# Capacitance in Series

1. Jun 21, 2006

### meldave00

have a question in regards to capacitors in series. We are taught that the charge must be equal on all plates of the capacitors in the case of a series connection. However, we also know that Q = CV. If we increase V, then Q must increase. How is this possible for the node between the two capacitors. Does it not have a finite amount of charge? Also, does not all the charge at the middle node just immediately rush to the perspective plates no matter what voltage is applied? Or does just some charge rush to the plates?

Hope this isn't too confusing. I will clarify more if people respond.

regards,

David

2. Jun 21, 2006

### Gokul43201

Staff Emeritus
You'll have to explain what you mean by "node".

What happens with a typical capacitor is that when you apply a potential difference (connect a battery) across the plates (or electrodes) of a capacitor you are providing a path and a reason for charge to travel from one plate to the other. The total charge is conserved - all that happens is a kind of rearrangement that minimizes energy (or balances forces).

3. Jun 21, 2006

### benzun_1999

Your resoning might seem logical but not scientific though Q=CV , the charge being withdrawn from the battery is due to the net capacitance due to the entire series combination. Now the charge on all plates remain same due to conservation of charge. So even if u change the capacitance of a capactior this equality remains ie the charge on all the plates will be same.

-Benzun

4. Jun 21, 2006

### meldave00

Let me reiterate.

Imagine two caps in series and a voltage source connected across both of them. In order to calculate the individual voltage drops across each cap you need to use the Q=CV relationship for each. The first thing that you need to do is assume that the charge Q is the same on both caps in order to solve for the individual voltage drop across each cap. I made this assumption and was able to solve the voltage drops across each cap correctly. However, it got me thinkin. How does the charge on the connection between the caps increase. If the caps remain the same value, but you just increase V then the charge Q must increase as well.

But how does it increase on the conductor between the caps??

regards

David

5. Jun 21, 2006

### Gokul43201

Staff Emeritus
It does not. What increases is the absolute value of the charge. To reiterate what I said above, increasing the voltage only takes more charge from cap#1 and transfers it across the node to cap#2 . As a result this plate of cap#1 becomes more negatively charged and the plate on cap#2 (that it is connected to) becomes more positively charged. But the total charge is still zero.

Remember that with the equation you wrote down, Q does not represent the total charge in the capacitor. It represents the absolute value of the charge on either of the electrodes of the capacitor. If one electrode has charge Q, the other has charge -Q.

6. Jun 22, 2006

### meldave00

First of all. Thank you for the reply. You have no idea how much I have been thinking of this problem.

Second. Please bare with me during this discussion. Please be patient with me on this one. I may take me awhile to understand this one.

So. Let me get a couple of things clear.
1). Are you saying that this is a false statement?
abs(Q1pos) = abs(Q1neg) = abs(Q2pos) = abs(Q2neg), where pos represent the top plates charge and neg represent the bottom plates charge.

2). I'm assuming that negative charge is electrons and positive charge is lack of electrons?

3). If #2 is true. Does not just all the electrons move to the bottom plate of Cap#1 with the application of any voltage across the caps? If the supply voltage increases, how can more electrons move towards the bottom plate of cap#1 from top plate of cap#2. The top plate of cap#2 does not have any more electrons to "transfer" as you call it.

Last edited: Jun 22, 2006
7. Jun 22, 2006

### Tom Mattson

Staff Emeritus
The charge doesn't increase on the node between the caps. I think you're missing the idea of displacement current here. You have a single conduction current $i$ that enters the first cap. That conduction current stops dead at the positive plate as positive charge builds up there (it's really negative charges leaving, but whatever). That conduction current doesn't flow across the capacitor plates, but rather is replaced by the displacement current $i_d=\epsilon d\Phi/dt$, which is exactly equal to the conduction current. Thus, the current is continuous.

On the opposite plate of the same capacitor (the plate that is connected to the node you speak of), negative charge builds up and pushes the positive charge away from it. Thus, the conduction current resumes. Observe that there is no charge stored here: the charge flows until it hits the near plate of the other capacitor, at which point another displacement current builds up, and the above process repeats itself.

8. Jun 22, 2006

### nrqed

By cap I guess you mean what I would call a plate of the capacitor. I don't know what you mean by the "value of the cap". I guess you mean the capacitance, which depends on the size of the plates, the distance between them and the material between the plates (for parallel plate capacitors, there are also a lot of different types of capacitors).

