive electrometer to detect it. I'm not familiar with experimental devices, so I won't be able to help you here, sorry.

pzlded
Placing a conductor between two charges does not change the electrostatic force between the charges. Adding capacitor plates to each of the two isolated charged insulators can reduce the voltage gradient between the two insulators to zero, w/o changing the force between the two charged insulators.

Point charges do not produce a force w/o a voltage gradient. Is there a theoretical explanation of how a conductor can reduce a voltage gradient w/o reducing force between two charges?

Mentor
Placing a conductor between two charges does not change the electrostatic force between the charges. Adding capacitor plates to each of the two isolated charged insulators can reduce the voltage gradient between the two insulators to zero, w/o changing the force between the two charged insulators.
Your first sentence is correct, but not the second sentence. The definition of the voltage field is that it is the gradient of the voltage is the e-field, and the definition of the e-field is that it is the force on a test charge. So if the voltage gradient is reduced at the insulator then the force is necessarily reduced also, by definition.

Note that a conductor in the middle will have a zero e-field inside the conductor, but will not change the e-field outside the conductor (for an infinite sheet charge and conductor at least, I am not sure about more complicated geometries).

pzlded
Hi DaleSpam,

Thank you for your thoughtful opinion.

The conductors in an electronic vacuum tube are tube elements that control (not short) the vacuum voltage gradients between elements. There is no current in the conductor inserted between to charged insulators within a vacuum, because there are no point charges in the vacuum. W/O current, the conductor does not convert vacuum volts to thermal energy or magnetism.

Perhaps you could more precisely explain the interaction between the conductor and the volts between the two-isolated (open circuit) charged insulators.

Mentor
The conductors in an electronic vacuum tube are tube elements that control (not short) the vacuum voltage gradients between elements. There is no current in the conductor inserted between to charged insulators within a vacuum, because there are no point charges in the vacuum. W/O current, the conductor does not convert vacuum volts to thermal energy or magnetism.
OK, that is true except for the small transient currents when you first insert the conductor. So what?

Perhaps you could more precisely explain the interaction between the conductor and the volts between the two-isolated (open circuit) charged insulators.
Are you familiar with the Gauss' Law derivation of the E-field between two infinite sheets of opposite charge? If so, then you can follow the same procedure to determine the E-field when you have a conductor between them. You will find that the E-field between the charge sheet and the conductor is unchanged. You can even use Gauss' Law to determine the amount of charge on each face of the conductor.

Mentor
The conductors in an electronic vacuum tube are tube elements that control (not short) the vacuum voltage gradients between elements. There is no current in the conductor inserted between to charged insulators within a vacuum, because there are no point charges in the vacuum. W/O current, the conductor does not convert vacuum volts to thermal energy or magnetism.
OK, that is true except for the small transient currents when you first insert the conductor. So what?

Perhaps you could more precisely explain the interaction between the conductor and the volts between the two-isolated (open circuit) charged insulators.
Are you familiar with the Gauss' Law derivation of the E-field between two infinite sheets of charge? If so, then you can follow the same procedure to determine the E-field when you have a conductor between them. You will find that the E-field between the charge sheet and the conductor is unchanged. You can even use Gauss' Law to determine the amount of charge on each face of the conductor.

pzlded
Isn’t there a problem with voltage gradients bound by conductors?

The amount of energy above a horizontal charge sheet equals the energy below the charge sheet. Placing a conductor on one side of a charge sheet converts (shorts) half the charge sheet’s energy. The short converts voltage's energy to magnetic or thermal energy.

Because of the short, devices that contain a linear voltage gradient bounded by conductors (including capacitors, vacuum tubes and resistors), are missing half their voltage’s energy. In an LC oscillator, when moving electrons from magnetism completely convert to a sheet of charge: the plates short out 50% of the resulting voltage’s energy.

The linear voltage gradient within resistors is another problem. The only configuration of point charges that produce a linear voltage gradient is an infinite plane. A resistor does not normally contain an infinite sheet of charge.

Mentor
Isn’t there a problem with voltage gradients bound by conductors?
The E-field inside a conductor is 0. I don't see why that is a problem.

Placing a conductor on one side of a charge sheet converts (shorts) half the charge sheet’s energy. ... Because of the short, devices that contain a linear voltage gradient bounded by conductors (including capacitors, vacuum tubes and resistors), are missing half their voltage’s energy.
This is incorrect. The E-field is only 0 inside the conductor itself. The E-field outside the conductor it is unaffected. You can derive that using Gauss' law.

