Capacitors connected in series: Why is the voltage the same?

[hutchphd]

I enjoyed a simulation of the equipotential sketch as attached. Battery right side and two condensers left side. The closer pair plate charges are put, the weaker electric field between the two condensers become, I see. If you are interested in join me in Kobe university site in Japanese,

That is a nice addition. I particularly enjoy the V=0 line which threads its way from infinity, through the center plane of the devices, and out to infinity again.

 

[etotheipi]

I enjoyed a simulation of the equipotential sketch as attached. Battery right side and two condensers left side. The closer pair plate charges are put, the weaker electric field between the two condensers become, I see. If you are interested in join me in Kobe university site in Japanese,
http://www2.kobe-u.ac.jp/~wakasugi/mats/elecflux/ .

I tried to draw a smiley face but it just ended up looking like a demonic frog...

 

[jbriggs444]

I enjoyed a simulation of the equipotential sketch as attached. Battery right side and two condensers left side. The closer pair plate charges are put, the weaker electric field between the two condensers become, I see. If you are interested in join me in Kobe university site in Japanese,
http://www2.kobe-u.ac.jp/~wakasugi/mats/elecflux/ .

I like the drawing, but should not the field lines and the equipotential lines always be at right angles?

 

[STEMucator]

When you connect an appropriate energy source, each series capacitor will store instantaneous charge. The charge for each capacitor is equal for every series capacitor if the capacitance for each capacitor is equal.

A single equivalent capacitor $\frac{1}{Ceq}$ will have a larger plate separation $d$. The single equivalent capacitor will span the same distance as the smaller series capacitors, and have a capacitance smaller than any of the series capacitors.

The total voltage difference from $V_1$ to $V_6$ is divided across each capacitor depending what the inverse of its capacitance is. More capacitance $\rightarrow$ less voltage $(V = \frac{q}{C})$.

The voltage loss across the wire separation between the capacitors is negligible if you have a superconducting wire (no electric field within an “ideal” wire). Hence:

$V = - \oint_C \vec{E} \cdot d \vec S = - \oint_C \vec{0} \cdot d \vec S = 0$​

 

[hutchphd]

I like the drawing, but should not the field lines and the equipotential lines always be at right angles?

If you look closely the field lines are brown and the equipotentials purple. There are both (and arrows!)

 

[jbriggs444]

If you look closely the field lines are brown and the equipotentials purple. There are both (and arrows!)

Yes. I get that. But they are not perpendicular. And they should be.

 

[DaveE]

I really dislike framing this problem with images that are labelled like these:

I think it is confusing to have defined (and redefined) the 0 charge location in multiple different places. It implies some equivalence between a + on one side vs. a + on the other side of the drawing, when in fact you have some undefined redefinition of the meaning of the charge magnitude at these different places. It also makes it essentially impossible to think about more than one capacitor at a time.

Much better to label the charge at each location as Q₁, Q₂, Q₃... Then, if you want, you set one to 0, not that that definition matters much, it's all about charge difference. My point is you only get to do it once.

Then it's easy to visualize some magnitude relationship between them. Something like Q₁ > Q₂ > 0 > Q₃.

 

[etotheipi]

I think it is confusing to have defined (and redefined) the 0 charge location in multiple different places. It implies some equivalence between a + on one side vs. a + on the other side of the drawing, when in fact you have some undefined redefinition of the meaning of the charge magnitude at these different places. It also makes it essentially impossible to think about more than one capacitor at a time.

Much better to label the charge at each location as Q₁, Q₂, Q₃... Then, if you want, you set one to 0, not that that definition matters much, it's all about charge difference. My point is you only get to do it once.

Then it's easy to visualize some magnitude relationship between them. Something like Q₁ > Q₂ > 0 > Q₃.

I don't understand what you mean about defining the zero of charge. The first capacitor has plates $Q_1$ coulombs above neutral and $Q_1$ coulombs below neutral respectively, and likewise for the other plates.

Aren't charges raw values? Potentials exist in an $\mathbb{R}$-torsor whilst charges exist in $\mathbb{R}$.

The charges feed directly into Maxwell's equations so absolute values matter for things like electric fields; not charge differences. Please do correct me if I am missing something!

 

[hutchphd]

Much better to label the charge at each location as Q1, Q2, Q3... Then, if you want, you set one to 0, not that that definition matters much, it's all about charge difference. My point is you only get to do it once.

I have already aired my annoyance with much of this discussion, but I don't see your point here. Typically a circuit element is presumed inactivated when placed in a circuit. Writing the Q as equal merely represents the fact that they are connected. Any overall charge is irrelevant to circuit.
Similarly each wire segment is not required to have a potential defined at each end, so that one can put in R=0 and show ΔV=0. Am I missing something?

Otherwise it has already been said

I agree. The whole of this thread rambles between elementary circuit theory and advanced EM concepts.

