Charge in RLC circuit when starting charges are 2Q and Q

[phantomvommand]

Hi all,

I understand the standard solution where charge in an RLC circuit decreases from +Q to 0, for a capacitor with starting charges +Q and -Q. May I know what the terminal charge on a capacitor in an RLC circuit is, when it’s original charges are 2Q and 0?

I am thinking it will be +Q and +Q on both plates, resulting in no net flow of charge.

Thank you :)

 

[kuruman]

Yes. Charge is conserved and eventually the voltage on the capacitor will go to zero.

 

[Baluncore]

May I know what the terminal charge on a capacitor in an RLC circuit is, when it’s original charges are 2Q and 0?

The energy is stored in the electric field between the plates. If you instantly disconnect the capacitor from the circuit, how can it have an unbalanced charge?

 

[kuruman]

The energy is stored in the electric field between the plates. If you instantly disconnect the capacitor from the circuit, how can it have an unbalanced charge?

I interpreted the situation as follows. You put charge 2Q by induction (or some other method) on an isolated plate connected to a wire and an open switch. You bring an identical uncharged plate close to the first one. The other end of the second plate is connected to a series RL. You connect the free end of the RL to the free terminal of the open switch. You close the switch and see what happens.

 

[alan123hk]

May I know what the terminal charge on a capacitor in an RLC circuit is, when it’s original charges are 2Q and 0?

Do you mean that the initial charge of the capacitor is +2Q on one plate and 0Q on the other plate?

I don't think we can define the charge of a capacitor in this way. For the charge of the capacitor itself at any time, if +Q on one plate, then the -Q must be on the other plate.

If the amount of charge on the two plates of the capacitor is not equal and the polarity is opposite, the excess charge will not belong to the capacitor.

 

[phantomvommand]

Do you mean that the initial charge of the capacitor is +2Q on one plate and 0Q on the other plate?

Yes.

I don't think we can define the charge of a capacitor in this way. For the charge of the capacitor itself at any time, if +Q on one plate, then the -Q must be on the other plate.

If the amount of charge on the two plates of the capacitor is not equal and the polarity is opposite, the excess charge will not belong to the capacitor.

I was thinking that the charge on the capacitor is found by taking (Q1-Q2)/2, where Q1 and Q2 are charges on the plates. So in the normal case where Q1=+Q and Q2=-Q, charge of capacitor is 2Q/2 = Q, while for this case where one plate has +2Q and the other plate is at 0 charge, the charge would still be Q.

 

[phantomvommand]

The energy is stored in the electric field between the plates. If you instantly disconnect the capacitor from the circuit, how can it have an unbalanced charge?

What do you mean by disconnect the capacitor? I meant that the capacitor starts with 1 plate having charge +2Q, and another with 0 charge. Then let the RLC circuit run for a very long time, and my question is what the final charge on the 2 plates of the capacitors are.

 

[Baluncore]

What do you mean by disconnect the capacitor?

I mean exactly that. Disconnect the capacitor from the R and L.
The capacitance C, is defined as the ratio of charge to voltage. C = Q / V. You can calculate the charge by knowing the capacitance and voltage on your hypothetical capacitor.

You appear to be hypothesising a displacement current that splits somewhere and flows to a third terminal. If that was the case there would need to be one or more additional capacitors, but you have only one so the charges must be equal and opposite.

 

[kuruman]

The capacitance C, is defined as the ratio of charge to voltage. C = Q / V.

The capacitance is strictly a geometric property and depends on the shape and the disposition of the conductors relative to each other. In any case, I proposed a scheme of how to view this in post #4. Is it not appropriate?

 

[Baluncore]

In any case, I proposed a scheme of how to view this in post #4. Is it not appropriate?

You hypothesise two isolated plates being charged relative to an unknown third conductor. You then change the isolation and therefore change the capacitance. Charge is conserved during the change in geometry, so the voltage must change. Then you connect only one R and only one L, to discharge the three capacitors you formed during the construction. What is the schematic for that circuit ?

