How does this capacitor circuit not violate conservation of charge?

[lightlightsup]

Homework Statement

See image below for the question: I was able to obtain the answers of 32uC, 16uC, and 16uC. But, wouldn't this total charge exceed the original charge provided to C1 by the battery (48uC)?

Relevant Equations

See image below for the question: I was able to obtain the answers of 32uC, 16uC, and 16uC. But, wouldn't this total charge exceed the original charge provided to C1 by the battery (48uC)?

There must be something I'm not understanding about capacitors in series.

I know that we can treat them as one equivalent capacitor with:
(1) with 1/Ceq,
(2) same Q as anyone of the capacitors,
(3) and add up the Vs for the sum total V across them.

If the equivalent capacitor (Ceq) would have the same Q as anyone of the capacitors, how would I know how much Q is held by anyone of them and the sum total Q held by them?

 

[haruspex]

If you connect a battery to both sides of an uncharged capacitor, what will happen?

 

[Merlin3189]

First, it looks as if you worked it out correctly. So it's just your niggling worry to address.

Seems to me to be the same underlying idea as Kirchhoff's law.
In circuits, as opposed to static electrics, charge does not accumulate. Charge into a node equals charge out.
Here treating each component as a node.

Charge flows through C1 until the charge gained by one plate and lost by the other is enough to balance the charging voltage. No net charge is gained or lost. You calculate the shift of charge from one plate to the other.

When the switch is changed, charge flows from one plate to the other, through both other capacitors. No charge accumulates on either capacitor. As charge flows into C2, it equally flows out of C2, then into C3. Neither can it stay there, but flows equally out of C3 and back to C1. Nobody gains any net charge. It's the same amount of charge going round the circuit at all places.

If you had resistors in series which charge flowing through driven by a battery, the same charge flows in all parts of the circuit. If 1 Cb flows out of the battery, then 1 Cb flows through each series resistor, no matter how many there are! If you have ten resistors in series, each still has 1 Cb flow through it. 1 Cb out of the battery but a total of 10 Cb through the resistors?
==============================================End of post!
At the risk of being banned for life from PF, a little hydraulic picture that has nothing whatsoever to do with electricity.
The curved lines are elastic membranes and the values by the boxes show how much liquid has to flow into stretch the membrane enough to raise the pressure difference by 1 Pa. (By magic, my membrane behaves linearly, like a piston and Hooke spring, which would have been harder to draw.)

Image edited
At the start I've pumped 48 ml into the LH device. The other two have unbent membranes.
When the valve is opened, 16 ml flows out of the left device and both the right devices gain 16 ml.

In losing 16 ml the pressure difference across the first device fell from 12 to 8 Pa. In gaining 16 ml, the pressure across the other two devices rose by 5.33 and 2.66 Pa respectively.

 

[lightlightsup]

If you connect a battery to both sides of an uncharged capacitor, what will happen?

+Q will accumulate on one side, -Q on the other.
Net charge will be zero.

Let's say I have two capacitors in series.
If I am asked what the total charge accumulated on any 1 plate on both capacitors is, what would the answer be?
Q+Q = 2Q?
Q of the equivalent combined capacitor = Q on any of the individual capacitors.

 

[lightlightsup]

https://www.physicsforums.com/attachments/259304

So, it may seem like we went from 48mL excess to 64mL excess and violated conservation of mass.
But, in reality, the water is in the pipes as well? Just at a different pressure?

There's some misconception I have that's causing the math to not add up:
LH: 48mL -> 32mL (-16mL)
RH: 0mL -> 32mL (+32mL)
Where did the extra 16mL come from?

Or is the LH device also generating a potential difference for us?

 

[lightlightsup]

At the start I've pumped 48 ml into the LH device. The other two have unbent membranes.
When the valve is opened, 16 ml flows out of the left device and both the right devices gain 16 ml.

In losing 16 ml the pressure difference across the first device fell from 12 to 8 Pa. In gaining 16 ml, the pressure across the other two devices rose by 5.33 and 2.66 Pa respectively.

Voltage drop on C1 led to Q gained on C2 and C3?

 

[lightlightsup]

Net charge is still zero for the circuit before and after.
It's just the potential difference caused by the charge separation on C1 that leads to the different Qs you see in the end state?