Let's sya there are two parallel plate capacitors (with air between the plates) in series.

You have

wire - plate - air - plate - wire - plate - air - plate -wire

And you connect this to a battery. If you increase the potential difference applied to the batteries, and you focus on the wire conencting the two capacitors, what happens is that some positive charges will move to one of the plates and and equal amount of negative charge (in absolute value) will flow to the other plate. The net charge of the wire conencting the two capacitors does not change (it remains zero). The distribution of the charges has changed. There is more charge on the two capacitors but if you look at the polarities you will see (say)

positive plate-air-negative plate-wire-positive plate-air-negative plate

That is, the negative plate (cap) of one capacitor is connected to the positive plate of the next capacitor and so on if there are several capacitors.

9. Jun 22, 2006

### meldave00

Capacitor in Series

Again thank you for your patience in this matter. I hope you stick with me on this matter. I'm getting close to understanding. By the way, what is your name or handle? (Don't have to give it to me if you don't wish)

I like your word picture of the capacitor circuit. I've added GND and Vsupply to complete the circuit. I've apppended numbers for each plate and wire. I will be referring to each plate/wire/air numbers throughout the message.

GND - wire1 - plate1 - air1 - plate2 - wire2 - plate3 - air2 - plate4 -wire3 - Vsupply

where : plate1 and plate2 make capacitor1
and plate3 and plate4 make capacitor2

Here is what I understand (Stop me if I'm incorrect):
1). When Vsupply is initially applied to this circuit. Electrons will build up on plate1 which in turn repels electrons away from plate 2 thus leaving positive charge built up on plate2. The electrons leaving plate2 causes electron to build up on plate3 which in turn repels electrons away from plate4(leaving behind positive charge) and the process is done and the circuit settles into equilibrium when Vcapacitor1+Vcapacitor2 = Vsupply. There is no current flowing thru the capacitors per se, but just displacement currents when Vsupply is first applied. No currents or displacement currents flow thru the circuit after the circuit settles into equilibrium.

2). Quote from my physics book. "For a series combinations of capacitors, the magnitude of the charge must be the same on all the plates when in equilibrium." That is Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). By the way, you can look this up in any physics book that explains two capacitors in series.

Here are my questions again:
It's obvious that I'm interested in the region that consists of plate2 - wire2-plate 3

1). When VSupply is increased then the voltage across each capacitor will increase which in turn will increase all the charge on each capacitor according to the Q = CV equation where Q = Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). This must mean that more additional electrons must leave plate2 and migrate over to plate3? My question is: How can there be more electrons to move from plate 2 onto plate3 with the addition of more voltage. Why doesn't "all" the electrons move over to plate3 with any small amount of Vsupply. Since the plate2 - wire2-plate 3region is a conductor (very low resistance). I would think that all the electrons would flow to plate 3 with any small amount of Vsupply voltage and then not change with any additional Vsupply voltage because "all" the electrons would have already migrated over to plate3.

1a) According to Gauss's Law, there is no "net" charge stored inside a conductor when in equilibrium and there is no electric field inside a conductor when in equilibrium. Also, all "net" charge must reside on the surface of a conductor when in equilibrium. So in the case of plate2 - wire2-plate3 region (which is a conductor), plate3 must borrow more additional electrons when Vsupply is increased. Is it borrowing the additional electrons from the surface of plate2 or from within the wire2? In which case if it is borrowing more electrons from wire2 then the electrons would not be considered "net" charge, but some other type of charge. Is there a difference? You stated earlier that the "net" charge within the plate2 - wire2-plate3 region is zero.

2). You stated that the net charge is zero in the plate2 - wire2-plate3 region. However, there is negative charge stored on plate3 and positive charge stored on plate2. I am wondering why this does not create potential difference form plate2 to plate 3? I know that it does not because its a conductive wire in the circuit that has very low resistance. However, I was under that assumption that charge imbalance would create an Efield which would create a voltage drop. Obviously I am wrong, but can you explain?

3). Last but not least. This is a doozy!. What is voltage? I know that voltage is a scalar quantity the equals the amount of work done to move a positive point charge from A to B within an electric field. It the line integral of the efield across some distance ds. This is the physics book definition, but what does it mean from a charge perspective. Say you have a node A in a circuit that is a very long distance away from node B. The nodes are far away from each other as to not be effected by any Efield produced by the charges that resides in the nodes. Say NodeA is 5V and NodeB is 3 Volts. What is the difference between the nodes from a charge standpoint? Is there more electrons in NodeB than NodeA? Or is there a greater charge density in NodeB than NodeA?

