Series capacitors and the amounts of charge on the plates.

[Kerrigoth]

First in Circuit A:
I hook up a battery with a voltage of V up to a capacitor C1. Once C1 is fully charged it has 6+ on its positive plate and 6- on the negative plate. Given it's fully charged, the +terminal of the battery and the +plate are in electrostatic equilibrium. Since they are both in electrostatic equilibrium AND are connected together by a conducting wire, they can be considered to be one large conductor with the same electric potential of V

Now in circuit B:
There are two capacitors in series; C2 and C3. Once fully charged, the +plate of C2 and the the +terminal of the battery are in electrostatic equilibrium and share the same electric potential because they are connected by the conducting wire. C2 and C3 duplicate capacitors that share the same specs as C1

So I'm struggling with two things.

1) Capacitors in series have the same magnitude of charge on all of the plates. This number comes from limited amount of charge that is available on the middle plates of the capacitors that are circled with purple. I understand that no charge can be added or subtracted from this area because it's isolated. What I don't understand is: Why is the amount of amount of charges inside the purple area able to limit how much charge the battery dumps onto the +plate of C2. Why can't +6 be added onto the +plate (like we see in circuit A) instead of only being limited to +3 (like we see in circuit B).

2) How can the +plate of C1 and the +plate of C2 share the same specs BUT have different amounts of charge AND be the same electric potential when hooked up to the same battery.

I can perform the math get the correct results, but I'm having a hard time coming up with the intuition to satisfy my answers. Thanks to anyone who responds. It's much appreciated.

 

[davenn]

hi there

Welcome to PF

because you are going to get a voltage drop across each capacitor. and totalling up the voltage drops will equal the supply voltage the same as for resistors in series

as an example ...

Dave

 

[Kerrigoth]

hi there

Welcome to PF

because you are going to get a voltage drop across each capacitor. and totalling up the voltage drops will equal the supply voltage the same as for resistors in series

as an example ...

View attachment 85780

Dave

Hey daaven,

Thanks for welcoming me to PF and thanks for the response! Like you showed in your post, the math very much so works out. But I'm still unclear about what phenomena is going on under the hood that address my original question(s).

I'm trying to figure out the phenomena that explains why the battery will only add a quantity of charge onto the +plate of C2 that is the same magnitude as the quantity of charge found on the middle plates that are in the purple region of my circuit diagram.

That brings me to my confusion about comparing of the +plates on C1 and C2 when they are fully charged. When I look at it from one way,their electric potential must be the same because they are connected to those same batteries by wires and are in a state of electrostatic equilibrium. This means that no point within the conductor has an electric field (which means no electric potential); This makes sense. But when I look at it from another angle then I am confused because they are essentially the same plate with different amounts of charge on them, yet they wield the same electric potential. This doesn't seem right with me.

 

[nsaspook]

I'm having a problem seeing what the 'other angle' is. B is electrically A with a center plate added.

 

[Kerrigoth]

I'm having a problem seeing what the 'other angle' is. B is electrically A with a center plate added.

This is true. But in circuit A the battery is capable of delivering more charge on the +plate of the capacitor attached to it.

In circuit A the +plate of the capacitor will hold +6. In circuit B, after you add the center plate, the +plate of the capacitor will only hold +3. What does the center plate do that prevents the battery from delivering more charge to the +plate of the capacitor?

 

[TurtleMeister]

You are taking your voltage readings from two different points. The 6 volts is from one end plate to the other end plate, and the 3 volts is from one end plate to the center plate. As Dave pointed out, you would get the same thing for two resistors in series. This is just a voltage divider.

 

[nsaspook]

This is true. But in circuit A the battery is capable of delivering more charge on the +plate of the capacitor attached to it.

In circuit A the +plate of the capacitor will hold +6. In circuit B, after you add the center plate, the +plate of the capacitor will only hold +3. What does the center plate do that prevents the battery from delivering more charge to the +plate of the capacitor?

You should use the idea of charge separation and energy storage in the electric field between the plates when you think of 'charging' a capacitor. When you say the plate changed from +6 to +3 what is the voltage reference for those values?

 

[Joe Ciancimino]

You are taking your voltage readings from two different points. The 6 volts is from one end plate to the other end plate, and the 3 volts is from one end plate to the center plate. As Dave pointed out, you would get the same thing for two resistors in series. This is just a voltage divider.

