Basic questions regarding theory of electrical capacitance

[HCB]

TL;DR Summary

I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

Hello, all. First, I want to apologize if this is not the correct forum or area of the forum for this question. Please direct me if I should be posting this somewhere else.

I have some questions regarding what I believe is best described as the "theory" of electrical capacitance. As my questions are about what I believe are the physics involved, and not about electronic components, I felt that a physics forum was the place to ask.

My examples are intentionally simplistic as I wish to understand the fundamentals. This is not about a particular application.

It is my understanding of electrical capacitance that if two pieces of electrically-conductive material are connected to a DC power supply, one to the negative terminal and one to the positive terminal, that it is possible to imply a durable charge to the two pieces (leaving an electrical potential between them), even if the power supply is disconnected from them. For an example: if two pieces of identical volume and shape copper are placed on a non-conducting surface and surrounded by air and then connected to a 12 volt DC power supply, after the power supply is disconnected from them, they should still have a 12 volt potential between them. For this example, we'll say the two pieces are 1/2 inch apart. As 12 volts is insufficient to arc across 1/2 inch of air, and the surface the two pieces are sitting on is non-conductive, theoretically speaking, would the 12 volt potential exist indefinitely? If the two pieces are moved to 1/4 inch distance from one another and again connected to the 12 volt DC power supply, would this increase the effective capacitance between them? I *think* that as the two pieces are moved closer and closer to one another, the capacitance between them will increase so long as the electrons are not able to arc between them.

When connecting two conductive pieces to a 12 volt DC power supply, let's continue with copper as the material, I *think* it is the electrons flowing from the negative terminal of the power supply which saturate the conductive material connected to it. Is that correct? Furthermore, *think* it is the presence of those extra electrons in the negatively-charged material which repel electrons in the conductive material connected to the positive terminal. Is that correct?

If these two pieces are electrically charged by such connection to a DC power supply, one has excess electrons (negative charge) and one has had electrons driven from it by the repelling force of the negatively-charged piece (positive charge). Is that correct?

Now, if these two pieces are electrically charged thusly, does this impart any actual physical attraction between them? This is where things get murky for me the most: I recall from physics classes (and chemistry, I believe, too) from many years ago that "opposites attract", and I recall that some atoms are attracted to other atoms based on their dissimilar charges (if I recall correctly). So, too, molecules I *think*. In this example of two pieces of copper which have been electrically charged, one positive and one negative, will there be any physical force between them caused by their different polarity charges?

Continuing in that vein, if these same two pieces of copper are charged as mentioned above, will they have any effect on charged particles around them? For instance, if these two pieces are charged, will they physically attract or repel ions in the air?

I'm not sure I'm knowledgeable enough to ask coherent questions. My apologies if these are dumb questions.

Thank you for your time.

--HC

 

[Drakkith]

When connecting two conductive pieces to a 12 volt DC power supply, let's continue with copper as the material, I *think* it is the electrons flowing from the negative terminal of the power supply which saturate the conductive material connected to it. Is that correct? Furthermore, *think* it is the presence of those extra electrons in the negatively-charged material which repel electrons in the conductive material connected to the positive terminal. Is that correct?

That's mostly correct. The power supply itself also pulls negative charges off of the positively charged material and into its positive terminal. It's not solely because the negative material repels electrons from the positive material.

Now, if these two pieces are electrically charged thusly, does this impart any actual physical attraction between them?

Absolutely. The two pieces will indeed attract each other when oppositely charged.

Continuing in that vein, if these same two pieces of copper are charged as mentioned above, will they have any effect on charged particles around them? For instance, if these two pieces are charged, will they physically attract or repel ions in the air?

Certainly. This ability to attract oppositely charged ions from the air is one reason air is an imperfect insulator. They will also polarize nearby molecules in the air if those molecules are able to be polarized.