When you charge a fixed capacitor, the virtual displacement current flows through the capacitor. The charge you deliver to one plate must be removed from the other. The charges on the two plates must therefore always be equal and opposite.

 

[Baluncore]

The capacitance is strictly a geometric property and depends on the shape and the disposition of the conductors relative to each other.

Physically yes. But the "concept" of capacitance is defined as the ratio of charge to voltage.

 

[kuruman]

You hypothesise two isolated plates being charged relative to an unknown third conductor. You then change the isolation and therefore change the capacitance. Charge is conserved during the change in geometry, so the voltage must change. Then you connect only one R and only one L, to discharge the three capacitors you formed during the construction. What is the schematic for that circuit ?

When you charge a fixed capacitor, the virtual displacement current flows through the capacitor. The charge you deliver to one plate must be removed from the other. The charges on the two plates must therefore always be equal and opposite.

It seems I didn't explain myself well. Here is the schematic.

 

[phantomvommand]

You hypothesise two isolated plates being charged relative to an unknown third conductor. You then change the isolation and therefore change the capacitance. Charge is conserved during the change in geometry, so the voltage must change. Then you connect only one R and only one L, to discharge the three capacitors you formed during the construction. What is the schematic for that circuit ?

Series. The +2Q plate is connected to an open switch from what I understand, so no charge is lost. The uncharged plate is connected in series to R, L, and the other end of the open switch. Close the switch.

When you charge a fixed capacitor, the virtual displacement current flows through the capacitor. The charge you deliver to one plate must be removed from the other. The charges on the two plates must therefore always be equal and opposite.

With this, you should get 2 plates with charge +2Q and -2Q. Use a 3rd, uncharged plate, and bring it near the plate with charge 2Q to form the "+2Q, 0" charge capacitor.

Assuming the 2Q-0 charge capacitor can be made and can be placed in a series RLC circuit, what do you think would be the final charges on the 2 plates?

 

[DaveE]

OK, so this is how I understand your question: A series RLC circuit with a (series) switch is at rest with the switch open. Let's say there is no net charge in the circuit an no charge on either side of the capacitor. Every part of the circuit starts with 0 charge. By some magic process (cosmic ray absorption?) some amount of charge is deposited on each plate, perhaps different amounts. Then the switch is closed and you want to know how much charge is on each capacitor plate after the initial transients have died away to (almost) nothing when you again open the switch*; i.e. steady state. Correct?

As others have said, in steady state the final charge on each plate must be equal. Otherwise there would be a voltage across the cap which would cause the charge to redistribute through the circuit. Also, I'm assuming charge can't leave the circuit and go somewhere else. So the final charge distribution will be half of the (net) added charge on each plate. Or the average value of the initial charge distribution.

So yes, you are correct the final charge for the (2Q,0) initial condition will be (Q,Q).

You might want to look into the superposition theorems of linear systems for this sort of question. Having solved it for one set of conditions, the answers for other values can be really easy to find that way. You could view the (2Q,0) case as the sum of the responses to the (Q,-Q) case and the (Q,Q) case. The former settles to (0,0) and the latter doesn't change.

*opening the switch let's me ignore questions about where is the charge located in the circuit. Is there charge on the inductor? If the plates are connected through other components the charge distribution question between plates confuses me. The charge ends up everywhere, equal and at rest.

 

[alan123hk]

It seems I didn't explain myself well. Here is the schematic.

Yes, I think it is not difficult to place any amount of charge on the two plates of the capacitor. For example, we only need to place each plate very close to the ground one by one, and then apply the appropriate voltage between the plate and the ground, so that each plate is charged to the required amount of charge, and then move the two plates close to each other to form a capacitor. Now we can move the capacitor far away from the ground and connect any form of circuit to this capacitor according to our interest.

Of course, this process is to add any amount of charge (positive or negative) on the two plates of the capacitor. As for the final voltage between the two plates of the capacitor, that is another matter.