 

[etotheipi]

I guess for any expression that involves a difference of terms that are linear in charge, you can get away with translating the values of the charges. If we take a single capacitor with $\sigma_1 \, \text{Cm}^{-2}$ on one plate and $\sigma_2 \, \text{Cm}^{-2}$ on the other plate, then the electric field between the plates is $\frac{\sigma_1}{2\epsilon_0} - \frac{\sigma_2}{2\epsilon_0}$. So $E = \frac{1 \, \text{Cm}^{-2}}{2\epsilon_0} - \frac{3 \, \text{Cm}^{-2}}{2\epsilon_0}$ is the same as $E = \frac{6 \, \text{Cm}^{-2}}{2\epsilon_0} - \frac{8 \, \text{Cm}^{-2}}{2\epsilon_0}$, for instance.

Maybe that covers all of CA purposes. Where you get into trouble is for anything that is not an explicit difference / not linear in charge, which is still quite a broad designation.

That's why I'm pretty skeptical of "setting the zero of charge". It's not like potentials which are truly elements of an R-torsor due to that constant of integration; generally it's incorrect to set a zero of charge.

 

[gleem]

The original problem was that the OP did not agree with the statement that the voltages across the three capacitors in series were the same. We saw line integrals and differential equations. We discussed fields outside the capacitors and we are at the 60 the posts with no explanation as to why the voltages are the same.

Capacitors connected in series no matter the capacitances will have the same charge on their plates.

Post 3 basically answered the question but did not go far enough and state V_i= Q/C_i which was written in posts 44 and 54 but did not explicitly state that :

Since the C_i's are assumed to be different then the voltages are not the same.

This thread should have been put to rest in post 5. We can do better, you think?

 

[hutchphd]

Absolutely.
When you say "I have a capacitor", you are tacitly announcing the idealized linear component. If you have a non-ideal component, that requires a different question with particular specification.
Such considerations are very interesting but can we please answer one question at a time, decide on an answer, and then proceed. Otherwise we end up with a diffusion process

 

[DaveE]

I don't understand what you mean about defining the zero of charge. The first capacitor has plates $Q_1$ coulombs above neutral and $Q_1$ coulombs below neutral respectively, and likewise for the other plates.

Aren't charges raw values? Potentials exist in an $\mathbb{R}$-torsor whilst charges exist in $\mathbb{R}$.

The charges feed directly into Maxwell's equations so absolute values matter for things like electric fields; not charge differences. Please do correct me if I am missing something!

Yes, I see your point. I just think there is a case where your various "neutral" points may have a difference in net charge; i.e. current would flow if you connected them. This would depend on where you define the neutral points and what the initial charge distribution is. If you care about that relationship then there is a place for a more global definition.

I suppose I am worried about non-trivial initial conditions. Perhaps because there are three capacitors, the single capacitor solution repeated three times seems too simple otherwise. But with superposition, I suppose all of the work is in solving the dynamic case of what happens as a battery is connected to the "zero voltages" initial condition. Then you can worry about the other stuff in the other solution.

In any case, the total charge does matter in the complete solution, but changing the problem by adding equal charge everywhere is a trivial change in the solution. I guess I meant that the total charge in the entire circuit isn't interesting.

Plus, I may have missed the assumption of simple initial conditions, in which case I'm just wrong.

In any case, I think I'm sorry I posted. I agree with @sophiecentaur, the thread is now circular pedantic rambling or maybe just semantic arguments. Really, it doesn't matter how you name things if we are solving the same problems and all getting the correct result. The fact that I dislike setting things equal by giving them the same name (or label?) in diagrams is quite irrelevant.

 

[vanhees71]

This looks overcomplicated to me. Let's argue from the conservation laws. We know that the left plate of $C_1$ carries a charge $Q$ and the right-most plate carries charge $-Q$. Then charge conservation tells you that all capacitor plates carry the same charge $\pm Q$ as given in the figure. Now the voltages across the capactors are $U_j=Q/C_j$. Now integrating along the circuit and using the fact that there are no magnetic fields in this electrostatic situation you get
$$U_{\text{Battery}}=U_1+U_2+U_3,$$
where $U_{\text{Battery}}$ is the electromotive force of the battery (often misnamed as "voltage of the battery"). this leads to
$$U_{\text{Battery}}=Q \left (\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right) =\frac{Q}{C_{\text{tot}}},$$
which leads to the well-known rule for the capacity of a series of capacitors
$$\frac{1}{C_{\text{tot}}}=\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}.$$

 

[Leo Liu]

This looks overcomplicated to me. Let's argue from the conservation laws. We know that the left plate of $C_1$ carries a charge $Q$ and the right-most plate carries charge $-Q$. Then charge conservation tells you that all capacitor plates carry the same charge $\pm Q$ as given in the figure.

I think I can provide an intuition for why $\Delta Q$ is the same across all the capacitors in series.