The linear voltage gradient within resistors is another problem. The only configuration of point charges that produce a linear voltage gradient is an infinite plane. A resistor does not normally contain an infinite sheet of charge.
How is that a problem? Consider an analogy: an infinite sheet of mass generates a uniform gravity field (the force on a mass is constant), a kinetic friction force is also constant. Is there a problem? Why not?

I don't know what your real question is, but I get the feeling that you are not asking it. Every time I answer a question it results in two more almost unrelated questions. Can you please take some time and think about what your underlying point of confusion is? All of these things that you list as problems are obviously not problems since these principles work and basic electronics depend on them. Therefore the problem is simply one of understanding, so please think carefully about the real question that you have and try to distil it into one or two fundamental points.

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pzlded
Hi DaleSpam

I am trying to reconcile the difference between shorting a capacitor and shorting the voltage between capacitors. I suppose I should first ask about ‘sheets of charge’.

A voltage gradient within a charged peanut shaped piece of Styrofoam does not require conductive plates. Therefore, a voltage gradient can exist within a vacuum w/o conductive plates. Both Styrofoam and a vacuum are insulators that can contain a voltage gradient. How can the conductor-less vacuum contain ANY point charges or sheet of point charges?

Shorting any portion of a voltage gradient reduces only that portion of the voltage gradient to zero. Therefore, shorting a portion of a voltage gradient within a vacuum reduces only that portion of the vacuum voltage gradient to zero. When a portion of a voltage gradient within a vacuum is shorted, there is no release or flow of point charges from the vacuum to the conductor.

In the original problem, the two isolated insulators could have been composed of vacuum dielectric material; the area between the insulators could also have been composed of vacuum dielectric material.

Homework Helper
A voltage gradient within a charged peanut shaped piece of Styrofoam does not require conductive plates. Therefore, a voltage gradient can exist within a vacuum w/o conductive plates. Both Styrofoam and a vacuum are insulators that can contain a voltage gradient. How can the conductor-less vacuum contain ANY point charges or sheet of point charges?
The "voltage gradient" you speak of, it's actually grad(V) where V is the electric potential function defined in 3D space is it not? That is equivalent to the electric field evaluated at the point. I don't know if this is what you are asking, but you seem to be asking how is it possible for an electric field to exist in a dielectric/insulator if there is no electric charge in the dielectric itself. Well, this is analogous to how a gravitational field may exist in a given region of space even if there is no mass within that specified region.

Shorting any portion of a voltage gradient reduces only that portion of the voltage gradient to zero. Therefore, shorting a portion of a voltage gradient within a vacuum reduces only that portion of the vacuum voltage gradient to zero. When a portion of a voltage gradient within a vacuum is shorted, there is no release or flow of point charges from the vacuum to the conductor.
I don't know what you mean by "shorting". In fact I would say that it is best if you define in advance any key words you use such as "voltage gradient", "shorting", and until recently, "charged". Sometimes a lot of misunderstanding can result over misinterpretation of these key terms. Although I can't be sure, the way in which the term "shorting" is used in this context appears to have the same meaning as "grounding". Perhaps you can clarify this point.

In the original problem, the two isolated insulators could have been composed of vacuum dielectric material; the area between the insulators could also have been composed of vacuum dielectric material.
Yes it is possible that you could use dielectric material instead of conductive capacitor plates. However, such a setup would be of limited practical usage, since charges cannot flow freely in insulators, the "capacitor" in a such a setup simply cannot function as one even though roughly speaking, it is similar to the ubiquitous parallel plate capacitor setup we see in first-year physics textbooks.

pzlded
Hi Defennder,

The existence of a voltage field within a vacuum is one of the problems that James Maxwell broached through the use of 'displacement current'. I suppose my 'no point charge' approach to the voltage within a vacuum problem is controversial, but I found this approach convenient for explaining my problems with sheet charges.

The "voltage gradient" you speak of, it's actually grad(V) where V is the electric potential function defined in 3D space is it not? That is equivalent to the electric field evaluated at the point. I don't know if this is what you are asking, but you seem to be asking how is it possible for an electric field to exist in a dielectric/insulator if there is no electric charge in the dielectric itself. Well, this is analogous to how a gravitational field may exist in a given region of space even if there is no mass within that specified region.
Evaluation of a voltage 'gradient' requires two points; a single point does not contain a voltage difference. Voltage is a property of voltage’s energy; voltage is not voltage’s energy. For example, a high capacitance capacitor charged with one volt contains more voltage’s energy than a low capacitance capacitor charged with one volt.