 

[kuruman]

The OP is interested in only one plate having charge while the other has zero charge. That is why the plate on the right is grounded. The plate on the right can be charged by induction.

 

[alan123hk]

I was thinking that the charge on the capacitor is found by taking (Q1-Q2)/2, where Q1 and Q2 are charges on the plates. So in the normal case where Q1=+Q and Q2=-Q, charge of capacitor is 2Q/2 = Q, while for this case where one plate has +2Q and the other plate is at 0 charge, the charge would still be Q.

I don't think it is necessary to define the amount of charge of the capacitor as $~\frac {Q_{p1}-Q_{p2}} {2}~$, and it may cause misunderstanding. Please see the following example of two capacitors in series.

Since the net charge of the middle plate is zero, the charge of C1 may become $~\frac {+2Q-0Q} {2} =Q ~$, but in fact the charge of C1 should still be regarded as $2Q$.

 

[Baluncore]

The OP is interested in only one plate having charge while the other has zero charge. That is why the plate on the right is grounded. The plate on the right can be charged by induction.

Grounding one plate of a capacitor does not discharge that plate, nor the capacitor.

I take a one farad capacitor and charge it to one volt. C = Q / V. It then has a positive charge excess of one coulomb on one terminal, and a charge of negative one coulomb on the other. The energy stored in the capacitor is; E = ½·C·V² = 0.5 joule. If I then connect one terminal of that capacitor to ground, the energy stored within the capacitor and the charge imbalance between the plates remain unchanged. The voltage between the plates of the capacitor remains the same, as does the balanced charge on each plate.

When I ground one terminal, it does not mean that plate has zero charge. The voltage of that plate becomes zero relative to ground, but the charge on that plate is temporarily connected to ground. If I then disconnect the terminal from ground, electrostatic induction will make sure that the charge on the previously grounded terminal remains constant, equal and opposite to the charge on the floating plate.

If you have two conductors, joined at a closed switch, then induce a distortion of the surface charge through the use of an external electric field, charge will still be conserved on the connected conductors. If you then open the switch, isolating the induced charge imbalance on either side of the switch, you have formed a charged capacitor. But whatever excess charge you have at one end, must be balanced exactly by the deficit at the other.

An RLC circuit is a closed electrical model. It obeys Kirchhoff's current law. What flows in through one terminal of a capacitor must flow out through the other. Isolated conductors, monopoles and electrostatic induction are not part of the RLC circuit model.

A capacitor that has a permanent charge built into the dielectric is called an electret. It is the electrostatic equivalent of a permanent magnet.
https://en.wikipedia.org/wiki/Electret

 

[kuruman]

Grounding one plate of a capacitor does not discharge that plate, nor the capacitor.

If the plates are far enough apart so that what is done electrically to one does not affect the other, then the charge on the grounded plate will be zero. That's what I meant when I specified in the schematic (post #12) that the initial separation is "Far apart".

Even if there were some residual charge on the grounded plate instead of zero, this does not change the answer to OP's initial question. When the plates are brought close together and connected through the RL circuit, the final charge on each plate would be $Q=\frac{Q_1+Q_2}{2}$. OP's question is a special case of this with $Q_1=2Q$ and $Q_2=0$.

BTW, this is why I prefer to think of capacitance as a geometrical quantity. The definition $C=Q/V$ comes with a procedure that is routinely omitted, but is nevertheless important: "One starts with two arbitrarily shaped conductors in close proximity. Then one removes charge $+Q$ from one conductor and places it on the other. The ratio of the charge $+Q$ to the absolute value of the potential difference (voltage $V$) between conductors is the capacitance."

Now consider the case of two isolated parallel plates of area $A$ and separation $d$. Before the plates are brought together, charges $Q_1$ and $Q_2$. How does $C=Q/V$ work as a definition in this case? The geometric expression $C=\epsilon_0A/d$ works just fine.