10. Jun 22, 2006

### nrqed

Hi .Myname is Patrick
Ok
Right
Right
As the electrons move away from plate 2, plate 2 becomes more and more positive obviously. That means that it gets harder and harder to move electrons away from plate 2 (more positively charged plate 2 is, the more the electrons are attracted by it!).

The key point is that there are two effects: the electrons of plate 1 are repelling the electrons of plate 2, pushing them toward plate 3. But at the same time, thenet positive charge of plate 2 are *attracting* electrosn toward it. Equilibrium is reached when the two effects cancel (that depends on how much charge is accumulated on plate 1, which depends on the capacitance of the first capacitor and on how "strong" the battery is.)
Well, it would be hard to follow individual electrons so it's hard to say But what is most likely to be happening is that electrons are pushed away from plate 2 and these push away some nearby electrons which then push away nearby electrons and so on in a domino-type effect.
The net charge is simply= (e times the number of protons - e times the number of electrons). No net charge means that the number of electrons is equal to the number of protons. For an isolate system the net charge can never change. The net charge of the system plate 2+ wire 2+ plate 3 does not change.
There is also an E field created by the charges on plate 1. It's the total E field which cancels out, at equilibrium.

More in a minute

Patrick

11. Jun 22, 2006

### nrqed

It has nothing to do with the amount of charge at the nodes or with charge density. And notice that there is no accumulation of charges at nodes (unless by nodes you would mean the plates of a capacitor). The charges are constantly flowing.

The electric potential is not a measure of how much charge is at a point.
The best analogy is in terms of gravity. Imagine skiers in a ski resort, taking a ski lift, going uphill , skiing down etc. Then the potential at a point is the analogue of the "height" (or altitude if you will) of a point.The larger the potential difference between two points is, the more work is required to move a charge between those two points (if it is on the way "up") or the more energy it will give away between those two points (if it moves on the way "down").

(here up means in the direction of increasing potential if the charge is positive and in the direction of decreasing potential if the charge is negative. A negative charge is like a "negative mass" in this analogy)

12. Jun 23, 2006

### meldave00

Capacitor in Series

So in the plate2 - wire2 - plate3 region there are multiple efields occuring.
One from plate 2 to plate 1 and one from plate2 into the wire 2 toward plate 3. These two Efield cancel out so that there is no potential drop across theplate2 - wire2 - plate3 region.

So there is either a step boundary of positive charge to negative charge going from plate2 toward wire2 and plate3. Or there has to be a gradual gradient of electrons that are at a minimum closest to plate2 and a maximum on plate3? Any idea?

I'm invisioning that all the electrons are not accumulated on plate 3, but is spread out across the wire2 to plate 3 connection where the electrons are maximum on plate 3.

Also, for the Voltage question. I've heard the analogy for uphill mountain potential comparision for voltage. However, no matter where you are on the hill there is still a constant force being exerted on an object no matter where they are on the hill.

However, in a electric circuit. I believe that it is different. I believe that it is more of a water analogy. The pressure of water is analogous to voltage. The greater the pressure the greater the voltage. This means that that there is more force on the electrons at the lower potential V and less force being exerted on the electrons at higher potential V. This is different than a hill analogy because an object on a hill always has the same amount of force on it no matter where its at on the hill. Granted the higher the object is the more work it can do. But still if has the same force of gravity on it no matter what.

13. Jun 23, 2006

### meldave00

Capacitors in Series

Sorry for the multiple threads. I'm new to this forum. I'll be better from here on out. I will consider this thread the main thread. Again sorry for the multiple threads. I don't want to scare people away.

14. Jun 24, 2006

### nrqed

That is not true. The force depends on how steep the slope is. The slope is not necessarily of equal steepness everywhere! This allows for a varying force. There may even be sections which are flat so that the force is zero (here by force I obviously mean the net force)
But it is NOT true that the lower the potential, the stronger the force on an electron is!!!!!!!!

An electron could be at a point where the potential is one gazillion volts and feel no force at all! Another electron could be at a point where the potential is zero and feel a huge force!

One way to see that the value of the electric potential at one point is unrelated to the force is that the value of the potential at any given point is arbitrary. By changing the choice of ground, the value of the potential will change. But of course the value of the force can't change.
(in a circuit, the potential at a given node can be defined to be anything you want, for example).