The voltage gradient I speak of is the difference in voltage between two points divided by the distance between them. Gravitational acceleration of mass does not deplete gravity or mass. A point charge traveling through a voltage gradient depletes voltage and voltage’s energy. Voltage’s energy resides in the volume (even if the volume is a vacuum) that contains the voltage gradient.

I don't know what you mean by "shorting". In fact I would say that it is best if you define in advance any key words you use such as "voltage gradient", "shorting", and until recently, "charged". Sometimes a lot of misunderstanding can result over misinterpretation of these key terms. Although I can't be sure, the way in which the term "shorting" is used in this context appears to have the same meaning as "grounding". Perhaps you can clarify this point.

My response to a claim my post is ambiguous because some people might think shorting is grounding, is my claim that there should is no need to define electrical ‘short’ to an audience with a knowledge of physics 101. If you do not understand my post, then explicitly say so. If you disagree with it, then explicitly state your disagreement.

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drxyxy
Thanks to match

Homework Helper
The existence of a voltage field within a vacuum is one of the problems that James Maxwell broached through the use of 'displacement current'. I suppose my 'no point charge' approach to the voltage within a vacuum problem is controversial, but I found this approach convenient for explaining my problems with sheet charges.
What exactly is your "no point charge approach" here? All I have said is that it is not necessary for any region of space to contain a point charge or a net charge in particular in order to for there to be an electric field within it, though by Gauss law, one could see that the flux through that enclosed surface is zero.

Evaluation of a voltage 'gradient' requires two points; a single point does not contain a voltage difference. Voltage is a property of voltage’s energy; voltage is not voltage’s energy. For example, a high capacitance capacitor charged with one volt contains more voltage’s energy than a low capacitance capacitor charged with one volt.
You appear to be stating facts which are not under dispute and irrelevant to the discussion. In fact this appears to be part of a disconcerting pattern of your behaviour in this thread; you would post a few questions using unfamiliar terms, and then when someone points out a misunderstanding you respond by stating a fact which is not disputed and seemingly irrelevant. If they are indeed relevant, you have not endeavoured to make it appear so. How does this help you?

The voltage gradient I speak of is the difference in voltage between two points divided by the distance between them. Gravitational acceleration of mass does not deplete gravity or mass. A point charge traveling through a voltage gradient depletes voltage and voltage’s energy. Voltage’s energy resides in the volume (even if the volume is a vacuum) that contains the voltage gradient.
How is "voltage gradient" as you have defined it different from the definition of the E-field in one dimension? When a point charge moves under the influence of an E-field, the field does work on the point charge. This is similar to the idea of a gravitational field doing work on a point mass when it moves under the influence of a gravitational field. The gravitational potential energy of the point mass decreases when work is done on it by the gravitational field, similarly the electric potential energy of a charged particle decreases when it accelerates under the influence of an electric field. In the case of gravitational field, you are conflating the change of gravitational potential energy of a system with the change in mass of the system. The former does not require the latter.

I believe that the problem you have is that you are thinking of energy storage in an electric field as something which is distinct from electric potential energy, as may be seen from the fact that you repeatedly reference "voltage gradient" instead of the more commonly used and understood "electric field". This simply isn't the case. I quote the following from my freshman physics textbook:

My response to a claim my post is ambiguous because some people might think shorting is grounding, is my claim that there should is no need to define electrical ‘short’ to an audience with a knowledge of physics 101. If you do not understand my post, then explicitly say so. If you disagree with it, then explicitly state your disagreement.
Your post is unclear, and I have asked you to clarify it as before. Your insistence in bringing up irrelevant facts which are true but which do not relate to your situation is both irritating and unhelpful, to say the least. I can neither disagree nor agree with something which is not properly articulated. Perhaps you can start by answering the question I posed as to whether "shorting" as you have used it is actually the same as "grounding". Otherwise we can spend the next few posts fruitlessly debating over a mere misunderstanding of your terminology.

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pzlded
I was hoping to keep this short and simple. Basically this is a disagreement over whether or not a voltage gradient must be nearby point charges. Perhaps the following scenario will settle the disagreement.

A voltage gradient will always accelerate a point charge within the voltage gradient, away from the voltage gradient. I assume you think it is plausible that a single electron and a positron are enough to form a voltage gradient.

A point charge released near a stationary positron will either collide with the positron, or oscillate back and forth across the original distance from the positron. This is similar to gravity causing a comet to either orbit the Sun or crash into the Sun. A point charge released between a stationary electron and a stationary positron will either crash or oscillate a shifted distance from the stationary positron. Neither case provides enough energy for the injected electron to escape.