First, let's move a first electron to the negative side of a capacitor. As you can imagine, it begins to repel a negatively charged electron on the other plate. The electron repelled by the electron we moved will reside on the plate of a capacitor and repel an electron on the other plate, and so on.

Then we (electromotive force) will continue to move the 2nd, 3rd and nth electron. Our action will make the same number of electrons being released from the positive side of the last capacitor.

However, as we try to move the next electron, although it is attracted by the other plate which is positively charged, it becomes harder because the excessive negative charge on the plate repels the electron we are moving.

After a while, the amount of charges on the negatively charged plate is so high that it is impossible for us to move an electron to the plate.

As a result, the circuit becomes stable and the charge entered the first capacitor is equal to the charge released by the last capacitor.

Also, you might be interested in this video.

 

[etotheipi]

It is true that the equal positive and negative charges on an ideal capacitor follows from the continuity condition for current (in physical terms, like you mention, electron in one side, electron out the other side).

I think we must be careful when talking about "how hard it is" to move an electron onto the negative plate of the capacitor from the battery. E.g. if the negative pole is connected to the negative plate with an ideal wire then no work needs to be done on the electron to move it onto the negative plate from the negative pole.

The current in the circuit is determined from the voltage already across the capacitor and the resistance of the circuit (along with constant EMF from the battery), so the current does indeed decrease as the capacitor charges. It means the voltage drop across the resistance gets progressively smaller and the capacitor voltage approaches the cell voltage. So we just end up allocating a greater proportion of our "voltage budget" to the capacitor and less to the resistance. But the electric field still does the same total amount of work.

Also, neat animations

 

[Leo Liu]

I think we must be careful when talking about "how hard it is" to move an electron onto the negative plate of the capacitor from the battery.

I agree with you but I just want to show that the potential energy of the plate becomes higher due to the accumulation of electrons, which lessens the potential energy difference between the battery and the capacitor.

The current in the circuit is determined from the voltage already across the capacitor and the resistance of the circuit (along with constant EMF from the battery), so the current does indeed decrease as the capacitor charges. It means the voltage drop across the resistance gets progressively smaller and the capacitor voltage approaches the cell voltage. So we just end up allocating a greater proportion of our "voltage budget" to the capacitor and less to the resistance. But the electric field still does the same total amount of work.

Good and rigorous explanation. Yet again I just want to make this process more intuitive.

 

[etotheipi]

I agree with you but I just want to show that the potential energy of the plate becomes higher due to the accumulation of electrons, which lessens the potential energy difference between the battery and the capacitor.

You must be careful here. There is an energy $\frac{Q^2}{2C}$ associated with the capacitor as a whole which does increase with charge, and there exists a potential difference between the two plates that also increases with charge.

The potential difference between either plate and the battery depends on whatever is between them in the circuit. So if one end of the capacitor is connected to one terminal of the battery by an ideal wire, they will be at equal potential always and no work would be done by the electric field moving a charge from the terminal onto that plate. If there's a resistor between the plate and the pole of the battery, then the p.d. across that resistor will decrease with time.

 

[hutchphd]

The potential difference between either plate and the battery depends on whatever is between them in the circuit. So if one end of the capacitor is connected to one terminal of the battery by an ideal wire, they will be at equal potential always and no work would be done by the electric field moving a charge from the terminal onto that plate. If there's a resistor between the plate and the pole of the battery, then the p.d. across that resistor will decrease with time.

I will ask the obvious question. If you perform this operation with just a wire the Battery will supply a total amount of energy equal to $VQ$ where Q is the charge supplied. Yet the capacitor stores energy $\frac {Q^2} {2C}=\frac {VQ} {2}$, exactly half as much. Just to clear this up, where does this energy go?

 

[etotheipi]

I will ask the obvious question. If you perform this operation with just a wire the Battery will supply a total amount of energy equal to $VQ$ where Q is the charge supplied. Yet the capacitor stores energy $\frac {Q^2} {2C}=\frac {VQ} {2}$, exactly half as much. Just to clear this up, where does this energy go?

Exactly half is dissipated in the resistance.

It does not matter on which side of the capacitor we put the resistance. If it is on the side of the positive pole, then the negative pole is always at the same potential as the negative plate of the capacitor. If it is on the side of the negative pole (as I have drawn above), there will be a voltage drop between the negative pole and the capacitor, and instead the positive pole is at the same potential as the positive plate of the capacitor.

The current is $I = I_oe^{-\lambda t}$. The energy dissipated in the resistor,

$$E = \int P dt = \int_0^T I_0^2 e^{-2\lambda t}R dt = -\frac{I_0^2 R}{2\lambda} [e^{-2\lambda t}]_0^T = \frac{I_0^2 R}{2\lambda}(1-e^{-2\lambda T})$$We take $T \rightarrow \infty$, i.e. $E = \frac{I_0^2 R}{2\lambda} = \frac{\varepsilon^2 R}{2\lambda R^2} = \frac{1}{2}C\varepsilon^2$.

