Charge conservation in capacitors

[Muthuraj R]

In the circuit, all elements are ideal.
For time t<0, S1 remained closed and S2 open.
At t=0, S1 is opened and S2 is closed.
If the voltage Vc2 across the capacitor C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ ??

I tried this problem.
Obviously, Vc1(0) = 3V ; Given that Vc2(0) = 0V.
Then C1 = 1μF and C2 = 2μF are connected at t=0.
So, Q1(0) = 3 C. Q2(0) = 0 C.
At t=0+ how charges are distributed and hence the voltage ??
I was thought (according to charge conservation) that, charges are distributed to two cap. such that voltage across them is equal. So, Q1 = 1 C , Q2 = 2 C. Hence Vc1 = Vc2 = 1 V.
All these things happen at steady state. How about at t=0+ ?

 

[Simon Bridge]

charge one cap to 3V, then use it to charge another cap - what is the voltage?
that the problem?

see:
https://www.physicsforums.com/showthread.php?t=234221

you have no resistances (ideal components remember) so the steady state will be all-but immediate. What is your question?

 

[technician]

You are correct to say that the total charge on the 2 combined capacitors is the same as the charge on C1 at t=0.
when S2 is closed you now have 2 capacitors connected in parallel so you should calculate the new capacitance, then you should be able to calculate V and the charge on each of the capacitors.

 

[Muthuraj R]

Dear Simon,
C1 is charged to 3V before t = 0.
C1 and C2 are connected at time t = 0.
As you say, C2 is to be charged by C1 (having a voltage of 3V).
Now, pls tell me
what will happen to the voltage across C1 and C2 at t=0+ (immediately after t=0)

 

[Muthuraj R]

Dear technician,
At time t = 0, the voltage across C1 is 3V ; while the voltage across C2 is 0V.
Different voltages. How do u say that they are in parallel ??

 

[Simon Bridge]

Since you have ideal components, and no resisters, then at the exact time the switch is closed, the voltage drops in an instant to 1V. It is a step function. v(t)=3-2u(t) where u(t) is the unit step function.

If you have some resistances in the circuit, i.e. the caps are not ideal, then v(t) drops exponentially to 1V ... see the link in my post for an example of how to do the calculation.

 

[sophiecentaur]

If you charge the first C then use it to charge the second, you have an insoluble problem unless you include some finite resistance and you 'admit' to some series Inductance in the connecting wires.
Without resistance in circuit, you have energy conservation as well as charge conservation. The energy must exist as magnetic energy in the L, giving you a continuous oscillation of energy from C to C and back.
If there is resistance, the 'excess' energy is dissipated in a simple exponential charge transfer.

 

[Simon Bridge]

Oh I forgot about the oscillation ... lone ideal reactances in a circuit - one of those classical fudges like "rigid" bodies.

 

[sophiecentaur]

Oh I forgot about the oscillation ... it's one of those classical fudges like "rigid" bodies.

It can be a serious problem in some applications. Ringing involves over-voltage and EM interference.

Not a "classical fudge" but a result of inadequate 'Classical" modelling by some teachers.

 

[Simon Bridge]

Yeh - we don't expect models to give good pictures in some situations.
I have to be alert for that in questions.

 

[sophiecentaur]

The 'two capacitor' paradox is a good one for students at Christmas / fun times, just when they are getting too damned smart!

 

[technician]

This is an easy problem to solve in practice by doing the experiment! You may not be able to get your hands 1F and 2F capacitors but it is not difficult to get capacitors that will give answers.
Are these capacitors in parallel ?...yes.
The symbol for a capacitor represents the parallel plates of a capacitor. When S2 is closed the 'top plates ' of C1 and C2 are connected together to give a plate of larger area. The bottom plates are already connected as shown in the diagram.
The combined capacitance of this arrangement is 3F
The charge on C1 at the start is Q = CV = 1 x 3 = 3Coulombs
The same charge is there when the switch S2 is closed so now a capacitor of 3F has a charge of 3Coulombs. The voltage across the combination = 1Volt.
The charge on C1 is now 1Coulomb and the charge on C2 is 2Coulombs

 

[technician]

Another good 'paradox' for students is to ask what happens to energy stored on capacitors when they are connected.

 

[jim hardy]

""How about at t=0+ ?
""
i think OP is asking what happens when you divide by zero.
In world of math thinking, you have a discontinuity and it's indeterminate.
Its limit approaches infinity . But "it" is not real anymore.

3 volts across zero ohms would be infinite amps , were it not for Sophie's real world inductance and resistance of the wires. When you force them out you make "it" not real anymore.

