[SOLVED] Initial and final energy stored in a capacitor

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sassyboy

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Hi. I thought the energy stored calculated in part (a) would be the final energy stored.

However, the textbook's answer indicate part (a) is 11mJ and part (c) final energy stored = 7.6mJ.

Please help! What is my misconception in this case?

 

Conservation of energy doesn't apply here, part of the energy is dissipated during the parallel connection. Instead use conservation of charge to calculate capacitor voltages as in the other capacitor exercise problem.
 
It is important to understand that the "missing" energy doesn't disappear immediately when the capacitors are connected.
An oscillation starts, and the energy disappears gradually via resistive losses and electromagnetic radiation.
The available charge (= the number of electrons) is constant, so conservation of charge will give the correct answer.
 
Good question. What happened to the missing energy?

Consider ideal case (no resistance etc) for simplicity.

Remember that the actual energy is stored in the dielectric (it can be a vacuum) in the electric field.

You are shorting a charged capacitor with another uncharged capacitor. This is an irreversible process.

Charges flow during this process and voltages are not equal (to start with).

That involves some work done (charge moving in an electric field) and that is missing energy in the calculation.

Of course total energy is conserved but what we are dealing with is actually the free energy (that has an entropy component).

In this example the free energy before and after are different. The missing energy appears as entropy in the dielectric (how?).

The polarized dielectric (even it is a good old vacuum) contributes to some entropy (and is associated with some energy)
 

You are asking the capacitor to supply an infinite amount of energy for an infinitesimal amount of time. That gives an indefinite answer, like 0/0. In the practical world, you will see the final energy to be about 1/2 the starting energy if both capacitors have equal values. What is happening is half the energy is wasted as heat from the connection resistance and dielectric resistance. Also there is electromagnetic radiation losses if the energy change is quick enough. If you put a 22 Meg resistor between the capacitor leads, then you would recover almost all the energy you started with. However, it might take a month to stabilize the circuit. Twice the resistance cuts the current in half. In total, twice the resistance means that energy dissipation if cut in half. So, use as high a resistance as you can tolerate when you de-energize one capacitor by energizing another one. Use the alternate formula of 1/2* Q^2/C to solve your energy problem.

The dielectric of a capacitor does not store energy. The energy is stored in the electric field that occupies the same volume as the dielectric. The dielectric allows the electric field between the plates to have a higher flux density so as to increase its energy capacity.

Ratch
 
I don't understand what the present discussion is all about, std_match has completely answered the question. No matter how fast the charge transfer between capacitors takes place, if the energy is dissipated in parasitic or intentional resistors, or radiated as electromagnetic wave, the final voltage is ruled by conversation of charge law.
 

Thanks for your help. I enjoyed reading the various answers, some suggestions are clearly beyond my level. My misconception is that I did not know about the conservation of charge law. It is not in the A-level text and therefore, I assumed the conservation of energy law. I will need some time to digest what others are suggesting, but I am thankful for the detail inputs.

I will now try to solve the problem in several ways that people have suggested. I am a visual learner and I want to try and imagine how the charge move from one fully charged capacitor to an uncharged capacitor, and identify the critieria that lead to the charges to stop flowing to the uncharged capacitor. This is A-level and I hope I can untangled this problem using very basic knowledge or the prior knowledge from the textbook. Thanks again!
 

An oscillation starts, and the energy disappears gradually via resistive losses and electromagnetic radiation.
Click to expand...

A circuit containing only C (or even and RC element) cannot produce oscillation. However, a single current pulse can produce some electromagnetic radiation.

Energy does not disappear just like that; it does get converted into heat and is dissipated (adds to entropy). But what happens if there is no resistance present at all?

Are you suggesting that the principle of conservation of energy does not apply to electronic circuits?

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The dielectric of a capacitor does not store energy.
Click to expand...

Please consider an elementary gedankenexperiment:

1. You have an isolated charged parallel plate capacitor with vacuum as the dielectric: it has some energy determined by the capacitor geometry and the voltage on the plates.

2. Insert one material dielectric (say a thin mica foil) in the space in between the parallel plates. Mica has a different dielectric constant (compared to vacuum).

3. Does the energy of the capacitor stay the same? If not, where the extra energy comes from? What has actually changed between case 1 and case 2?

The dipoles in the dielectric are in random orientation before the electric field is applied; they are oriented after the electric field is applied. This contributes to entropy reduction and consumes energy.
 
