# One capacitor charging another capacitor

likephysics

## Homework Statement

You have one capacitor C1 (1uF) charged to 10V.
Now the capacitor is switched to charge another capacitor C2 (0.25uF),
Whats the voltage on C2?
Everything's ideal. No loss caps, no wire resistance.

## The Attempt at a Solution

I tried,
Q1=C1V1
Q2=C2V2

total charge will be constant, so Qtotal, Qt=CtV
Both caps are in series, so Qt= (C1+C2)V

(C1+C2)V = C1V1+C2V2

V2 is 0v

V = C1V1/(C1+C2)

So, V=8V.

But when I do E = 0.5*C*V^2 for C1. (80uJ)
and then for both the caps(32uJ and 8uJ), the numbers don't add up.

## The Attempt at a Solution

Mentor
Was there a question to go with this?

likephysics
Sorry, the question is to find the voltage at capacitor C2.

Mentor
Sorry, the question is to find the voltage at capacitor C2.

It looks like you found it when you calculated the 8V.

You could also have found it by interpreting the capacitors as being in parallel, and placing the total charge on the equivalent capacitor C1+C2.

Don't fret about the loss of energy between before and after; When the charges were redistributed they had to move. Anytime something moves or changes spontaneously in nature (that is without some external energy source to drive it) it's because the system is seeking a lower energy state and one is available to be filled. If the components were truly ideal, with zero resistance, then the charge would continue sloshing back and forth between the capacitors (even a perfectly straight, 1cm length of wire has some inductance!). In real life the oscillations would eventually bleed away the energy via electromagnetic radiation and the system would settle down to the quiescent steady state.

Homework Helper
Gold Member
But when I do E = 0.5*C*V^2 for C1. (80uJ)
and then for both the caps(32uJ and 8uJ), the numbers don't add up.

Your capacitor leads have finite resistance. If you connect C1 to C2 with a resistor, no matter how small, you will find by the usual analysis that the missing energy is wholly disspiated by that resistor.

Staff Emeritus
Your capacitor leads have finite resistance. If you connect C1 to C2 with a resistor, no matter how small, you will find by the usual analysis that the missing energy is wholly disspiated by that resistor.
And in the dielectric losses. (unless the dielectric is vacuum)

likephysics
I know the losses- ESR, dielectric loss, wire resistance.
The Q=CV equation does not contain a loss term, but the result contains loss of energy. How is this possible?
Let's assume ideal capacitor, idea wire (zero resistance).

Homework Helper
Gold Member
I know the losses- ESR, dielectric loss, wire resistance.
The Q=CV equation does not contain a loss term, but the result contains loss of energy. How is this possible?
Let's assume ideal capacitor, idea wire (zero resistance).

You have to accept that there is SOME resistance. It can be as small as you like, but not zero. ∫i^2*Rdt from 0 to ∞ will always be the difference between energies before & after connection is made, for ANY value of R.

This is a well-known conundrum. It is always dangerous to deal with singularities. So just accept that R cannot be absolute zero.

cmb
You have one capacitor C1 (1uF) charged to 10V.
Now the capacitor is switched to charge another capacitor C2 (0.25uF),
Whats the voltage on C2?
Everything's ideal. No loss caps, no wire resistance.

I would say that the reason you are getting caught in a quandry over how to answer this is because it cannot be answered. The question is, itself, 'wrong'.

This, as posed, would lead to the condition well described in #4 above:

If the components were truly ideal, with zero resistance, then the charge would continue sloshing back and forth between the capacitors

That is to say, there would be no static value of voltage on C2 that could be defined. If everything was perfect and zero resistance with no inductance, the frequency of VACC2 would be infinite!

Think of it as a Newton's cradle with two balls of different masses. The little one is stationary and the big one is dropped. There are no losses. How high is the smaller ball? You see, it is a time-dependent thing and the energy just swings one into the other, then back again.

To answer the question, you must pragmatically include an assumption about energy losses, else the answers you get are meaninless. If you like, you could work out the max and min volts on each capacitor during you theoretically perfect oscillation, because those would be finite values. I'll start you off; the max volts on C1 is 10V, and the min on C2 is 0V. ;)

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Staff Emeritus
The smaller the resistance, the greater the current surge, therefore the higher the i2R losses. With no wire resistance, the L-C ringing and associated radiated electromagnetic energy (radio waves) would be the principal energy loss from the system.