 

[Merlin3189]

Ok. It looks like my diagram wasn't helpful. I've updated it a bit to see if that helps.
The absolute pressure in the pipes is unimportant: water is roughly incompressible. The pressure is the difference across the two sides of a device.

============================================Start of post!
What you are misunderstanding about the electric situation, is that the 48 Cb is conserved - it was there all along and will be there long after you and I are dead! In an uncharged capacitor both plates have zillions of positive charges (protons) and an equal number of negative charges (electrons).
What the battery did was simply to move 48 Cb from one side of the capacitor to the other. (in our case, electrons move from top to bottom, leaving an excess of protons on top and excess electrons on bottom.)

When it was connected to the other capacitors, 16 Cb managed to flow back to the other side.
But to get there it had to go "through" the other two capacitors. Now, you know that current can't flow through the gap between the plates (*). So what happens is that as 16 Cb moves onto the top plate, 16 Cb moves away from the bottom plate. And that 16 Cb moves onto the top plate of the next capacitor and 16 Cb moves off the bottom plate, returning to the C1 lower plate.

In summary: the total charge on everything in the circuit never changes.
The battery moves 48 Cb on C1.
Later 16 Cb moves around the circuit, changing the location of charge distribution.
Capacitors always stay electrically neutral - charges just get moved from one side to the other.(**)
-----------------------------------------------------------------------
* If you're unlucky enough, one day you'll hear about displacement current through dielectrics. But do not even give it moment's worry until you are much more advanced.
** From static electricity you may come across charged objects, like metal spheres on insulating rods. You may learn that they have capacitance. They behave a bit like a capacitor with one plate being the sphere and the other (often) being the Earth. Because there isn't a circuit of good conductors linking everything, it gets harder to track where the charges come from and go to.

 

[lightlightsup]

But to get there it had to go "through" the other two capacitors. Now, you know that current can't flow through the gap between the plates (*). So what happens is that as 16 Cb moves onto the top plate, 16 Cb moves away from the bottom plate. And that 16 Cb moves onto the top plate of the next capacitor and 16 Cb moves off the bottom plate, returning to the C1 lower plate.

JACKPOT.
That's what I was missing.
Thank You!
I should have thought of it as a displacement of Q.

 

[kuruman]

I find that questions of this kind can be answered directly by assuming charge conservation without recourse to what's in parallel and what's in series. The "H"-shaped conductor piece between $C_2$ and $C_3$ is and remains neutrally charged. This means that when $C_1$ is connected, $C_2$ and $C_3$ have the same charge. This is the charge on $C_1$ before the switch is moved minus the charge on $C_1$ after the switch is moved. $Q_2=Q_3=C_1V_0-Q_1$, where $Q_1$ is the final charge on $C_1$. Since the final voltage across $C_1$ is the sum of the voltages across the other two,$$\frac{Q_1}{C_1}=\frac{C_1V_0-Q_1}{C_2}+\frac{C_1V_0-Q_1}{C_3}$$which can be easily solved for $Q_1$.

 

[chyiLee]

Homework Statement:: See image below for the question: I was able to obtain the answers of 32uC, 16uC, and 16uC. But, wouldn't this total charge exceed the original charge provided to C1 by the battery (48uC)?
Relevant Equations:: See image below for the question: I was able to obtain the answers of 32uC, 16uC, and 16uC. But, wouldn't this total charge exceed the original charge provided to C1 by the battery (48uC)?

View attachment 259288

There must be something I'm not understanding about capacitors in series.

I know that we can treat them as one equivalent capacitor with:
(1) with 1/Ceq,
(2) same Q as anyone of the capacitors,
(3) and add up the Vs for the sum total V across them.

If the equivalent capacitor (Ceq) would have the same Q as anyone of the capacitors, how would I know how much Q is held by anyone of them and the sum total Q held by them?

same Q, but total Q is not Q2+Q3 in capacitors series. same as series's current is not equal i2+i3 , just equal i2 or i3.

 

[SammyS]

same Q, but total Q is not Q2+Q3 in capacitors series. same as series's current is not equal i2+i3 , just equal i2 or i3.

Hello, @chyiLee .



The thread you are responding to is more than 2 years old. It has been answered well by @kuruman , but you do make a good point.