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    Ratch

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A detailed discussion can be found e.g. here https://en.wikipedia.org/wiki/Two_capacitor_paradox

The confusion is brought up by not considering real circuit properties, a paradox exists only in the fictive world of ideal electric circuits. Any real capacitor involves parasitic serial inductance and resistance (ESL and ESR). The transient charge/discharge process after parallel connection of both capacitors involves a damped oscillatory or exponential current flow which dissipates part of the initial energy. Conservation of energy law is in so far satisfied, but energy balance isn't well suited to calculate the final voltage.
 

A detailed discussion can be found e.g. here https://en.wikipedia.org/wiki/Two_capacitor_paradox...
Click to expand...

Pardon my ignorance, I did not know that it is a burning issue. Let me present my solution in brief, without equations and complex solutions (sometimes descriptive ideas catch the mind easily):

1. Simplify the problem and reduce to basic essentials: consider an ideal capacitor charged to a voltage V being shorted by a wire of negligible resistance.

2. To get some idea fixed, consider some small but finite resistance: all the energy is dissipated in the resistance. After all, this is how capacitative discharge is used in cars and also in welding!

3. Consider now the resistance of the wire is being reduced to zero. What happens to the energy now?

4. There is some displacement current flowing through the capacitor dielectric. When the capacitor is charged, the dielectric is polarised (low entropy) and the external current is infinite and the dielectric is now returns to unpolarised state (high entropy)

5. Hence when a capacitor is shorted, the dielectric dissipates all the energy.

6. Consider a parallel plate capacitor with vacuum as the dielectric. There is no dipole to orient or disorient now.

7. The electric field collapses rapidly and that produces a burst of electromagnetic radiation.

8. The displacement current (during the discharge of the capacitor) is not infinite because Maxwell's law. The electromagnetic waves can be detected with a shortwave radio.

9. This remains valid even in absence of parasitic elements.

Thermodynamically speaking, this is an irreversible process and energy is lost to entropy. However, if you connect an ideal motor in the discharge circuit, you can recover all the "lost" energy.
 

    V

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Hi,

Actually, the textbook is correct.

The energy here is a function of two factors: the square of the voltage and the capacitance. Remember (1/2)*C*V^2?

The initial energy is the energy in the 220uF capacitor at 10V and that energy is W = (1/2)*C*V^2 = 0.5*220u*10^2 = 11mJ. That is the solution to the (a) part. Here the 220uF-capacitor is fully charged so the capacitance is 220uF.

For the (c), the energy drops not because of anything else but because the charges that 220uF-capacitor stored at 10V is now contained in a larger vessel (ie 220uF+100uF = 320uF-capacitor). The misconception here is due to the fact that we often are not able to differentiate between capacitance and capacitor.
Capacitance is actually a measure of the amount of charges that a capacitor can store per unit of voltage (C = q/V) whereas a capacitor is just a vessel for storing charges. What (c) is really asking of you is what the energy would be if this same quantity of charges were contained in a larger vessel. Realize that when the 220uF-capacitor is expanded to a 320uF-capacitor (ie 220uF+100uF) no charge is lost or gained but the voltage drops and that the energy W is a function of a square of this voltage.

To get the energy is quite simple:
W/11mJ = 220uF/(220uF+100uF)
:. W = 11mJ*220/320 = 7.5625mJ.

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Let me explain the solution to part (c) a little further, mathematically.

Given:
V1 = 10V
C1 = 220uF
C2 = 220uF + 100uF = 320uF

Now,
C1*V1 = Q = C2*V2
Q = C1*V1 = 220uF * 10V = 2200uC
Q = C2V2
2200uC = 320uF * V2
V2 = 2200uC/320uF = 6.875V
W2 = (1/2)*C2*V2^2 = 0.5 * 320uF * 6.875^2
W2 = 7.5625mJ.
 
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    V

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I see a lot of technological slang thrown around in these discussions instead of precise descriptions of what things are. Slang should not be used in technical discussions because different folks might have different ideas of what slang means.

A) Current flow-----Current is charge flow, so current flow is charge flow flow. That is redundant and ridiculous. One should instead say, "charge flow", "current', "current is present", or "current exists." Use good English when describing something.

B) Charged:-----When you say a cap is charged, what is it charged with? For every electron that enters a capacitor lead at one end, a matching electron exits the opposite lead. Therefore, a cap with a 1000 volts across it has the same net number of electrons as a cap with no voltaqe across it, namely zero net change of charge. The cap is being "charged" with energy when a voltage is applied across it, so you might as well say the cap is being energized.

C) Displacement current-----This is a misnomer. due to Maxwell. He corrected one of his equations with a term that had units of amperes. But the term was not really current (charge flow) even though it he named it "displacement current". That is analogous to torque (newton-meters) which has the same units as work (newton-meters). Yet torque and work are two different physical entities. Same with Maxwell's correction term. A perfect cap does not permit current through its dielectric no matter how it becomes polarized.

