Energy in a Capacitor: Explained

[BillyT]

Yes. I completely agree with post 30. I'll add from my youth as a radio "ham" that the impedance of the popular center feed half wave length antenna is about 50 ohms. So that is the impedance of most twin lead cables available and the finnal "tank coil" is an air core step down transformer. (at least back in the days of tubes).

 

[AlejandroDes]

why is the formula of energy in a capacitor E= C x U^2 /2
I understand it mathematically, but I do not understand it if you apply it to a real situation. Is there anyone who can explain that?

I think someone already posted the link but this video might answer your questions:

https://www.khanacademy.org/science...c/circuits-with-capacitors/v/energy-capacitor

The total stored energy is half the potencial energy because you lose voltage while discharging the capacitor.

*http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c2

Cheers

 

[Mark Harder]

I see by one site you lose half the energy used to charge a cap because of resistance. And it says if you lower the resistance it doesn't help. What if the energy for the charge is sent by a superconductor and the plates are superconductive, would there still be a loss of half the charging energy?

That statement makes no sense to me. The energy stored in a capacitor is the energy in the capacitor when it is in a steady state, when the voltage across it is constant. Ideally, one can remove the charging voltage source and a cap will keep its charge and therefore it's voltage forever. In the real world, however, the insulators in the cap leak current, so a capacitor cannot store charge indefinitely. Therefore, in order to derive a formula for the stored energy, ideal conditions are probably assumed. Thus, no current is flowing in or out of the device, so dissipation of energy is not a factor needed to create a model to use when deriving the energy formula. So, we are left with that factor of 1/2 to explain. As for me, I don't understand why the charge Q= C*U is multiplied by U to arrive at the energy. I suspect it has something to do with the definition of voltage, which is a term for electrical potential energy. Is voltage the potential energy per charge? If it is, then Energy volts*charge = U*(C*U) = C*U^2. That still doesn't explain that factor of 1/2 however. I'm curious about the answer myself. I'll think about it some more.

 

[sophiecentaur]

The energy stored in a capacitor is the energy in the capacitor when it is in a steady state,

Not really. The formula CV2/2 applies at any time in the process.
When you say that "dissipation is not a factor" you need to realize that the dissipation takes place whilst the charges are moving and not in the final state.

I'm curious about the answer myself.

If you are curious then go to a Physics textbook or, if you want a brief explanation, go to this hyperphysics link. You will not (cannot) find an explanation that doesn't involve some Maths. After all, you are asking for an explanation of a mathematical formula in the first place.

 

[Mark Harder]

Not really. The formula CV2/2 applies at any time in the process.
When you say that "dissipation is not a factor" you need to realize that the dissipation takes place whilst the charges are moving and not in the final state.

If you are curious then go to a Physics textbook or, if you want a brief explanation, go to this hyperphysics link. You will not (cannot) find an explanation that doesn't involve some Maths. After all, you are asking for an explanation of a mathematical formula in the first place.

Sophie, thank you for your response. You are correct when you say the energy formula applies to any point on the charging curve - as long as you are referring to the energy of the STATE, not the energy required to attain that state. In other words, we need to make a distinction between the energy content of a charged capacitor, which does not include the energy that must be expended to put that energy there, and the energy required to charge the capacitor, which must include the energy dissipated during the process. I believe the latter is a path-dependent process, since faster charging requires more current, hence more heating due to the finite resistance in the internal conducting path. There is also the matter of leakage current. I haven't tried to calculate its contribution to dissipation during charging. But I think we can say that there are 2 competing processes taking place during charging: adding charge to the cap and losing charge through leakage. At some point, a steady state is attained, in which the rate of adding charge equals the rate of loss through leakage. Thus, even maintaining constant voltage in the fully charged state requires some current to replace the charge lost through leakage.

The formula E = C*V^2 /2 is derived in the online physics tutorials (I no longer own a general physics text.) by calculating the work required to move charge into the cap against the growing electrical potential. When one calculates work against a field this way, dissipation is ignored. The assumption that the field against which work is done is conservative is typically made in physics texts. The textbook transition to systems with dissipation is made when they discuss thermodynamics. In my day, Physics 101 textbooks didn't make this transition explicit (which is a shame since a teachable moment, the historical co-evolution of the science of thermodynamics and the design of heat engines during the industrial revolution, is passed over.)

You are also correct when you point out that the problem can't be understood without math. Believe me, I'm trying to understand the math and the physics together.

 

[sophiecentaur]

When one calculates work against a field this way, dissipation is ignored.

Of course it is. It's a definition. It also represents the energy that you could get out, which assumes an ideal capacitor, of course.

hence more heating due to the finite resistance in the internal conducting path.

You are introducing more 'practicalities than are appropriate in a definition. However, if you want a series internal resistance, then this only needs to be added to the series source resistance. If you look at your on line tutorial and extend it to work out the energy dissipated in the series R, you will find that the value of R cancels out and you are left with CV²/2. Charging the C slowly or fast makes no difference. Believe the Maths.

 

[gleem]

f you look at your on line tutorial and extend it to work out the energy dissipated in the series R, you will find that the value of R cancels out and you are left with CV2/2. Charging the C slowly or fast makes no difference. Believe the Maths.