We notice also that the energy gained by the capacitor in this time is $\frac{1}{2}C\epsilon^2$. The total work done by the battery on charges moving through it is $C\varepsilon^2$. These two results can be obtained via similar integrals.

Now in the arrangement I have drawn, an ideal wire connects the positive pole of the battery with the positive plate of the capacitor. This means that they are at equal potential, and the electric field does no work transporting charges from the positive terminal to the positive plate of the battery.

Now @hutchphd asks an interesting question; what if we remove the resistance? I don't think that it is possible to do so in our model. Indeed physically there will always be resistance in our wires even if we do not connect one up explicitly. Hypothetically, though, our time constant would go to zero, the capacitor would charge instantly. The integrals I have shown above would not work.

 

[hutchphd]

And of course in the real world all batteries will have some finite internal resistance. I very much appreciate the complete answer and plead laziness for not supplying same. Also yours was better

 

[Delta2]

I think in the case of zero ohmic resistance, the capacitor is charged with an infinite current in zero time, and the energy lost is of the form $\int I^2 R dt=\infty\cdot 0=\frac{VQ}{2}$ we have to impose this for the sake of conservation of energy.

 

[etotheipi]

I think in the case of zero ohmic resistance, the capacitor is charged with an infinite current in zero time, and the energy lost is of the form $\int I^2 R dt=\infty\cdot 0=\frac{VQ}{2}$ we have to impose this for the sake of conservation of energy.

It is an interesting take. It does beg the question of the mechanism by which that energy is dissipated! But I think you must be correct, especially since we can make $R$ arbitrarily small with the theoretical model and it still comes out to one half.

 

[Leo Liu]

The potential difference between either plate and the battery depends on whatever is between them in the circuit.

Are we really talking about the same difference :D? I am talking about the difference of potential energy $U$.

The potential difference between either plate and the battery depends on whatever is between them in the circuit. So if one end of the capacitor is connected to one terminal of the battery by an ideal wire, they will be at equal potential always and no work would be done by the electric field moving a charge from the terminal onto that plate.

I don't agree with this, but I could be wrong.

First, even if the resistance is 0, there is still an electrostatic force opposing the motion of the electron after $\Delta t$. As shown by the diagram below, although the electron is being attracted to the positively charged right plate, the negatively charged left plate (# of e- should be greater than the number of protons, I messed it up), which is closer to the electrons than the right plate, repels it. In this case, the net force exerted on the electron should point toward the anode which is the battery (the rectangle is not a battery--it is just for highlighting). Therefore, emf must do work to counteract this opposing force. Thank you.

 

[etotheipi]

First, even if the resistance is 0, there is still an electrostatic force opposing the motion of the electron after $\Delta t$.

Remember that inside the ideal (conducting) wire, we have $\vec{E}=0$. So there is no electrostatic force present there. We also note that the electric field of the ideal capacitor (i.e. that produced by the charges which line the plates) is confined to the capacitor.

In this case, the net force exerted on the electron should point toward the anode which is the battery. Therefore, emf must do work to counteract this opposing force. Thank you.

The EMF is a line integral $$\oint (\vec{E} + \frac{1}{q}\vec{f}_{chem}) \cdot d\vec{x}$$ around the whole circuit. The $\frac{1}{q} \vec{f}_{chem}$ exists only inside the battery, whilst the electric field (which is conservative, here) is not confined to the battery.

Inside the battery the electric field perfectly balances the chemical force (per unit charge). In an ideal wire, the electric field is zero. In a component like a resistor, we have a non-zero electric field which results in a voltage drop; the difference in energy here is that which is dissipated in the resistance.

Are we really talking about the same difference :D? I am talking about the difference of potential energy $U$.

I am not sure what the potential energy of a battery is . I don't know if this makes sense. But we can speak of the electric potential fields and chemical potential fields, for instance?

 

[Leo Liu]

Remember that inside the ideal (conducting) wire, we have E→=0. So there is no electrostatic force present there. We also note that the electric field of the ideal capacitor (i.e. that produced by the charges which line the plates) is confined to the capacitor.

I see. You are talking about an infinitely large capacitor, but my example was a realistic case. The image below shows the electric field around a real capacitor.

(Credit to Kogence)
If it was an ideal capacitor, you were right.

The EMF is a line integral ∮(E→+1qf→chem)⋅dx→ around the whole circuit. The 1qf→chem exists only inside the battery, whilst the electric field (which is conservative, here) is not confined to the battery.

Inside the battery the electric field perfectly balances the chemical force (per unit charge). In an ideal wire, the electric field is zero. In a component like a resistor, we have a non-zero electric field which results in a voltage drop; the difference in energy here is that which is dissipated in the resistance.