What TurtleMeister said is the correct textbook answer but I think I understand the question you are trying to ask. To visualize the circuit imagine a water pump pushing 6lbs of pressure and a pipe fitted with a a flexible diaphragm. No water can pass through the diaphragm, but as the diaphragm stretches it pushes the water on the other side. Adding a second diaphragm, this only expands from the pressure of the first diaphragm, not from any water flowing through the pipes as the passage is blocked. So with 6lbs of force being divided between the two diaphragms, each diaphragm divides the force between them making each one hold back 3lbs of pressure.

 

[sophiecentaur]

If you are prepared to accept and to start with Q = CV and, if you accept that the excess + charge on the bottom of the upper plate has to be the same as the excess - charge on the bottom plate (there's nowhere for charges to arrive on those plates but the plates themselves) so the Volts will be shared inversely with the Capacitance values. It isn't much of a step to let the Maths take over from that point to give you the 'right' answer. If you want to consider the Energy situation, you have E=QV, so the stored energy will be shared according to the PD across each Capacitor. The smaller value capacitor will have more energy - which may be a bit counter intuitive.

There's another analogue, as well as the pump / diaphragm. Consider what happens with two springs with different k values, connected in series, where the Tension is the same in both. It's the weaker spring that holds nearly all the energy. (A bit of mental gymnastics with choosing the appropriate analogue quantities will prove it). Alternatively, you can do two springs in parallel, with a weight hanging on the bottom. It's the other way round, in that case. That Maths thing is such a clever tool and language.

 

[Kerrigoth]

I'm not disputing the math adds up. What I'm struggling with is identifying what happened after I slid the center plate in between the capacitor. There must have been some transient event that caused the +plate of the capacitor from being able to hold 6+ units of charge, to only 3+ units of charge. As far as I can understand, 3+ units charge just disappeared (which obviously isn't true)

Like circuit A in my picture. Let's say one single capacitor is hooked up to battery; Instead of a voltage rating of V I will say the battery has 9V. When the capacitor is charging, + units of charge are taken from the bottom plate and put onto the top plate (this is why the magnitude of charge on the top and bottom plates are equal). After a while, the repulsion force of the top plate becomes so strong that no more charges can be added to it and the system reaches steady state. Once in steady state, I observe there is +6 units of charge on the +plate of the capacitor.

As it stands I have a capacitor that is 9V with 6 units of charge on the positive and negative plates.
Q = 6, V = 9
Q/V = C ----> C = 2/3

But now I slide a center plate between the two plates of the capacitor. The center plate has only 3+ and 3- units of charge on it. When I look back at the positive plate of the capacitor, to my surprise I see that there is not 6+ units of charge on the +plate, but only 3+ units of charge on it. Same goes for the -plate.
So at this point I essentially have two capacitors in series. Both of which have:

Q=3
C = 2/3 (as we found in this first situation)
V = Q/C
V = 4.5 (for one capacitor)
V * two capacitors = 9V equals the voltage rating of the battery.Why does the excess + charge on the positive plate have to be the same as the excess - charge on the middle plates AFTER you slide the middle plate in between the two original plates.

 

[Joe Ciancimino]

I think what you are doing is confusing the difference of amperage and voltage. From your own terminology you are trying to describe how much charge the capacitor can hold. This problem being that it is not a question of how much charge it can contain, but how much charge the battery can give. Voltage is pushed from the battery, and it can push no more, so there is no more to put on the plates. It is amperage that is the force derived from the capacitor rather than voltage.

 

[nasu]

I'm not disputing the math adds up. What I'm struggling with is identifying what happened after I slid the center plate in between the capacitor. There must have been some transient event that caused the +plate of the capacitor from being able to hold 6+ units of charge, to only 3+ units of charge. As far as I can understand, 3+ units charge just disappeared (which obviously isn't true)

.

There is no transient. You forget that in order to get two capacitors identical with the capacitor in A, the distance between plates should be double. So before inserting the middle plate you have a capacitor with half the capacitance of the one in A. So it has half the charge. Inserting the middle plate does not change anything. The capacitance stays the same (assuming thin plate) and the charge on the external plates stays the same (the 3 units). With the plate in you have two capacitors with capacitance C each which in series give C/2 again. ( C is the capacitance of the original capacitor).

 

[sophiecentaur]

What I'm struggling with is identifying what happened after I slid the center plate in between the capacitor.