What *does* determine the force is the rate of change of the potential with respect to the position. So the potential is really analogue to the *height* of a point on a hill (you can put the origin of your coordinate system at the top of the hill, 100 m above the top of the hill, at the center of the Earth or, as people usually do, at sea level so the height could be anything). If you want to determine the force on a ball placed at some point on the hill, just knowing the height of that point is useless. what you need to know is the gradient of the height at that point, basically how steep the hill is. It's the same thing for the electric potential. You eed to know how it *varies* near the point where your charge is placed.

15. Jun 26, 2006

### meldave00

Capactitor in Series

Nrqed,

I started a thread in "Electrical Engineering" under the subject of "Voltage". I'm going to reply to your response there. I felt the subject of voltage deserved its own thread. m Also, please read the history of replies prior to my response to you.

regards,

David

16. Jun 30, 2006

### Firefox123

Hi David...

Since this appears to be the active thread I will post it here. Hopefully you will find something useful in my reply.

Keep in mind that plate3 is negative with respect to other specific points or nodes in the circuit and the same applies to plate2. We put a '+' symbol on plate2 because we are denoting a voltage drop over plate2-air1-plate1. We put a '-' symbol on plate3 because we are denoting a voltage drop over plate4-air2-plate3.

Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1.

Keep in mind that when you say “NodeA is 5 V and NodeB is 3 Volts” you have to have some kind of a point of reference…..Node A is 5 Volts and Node B is 3 volts with respect to what? Lets say we have a third “node”…Node ‘C’ which is ground.

What are we really saying? We are saying that Node A is 2 volts higher in potential (which is potential energy per unit charge) than Node B is and if we put a “test” charge in the potential field of Node A and Node B we would notice a difference of 2 Volts in the potential energy per unit charge.

Russ

17. Jun 30, 2006

### meldave00

Russ,

Thanks for the reply. By the way, how do you quote other peoples words?
I haven't figured that out yet.

Any in response to your response. You said "Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1."

What I'm saying is that there is a charge inbalance in the node between the two capacitors. So why is there not a voltage drop? I know that the E-Field inside a conductor is zero when in equilibrium.

regards,

David

18. Jul 1, 2006

### Firefox123

Ill put an extra character that you would not include in the example by using a '*'

To quote someone type...[*quote=whoeveryouarequoting]*Insert the persons text here*[*/quote]

So the commands are [operation]thing to operate on [/operate] where the '[]' signify the operation and the '/' is the close for that operation.

To make text bold type [*b]*Insert text here[*/b]

Okay...Im going to give this a shot. Someone please correct me if I am wrong.

Here is what I believe happens....

Lets focus on the node in between the two capacitors. So we will look at both plates plus the wire conductor in between.

When the caps are charging we have one plate that basically deposits electrons onto the other plate via the wire conductor.

We still have the same number of charges in the plate-wire-plate system, but the charges are now distributed differently than they were before.

On one plate we have a '+' charge because electrons were removed and on the other plate we have a '-' charge because electrons were "added".

But what about the wire in between? I would imagine that the free electrons in the wire are not evenly distributed along the wire but are displaced the exact opposite of the plates.

In other words, I would assume a higher electron density near the positive plate than near the negative plate. So an opposite electric field is created in the wire so the net field is zero.

Some might object at this point and ask why the electrons dont flow onto the positive plate and replace the charge deficiency there....I would answer that the electric field across the capacitor prevents this from occuring on both capacitors.

So we have four different fields in this situation.

Fields #1 and #2: The two fields across both capacitors repel and attract charges to and from the plates and create a potential energy between the plates. If we put a test charge in this region, it will flow from one point to another.

Fields #3 and #4: These two fields are set up by the charge distributions in between the two capacitors.

Field #3 is the field between the plates caused by the local distribution of extra electrons on one plate and missing electrons on the other plate.

Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that a test charge placed anywhere in this region will experience no net electric field acting on it.

So my short answer is that the reason there is no voltage drop is because the charge distribution in the wire cancels out the electric field created by the charge imbalance. Since there is then no net electric field, there is no voltage drop.

I hope that helps David....this is the best I can do for a quick answer. Someone please correct me if I am way off here.

Russ

Perhaps it is not the charge distribution in the wire, but the charge distribution on the plates themselves that causes this opposite electric field.