I don't know what you meant by this. I talked about the direction of the net force which has nothing to do with the battery. What I really mean is that you must do a bit work to counteract the electric field generated by the capacitor in reality. Can you elucidate you answer?

Nevertheless, thank you for writing that equation. I just learned something new today.

I am not sure what the potential energy of a battery is . I don't know if this makes sense. But we can speak of the electric potential fields and chemical potential fields, for instance?

I meant the energy released by the redox reaction occurring within the cells of the battery.

 

[etotheipi]

I see. You are talking about an infinitely large capacitor, but my example was a realistic case. The image below shows the electric field around a real capacitor.

It could be an infinitely large capacitor, but it is also an assumption of circuit theory that the electric field of a capacitor is confined to within the capacitor. But even if we did allow for fringing, the electric field inside a perfect conductor is still constrained to be exactly zero.

I don't know what you meant by this. I talked about the direction of the net force which has nothing to do with the battery. What I really mean is that you must do a bit work to counteract the electric field generated by the capacitor in reality. Can you elucidate you answer?

If the wire is ideal, there will be no such "electrostatic force" opposing the motion of the electrons inside the ideal wire. There is also no other source of EMF outside the battery which could oppose such an "electrostatic force" even if it were there.

Outside of the wire in the region of a physical capacitor you will get non-zero electric fields, as has been mentioned by a few others earlier in the thread. But we are not too interested in them here .

In reality, neither the wire nor the capacitor will be ideal. Perhaps yes in reality you will obtain some electric field due to the capacitor in the wire, that does some negative work on the electrons. But I don't know how you would go about analysing this. It's certainly outside the realm of circuit theory.

I meant the energy released by the redox reaction occurring within the cells of the battery.

This is a slightly different ball-game, and is really more a question of the operation of batteries. The work you can extract from the battery can be expressed in terms of the (negative) of the free energy change, $nFE^o_{cell}$, where $E^o_{cell}$ is a difference of electrode potentials (which are themselves, loosely, the voltage across the respective double layers at each electrode). But I would refrain from trying to weave this into circuit theory, and I wouldn't identify it as a "potential energy of the battery" in any case.

 

[hutchphd]

The potential energy of a battery is $V*(Amp-hrs)*3600$ and it is a chemical number.! No fictitious fields please.
I will note in passing that this same factor of 1/2 occurs with a mass on a spring having gravitational potential. If the mass is allowed to settle to new equilibrium, the spring will end up with exactly mgh/2 more energy and the other half goes to friction to settle the system.
Absent any resistance in the electrical case, there must be some inductance in order to make a topological circuit and the circuit would simply oscillate forever. The factor of exactly 1/2 for both I think comes from the fluctuation-dissipation theorem but that may be an overreach. Needs more thought, perhaps not by me.

 

[etotheipi]

I will note in passing that this same factor of 1/2 occurs with a mass on a spring having gravitational potential. If the mass is allowed to settle to new equilibrium, the spring will end up with exactly mgh/2 more energy and the other half goes to friction to settle the system.
Absent any resistance in the electrical case, there must be some inductance in order to make a topological circuit and the circuit would simply oscillate forever. The factor of exactly 1/2 for both I think comes from the fluctuation-dissipation theorem but that may be an overreach. Needs more thought, perhaps not by me.

That's interesting, hadn't heard of this theorem before! I found this which seems to give a good outline of the theorem. It certainly looks like it's along the right lines...

Anyway it is now nearly 2AM so I'm going to go and watch netflix. Perhaps someone will have provided some extra insight by tomorrow morning . It's a good question!

 

[Leo Liu]

It could be an infinitely large capacitor, but it is also an assumption of circuit theory that the electric field of a capacitor is confined to within the capacitor.

It would be great if you can share your insight into the mechanism of capacitor! I am a bit confused now, haha.

Besides, does circuit work like water pipe? If it does then I guess once an electron enters the electric field of a capacitor, it will feel a force and will subsequently pass it on to the adjacent electron in the circuit. Thus, I think emf does work to counteract this force, rather than the field outside the capacitor (which DNE ideally). Please correct if I am wrong!

 

[vanhees71]

I think in the case of zero ohmic resistance, the capacitor is charged with an infinite current in zero time, and the energy lost is of the form $\int I^2 R dt=\infty\cdot 0=\frac{VQ}{2}$ we have to impose this for the sake of conservation of energy.

Well, and I thought $0 \infty=42$ ;-)). The correct answer is of course that, if you want to make the unphysical assumption that $R=0$ for the entire circuit you have to take the limit of the physical situation with $R \rightarrow 0^+$.

 

[etotheipi]

It would be great if you can share your insight into the mechanism of capacitor! I am a bit confused now, haha.