Consider a capacitor being charged by a battery of voltage V through a resistance R. The power produced by the battery is
$$ P = Ri^{2} $$
Thus the energy needed to be delivered by the battery is
$$E= \int_{0}^{\infty }Pdt $$
The current charging the capacitor is$$i = (V/R)e^{^{-t/RC}} $$
thus the total energy supplied by the battery is $$ E= R(V/R)^{^{2}} \int_{0}^{\infty }e^{-2t/RC}dt $$ $$ E = (V^{2}/R)(-RC/2)[e^{-2t/RC}]_{0}^{\infty }$$
$$ E = CV^{2}/2$$

thus the energy delivered by the battery is equal to the energy stored in the field of the capacitor.You can also show that a capacitor B charged by another capacitor A through any resistance produces a similar result.

Starting with KVL V_A = iR + V_B

If Q₀ is the initial charge on capacitor A and q is the charge flowing into capacitor B then we can write the differential equation
$$(Q_{0} - q)/C_{A} = dq/dt + q/C_{B}$$ Solving for q , determining the current flowing and Using the same method as above we get

$$E = (\frac{C_{A}C{_{B}}}{C_{A} +C_{B}})(\frac{V_{0}^{2}}{2})$$
and V₀ = Q₀/C_A

if C_A = C_B

Then $$E = V_{0}^{2}/4$$

which is the energy stored in the fields

Note no need to invoke em radiation emission. The energy used by the charging source is just the energy used to transport the charge in the process and equal to the energy stored in the field(s).

Believe the Maths.

Yes and the Math will show you the way.

 

[sophiecentaur]

The Resistor will have to be dissipating Power as the current flows through it so the Energy supplied has to be more than what is stored in the Capacitor. From your post, the Energy dissipated in the Resistor is :

thus the total energy supplied by the battery is

E=R(V/R)2∫∞0e−2t/RCdt​

E= R(V/R)^{^{2}} \int_{0}^{\infty }e^{-2t/RC}dt

E=(V2/R)(−RC/2)[e−2t/RC]∞0​

E = (V^{2}/R)(-RC/2)[e^{-2t/RC}]_{0}^{\infty }

E=CV2/2​

(I apologise for the way your code seems to have got through my attempted quote)
In addition, the Energy in the Capacitor is also CV²/2 so the total energy supplied is CV²
So the Capacitor has half of the total Energy that was supplied. (The two expressions are the same, which may be confusing you.)

In the case of the two Capacitors, you have shown that the original stored Energy is CV²/2 and that the total Energy in the two Capacitors is CV²/4 so there is an energy deficit. How do you account for it? It just has to go somewhere. Dissipation in the internal resistances in the circuit or radiation are the only possibilities. In fact, you cannot deny that EM radiation will always take place in any circuit with changing current.
It seems that you are invoking a different way in which Energy can exist i.e. "used in charging". But that is not a valid idea. Can you supply a proper reference to it?

 

[nasu]

@gleem
1. The energy delivered by battery is not that in your post but
$E_{battery} = \int_0^\infty E i(t) dt$
where E is the emf of the battery. E is a constant whereas the voltage across the resistor is a function of time.
The integral of Ri^2 is just the energy dissipated in the resistor.
Taking
$i(t)=E/R e^{-t/RC}$
you get the energy as $E_{battery} =CE^2$, as expected.

2. You assumed that the circuit has no inductance so of course there is no inductive effect or radiation.
Taking a more realistic model you will get all (or some) of the extra effects.

So you either assume a purely RC circuit and then the dissipation is all resistive or you take a more complex model and you have other way to dissipate energy.

 

[gleem]

In fact, you cannot deny that EM radiation will always take place in any circuit with changing current.

I do not deny some possibility of em radiation. EM radiation is produced by the accelerations of charge which is minimized by the resistance.

An yes @nasu In a perfect battery of emf V you move a total of Q coulombs through V volts so QV work was done by the battery. Half of the energy was dissipated in the resistor and half used to charge the capacitor.

In the situation where one capacitor charges another one might expect that again only one half of the energy available would be used to charge the capacitor connected to it. In this case however the charging emf is not constant.

It seems that you are invoking a different way in which Energy can exist i.e. "used in charging".

So like the battery we see the net energy stored in the new arrangement ( if both capacitors are equal) is one half the energy available (from the charging capacitor). But that energy is shared between the two capacitors. so each one has 1/4 of the initial energy. So the initially charged capacitor had to have delivered 3/4 of its energy. But 1/2 of this is dissipated in the resistor with 1/4 going to the other capacitor. Thus all the energy is accounted for.

Do I have it straight now?

 

[sophiecentaur]

Yes, I would say that you are increasingly getting it "straight".

I do not deny some possibility of em radiation.

Although it's always gong to be there, it is going to be very small unless the frequency of any ringing is high enough for the wavelength to be of the order of the size of the circuit. That's the only thing that's going to make a significant amount of radiation resistance affect the circuit. Once you have 'common' values of R and C (and L) , the resonant frequency will be in the MHz or even hundreds of MHz, which means that the structure needs to be pretty big. The point is that, when a question requires having 'zero' Resistance, the ringing frequency will be so high that the radiation resistance will be the only resistance that needs to be considered.