Could this be your problem, perhaps? It happens whilst you are moving the dielectric between the plates. Energy transfer is involved as the charges flow out of the battery to re-establish the equilibrium; the battery volts pull the dielectric into the middle position. Inserting either a pair of conducting plates, as in the OP or inserting a Dielectric will allow more charge imbalance for a given battery voltage. (i.e. increasing Capacitance).
If you do the before and after Energy situations, there will be an error and that is because, without some mechanical friction involved in the mechanism that inserts the plate. Without some friction, the plate will actually be pulled into the middle, carry on to the other side and back again and then keep on oscillating that way for ever. It turns out that the mechanical energy involved is half the change in stored energy.
This is yet another one of the 'where does the Energy go?' questions that students get asked when learning about Capacitors.

 

[Kerrigoth]

You forget that in order to get two capacitors identical with the capacitor in A, the distance between plates should be double. So before inserting the middle plate you have a capacitor with half the capacitance of the one in A. So it has half the charge.

Wow, I can't believe I was overlooking that distance factor the whole time...When I was drawing out my situations on paper I wasn't giving the physical distance the right attention because it was just a sketch. You are correct. Thanks!

 

[davenn]

...... Voltage is pushed from the battery, and it can push no more, so there is no more to put on the plates. It is amperage that is the force derived from the capacitor rather than voltage.

Joe,
you were making this same mistake in another recent thread
Voltage is pushed anywhere. Voltage doesn't move
please take care with your wording

 

[sophiecentaur]

I think what you are doing is confusing the difference of amperage and voltage. From your own terminology you are trying to describe how much charge the capacitor can hold. This problem being that it is not a question of how much charge it can contain, but how much charge the battery can give. Voltage is pushed from the battery, and it can push no more, so there is no more to put on the plates. It is amperage that is the force derived from the capacitor rather than voltage.

I would strongly advise you to avoid using these terms in the way you are doing. What you write is not accurate and can only confuse you and other readers. You really should go back to the start and get your basics right. This stuff is not just a matter of personal opinion but well established fact.

 

[sophiecentaur]

it was just a sketch

Very risky to use sketches when the Maths is available. This applies even to putting up shelves.

 

[Let'sthink]

Kerrigoth, I appreciate your persistence. Here I am presenting another generalized view to the questions you are raising. Somewhere I think you are equating a battery and capacitor may be in a limited manner. But let me point out to you the differences. The fundamental and individual property of a given capacitor or a given arrangement of plates which makes a capacitor or capacitor system of capacitors, is its capacity it is neither voltage nor maximum charge it can store for a given voltage both vary in a proportionate manner. individual plates do contain residual positive or negative charges, which gives rise to the potential difference the plates show. Now look n which way a battery is different from a capacitor the two terminals of a battery do have a maintained potential difference between them due to a chemical reaction or may I say chemical emf. The potential difference between them is not because of the separated opposite charges. But as a dynamic equilibrium of chemical forces and the current which the battery supplies. You may treat my answer as my question put to you.

 

[sophiecentaur]

Somewhere I think you are equating a battery and capacitor may be in a limited manner.

You are right BUT it is still true to say that there is no net gain or loss of charges for either. A battery a charged capacitor both have a displacement of charge; extra electrons turning up at the negative terminal and a lack of electrons at the positive terminal. The imbalanced charge on a capacitor is just due to temporary polarisation where the 'available' imbalanced charge on a battery is maintained chemically, more charges being supplied, once the PD across it starts to drop.
In some ways, a battery can 'behave like' a big capacitor. It is not uncommon to use the two in parallel, to get the best of both worlds with low audio frequency source impedance and a good. constant DC voltage (lunatic high power audio systems in cars are an example).

 

[Let'sthink]

Arrangement of electric field for capacity and battery outside and inside is vastly different. Then the fundamental individual property of battery is its emf and internal resistance not its capacity.

 

[nsaspook]

Arrangement of electric field for capacity and battery outside and inside is vastly different. Then the fundamental individual property of battery is its emf and internal resistance not its capacity.

Say again. I'm not sure what you mean by that.

 

[Let'sthink]

For Capacitor electric field outside the plate is zero inside normal and uniform. Not so with battery terminal.
Electric field produced in capacitor is because of the held up and accumulated charges. In battery it is produced due to motional emf, whose cause is chemical reaction. As in case of a closed conducting loop, non-conservative electric field is produced due to changing of flux linked with it with time. So [hysically there is no accumulation of charges at the battery terminals.

 

[sophiecentaur]

Arrangement of electric field for capacity and battery outside and inside is vastly different. Then the fundamental individual property of battery is its emf and internal resistance not its capacity.

OF course. What goes on inside is very different. But, to someone with a resistor and a meter, the two are sometimes hardly distinguishable because they both present a charge imbalance (with PD) which will make current flow. Under many circumstances and for small changes, they can be considered as equivalent.