Besides, does circuit work like water pipe? If it does then I guess once an electron enters the electric field of a capacitor, it will feel a force and will subsequently pass it on to the adjacent electron in the circuit. Thus, I think emf does work to counteract this force, rather than the field outside the capacitor (which DNE ideally). Please correct if I am wrong!

It might help to have a read through Feynman's bit on this for the full rundown, since I don't think I can offer much in the way of insight .

For electric currents, we have the continuity equation $\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t}$. If we take a surface completely enclosing the capacitor, then the total charge in this region must not change with time, and we deduce that the current in equals the current out (N.B. inside the capacitor we can also define a displacement current; we call the displacement field $\vec{D} = \epsilon \vec{E} + \vec{P}$ and the displacement current density $\vec{J}_D = \epsilon \frac{\partial E}{\partial t}$, so that the continuity equation remains valid everywhere). I think that in physical terms we can imagine that electrons arriving at the negative plate repel electrons on the positive plate, since in reality the E field will penetrate at least a little into the plates.

You mention the hydraulic analogy for current; it's good in some respects, but not completely. It's good in the sense that a flow of liquid is a good way to understand the continuity equation (i.e. volume of liquid into a pipe equals volume out, if it's incompressible).

However, in water pipes, each section of water exerts a contact force on adjacent sections of water. This does not happen with charge carriers in an ideal circuit! You can think of the sea of charge carriers a bit like an ideal gas, in that there is no interaction between them. The net E field inside the wire is zero, concretely because of Gauss' theorem but loosely because the sum of the forces on anyone charge carrier from the mobile charge carriers and positive cores in the wire is zero.

This just means that in an ideal wire, the charge carriers maintain their kinetic energy and no work is done on them. In some components, however, there is a non-zero electric field. In a resistor, the current density is related to the electric field by $\vec{J} = \sigma \vec{E}$. We note that if there is any current flowing through, there will be a voltage drop across the terminals. Where does this electric field come from physically? It is due to surface charges on the wire. The charge carriers are crashing into ions in the resistor, transferring and losing their energy. This is made up for by the positive work done by the electric field, so that they maintain loosely a constant kinetic energy.

And in a battery, you also have an electric field from the positive to negative plate. There $-\int \vec{E} \cdot d\vec{x}$ is the voltage across the terminals and the chemical force does work against the electric field in order to raise the electric potential of the charge carriers.

 

[Delta2]

Well, and I thought $0 \infty=42$ ;-)). The correct answer is of course that, if you want to make the unphysical assumption that $R=0$ for the entire circuit you have to take the limit of the physical situation with $R \rightarrow 0^+$.

That is absolutely correct, we have to take that limit. Taking that limit proves that the energy is independent of R and equal to $\frac{VQ}{2}$. Indeed continuing from the nice post #70 from @etotheipi we see that the energy dissipated in the ohmic resistance is $$E=\frac{I_0^2R}{2\lambda}$$ where $\lambda=\frac{1}{RC}$, and taking the limit of $E$ as $R$ tends to zero we have :
$$\lim_{R\to 0}E=\lim_{R\to 0}{\frac{I_0^2R}{\frac{2}{RC}}}=\lim_{R \to 0}\frac{I_0^2R^2C}{2}=\lim_{R\to 0}\frac{1}{2}\frac{V_0^2}{R^2}R^2C=\frac{CV_0^2}{2}=\frac{Q_0V_0}{2}$$

 

[etotheipi]

Damn, I was really hoping it was going to be 42!

 

[sophiecentaur]

I think I can provide an intuition for why ΔQ is the same across all the capacitors in series.

Yes. There are a number of approaches to giving very reasonable intuitive explanations and the OP should try to get into the topic from that direction before subjecting the problem to a higher level of thought. I suggest asking oneself "if it were not true, where would the extra charge come from?" (i.e. think of the equilibrium condition)
Yards of Maths may not help with getting over an initial conceptual problem. The reason for most misconceptions doesn't lie with the Maths (which is, of course, necessary to give formal confirmation) but with basic ideas like conservation principles.
No paradoxes, no internal contradictions involved. A single post reply could have sufficed to point out the Field vs Potential mix-up (There is plenty of scope for the rich PF discussions to take place where applicable.)

 

[STEMucator]

I see. You are talking about an infinitely large capacitor, but my example was a realistic case. The image below shows the electric field around a real capacitor.
View attachment 265146
(Credit to Kogence)
If it was an ideal capacitor, you were right.

I don't know what you meant by this. I talked about the direction of the net force which has nothing to do with the battery. What I really mean is that you must do a bit work to counteract the electric field generated by the capacitor in reality. Can you elucidate you answer?

Nevertheless, thank you for writing that equation. I just learned something new today.

I meant the energy released by the redox reaction occurring within the cells of the battery.

An insulator surrounds the wire that the current is flowing through. The electric field emanating from the capacitor would not affect the wire inside the insulation.