 

[Baluncore]

Now consider the case of two isolated parallel plates of area A and separation d. Before the plates are brought together, charges Q1 and Q2. How does C=Q/V work as a definition in this case? The geometric expression C=ϵ0A/d works just fine.

The voltage across the capacitor and the charge are needed for RLC circuit modeling. KIL and KVL work well. The dielectric constant, area, and separation of the plates is quite irrelevant.

Capacitors are designed to contain their internal electric fields, so they should not behave like antennas. The suggestion that individual parts of a component be transported from zero to infinity and back, while assembling something that is physically impossible, gets more ridiculous by the post.

This thread is based on the absurdity that the two opposite plates of the same capacitor can have different charges. That requires there be three capacitors, not one. You will then find that the unbalanced charge is actually on the plates of another capacitor, not on the plate of the capacitor you expect.

 

[DaveE]

This thread is based on the absurdity that the two opposite plates of the same capacitor can have different charges. That requires there be three capacitors, not one. You will then find that the unbalanced charge is actually on the plates of another capacitor, not on the plate of the capacitor you expect.

OK, for practical solutions, I agree. However, charged particles are real, countable, things. The entire circuit can have more protons than electrons, for example. This combined with some induced charge redistribution can result in different charge quantities on the plates. I guess you would refer to that as your 3rd capacitor, but I'm not sure you need to invoke a remote reference point for countable objects; I don't think it usually matters where the extra charges came from. If it does matter for the analysis, then that would have to become part of the circuit definition. This is why in a previous post I referred to decomposition into charge difference and common charge. Of course the only interesting part of the solution is the symmetric charge difference you refer to.

 

[Baluncore]

However, charged particles are real, countable, things. The entire circuit can have more protons than electrons, for example.

How would you model such circuit features with SPICE ?
Can you buy the components and assemble the circuit ?

 

[kuruman]

The voltage across the capacitor and the charge are needed for RLC circuit modeling. KIL and KVL work well. The dielectric constant, area, and separation of the plates is quite irrelevant.

Capacitors are designed to contain their internal electric fields, so they should not behave like antennas. The suggestion that individual parts of a component be transported from zero to infinity and back, while assembling something that is physically impossible, gets more ridiculous by the post.

This thread is based on the absurdity that the two opposite plates of the same capacitor can have different charges. That requires there be three capacitors, not one. You will then find that the unbalanced charge is actually on the plates of another capacitor, not on the plate of the capacitor you expect.

I think we are talking past each other, so I will stop here.

 

[DaveE]

How would you model such circuit features with SPICE ?

The same way you would. The extra net charge is irrelevant to the circuit dynamics. It can be ignored unless you actually want to count the charged particles. SPICE, and circuits in general, are an abstraction of the real world. We only model what we want the computer to analyze. I wouldn't want to add a trivial bit of the analysis to my model.

Now, if I added some static charge imbalance by touching the circuit after I shuffled my feet on the carpet, and if you thought it was important to know how my finger affected the circuit dynamics, then you could add me and the carpet to your circuit model. That would be the third capacitor your referred to. But if those charges arrived as cosmic rays from Alpha-Centauri, I wouldn't imagine you would really care about the capacitive coupling from that star. That extra charge isn't ever going to go back where it came from. OTOH, you could easily add it; I'd choose a value of 0 farads, myself.

Can you buy the components and assemble the circuit ?

I could use ANY components or assembly you choose with regard to net charge balance. I do need a source of charged particles to add, like cosmic rays, an electron gun, or whatever.

The key point is nearly a semantic one. The "circuit" needs to be defined so we know what to include in the analysis. I suppose philosophically, every circuit includes Alpha-Centauri; but it's not ever included in the analysis.

 

[anorlunda]

The extra net charge is irrelevant to the circuit dynamics.

I agree. I've been making that point using other words. In circuit analysis, only voltage differences, not absolute potentials, play a role. I use these figures to illustrate. You can think of case #6 as the one with excess charge in the circuit.