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Staff Emeritus
That is to say, there would be no static value of voltage on C2 that could be defined. If everything was perfect and zero resistance with no inductance, the frequency of VACC2 would be infinite!
If there was no inductance, then what mechanism would cause the charge redistribution to overshoot?

Homework Helper
Gold Member
The smaller the resistance, the greater the current surge, therefore the higher the i2R losses. With no wire resistance, the L-C ringing would radiate electromagnetic energy (radio waves), this itself posing an energy loss from the system.

Incorrect. The smaller the resistance, the greater the current surge, but the shorter the duration of the surge. So the integral of i^2*R is the same over t=0 to ∞.

Work the math and forget radio waves.

Staff Emeritus
We can't simply "forget" EM radiation when in the limiting case it is the only existing loss mechanism.

cmb
If there was no inductance, then what mechanism would cause the charge redistribution to overshoot?

But we should still be able to theorise about idealised systems, and to address your question you should think again on the Newton's cradle example. You are asking why the falling ball does not instantly stop, but its energy 'overshoots' into the other smaller ball. (the analogy is - ball height is voltage).

There is no 'overshoot', there is merely the bouncing of energy back and forth. There are two competing, opposed, osillatory drivers (as there usually are with oscillators!): You have the capacitors trying to even out the charge such that the smaller capacitor takes proportionately more voltage, whilst you have the system trying to even out the voltage.

Nothing complicated in this, just think about the two germane elementary rules in electronics; the potential across serial capacitors will distribute itself proportionately to the capacitance, yet a differential voltage will cause a current.

cmb
We can't simply "forget" EM radiation when in the limiting case it is the only existing loss mechanism.
Yes we can. It is a theoretical device, and it is infinitely small. No radiating elements.

Staff Emeritus
There is no 'overshoot', there is merely the bouncing of energy back and forth.
If there is no overshoot, there is no oscillation--there can be no "bouncing back and forth".

cmb
If there is no overshoot, there is no oscillation--there can be no "bouncing back and forth".

I am not describing inductive overshoot, per se.

As I stated, this is not a real-world example for which there would always be some inductance. An infinite current (I) and a zero inductance (L)... what magnetic energy does the inductance formula [0.5xLxI^2] give you with those two values?

Don't stress over it... the question is answered. It is either unrealistic question, or you make an assumption of some teeny amount of ESR so it damps promptly.

Staff Emeritus
I am not describing inductive overshoot, per se.
Then what back and forth phenomenon are you talking about?
As I stated, this is not a real-world example for which there would always be some inductance. An infinite current (I) and a zero inductance (L)... what magnetic energy does the inductance formula [0.5xLxI^2] give you with those two values?
If there is no inductance, there can be no stored magnetic energy. That's not too complicated, is it?
or you make an assumption of some teeny amount of ESR so it damps promptly.
Damps? What exactly do you imagine we need to damp? With no inductance, it is not a second order system, there is oscillation, so there is nothing to dampen.

Homework Helper
Gold Member
Yes we can. It is a theoretical device, and it is infinitely small. No radiating elements.

Good point! No antenna! Hertz would have understood.

cmb
If there is no inductance, there can be no stored magnetic energy. That's not too complicated, is it?

It is complicated by the fact that it is an impossible question. You say there is no magnetic energy because there is no inductance. But there is an infinite amount of current. In fact, there is an infinite amount of current squared! I say my infinite current [squared] trumps your zero inductance! There WOULD BE magnetic energy there, in this dream-world fantasy!

This is quite a simple problem to sort out, even if you do not have all the info for the losses. All you need to recognise is that there will be energy losses. Then you treat the two capacitors as parallel (hence you can deduce a sum capacitance, C1+C2). Then you calculate the amount of charge present before the switch was thrown. The charge cannot be lost, as the + and - terminals of each capacitor, so to speak, are mutually isolated. So we then know charge, and total capacitance - from whence the voltage is defined!

