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

  • #1
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Summary:

We have three capacitors connected in series, my book says that plates which are connected are at same voltage. I don't agree with this because plates are equal and oppositely charged.

Main Question or Discussion Point

Here is a circuit diagram:
Screen Shot 2020-06-18 at 11.06.30 AM.png
.
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.
 
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Answers and Replies

  • #3
davenn
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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"
 
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  • #4
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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 ?
 
  • #5
etotheipi
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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!
 
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  • #6
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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.
 
  • #7
Ibix
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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?
 
  • #8
DaveE
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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.
 
  • #9
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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).
 
  • #10
etotheipi
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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.
 
  • #11
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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?
Screen Shot 2020-06-18 at 2.04.00 PM.png
 
  • #12
davenn
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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 :wink:
 
  • #14
etotheipi
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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:

1592469568883.png


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
 
Last edited:
  • #15
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@etotheipi @Ibix Why electric field is zero all the way through the purple path?
 
  • #16
etotheipi
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@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:

1592470280923.png
 
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  • #17
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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 ?
 
  • #18
anuttarasammyak
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@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.
 
  • #19
etotheipi
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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##.
 
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  • #20
Ibix
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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.
 
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  • #21
cnh1995
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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.
 
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  • #22
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But the E-field outside the wire is not necessarily 0.
That’s really what I’m saying.
 
  • #23
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@Ibix @etotheipi
1592470280923.png
.

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.
 
  • #24
cnh1995
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That’s really what I’m saying.
This does not mean A2 and A3 are not at the same electrostatic potential.:wink:
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.
 
  • #25
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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.:wink: 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) 😁
 

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