Looking at it another way. One can not combine field theory for how charges distribute themselves in a conductor, with circuit analysis. In circuit analysis, none of the ideal components in the circuit need to be physically realizable. I could replace all the wires, V, R, L, and C in the circuit with digital simulators of wires, V, R, L, or C and the circuit solution would be the same.

 

[alan123hk]

Now, if I added some static charge imbalance by touching the circuit after I shuffled my feet on the carpet, and if you thought it was important to know how my finger affected the circuit dynamics, then you could add me and the carpet to your circuit model. That would be the third capacitor your referred to. But if those charges arrived as cosmic rays from Alpha-Centauri, I wouldn't imagine you would really care about the capacitive coupling from that star. That extra charge isn't ever going to go back where it came from. OTOH, you could easily add it; I'd choose a value of 0 farads, myself.

I agree. Static electricity is inevitable in daily life. All electronic and electrical products may carry more or less static electricity. Unless the voltage and energy of these static electricity are quite high, it will generally not affect the operation of the electronic and electrical products. .
Because the product must pass anti-static safety standards, the problem of static electricity will be taken into account when designing the product, but it seems that not every design engineer believes that it is necessary to add additional capacitors to simulate the interaction between these static electricity and surrounding space objects.

 

[alan123hk]

Another thing worth mentioning is the final charges on the two plates of the capacitor. Of course, I also agree that the charge on each plate will be $~Q=\frac{Q_1+Q_2}{2}~$, but I think this only applies when the two conductors of the capacitor are completely symmetrical.

When the two conductors of the capacitor are not completely symmetrical, even if it is assumed that the charge will not accumulate on the LR circuit, the charge distributed to the two conductors may not be equal.

For example, in the following example, it is obvious that most of the charge will accumulate on the outer conductor.

 

[sophiecentaur]

This thread is based on the absurdity that the two opposite plates of the same capacitor can have different charges. That requires there be three capacitors, not one. You will then find that the unbalanced charge is actually on the plates of another capacitor, not on the plate of the capacitor you expect.

I think this comment should have appeared in post #2 and we could have given the question a peaceful death. To set up the scenario in the OP, you would have had to connect the 'charged' (2Q) plate to, say 10kV and leave the uncharged plate a fraction of a mm away from it (with a good insulator, of course but that's high voltage capacitor). The Capacitance of the Capacitor (say 1uF) would be around 100,000 times the Capacitance between the charged plate and Earth ( say 10pF). That would give a PD across the plates of the Capacitor or 0.1V. (Actual numbers may be worth checking but the idea is there
'Doable' but for what purpose? You would still have an RLC circuit with 0.1 V and a tiny oscillating charge but the whole experiment would need to be on an insulated platform.

 

[vanhees71]

I think it's not so absurd a question, though somewhat academic.

You can in principle have such an unusual initial condition and why not asking for the dynamics of this unusual "circuit", but I think it's not as easy to solve as it looks at the first glance. I'm not so sure, whether one can use the standard AC circuit theory in terms of the parameters R, L, and C; particularly the latter, which has a well-defined meaning for the case that the capacitor is connected to a (DC) voltage source, which implies that one plate carries a charge Q and the other a charge -Q, and then you use the quasistatic approximation for the AC case too, which is legitimate if the typical wavelength of the em. waves involved is very large compared to the extension of the circuit. I guess one has to carefully think about the problem using a concrete model for the capacitor (I guess a spherical capacitor is doable without to much effort) with these initial conditions and then closing the circuit with a series of resistor and inductance. I hope I find the time to think about it over the weekend (and get something useful out ;-)).

 

[sophiecentaur]

I think it's not so absurd a question, though somewhat academic.

No. Not absurd 'as such' but it makes a lot of assumptions about the common sense of any reader. If they don't realize that it's a crazy scenario then they can spend a lot of time worrying about it and coming to some pretty pointless conclusions.
I think the model I introduced higher up, of taking two plates, one charged and one uncharged, and bringing them together is a good way into getting what's going on. That is easy to achieve but measurements could be difficult.