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

[Adesh]

Here is a circuit diagram:

.
We have three capacitors, with capacitances $C_1$, $C_2$ and $C_3$. Plates are labelled as $A_1, A_2, A_3 ... A_6$. Point P is connected to the positive terminal of the battery and point N is connected to the negative terminal of the battery, positive terminal of the batter will provide a charge of $+Q$ to the plate $A_1$ and a charge of $-Q$ to the plate $A_6$. Due to mutual induction we will get a charge distribution as shown in the diagram. (Sorry if my diagram contains too much letters in a small space). Now, my book says that

let the point N be at potential 0, as point N is connected to the plate $A_6$ therefore plate $A_6$ will also be at potential 0. Assume point P to be at potential $V$, so the plate $A_1$ will also be at potential $V$. Let plate $A_2$ be at potential $V_1$, since it is connected to plate $A_3$ we will have the potential of plate $A_3$ also as $V_1$. Similarly, assume plate $A_4$ to be at potential $V_2$ so will be the plate $A_5$ as they are connected.

Now, the problem is since the plate $A_2$ and $A_3$ are oppositely charged how can they be at same potential? My intuition tells me that there will be a net field going from $A_3$ to $A_2$, but I also know that they are connected by a conducting wire and no field exist inside the conductor. I'm just unable to connect the dots, can someone please clear this doubt.

Thank you.

 

[anuttarasammyak]

Please find attached my sketch.

 

[davenn]

Summary:: We have three capacitors connected in series, my book says that plates which are connected are at same voltage.

if the capacitor values are the same, then the voltage across each cap will be the same and will add up to the power supply voltage
if your example of 3 cap's and if equal value, the voltage across each one will be 1/3 of the total voltage.
This is Kirchhoff’s voltage law and it applies to all series circuits.

If the values are unequal, then you can work out the values of the voltage drops across each cap. and their values will still
add up to the supply voltage

I don't agree with this because plates are equal and oppositely charged. Thank you.

A bold thing to say in the face of proven theory, you would have been better to say ... " I don't understand how"

 

[Adesh]

A bold thing to say in the face of proven theory, you would have been better to say ... " I don't understand how"

Yeah, you’re right I should be careful with my writings. But can you please help me with that part? Why they are at same voltage ?

 

[etotheipi]

Yeah, you’re right I should be careful with my writings. But can you please help me with that part? Why they are at same voltage ?

The voltage between two points is $V = -\int_a^b \vec{E} \cdot d\vec{x}$. Inside the ideal wire between the two (oppositely charged) plates, in a steady state, we take $\vec{E} = \vec{0}$ which also means $V =0$. Any two points in a circuit that are joined by an ideal wire (and nothing else in between) are at the same potential!

 

[Adesh]

The voltage between two points is $V = -\int_a^b \vec{E} \cdot d\vec{x}$. Inside the ideal wire between the two (oppositely charged) plates, in a steady state, we take $\vec{E} = \vec{0}$ which also means $V =0$. Any two points in a circuit that are joined by an ideal wire (and nothing else in between) are at the same potential!

I too thought that, but we can follow any other path and hence integral will not be zero. In your integral you took the path totally inside the wire but Voltage should path independent and hence it’s value will not be zero for some other path.

 

[Ibix]

I too thought that, but we can follow any other path and hence integral will not be zero.

Can you specify what path you are thinking of where you think this statement is true?

 

[DaveE]

if the capacitor values are the same, then the voltage across each cap will be the same and will add up to the power supply voltage
if your example of 3 cap's and if equal value, the voltage across each one will be 1/3 of the total voltage.

The voltages do have to add up to the battery voltage, but they don't have to be equal. The relative voltages on the capacitors depends on the initial state of charge distribution which doesn't have to be equal. Of course, everyone assumes that they are, since there doesn't appear to be any mechanism shown to "set up" the initial state except for connection to the battery. However, there is nothing in the math that says you can't have different capacitor voltages.

 

[Adesh]

Can you specify what path you are thinking of where you think this statement is true?