 

[DaveE]

I think your questions are good ones.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

It is my understanding of electrical capacitance that if two pieces of electrically-conductive material are connected to a DC power supply, one to the negative terminal and one to the positive terminal, that it is possible to imply a durable charge to the two pieces (leaving an electrical potential between them), even if the power supply is disconnected from them.

yes.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

would the 12 volt potential exist indefinitely?

yes. Unless other charges are somehow allowed to move on/into the capacitor plates.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

If the two pieces are moved to 1/4 inch distance from one another and again connected to the 12 volt DC power supply, would this increase the effective capacitance between them? I *think* that as the two pieces are moved closer and closer to one another, the capacitance between them will increase so long as the electrons are not able to arc between them.

Yes, if they were farther apart than 1/4" before. However, the capacitance between the plates is a feature of their construction; the geometry. It has nothing to do with whether there is a net charge on the plates or a voltage across the capacitor. In the same way that an airplane is an airplane whether it has passengers on it or not.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

When connecting two conductive pieces to a 12 volt DC power supply, let's continue with copper as the material, I *think* it is the electrons flowing from the negative terminal of the power supply which saturate the conductive material connected to it. Is that correct? Furthermore, *think* it is the presence of those extra electrons in the negatively-charged material which repel electrons in the conductive material connected to the positive terminal. Is that correct?

If these two pieces are electrically charged by such connection to a DC power supply, one has excess electrons (negative charge) and one has had electrons driven from it by the repelling force of the negatively-charged piece (positive charge). Is that correct?

It's 1/2 correct. The battery will cause electrons to be removed from one side and sent to the other side. It is a symmetrical process. Whatever argument you use to explain extra electrons on one side is the same for the depletion of electrons on the other side. Batteries don't make electrons, they move them.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

Now, if these two pieces are electrically charged thusly, does this impart any actual physical attraction between them? This is where things get murky for me the most: I recall from physics classes (and chemistry, I believe, too) from many years ago that "opposites attract", and I recall that some atoms are attracted to other atoms based on their dissimilar charges (if I recall correctly). So, too, molecules I *think*. In this example of two pieces of copper which have been electrically charged, one positive and one negative, will there be any physical force between them caused by their different polarity charges?

yes. there is an electrostatic force pushing them towards each other.

Summary:: I want to better understand the theoretical functioning of electrical capacitance. This is not about discrete electronic components.

Continuing in that vein, if these same two pieces of copper are charged as mentioned above, will they have any effect on charged particles around them? For instance, if these two pieces are charged, will they physically attract or repel ions in the air?

Yes. Free charges will be attracted, or repelled, by the electrostatic force of the net charge on each plate. But not from far away. The attractive force from one plate will be balanced by the repulsive force from the other when the distance is large.

 

[jbriggs444]

It is my understanding of electrical capacitance that if two pieces of electrically-conductive material are connected to a DC power supply, one to the negative terminal and one to the positive terminal, that it is possible to imply a durable charge to the two pieces (leaving an electrical potential between them), even if the power supply is disconnected from them. For an example: if two pieces of identical volume and shape copper are placed on a non-conducting surface and surrounded by air and then connected to a 12 volt DC power supply, after the power supply is disconnected from them, they should still have a 12 volt potential between them.

Yes. Give them a 12 volt potential difference, remove the power supply, do not let any charge leak away and the potential difference will remain.

For this example, we'll say the two pieces are 1/2 inch apart. As 12 volts is insufficient to arc across 1/2 inch of air, and the surface the two pieces are sitting on is non-conductive, theoretically speaking, would the 12 volt potential exist indefinitely? If the two pieces are moved to 1/4 inch distance from one another and again connected to the 12 volt DC power supply, would this increase the effective capacitance between them? I *think* that as the two pieces are moved closer and closer to one another, the capacitance between them will increase so long as the electrons are not able to arc between them.

Yes. The closer the two plates are to one another the more charge they can hold for a given potential difference. That is to say the capacitance increases.

When connecting two conductive pieces to a 12 volt DC power supply, let's continue with copper as the material, I *think* it is the electrons flowing from the negative terminal of the power supply which saturate the conductive material connected to it. Is that correct? Furthermore, *think* it is the presence of those extra electrons in the negatively-charged material which repel electrons in the conductive material connected to the positive terminal. Is that correct?

