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

[hutchphd]

Sometimes the differential form of Maxwell is more useful. Clearly the curl of the E field you drew is manifestly nonzero at the edge. Alarms should go off in your head ( you need to train them well !)

I'm absolutely certain it will not happen again.

 

[Adesh]

There may still be an E field outside the wire between the plates. There is only no E field inside the conductor.

If there is a field outside the wire and between the plates, then how the work done (Voltage difference) will be zero for the purple path in post #11?

 

[cnh1995]

If there is a field outside the wire and between the plates, then how the work done (Voltage difference) will be zero for the purple path in post #11?

If you drew the electric field map, the two plates along with the conducting wire will be an equipotential region.

 

[Adesh]

Clearly the curl of the E field you drew is manifestly nonzero at the edge

I would really like to enrich my next to nothing knowledge by asking how that diagram is manifesting non-zero curl at the edges? I know there will be some field at the edges which fill follow a curvy path but they will begin and end on the plates. Can you please tell me where my friend, @etotheipi, assumed a non-zero curl? (I know he missed the field at the edges)

 

[hutchphd]

The curl is the infinitesimal version of the closed line integral (at least in my head). So consider a very small circular line right at the edge of the field...half the integral will be inside and entirely positive (or entirely negative) and the rest will be zero. Absent $\frac {\partial B} {\partial t}$ that nonzero result is impossible.

 

[Dale]

If there is a field outside the wire and between the plates, then how the work done (Voltage difference) will be zero for the purple path in post #11?

The point at the corner may be at a higher potential so whatever potential difference you go up getting to it you will go down the same amount going from it.

 

[Adesh]

The point at the corner may be at a higher potential so whatever potential difference you go up getting to it you will go down the same amount going from it.

For that to happen we have to have a non-perpendicular field, so does it imply that conducting wire would have distorted the field in a weird way?

 

[Dale]

For that to happen we have to have a non-perpendicular field, so does it imply that conducting wire would have distorted the field in a weird way?

Yes, defnintely. A conductor will distort the E-field outside the conductor. That is, in fact, how lightning rods work.

 

[hutchphd]

I think one other piece of realism missing from this picture should be touched upon. I am unsure that it will affect anyone's thought processes, but we shouldn't be too enamored of our sketches.
The equilibrium charge distribution in the plates will always favor the outer edges. For a cylindrical plate it will look like you took a uniform spherical shell of charge and squashed it flat, for instance. This will always make the edge effects worse than the pretty drawings. Square plates have corners!
And one more minor semantic question

An ideal capacitor has a uniform electric field that exists only between the two of its plates. Anywhere outside the cuboid bounded by the ideal capacitor the electric field is still of zero magnitude, not just inside the wire.

My definition of an ideal capacitor is $C=Q/V$ and nothing else (no inductance, no resistance, no hysteresis). Not a question of fringing fields. Perhaps a specific idealized capacitor is being described.
I am not trying to be pedantic on a whim here...much of this discussion and several previous about batteries and capacitors have been ill served by imprecise language InMyHumbleOpinion.

 

[sophiecentaur]

Well, why there is no electric field between the connected plates? We have a negative charge on one side and positive charge on the other Why why would there be no field ?

If you are dealing with ideal wires, there can be no potential between the ends. Field is the gradient of the Potential so it is zero.

My definition of an ideal capacitor is C=Q/V and nothing else (no inductance, no resistance, no hysteresis). Not a question of fringing fields. Perhaps a specific idealized capacitor is being described.
I am not trying to be pedantic on a whim here...much of this discussion and several previous about batteries and capacitors have been ill served by imprecise language InMyHumbleOpinion.

I agree. The whole of this thread rambles between elementary circuit theory and advanced EM concepts. If the OP wants, as he says,

I would really like to enrich my next to nothing knowledge

then we should insist the the first pass through this question uses the very basic circuit rules. Only when it becomes 'obvious' how ideal components behave that is worth getting involved with Maxwell; that approach only adds confusion. Its can be really unhelpful to introduce ideas that are too sophisticated when trying to answer what is really a basic question.

 

[gleem]

Its can be really unhelpful to introduce ideas that are too sophisticated when trying to answer what is really a basic question.

Amen.

The figure in post 23 is misleading. The conductor between the two capacitors is an equipotential surface. Near the surface of the conductor, equipotential lines follow the outline of the conductor gradually departing from that outline as you farther away. This behavior prevents or reduces to a great extent a field between the two capacitors. Below is my guestimate of the equipotential lines between the capacitors. EDIT: I revised my original equipotential lines for the original post.

Most of the electric field outside of the capacitors should only be extending to or from infinity.

Also, keep in mind that the electric field is ∇V/Δx this being many orders of magnitude greater in the capacitor then outside of it since most likely Δ x < 0.1 mm

 

[Adesh]

Yes, defnintely. A conductor will distort the E-field outside the conductor. That is, in fact, how lightning rods work.

So, what we do is to generally accept that the field gets distorted so that the line integral $\int_{L} \mathbf E \cdot d \mathbf l$ is zero for any path $L$? I think the distorted field will be unimaginative, beacuse if for purple path the integral zero then it becomes natural to ask will it be zero even if we follow a straight perpendicular path from A2 to A3 (outside the wire).

 

[Delta2]

So, what we do is to generally accept that the field gets distorted so that the line integral $\int_{L} \mathbf E \cdot d \mathbf l$ is zero for any path $L$? I think the distorted field will be unimaginative, beacuse if for purple path the integral zero then it becomes natural to ask will it be zero even if we follow a straight perpendicular path from A2 to A3 (outside the wire).

You are worrying too much, and try to find every possible explanation for every possible path , why the line integral becomes zero. It should be enough that in circuit theory we do the approximation that the E-field is conservative and hence if we find the line integral zero for one path, then "the universe will conspire in view that the E-field is conservative" and there will be an explanation on why the line integral becomes zero for any other path with same end points. You don't have to know every possible explanation for every possible path, in my opinion.
But you are right deep down that the E-field is not conservative, even in the case of charging or discharging capacitors with a DC source, there is a time varying current, hence a time varying magnetic field (or time varying vector potential) and it will be $$\nabla\times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}\neq 0$$ and hence the e-field will not be conservative and hence the line integral might be different for different paths.

 

[anuttarasammyak]

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/ .

 

[sophiecentaur]

So, what we do is to generally accept that the field gets distorted so that the line integral $\int_{L} \mathbf E \cdot d \mathbf l$ is zero for any path $L$? I think the distorted field will be unimaginative, beacuse if for purple path the integral zero then it becomes natural to ask will it be zero even if we follow a straight perpendicular path from A2 to A3 (outside the wire).

I think you are making assumptions about what the line integral implies. Take an entirely different situation, involving Gravitational potential. To travel on a horizontal roadway one side of a mountain to a point with the same altitude on the other side involves NO CHANGE of potential; the line integral is Zero. If your path takes you up one side and down the other, there is still NO CHANGE in potential. The height you went to, on the way makes no difference. Similarly, whatever fields you pass through on the way from A2 to A3, the line integral will be Zero - just as it is when you travel through the wire.
Look at the black Potential contours in the above post from @gleem. You can see the equivalent of the mountain's height contours.
You need to stamp on that little monster in your brain that's saying it doesn't feel right!

 

[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.