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

  • #76
113
47
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.
1592870654122.png

(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 :wink:. 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.
 
  • #77
etotheipi
Gold Member
2019 Award
2,167
1,152
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 :wink:.

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.
 
  • Like
Likes Leo Liu and hutchphd
  • #78
1,613
975
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.
 
  • Like
Likes etotheipi
  • #79
etotheipi
Gold Member
2019 Award
2,167
1,152
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!
 
  • #80
113
47
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!
 
Last edited:
  • #81
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,373
6,756
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^+##.
 
  • Like
Likes etotheipi
  • #82
etotheipi
Gold Member
2019 Award
2,167
1,152
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 any one 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.
 
Last edited:
  • Like
  • Informative
Likes Leo Liu and vanhees71
  • #83
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
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 :
$$\lim_{R\to 0}E=\lim_{R\to 0}{\frac{I_0^2R}{\frac{2}{RC}}}=\lim_{R \to 0}\frac{I_0^2R^2C}{2}=\lim_{R\to 0}\frac{1}{2}\frac{V_0^2}{R^2}R^2C=\frac{CV_0^2}{2}=\frac{Q_0V_0}{2}$$
 
  • Like
  • Informative
Likes Leo Liu, vanhees71 and etotheipi
  • #84
etotheipi
Gold Member
2019 Award
2,167
1,152
Damn, I was really hoping it was going to be 42!
 
  • Like
Likes hutchphd, vanhees71 and Delta2
  • #85
sophiecentaur
Science Advisor
Gold Member
24,844
4,639
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.)
 
  • Like
  • Love
Likes hutchphd, Leo Liu and gleem
  • #86
STEMucator
Homework Helper
2,075
140
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.
 
  • #87
sophiecentaur
Science Advisor
Gold Member
24,844
4,639
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.
 
  • Like
Likes Delta2, vanhees71 and etotheipi
  • #88
113
47
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.
 
  • #89
etotheipi
Gold Member
2019 Award
2,167
1,152
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.
 
  • Like
Likes Delta2
  • #90
sophiecentaur
Science Advisor
Gold Member
24,844
4,639
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.
 
  • Like
Likes vanhees71 and etotheipi
  • #91
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
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.
 
  • Informative
  • Like
Likes Leo Liu and etotheipi
  • #92
113
47
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?
1593001243041.png

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.
 
  • #93
113
47
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.
 
  • #94
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
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.
 
  • Like
Likes etotheipi and Leo Liu
  • #95
Delta2
Homework Helper
Insights Author
Gold Member
2,987
1,043
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.
 
  • #96
etotheipi
Gold Member
2019 Award
2,167
1,152
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.
 
  • Like
Likes Delta2 and Leo Liu
  • #97
1,613
975
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"
 
  • Haha
  • Like
Likes Adesh and etotheipi
  • #98
etotheipi
Gold Member
2019 Award
2,167
1,152
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 :wink:.

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.
 
Last edited:
  • Like
Likes hutchphd
  • #99
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,373
6,756
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##.
 
  • Like
Likes Delta2, weirdoguy and etotheipi
  • #100
sophiecentaur
Science Advisor
Gold Member
24,844
4,639
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.
 
Last edited:
  • Like
Likes etotheipi

Related Threads on Capacitors connected in series: Why is the voltage the same?

  • Last Post
2
Replies
34
Views
2K
Replies
2
Views
807
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
2
Replies
28
Views
2K
Replies
3
Views
11K
Replies
19
Views
630
Replies
5
Views
7K
Top