Saturate? Not for plates at 1/2 or 1/4 inch separation. One coulomb is far less than Avogadro's number of electrons. And a capacitor with that kind of separation is likely going to have a capacitance in the microfarad range or worse. I do not have the relevant constants memorized to do a proper calculation. But the point is that the number of electrons that have to move to establish a 12 volt potential difference is quite miniscule compared to the number of electrons in a neutral plate.

If these two pieces are electrically charged by such connection to a DC power supply, one has excess electrons (negative charge) and one has had electrons driven from it by the repelling force of the negatively-charged piece (positive charge). Is that correct?

Yes. One will have a relative surplus and one will have a relative deficit. Whether this is because they are pushed away by the opposing plate or pulled away by the power supply is not worth arguing about. They move.

Now, if these two pieces are electrically charged thusly, does this impart any actual physical attraction between them?

Yes. The negative plate will attract the positive plate and vice versa.

Consider for a moment what happens if you charge up this capacitor with a 1/2 inch separation. The plates are attracting one another and you have 12 volts of potential difference.

You allow the plates to move together to a separation of 1/4 inch. The capacitance has halved. The charge is unchanged. So the potential difference must now be only 6 volts. Where did the energy go?

It manifested by the work that could be harvested from the attraction of the plates to each other over that 1/4 inch displacement.

The same thing can be done in the opposite direction. Take the 6 volt plates at 1/4 inch separation and force then back apart. The work that is done to force them apart manifests as a return to a 12 volt potential difference. You've converted mechanical energy into electrical potential energy.

Continuing in that vein, if these same two pieces of copper are charged as mentioned above, will they have any effect on charged particles around them? For instance, if these two pieces are charged, will they physically attract or repel ions in the air?

Yes.

Far from the capacitor, the net effect is roughly zero since any attraction from the one plate will be a nearly equal repulsion from the other. Near the capacitor, negative ions will tend to be repelled from the negatively charged plate and attracted to the positively charged plate.

 

[HCB]

That's mostly correct. The power supply itself also pulls negative charges off of the positively charged material and into its positive terminal. It's not solely because the negative material repels electrons from the positive material.

Absolutely. The two pieces will indeed attract each other when oppositely charged.

Certainly. This ability to attract oppositely charged ions from the air is one reason air is an imperfect insulator. They will also polarize nearby molecules in the air if those molecules are able to be polarized.

Drakkith, thank you for the reply and information. I understand what you've stated about the power supply and read that it's basically acting as an electron pump: pull from this side and push to that side.

--HC

 

[phyzguy]

. For this example, we'll say the two pieces are 1/2 inch apart. As 12 volts is insufficient to arc across 1/2 inch of air, and the surface the two pieces are sitting on is non-conductive, theoretically speaking, would the 12 volt potential exist indefinitely?

One thing to add to the other excellent answers. In practice, the charge will leak off fairly quickly. The solid table they are sitting on is not a perfect insulator, so small currents can flow across the surface between the two pieces of copper even if there is no arcing. Also, as you said later, the two pieces will attract ions from the air, and these will act to discharge the potential between the two pieces of metal.

 

[HCB]

I think your questions are good ones.

yes.

yes. Unless other charges are somehow allowed to move on/into the capacitor plates.

Yes, if they were farther apart than 1/4" before. However, the capacitance between the plates is a feature of their construction; the geometry. It has nothing to do with whether there is a net charge on the plates or a voltage across the capacitor. In the same way that an airplane is an airplane whether it has passengers on it or not.

It's 1/2 correct. The battery will cause electrons to be removed from one side and sent to the other side. It is a symmetrical process. Whatever argument you use to explain extra electrons on one side is the same for the depletion of electrons on the other side. Batteries don't make electrons, they move them.

yes. there is an electrostatic force pushing them towards each other.

Yes. Free charges will be attracted, or repelled, by the electrostatic force of the net charge on each plate. But not from far away. The attractive force from one plate will be balanced by the repulsive force from the other when the distance is large.