Interestingly, you could stick a megohm or a milliohm resistance in the way, a Henry's worth of inductance, with a big Q or a little Q, or none at all, but the energy lost would be the same. Intuitively a bit wierd, but there it is! The charge and capacitances are immutable values, so the final energy must also be immutable.

(It is so counter-intuitive, I am not even sure of this! Can anyone spot any flaws?)

Staff Emeritus
It is complicated by the fact that it is an impossible question. You say there is no magnetic energy because there is no inductance. But there is an infinite amount of current. In fact, there is an infinite amount of current squared! I say my infinite current [squared] trumps your zero inductance! There WOULD BE magnetic energy there, in this dream-world fantasy!

This is quite a simple problem to sort out, even if you do not have all the info for the losses. All you need to recognise is that there will be energy losses. Then you treat the two capacitors as parallel (hence you can deduce a sum capacitance, C1+C2). Then you calculate the amount of charge present before the switch was thrown. The charge cannot be lost, as the + and - terminals of each capacitor, so to speak, are mutually isolated. So we then know charge, and total capacitance - from whence the voltage is defined!

Interestingly, you could stick a megohm or a milliohm resistance in the way, a Henry's worth of inductance, with a big Q or a little Q, or none at all, but the energy lost would be the same. Intuitively a bit wierd, but there it is! The charge and capacitances are immutable values, so the final energy must also be immutable.

(It is so counter-intuitive, I am not even sure of this! Can anyone spot any flaws?)
So, that's a summary of the situation. But your back and forth oscillation? We're waiting to hear how you account for that, since you say your explanation doesn't involve inductance.

cmb
..your back and forth oscillation? We're waiting to hear how you account for that..

(Sigh!)

I believe you are the only one waiting to hear. And there is nothing to hear. The question itself is a theoretical proposition, fraught with inconsistencies. The point was made simply to say that if there were NO energy losses, then the 'rest' condition (as per my definition above) would only be transitory as there would still be this other energy in the system. Electrostatic, magnetic, I don't care? :grumpy: Whatever that energy is, it would push the system away from that 'rest' condition. Then it would try to come back to that rest condition. Then it would move away. Then it would try to come back... etc.. It would never reach an end. But my posts here have!...

Staff Emeritus
cmb said:
Yes we can. It is a theoretical device, and it is infinitely small. No radiating elements.
Good point! No antenna! Hertz would have understood.
I have my doubts that Herr Hertz would have understood, actually. But since you at least seem to understand perhaps you can clear up this conundrum for the rest of us.

How are you going to pack a bunch electrons (remembering that 1 coulomb =6.24×10^18 electrons) into something infinitely small, when, for starters, we know that individual electrons have finite size?

Just wondering ...

Mentor
How are you going to pack a bunch electrons (remembering that 1 coulomb =6.24×10^18 electrons) into something infinitely small, when, for starters, we know that individual electrons have finite size?

Just wondering ...

Actually, as far as we can tell electrons are point particles -- no size! But what will kick into prevent such close packing is the Pauli Exclusion Principle. You just can't superimpose all those electrons in the same place with the same quantum state without a really, really good pair of vice-grips! Or the gravitation of a black hole...

Anyways, you can only throw out so many laws of physics for a gedanken experiment before you're no longer discussing physics or anything remotely like it. Resistance you can safely do away with since we know that superconductors exist. But you cannot wave your hands and eliminate inductance any more than you can banish inertia in a kinematics problem and still have it make sense. And accelerated charges WILL radiate.

technician
When current flows from one capacitor to another energy can be lost in a variety of ways
1) due to resistance in connecting wires... this can be made zero
2) sparking at the switch... this can be made zero
3) electro-magnetic radiation (could be radio waves) from the connecting wires. Whenever a current changes electro-magnetic radiation is produced. (That is what a radio transmitting aerial is). Current is changing during the transfer of charge from one capacitor to another... or from a battery to a capacitor. So there will always be some electro-magnetic radiation... it can be detected on a radio set as a crackling 'interference'
In mechanics there is an analogy with placing a weight on a spring, if the weight is lowered onto the spring then the spring achieves its equilibrium position and the energy stored is 0.5F x extension. If the weight is released then the energy given up by the weight is F x extension, the extra energy appears as oscillations... Kinetic energy.

technician
'Nascent Oxygen' post #13 is the most significant post in this discussion.
There will be electro-magnetic radiation when the current changes. This is undoubtedly energy lost from the circuit.

cmb
'Nascent Oxygen' post #13 is the most significant post in this discussion.
There will be electro-magnetic radiation when the current changes. This is undoubtedly energy lost from the circuit.