 

[sophiecentaur]

An insulator surrounds the wire that the current is flowing through. The electric field emanating from the capacitor would not affect the wire inside the insulation.

Why do you say that? A dielectric will reduce a field, locally but not eliminate it. The wire allows the charges to flow, removing any P D, in any case. The actual resistance of the wire will only effect the time taken for equilibrium to be reached (RC time constant) and we are dealing with the equilibrium condition, I believe.

 

[Leo Liu]

We note that if there is any current flowing through, there will be a voltage drop across the terminals. Where does this electric field come from physically? It is due to surface charges on the wire. The charge carriers are crashing into ions in the resistor, transferring and losing their energy. This is made up for by the positive work done by the electric field, so that they maintain loosely a constant kinetic energy.

I understand what you mean. However, why does the current in the capacitor gradually approaches 0? What is happening in the circuit? I know how to explain this using a differential equation, but I really want to know what is going on behind the math. Thank you.

 

[etotheipi]

I understand what you mean. However, why does the current in the capacitor gradually approaches 0? What is happening in the circuit? I know how to explain this using a differential equation, but I really want to know what is going on behind the math. Thank you.

I don't know if I can completely get rid of the maths; maybe someone else can jump in.

In the simple RC circuit, as charge accumulates on the plates, the electric field between the plates increases in magnitude ($E = \frac{\sigma}{\epsilon_0}$). That causes a larger voltage drop across the plates, which means you must have a lower voltage drop across the resistor. And the voltage drop across the resistor is related to the current in the circuit by $V=IR$. So we infer that as the capacitor voltage approaches the cell voltage, the current in the circuit must also approach zero.

I'm not sure if I can do much better than that. You can think of Kirchoff's voltage law for a circuit like this in terms of conservation of energy, i.e. carry a unit charge around the circuit and when you get back to where you started it must have the same potential energy it started with. Then you can deduce the current must go to zero, i.e. the previous paragraph.

 

[sophiecentaur]

I know how to explain this using a differential equation,

I think you mean describe ??
So what does the equation not say that a long verbal description does? The current flow into the capacitor will depend on the voltage across the resistor. The voltage across the capacitor is the supply voltage minus the volts across the resistor. The higher the capacitor volts, the lower the rate of charge flow. This leads to an exponential decay of the capacitor volts etc. etc. etc. And, even there, I used the word "exponential" because we all know and use the term.
Someone (several someones) went to the trouble of inventing a language (Maths) for describing these things. Merely expressing something efficiently doesn't devalue the explanation. Would you say that modern Science could be adequately described in Latin? Latin just doesn't have the ready made vocabulary to make things short and sweet.
They used to say the a picture is worth a thousand words. Well - Maths too and there's no compromise involved.

 

[Delta2]

In a despaired attempt to use as little math as possible:
The voltage source establishes an electrostatic field inside the resistor. This drives initially the current and charges the capacitor, but as the charge in the capacitor builds up, the capacitor establishes its own electrostatic field inside the resistor. The two electrostatic fields battle each other, with that of the voltage source prevailing and keep driving a decreasing current, but when the capacitor is fully charged, the two electrostatic fields eliminate each other and hence there is zero current inside the resistor.

 

[Leo Liu]

That causes a larger voltage drop across the plates, which means you must have a lower voltage drop across the resistor.

Thanks again for showing me how this works. But shouldn't the voltage difference between the plates increase, which causes the current in the circuit to drop?

You can think of Kirchoff's voltage law for a circuit like this in terms of conservation of energy, i.e. carry a unit charge around the circuit and when you get back to where you started it must have the same potential energy it started with. Then you can deduce the current must go to zero, i.e. the previous paragraph.

I understand this, yet it appears quite unintuitive to me. I use the same method to solve problems on the exams; nevertheless, I really want to know the movement an individual electron in the circuit because I think it is fun to think and makes grasping related concepts much easier.

 

[Leo Liu]

The two electrostatic fields battle each other, with that of the voltage source prevailing and keep driving a decreasing current, but when the capacitor is fully charged, the two electrostatic fields eliminate each other and hence there is zero current inside the resistor.

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

 

[Delta2]

Thanks again for showing me how this works. But shouldn't the voltage difference between the plates increase, which causes the current in the circuit to drop?

The voltage difference between the plates indeed increases as the capacitor charge build up and that is why (if we apply KVL) the voltage difference across the resistance decreases which is what @etotheipi says.

 

[Delta2]

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

Well the medium is the interior of the resistor which comprises of positive ions and free electrons.

 

[etotheipi]

I understand this, yet it appears quite unintuitive to me. I use the same method to solve problems on the exams; nevertheless, I really want to know the movement an individual electron in the circuit because I think it is fun to think and makes grasping related concepts much easier.