Mentor
My response to a claim my post is ambiguous because some people might think shorting is grounding, is my claim that there should is no need to define electrical ‘short’ to an audience with a knowledge of physics 101.
The standard term "short" refers specifically to an unintended conduction path between two parts of an electrical circuit. There is no circuit in the context of "shorting the voltage between capacitors". So although you are using a standard term you are using it in a very non-standard context and it is perfectly reasonable for someone to ask you to define it.

Your response to Defennder that it is a standard term that you should not have to define is not helpful, and it is not polite to imply (I hope unintentionally) that someone who is trying to help you lacks even a freshman-level knowledge of physics.

I am trying to reconcile the difference between shorting a capacitor and shorting the voltage between capacitors. I suppose I should first ask about ‘sheets of charge’.
Let me try to define the terms as I understand the question and you can agree or clarify:

Shorting a capacitor is placing a conductor (wire) such that it touches both plates.

Shorting the voltage between capacitors is placing a conductor (plate) such that it does not touch either capacitor plate, but that it is in located in the region between the capacitor plates.

If that is the question then the differences are:
1) there is no charge separation in the first case
2) there is no E-field anywhere in the first case

Both Styrofoam and a vacuum are insulators that can contain a voltage gradient.
Be careful in assigning material properties to vacuum, but yes a voltage gradient is an E-field, which can exist in a vacuum.

Shorting any portion of a voltage gradient reduces only that portion of the voltage gradient to zero. Therefore, shorting a portion of a voltage gradient within a vacuum reduces only that portion of the vacuum voltage gradient to zero. When a portion of a voltage gradient within a vacuum is shorted, there is no release or flow of point charges from the vacuum to the conductor.
Correct.

In the original problem, the two isolated insulators could have been composed of vacuum dielectric material; the area between the insulators could also have been composed of vacuum dielectric material.
Again, be careful in assigning material properties to vacuum, here if you want your charges to "stay put" then you will have to use a charged insulator not vacuum. Otherwise they will simply accelerate towards each other.

Mentor
I was hoping to keep this short and simple. Basically this is a disagreement over whether or not a voltage gradient must be nearby point charges.
It doesn't have to be a point charge (it could be any charge distribution), but yes, there must be some charge in order to get an E-field. I do not think this has ever been a point of disagreement.

A voltage gradient will always accelerate a point charge within the voltage gradient, away from the voltage gradient.
Yes, F = q E. For a negative q you get F in the direction opposite the E-field, and if there are no other forces acting on the charge F = m a.

Again, I feel like we are not getting to your core confusion. I haven't found any gross misunderstandings in your last couple of posts. Does your question have something to do with energy conservation? Or conservation of charge? Or the definition of capacitance? Please try to distill it down to your fundamental question. I think we are "not seeing the forest for the trees".

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pzlded
Hi Defennder

I appreciate the help you have been providing. I wish to apologize for making the following comment.
My response to a claim my post is ambiguous because some people might think shorting is grounding, is my claim that there should is no need to define electrical ‘short’ to an audience with a knowledge of physics 101.
I used ‘shorting’ to mean the short-circuit condition used in network theory to describe a general condition of zero voltage. For example, the term short-circuit admittance (or impedance) is used to describe a network condition in which certain terminals have had their voltage reduced to zero, for the purpose of analysis.

I realize that you were presenting thoughtful responses to my posts. Albeit I sometimes fail, I’ll make an extra effort to keep my posts within this forum professional. Please realize that this is not easy. It is difficult for me to present complex ideas as if they were intuitively simple. In this case, I forgot that both the words used to present the ideas and the ideas are not intuitively simple.

Homework Helper
Hi pzlded,

I guess it's normal for posters to get carried away at times just as I might have responded to your post more harshly than I intended to. It's something we all have to live and deal with I suppose.

In any case, I'm curious about the origin of the questions you're asking. Are they from some textbook you are reading or some lab assignment? I think it'll help a lot if you would be kind enough to tell us where these questions are from, that way it can help provide some perspective and context for us to answer your questions in a more productive manner.

pzlded
Hi Defennder
In any case, I'm curious about the origin of the questions you're asking. Are they from some textbook you are reading or some lab assignment? I think it'll help a lot if you would be kind enough to tell us where these questions are from, that way it can help provide some perspective and context for us to answer your questions in a more productive manner.
I’ve prepared a PowerPoint sideshow that is definitely 'draft' to provide an overview of my radical ideas related to point charge physics. The sideshow can be downloaded from the following website:

http://mysite.verizon.net/vzezz9ms/

Let me know if you have any problems accessing this material. Please be careful about referencing the material in the sideshow: The material within this sideshow is not peer reviewed and many people feel that there is no need to create an alternative to traditional theory.