I don't have any regard for that comment. Imagine we shield the interference in a closed box smaller than the near field, keeping all the energy coupled in the system. What is intellectually curious is that whether you were to examine such a system in a box, or not in a box, the energy lost would be exactly the same. That suggests the ways and means for radiative losses are immaterial. There is nothing of relevance to the question in discussing any energy loss mechanisms.

Radiative losses, or any come to that, are simply not part of the question. The nature of the losses are entirely immaterial.

Homework Helper
Gold Member
cmb - I think you're spinning your wheels here ...

Staff Emeritus
Actually, as far as we can tell electrons are point particles -- no size!
I'm uneasy with absolutes; too extreme. I'm more comfortable with "vanishingly tiny". Besides, it avoids the associated problem of explaining exactly what would be meant by the "size" of an electron, and how it would be measured.
Resistance you can safely do away with since we know that superconductors exist. But you cannot wave your hands and eliminate inductance any more than you can banish inertia in a kinematics problem
No need to eliminate resistance or inductance, either, to picture how near-ideal capacitors would behave. All we have to do is reduce L and R to the point that they play no predominate role in the subsequent operation. Again, I'm uneasy with absolutes, and the questionable relevance to reality. All we need do is minimize inductance, and go for low resistance, and we have near-ideal circuit operation.

Staff Emeritus
(Sigh!)

I believe you are the only one waiting to hear.
Alas, probably true. You lost most readers back when you wandered off into vague arm waving and anthropomorphic mutterings. But I stayed on to give you the benefit of the doubt, in case you had something insightful to say.
And there is nothing to hear.
So I found.

The point was made simply to say that if there were NO energy losses,
If pigs might fly.

cmb
Alas... I stayed on to give you the benefit of the doubt, in case you had something insightful to say.

I am absolutely dumbfounded that you do not find my conclusion above fascinating.

I suppose we live in an age where everyone wants a soundbite, spoon fed to them.

The conclusion, that it doesn't matter how the energy is lost, it is the same amount of energy, is fascinating. In fact, I suppose it is a form of 'Gibbs free energy' type calculation, in which we look at the 'before' and 'after' state and recongise that both are defined, so the route to the end-point is immaterial.

The example given above with the spring is similarly so. Imagine you drop a mass on a spring, through lossy fluids, magnetic fields, whathaveyou that might impede its decent, yet at the end, once equilibrium is achieved, exactly the same amount of potential energy would have been expended.

I think that is fascinating. You don't? Time for you to move on to graze for your next sound-bite...

technician
I would like to go right back to your first post and do what I think the question is asking.
I am going to assume that when C1 is connected to C2 they are in parallel... that is a sensible way to use one capacitor to put charge on another.
First of all C1 is 1μF charged to 10V so energy stored is 0.5 x 1 x 10^-6 x 10^2
This gives 5.0 x 10^-5J
also the charge on C1 = C x V = 1 x 10^-6 x 10 = 1 x 10^-5 C
When C1 is connected (in parallel) to C2 the combined capacitance = C1 + C2 = 1.25x10^-6 F
The charge in the combination remains the same (1x10^-5C)
therefore the Voltage across this parallel combination is V =Q/C = 1x10^-5/1.25x 10^-6 = 8V
The energy stored on the combination is 0.5xCxV^2 = 0.5 x 1.25x10^6 x 8^2 = 4 x 10^-5J
So there has been an energy loss of 5x10^-5J - 4x10^-5J = 1x10^-5J
In an earlier post I explained how energy is lost when charge flows on to or from a capacitor.
Resistance of any connecting wire, sparking at the switch and in the last resort electromagnetic radiation (probably radio waves).
This radiation can be detected on a radio in the vicinity, it can be shown on an oscilloscope with a pickup coil of a few hundred turns connected to the input of the CRO