 

[Mark Harder]

It's not central to the topic, but there's this subtopic running through this thread I want to comment on. It's the notion that all you have to do to solve problems like this is 'follow the math' and the like. There are problems with this approach to physics derivations. First, the symbols used are not abstract entities that can be pushed about they way they would in abstract algebra derivations, for instance. I believe this subconscious equivalence can lead one to use a symbol, say ' V ', to represent more than one physical entity. In reading the posts above, I see that this error has been made more than once. For instance, in one derivation V was used to stand for the voltage of the battery (or other power sources), which is assumed to be constant over time. Then it was used as the voltage over the capacitor, which increases as the cap is charged, or the voltage drop in the resistor, which decreases over the charging time, since the current is decreasing then. The math that followed lead to confusing and in some cases downright incorrect, results; because the capacitor voltage or resistor voltage cannot factored out of a time integral the way the battery voltage can. If the physical situation, in which battery voltage, resistive voltage drop and capacitor voltage are all different, is clear in one's head, such mistakes can be avoided. Furthermore, if all three voltages (designated by distinct symbols) are included in the integration in their proper places, the expressions for the energy lost to dissipation and the energy stored in the cap will both reveal themselves. In that case, the math will show how the 2 energies are equal, their sum equal to the energy delivered to the system by the battery.

Another mathematical assumption used in this problem, that the current around the circuit is equal everywhere, is very convenient. You can write expressions for the current through the resistor or into the capacitor and use the result anywhere. This is correct, but why? Again, picturing the physical set up shows you that this must be so. Pushing symbols around won't do this for you. You must bear in mind that all 3 elements in the circuit are in series. Then you can apply a physical law, which in itself is derived from the physical concept that charge is conserved throughout the circuit. That law is known as Kirchoff's Current Law. It states that the sum of the currents flowing into (positive sign) and out of (- sign) any circuit node must be zero. In a series circuit like this one, the junctions between the components have exactly one input and one output. There are no points where the current branches in more than one direction. Therefore, the currents flowing into and out of every component are the same. If you are not aware of this and you are presented with system with branching nodes, you may not realize that you need to compute the currents in the different branches according to the elements in them.

Of course I realize most of you already know the principles used in this problem. Rather, my point is that to do physics, you need to know physics, and physics cannot be reduced to mathematics alone. Otherwise, who needs experiments? Maybe you know that as well. If you do, then please stop saying it isn't so.

 

[sophiecentaur]

and physics cannot be reduced to mathematics alone.

True - you have to 'translate' from the physical situation to the mathematical model and back again. The point about using Maths is that it is so well suited to describing relationships between variables. It is an appropriate language which is often much better than mere words for describing patterns and for predicting the outcome of a new situation. When I say "follow the maths and believe the result" I am assuming that the original simple empirical based rules have been applied correctly in the first place. There is a terrible risk that a rule that is badly stated in words can be applied badly in an arm waving way and result on a nonsense conclusion. The rigour of simple algebraic manipulation is something that can usually be relied on to take you from A to B in a descriptive argument. Maths can be very hard, of course, but people often reject its use for that reason, rather than because of a genuine objection about its accuracy or applicability.
Verbal descriptions are actually no nearer to what is 'the truth' than a chain of mathematical steps. Quite the reverse, very often, as they cloud the issue due to mis understandings and the result of using numerical examples which happen to produce a one-off answer that looks right.
The acid test is to look at the number of people who have produced useful and verifiable theories in Physics without the use of Maths. I can't think of any, off hand.

 

[Mark Harder]

True - you have to 'translate' from the physical situation to the mathematical model and back again. The point about using Maths is that it is so well suited to describing relationships ...

Sorry I didn't select the most trenchant quotes. It's all worth repeating. You touch on a point made by Eugene Wigner in his famous article on the Unexpected(?) Effectiveness of Mathematics. As far as we know so far, math is the best, perhaps, language for describing reality. I'm deliberately vague because math applies to so many areas of quantitative thought in addition to physics. Many ideas in biology, for instance, are inherently mathematical. Take DNA, for instance. The structure and function of DNA and their intimate relationships is one of the most central ideas in biology. But why? For the longest time, beginning in the 19th C., DNA was known as a chemical component of all living things that were studied up to that time. It was not until Watson & Crick - and Franklin - that the notion that DNA carried "information" with it was universally accepted. Without this rather abstract property, without mathematics in other words, DNA (as I love to point out) is just so much snot you can find in living things. At the core of genetics, organismal and molecular, are the mathematical ideas of, first, information, and also formal language. My gut sense is that math still has much to contribute to the understanding of molecular genetics and therefore, all of life.

 

[Physics_Kid]

why do you lose energy? why do you say 1/2?

if i charge a cap very slowly the heat generated is much less than if i charged it at 2x the watts.

 

[sophiecentaur]

Do the sums, rather than relying on intuition. They tell you that the same energy is lost, however long the charging takes. There is no hand waving explanation afaik.