Ratch
 

Thanks Ratch for clarifying the term "displacement current". I think at this stage I comprehend Akanimo's explanation the most, and the others' perspectives are definitely worth further reading. So many A-level physics textbooks out there and they are all using the same terms. Otherwise, it will confuse students even more. I have been a member since 2003, this post is by far the most exciting read. I realise how much under par I am compare to you guys. Plenty of room for improvement in my part!

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Thanks for pointing out my conception. I shall use your analogy of capacitor as a vessel. This has definitely cracked it!
 


Please see the basic definition of a capacitor: the amount of charge needed to increase the potential of a body by one volt.

For every electron that enters a capacitor lead at one end, a matching electron exits the opposite lead.
Click to expand...

Surely you have seen a capacitor grounded at one end?

By the way, what is the technological slang that you found in the post? (OED: Slang; n. A type of language consisting of words and phrases that are regarded as very informal, are more common in speech than writing, and are typically restricted to a particular context or group of people.)

Yet torque and work are two different physical entities.
Click to expand...

Please look up generalized forces and displacements; Goldstein is a classic work.
 

c_mitra,

{Please see the basic definition of a capacitor: the amount of charge needed to increase the potential of a body by one volt.}

The correct wording is "the amount of charge imbalance between the plates ..." In other words, the cap only shows a voltage across itself when there is a difference in the amount of charge on its plates. The excessive charge on one plate is exactly balanced by a charge deficiency on the opposite plate. The net charge is still the same no matter what voltage is present across the cap.

{Surely you have seen a capacitor grounded at one end?}

Sure, but so what? Ground just a reference point and has no other special characteristic. Electrons travel to/from grounded leads like they would from any other conductive point.

{By the way, what is the technological slang that you found in the post? (OED: Slang; n. A type of language consisting of words and phrases that are regarded as very informal, are more common in speech than writing, and are typically restricted to a particular context or group of people.)}

See A, B, C of post #13 of this thread. NASA uses slang when they refer to their astros as "walking" in space. Would they walk away from their vehicle if their tether broke? It just doesn't seem like the right descriptive does it?

{Please look up generalized forces and displacements; Goldstein is a classic work.}

I don't know of what you are averring. Please supply a link and an explanation of how it applies to "current displacement".

Ratch
 

Hi,

How I see it:
* two ideal capacitors, different voltage
Now assume there is a resistor involved when both are connected.
There is charge flow through the resistor...causing energy to be transformed into heat. (We call this "loss of energy".... electrical energy)
One can calculate this.
Now reduce the resistor value.
You can still calculate the loss. It is the same as with the higher value resistor.
You can do this with all resistor values ... but not with 0 Ohms.

For ideal capacitors we have no loss...not in the wires, not in the plates, not in the dielectric, not causing electromagnetic waves.

This "ideal circuit" is not realistic.
And it's impossible to solve.
Thus the problem is called the "two capacitor paradox".

And I think this is the target of the task: To recognize ... You can't solve it with ideal parts, but you can solve it with real parts.

Klaus
 

If I am to measure the energy stored in a capacitor, but I have no joulesmeter (it broked), then how should I go about with this practical experiment.
 

Hi,

The energy is 0.5 × C × V^2.
V is relatively simple to measure.
C can be measured with a capacitance meter, some DVM include this function.

This how I'd measure the energy.

Klaus
 

sassyboy said:
If I am to measure the energy stored in a capacitor, but I have no joulesmeter (it broked), then how should I go about with this practical experiment.
Click to expand...

You need a recording ammeter or a recording voltmeter. For the ammeter, integrate the square of the current and multiply by the resistance. For the voltmeter, integrate the square of the voltage and divide by the resistance.

The energy difference before and after connection is caused by the dissipation due to the current in the conduction path. I will show that mathematically using Laplace transforms to represent the differential equations needed to solve this problem.

The loop equation representing two capacitors C1 and C2 with voltaqes V1 and V2 connected together with a resistor R is:



Now solving for I:



The inverse Laplace transform:



And, integrating the square of the current times R with respect to time, we get the general formula for two capacitors connected together.



As you can see, all values of R give the same answer because R cancels out in the resistive dissipation equation. That means I was wrong in post #5. Plugging in the values for your problem, we
get 3.44E-3 joules for the dissipation. Subtracting from the initial energy of 11E-3joules, we get 7.5625E-3 joules. This agrees with the alternate capacitor energy formula of (Q^2)/2C .

Ratch
 

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