We can go from $A_2$ in a cycloid manner to $A_3$ (cycling out of the wire).

 

[etotheipi]

We can go from $A_2$ in a cycloid manner to $A_3$ (cycling out of the wire).

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.

 

[Adesh]

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.

Can't we follow this purple path?

 

[davenn]

The voltages do have to add up to the battery voltage, but they don't have to be equal. The relative voltages on the capacitors depends on the initial state of charge distribution which doesn't have to be equal.

YES... I stated that

 

[Ibix]

Can't we follow this purple path?
View attachment 264809

Yes. As etotheipi said in the post you quoted, the electric field is everywhere zero along it.

 

[etotheipi]

Can't we follow this purple path?

You sure can, but the electric field is zero at all points along the path so the line integral is still zero. A slightly more sketchy path would be this one:

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.

Edit: Last part is incorrect, see post #35

 

[Adesh]

@etotheipi @Ibix Why electric field is zero all the way through the purple path?

 

[etotheipi]

@etotheipi @Ibix Why electric field is zero all the way through the purple path?

Take a single isolated (charged) capacitor, and surround it completely with a Gaussian surface. You have $Q=0$ which means that the net flux is zero. And since there are no external electric fields outside the capacitor as a whole, we deduce $\vec{E} = \vec{0}$ outside the capacitor too.

When you connect them up, there is still only an electric field between the plates. It looks like this:

 

[Adesh]

Take a single isolated (charged) capacitor, and surround it completely with a Gaussian surface. You have $Q=0$ which means that the net flux is zero. And since there are no external electric fields outside the capacitor as a whole, we deduce $\vec{E} = \vec{0}$ outside the capacitor too.

When you connect them up, there is still only an electric field between the plates. It looks like this:

View attachment 264811

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 ?

 

[anuttarasammyak]

@etotheipi @Ibix Why electric field is zero all the way through the purple path?

Put a wire on the purple path. Nothing change. So you know with or without real wire, E=0 on the purple path. ref.my sketch on post #2

We have a negative charge on one side and positive charge on the other Why why would there be no field ?

We have also a positive charge on one side far and a negative charge on the other side far.
All these charges should be considered.

 

[etotheipi]

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 ?

Try thinking about the superposition of electric fields at any point due to infinite sheets of charge. At any point outside of a capacitor, the total $x$ component of the electric field due to that capacitor is $\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$.

And if you have 3 capacitors in series, the electric field between the first and second capacitor (for instance) would be $\frac{3\sigma}{2\epsilon_0} - \frac{3\sigma}{2\epsilon_0} = 0$.

 

[Ibix]

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 ?

Because the field lines run from one plate to the other of the same capacitor. You are thinking of the connected plates and forgetting the other plates of the capacitors, which also have charge.

 

[cnh1995]

Referring to the diagram in the OP:
Shouldn't there be some surface charge buildup on the connecting wire between A2 and A3, which cancels the E-field due to capacitor plates inside the wire?

The E-field inside the wire is 0. But the E-field outside the wire is not necessarily 0.

 

[Adesh]

But the E-field outside the wire is not necessarily 0.

That’s really what I’m saying.

 

[Adesh]

@Ibix @etotheipi

.

Capacitor 1 will have no influence whatsoever outside their plates (thats what we get when we consider two oppositely charged plates of infinite dimensions we can see it here.) Now the positive plate of capacitor 2 and negative plate of capacitor 1 should have a electric field between them, the way I have drawn it.

 

[cnh1995]

That’s really what I’m saying.

This does not mean A2 and A3 are not at the same electrostatic potential.
Your book is still correct (though it is using the term "voltage" for what I understand to be electrostatic potential).

Now the positive plate of capacitor 2 and negative plate of capacitor 1 should have a electric

I believe you need to consider the field contribution from the surface charges on the wire too.

 

[Adesh]

I believe you need to consider the field contribution from the surface charges on the wire too.

That’s why I have drawn no arrows in the wire .