If you want to know more about the how charges move then I think you must look into the continuity equation and details to do with it. If $\rho$ is the charge density and $\vec{v}$ the drift velocity, then the current density $\vec{J} = \rho \vec{v}$. We know that the current through a surface is the surface integral of the current density. You can imagine, in the ideal case, a homogenous volume of charge drifting around the circuit at a drift speed that is the same everywhere in the series circuit at any given time.

Thank you for attempting at answering my question. I just want to know through what medium the two electric fields fight with each other.

The electric field in the battery does negative work on the charge carriers whilst the electric field in the capacitor would do positive work on the charge carriers (well, charge carriers don't move through the capacitor actually but we can ignore that, the voltage drop definitely still exists!). And when the voltage of the capacitor approaches the voltage across the cell, the work done by the electric field in the resistor must also decrease in order to satisfy $\oint \vec{E} \cdot d\vec{x} = 0$, and microscopically in the resistor $\vec{J} = \sigma \vec{E}$ so the current density needs to decrease in magnitude.

At any given point in the circuit, you only have one electric field. In the ideal (resistanceless) battery, the electric field balances the chemical field. In the ideal wire, there is no electric field. In the capacitor, there is an electric field of magnitude $\frac{\sigma}{\epsilon_0}$. And in the resistor, there is an electric field of magnitude $\frac{1}{\sigma}J$. Since there are no time-varying magnetic fields, the closed loop line integral of these will be zero.

 

[hutchphd]

In the ideal (resistanceless) battery, the electric field balances the chemical field

I very much dislike this terminology. WTF is a "chemical field"?
How about:
The ionic processes in the battery endeavor to maintain a fixed potential difference at its terminals. This open circuit potential is commonly called the battery "voltage".
Please no chemical fields. No morphogenic fields. No "fields of psychic energy"

 

[etotheipi]

I very much dislike this terminology. WTF is a "chemical field"?
How about:
The ionic processes in the battery endeavor to maintain a fixed potential difference at its terminals. This open circuit potential is commonly called the battery "voltage".
Please no chemical fields. No morphogenic fields. No "fields of psychic energy"

Okay yes that was sloppy. Let me define a vector field $\vec{f}_{chem}(\vec{r})$ s.t. $q\vec{f}_{chem}(\vec{r}) = \vec{F}_{chem}(\vec{r})$ is the chemical force on a charge carrier in the battery. That is my chemical field .

I guess this is fine for Physics purposes, and for calculating EMF, etc. But in reality like you say, the chemistry and processes at the electrodes is much more complicated than just a single tenuous field.

 

[vanhees71]

Instead of many words, simply do the math. As explained several times above a series of a battery (EMF $U$), resistor, $R$, and a capacitor, $C$, using Faraday's Law of induction in the quasistatic approximation, i.e., neglecting the self-inductance of the circuit, leads to
$$U_C+R i -U=0.$$
Now
$$U_C=Q/C, \quad i=\dot{Q},$$
where $U$ is the charge on one of the plates of the capacitor and thus
$$\frac{1}{C} Q + R \dot{Q}=U.$$
Initial condition: $Q(0)=0$. The general solution of this ODE is the superposition of a particular solution of the inhomogeneous equation, which obviously is given by
$$Q_{\text{inh}}(t)=C U=\text{const}.$$
and the general solution of the homogeneous equation, which is
$$Q_{\text{hom}}(t)=A \exp\left (-\frac{t}{R C} \right),$$
i.e.,
$$Q(t)=A \exp \exp\left (-\frac{t}{R C} \right) + CU .$$
The initial condition determines the integration constant $A=-CU$:
$$Q(t)=CU \left [1-\exp \exp\left (-\frac{t}{R C} \right) \right].$$
So the charge and thus $U_C=Q/C$ reaches exponentially its stationary limit (relaxation time $\tau=R C$). The current through the circuit is
$$i(t)=\dot{Q}(t)=\frac{U}{R} \exp \left (-\frac{t}{R C} \right).$$
So it starts at $t=0$ as if there's no capacitor in the circuit. In other words, in the first moment an uncharged capacitor acts like a short-circuit, but the current drops exponenitialle again with the said relaxation time $\tau$.

 

[sophiecentaur]

I just want to know through what medium the two electric fields fight with each other.

That's a bit too anthropomorphic for my liking!

Using the Fields approach is needlessly difficult, compared with looking at the Potential Energy situation. Nothing is fighting anything. The increasing charge in the capacitor is increasing the potential across the terminals so the PD across the resistor (its share of the available battery PD) is decreasing - sooooo the current through it also decreases.
There is absolutely no difference in the validity of a Field or Potential treatment so it's perfectly OK to choose the one which makes most sense. Many people seems to find it important to justify Electrical calculations by using Fields - why? It's just got to be harder when you think that a Vector has two values associated with it but a Potential has only one.
In the analagous situation of water flowing between two tanks, with different water levels, through a constriction, the flow is proportional to the Difference in Heights between the tanks. Nothing fights anything.