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

[jbriggs444]

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 ?

There is zero total charge on both sides. Look at the picture again. All of the plates matter.

 

[Adesh]

You are getting distracted by one part of the picture and not seeing the whole picture.

Consider a spherical conductor in a vertical (up) E field. You will get positive charge on the top and negative charge on the bottom. If you incorrectly consider only the field due to the surface charges then you would assume that there is a vertical (down) E field in the conductor. But that neglects the external field. The total field is the sum of the external and the field from the surface charges, which is 0 inside the conductor.

At equilibrium, a conductor produces a surface charge distribution that exactly cancels out the external field. The shape of the conductor doesn’t matter, nor does the presence or absence of an external field. The charge distribution cancels out the external field inside the conductor, which makes the voltage the same at all points in the conductor

This has really helped me. This is the way I have understood your reply: There is a conductor between the plates A2 and A3, since I’m absence of the conductor there would be a net field, but when the conductor is placed it’s free electrons distributed themselves as such so that there is no electric field anywhere between the plates.

Have I understood you correctly? Can you please shine some light on how connecting conductor and a conductor just between them (not touching them or joining them) would affect the field differently?

 

[Dale]

when the conductor is placed it’s free electrons distributed themselves as such so that there is no electric field anywhere between the plates.

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

 

[hutchphd]

Here you would get a non-zero line integral along the green path, because of the first segment between the plates. The resolution is, I think, that there are problems with doing the line integral over the discontinuity in the second segment.

This is absolutely wrong.
You cannot construct electric fields that look like your picture. In a static situation that line integral is zero because of fringing fields. Always. Always.

 

[etotheipi]

This is absolutely wrong.
You cannot construct electric fields that look like your picture. In a static situation that line integral is zero because of fringing fields. Always. Always.

Yes I realized this a bit later on. It's really a consequence of the CA assumptions. For a perfectly uniform field you would need infinite charged plates, and then the path I drew would still have the same line integral overall. Likewise if instead you accounted for fringing.

What I did was mix up two different sets of assumptions (no fringing + finite plates ), and that led to a wrong conclusion. Apologies!

I'll add a correction to that post.

 

[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.