DaveE, thank you for your reply and information. I sort of understand what you're saying about the capacitance being a function of their geometry. I was confused so I started questioning my understanding of the word "capacitance". I looked that up...my understanding of the term capacitance was incorrect. I was using it as a fixed unit of measure like "gallon" or "mile". It's not a fixed unit of measure, it's a ratio of charge stored to electrical potential. Kind of like trying to determine the applied force of "5 PSI"...we know the P, we need to know the SI. So, the "capacitance" is a ratio and the ratio doesn't change because of electrical potential. But would not the ratio (capacitance) change as the two pieces are moved closer or farther apart?

I very much understand the "net charge at a distance" statement you've made, thank you.

Thank you again.

--HC

 

[HCB]

Yes. Give them a 12 volt potential difference, remove the power supply, do not let any charge leak away and the potential difference will remain.Yes. The closer the two plates are to one another the more charge they can hold for a given potential difference. That is to say the capacitance increases.Saturate? Not for plates at 1/2 or 1/4 inch separation. One coulomb is far less than Avogadro's number of electrons. And a capacitor with that kind of separation is likely going to have a capacitance in the microfarad range or worse. I do not have the relevant constants memorized to do a proper calculation. But the point is that the number of electrons that have to move to establish a 12 volt potential difference is quite miniscule compared to the number of electrons in a neutral plate.Yes. One will have a relative surplus and one will have a relative deficit. Whether this is because they are pushed away by the opposing plate or pulled away by the power supply is not worth arguing about. They move.Yes. The negative plate will attract the positive plate and vice versa.

Consider for a moment what happens if you charge up this capacitor with a 1/2 inch separation. The plates are attracting one another and you have 12 volts of potential difference.

You allow the plates to move together to a separation of 1/4 inch. The capacitance has halved. The charge is unchanged. So the potential difference must now be only 6 volts. Where did the energy go?

It manifested by the work that could be harvested from the attraction of the plates to each other over that 1/4 inch displacement.

The same thing can be done in the opposite direction. Take the 6 volt plates at 1/4 inch separation and force then back apart. The work that is done to force them apart manifests as a return to a 12 volt potential difference. You've converted mechanical energy into electrical potential energy.Yes.

Far from the capacitor, the net effect is roughly zero since any attraction from the one plate will be a nearly equal repulsion from the other. Near the capacitor, negative ions will tend to be repelled from the negatively charged plate and attracted to the positively charged plate.

Jbriggs444, thank you for the reply.

DaveE had made a comment about the capacitance and change due to the distance between the pieces being charged. I thought, and you've stated, that the capacitance (ratio to potential voltage) increases as the pieces get closer to one another.

I used a technical term incorrectly with "saturation/saturate". I would think "saturation" would be that all electrons which could be moved onto or off of a material had been moved. That was not technically correct. My mistake.

That's very interesting regarding the change in physical separation and the impact on the electrical potential.

Thank you again.

--HC

 

[HCB]

One thing to add to the other excellent answers. In practice, the charge will leak off fairly quickly. The solid table they are sitting on is not a perfect insulator, so small currents can flow across the surface between the two pieces of copper even if there is no arcing. Also, as you said later, the two pieces will attract ions from the air, and these will act to discharge the potential between the two pieces of metal.

Phyzguy, thank you for the reply. I understand what you're saying. This is very useful to know as I had envisioned a test setup where I had hoped that the charges might be durable for a while (as in as much as several minutes). It sounds like there will be several variables in play so determining the amount of time for the charges to dissipate will not be easy.

Thank you again.

--HC

 

[phyzguy]

I'm not sure how long the charge will persist. It depends how clean the table is and the temperature and humidity of the air. It might last several minutes. It would be an interesting experiment to try.

 

[DaveE]

I'm not sure how long the charge will persist. It depends how clean the table is and the temperature and humidity of the air. It might last several minutes. It would be an interesting experiment to try.

This depends entirely on the construction. Real (encapsulated) caps can hold their charge for a really long time, at least months. Some memory ICs can hold charge nearly forever, but it's probably unfair to call those capacitors.

 

[phyzguy]

This depends entirely on the construction. Real (encapsulated) caps can hold their charge for a really long time, at least months. Some memory ICs can hold charge nearly forever, but it's probably unfair to call those capacitors.

Agreed, but the OP was talking about two pieces of copper laying on a table.