This does not mean A2 and A3 are not at the same electrostatic potential. Your book is still correct.

I know it’s correct, experimentally proven. You also had it during your IIT preparation (btw book is HC Verma’s)

 

[cnh1995]

@Adesh,
All textbooks use ideal circuit elements and lumped circuit abstraction for demonstrating the circuit-behavior(my lousy shortcut for 'behavior when connected as a circuit element') of the components.
In your diagram in the OP, the capacitors, wires and the voltage source are all ideal.
In case of an ideal capacitor, all the E-field exists inside the capacitor (i.e. no fringe field). So a capacitor as a circuit element is just a black box enforcing its v-i relationship across its terminals. The same holds true for all other circuit elements.
So if you apply Kirchhoff's voltage law to this circuit, all the components connected by a metal wire are at the same potential. If plates A2 and A3 weren't equipotential, there would be current flowing between them. In circuit theory (lumped circuit abstraction), things like geometry of the circuit, distance between two components, size, shape etc are irrelevant.

If you want to draw the field between the two equipotential plates, you need to consider things like shape of the wire, size and thickness of the metal plates and the distance between the two plates.

You might find this interesting.
https://books.google.co.in/books?id...hUKEwjUxcmLnIvqAhW_yzgGHaNABHkQ6AEwAHoECAQQAQ

 

[Adesh]

So if you apply Kirchhoff's voltage law to this circuit, all the components connected by a metal wire are at the same potential. If plates A2 and A3 weren't equipotential, there would be current flowing between them.

It may be off-topic, but I want to know why don’t we want any current to flow from A2 to A3? Is it like that current will flow initially until charge builds up on both the plates and finally we will get a equilibrium condition and hence no current will flow?

 

[cnh1995]

Is it like that current will flow initially until charge builds up on both the plates and finally we will get a equilibrium condition and hence no current will flow?

Sort of like that.
When you close a switch in an electric circuit, an EM wave (disturbance) flows around the circuit at nearly the speed of light. Maxwell's equations apply at each point in the circuit and fields, charges change accordingly. This all happens at an extremely small time scale.
For practical circuits, we make certain (valid and really useful) assumptions that make life much easier. These assumptions translate Maxwell's equations into the laws of Circuit Theory, which is what we extensively use in practice.
Check out this informative insights article here.
https://www.physicsforums.com/threa...circuit-analysis-assumptions-comments.930132/

 

[Delta2]

My thoughts on this:

1. In circuit theory we work with the quasi-static approximation for the E-field. This means that the E-field is everywhere conservative, that is $\vec{E}=-\nabla V$ and $\nabla\times\vec{E}=0$. This means that the integral $\int_C \vec{E}\cdot d\vec{l}$ is independent of the path C, so if we find one path C such that $\int _C \vec{E}\cdot d\vec{l}=0$ then it is $\int_F \vec{E}\cdot d\vec{l}=0$ for all paths F with the same endpoints.
2. In the quasi static approximation the potential across the region of any conductor with zero ohmic resistance is constant. Indeed the E-field inside the conductor with zero ohmic resistance is zero at all times (we can infer this from ohm's law $J=\sigma E$ since $\sigma$ is infinite(zero resistance) and $J$ is finite,$E$ has to be zero), and hence the $\int_C \vec{E}\cdot d\vec{l}=0$ for any path C that runs through the interior of the conductor (where e-field is zero) and connects any two points of the conductor. Hence the potential difference is zero and hence the two points are on the same potential. So if you treat the plates A2 and A3 and the connecting wire as one conductor with zero ohmic resistance, then the potential is everywhere the same(regardless if A2 is negatively charged and A3 is positively charged, we don't care about that).

 

[Dale]

Now, the problem is since the plate $A_2$ and $A_3$ are oppositely charged how can they be at same potential? My intuition tells me that there will be a net field going from $A_3$ to $A_2$, but I also know that they are connected by a conducting wire and no field exist inside the conductor. I'm just unable to connect the dots, can someone please clear this doubt.

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

 

[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!