Think about a single capacitor that is charged ...as one side becomes "negative" the other becomes "positive".....so the "positive" charged plate helps to hold the electrons in place on the "negative" charged plate.

Now disconnect the capacitor and think about the molecules and free electrons on both plates...

The negative plate has "extra" electrons on its surface that repel the remaining free electrons on the positive plate to the outside of the plate.

So if we looked at charge distribution from left to right going from negative plate to positive plate we get.... '-' '+' '-'

Perhaps this distribution is what creates the opposing electric field bewteen the two capacitors.

As a final thought maybe we can say that each individual situation is different depending on the value of the capacitors and how close they are together, but an opposing field is set up within the plate-wire-plate interface by the distribution of charges on the capacitor plates combined with the effect this has on the free electrons in the wire.

Does that make sense?

I would be surprised if the electron distribution in the wire in between the capacitors is uniform for all cases of two series capacitors.

Russ

Last edited: Jul 1, 2006
19. Jul 3, 2006

### meldave00

Russ,

I like what you are saying. I need to think about what you said a little longer.

But in the mean time here is something for you to think about. Gauss's Law keep poppin in my head when I think about your response. Gauss's Law says that all excess charge lies on the surface of a conductor in equilibrium and that there is no excess charge inside a conductor in equilibrium and the net E field inside a conductor is zero in a conductor in equilibrium. I'm wondering if I'm misintrepreting this law incorrectly or not. How does Gauss's Law stay valid for the case of the node between two capacitors in series. Again, I'm still thinking about your answer. You might be on to something.

bye for now,

David

20. Jul 3, 2006

### Firefox123

Hello again Dave...

It is true that

1. All excess charge lies on the surface of a conductor in equilibrium.

2.There is no excess charge inside a conductor in equilibrium.

3.The net E field inside a conductor is zero in a conductor in equilibrium.

Assuming that any one of these 3 statements are true, we can then use Guass's law to prove the other two.

Okay...Guass's relates the flux of an electrostatic field over some surface to the charge enclosed by that surface.

So Guass's law looks like this:
$$\iint_{\mathcal{S}}\vec{E}(x,y,z)\cdot{\vec{n}} dS = \frac{q} {\epsilon}$$

where $$q$$ is the enclosed charge and $$\epsilon$$ is the permittivity of free space.

So the "flux" or "amount of flow" (not to be taken literally, since the electric field really doesnt "flow" like a fluid does) of the electric field through the surface is
$$\iint_{\mathcal{S}}\vec{E}(x,y,z)\cdot{\vec{n}} dS$$

Suppose we have some conductor and we put a bunch of electrons inside the conductor....what will happen?

The electrons will repel each other and move as far apart as possible which means they will be on the outside of the conductor. So we have a bunch of free electrons just sitting on the outside of the conductor.

So why would the electric field be zero inside the conductor even though each electron is surrounded by an electric field?

The reason why it is zero is the same reason why the charges eventually stop moving...because a state of equilibrium has been reached. Each charge experiences a net force of zero from the other charges because they are distrubuted in such a way that the fields end up cancelling each other out.

Kind of like one person pulling your right arm while another pulls your left arm with equal force.

The same is true for the plate-wire-plate region. We have a case of equilibrium.

Picture the two capacitors plus the wire in bewteen....go from left to right with the negative side of the battery on the left..Ive exagerrated the spacing between the capacitor plates to make it easier to visualize. The charge signs in () are the induced charges on the outer edge of the plate away from the opposite capacitor plate.

CAP 1 ***wire in between*** CAP 2
- +(-) ______________________ - +(-)

Before equilibrium is reached the negative plate of CAP 1 (connected to the negative plate of the battery) forces electrons to leave the positive plate of CAP 1 and go onto the negative plate of CAP 2

As charge builds up on CAP 2 it increasingly resists this push from CAP 1 until equilibrium is reached and the "push" from CAP 2 equals the "push" from CAP 1.

I hope that makes sense.

***Curiousity question....what would happen is we removed the wire along with the two plates connected to it?

I would think that the part of the circuit still connected to the battery would not change but would just be an open circuit at battery potential....but what about the section no longer connected to the circuit?

What about the plate-wire-plate section? I would imagine that the induced charge on the plate of CAP 1 would no longer be seperated and the excess static charge on the plate of CAP 2 would flow through the wire to the CAP 1 plate as a static discharge....

Russ

Last edited: Jul 3, 2006