It's analogous to the thermodynamics experiment of a gas confined then the wall immediately removed. You can only define the end states.

observe with zero ohms, charge is concerved but energy is not.


energy in a cap is 1/2 (C*V^2)
when open S1, C1= 1uf at 3volts , 1/2C*V^2 = 1/2 X 1 X 9 = 4.5 joules
after close S2, C1 + C2 = 3 uf at 1 volt, 1/2 X 3 X 1^2 = 1.5 joules
Those are the end states. 3 joules are gone.

what happened at instant of closing S2? musta been some energy radiated into space .

EDIT ahh,, Tech was here already..

 

[technician]

I hope I have not mis-read the original post ! I thought that an explanation of the final voltage with some reference to charge conservation was required.
I hope there is no doubt or misunderstanding about what would happen if you got 2 capacitors and a battery and did this exercise for real!

 

[jim hardy]

i thought he was asking about instant of switch closure
but i could be wrong.
maybe i spoke to wrong question.

indeed charge is conserved as you pointed out
and energy goes someplace.

 

[technician]

Jim:
I think you could be right... his last sentence...'all these happen at steady state' suggests that.
It will be interesting to see what he thinks of this discussion.
Cheers
If this is the case then it is not a simple circuit with 2 capacitors and a battery !

 

[sophiecentaur]

Whatplace? ;-)

 

[technician]

Heat in any connecting wires ? Electro-magnetic radiation from connecting wires?
(this is way off the original OP and I hope it does not cause confusion for the OP)

 

[sophiecentaur]

Heat in any connecting wires ? Electro-magnetic radiation from connecting wires?
(this is way off the original OP and I hope it does not cause confusion for the OP)

I think it is bang on topic.
If you calculate the resulting charges and then the resulting energy, you find you've lost energy. It's a small step from that to the ensuing discussion and it's a salutary lesson about the shortcomings of 'ideal' models.

 

[Antiphon]

I think it is bang on topic.
If you calculate the resulting charges and then the resulting energy, you find you've lost energy. It's a small step from that to the ensuing discussion and it's a salutary lesson about the shortcomings of 'ideal' models.

I wouldn't call it a shortcoming of circuit theory; you'll get the right answer for the energy if you put resistors in series with the capacitors and take the limit as they go to zero.

In circuit theory you may not change the voltage abruptly on a capacitor. I'd call the original problem poorly formulated.

 

[sophiecentaur]

I'd call the original problem poorly formulated.

As are the majority.

 

[jim hardy]

Let us then alter the experiment to one that does not involve division by zero.

I am real simple minded and can understand mechanical things.

Take a 1 farad parallel plate capacitor. Let us run the experiment on surface of moon so our dielectric is free space.
Capacitance is ε* Area / Distance between plates

Charge the capacitor to 3 volts.
That takes 3 coulombs and the capacitor now holds (CV^2)/2 which is 4.5 joules in its dielectric.

Because the plates hold opposite charge they attract each other.
So let them move closer together until distance between them is one-third what it was at start of experiment.

Now capacitance is 3X what it was,
ε*Area /(D/3) = 3*ε*A/D

so voltage ix Q/C = 3 coulombs/3farads = 1 volt as before.
Energy in dielectric is (CV^2)/2 = 1.5 joules as before.
But now it is VERY CLEAR where the energy went - into the mechanical work done by plates on whoever was holding them apart as they moved toward each other, everyday F X Distance.

I can imagine feeling that force.
And we avoided division by zero.

I hope this helps.

 

[Noble3992]

Charge Conserved - Energy Lost

Sophie,

When you say "it's a salutary lesson about the shortcomings of 'ideal' models," I disagree.

There is a much more beautiful solution to this problem that involves solving the first order differential equation for energy lost in the flow of current from one capacitor to the other. Despite 'ideal' conditions, if you assume some real impedance in the connection between the capacitors and solve for the energy lost due to current flow through that impedance (integrating I²(t)R, you find that the energy loss is independent of the impedance. Therefore, even in an 'ideal' case, there is some energy lost as charge moves from one capacitor to another.

For the "mechanical minded," this situation can be best understood in the case of discrete charges on two sets of parallel plates being ideally joined at the edges into a single larger capacitor. The charges will ideally distribute to a uniform charge density on each plate. As this happens, the mean distance between charges increases resulting in a system with a lower potential energy.

This problem is a fantastic problem that is less about shortcomings of ideal models and more about a deeper understanding of the underlying physics. It's also a great way to get students to practice solving \frac{dq}{dt} = \alpha - \beta q.

 

[sophiecentaur]

@Noble
My basic point was that 'ideal' models tend to take the unwary into the realms of paradoxes because the ideal model often leaves out the factors that exclude a paradox.

Reality (over simplified) >> Maths >> Possible non-reality