Energy in a Capacitor: Explained

In summary: I cannot open the... link because it says it is not allowed.In summary, the formula of energy in a capacitor E= C x U^2 /2 is explained mathematically as how it applies the formula of charge to the formula of energy and converts Q to C. However, the application to a real situation is explained by looking at end of chapter problems. If you want to understand it applied, you would need to look at examples.
  • #1
naiasetvolo
22
0
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?
 
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  • #2
How do you understand it "mathematically"?
 
  • #3
nasu said:
How do you understand it "mathematically"?
Perhaps I used the wrong term, but I understand it from the point of Q= C x U in which it applies the formula of charge to the formula of energy and converts Q to C. So
Q= C x U --- > E= C x U^2 /2
But I want to understand it from an applied matter, right now I just see numbers and not the actual reason
 
  • #4
To what formula of energy do you apply the formula of charge?
 
  • #5
nasu said:
To what formula of energy do you apply the formula of charge?
To the potential electrical energy
 
  • #6
And that formula is...?
 
  • #7
nasu said:
And that formula is...?
I am not sure
 
  • #8
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?
 
  • #9
litup said:
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?
So have I understood it correctly, when the charge is passing the conductor there will be an electrical resistance . This resistance will cause it to loose half the energy used to charge a cap as you just referred. That is why we divide it by 2? But then why is the current multiplied by itself ( U^2)?
 
  • #10
naiasetvolo said:
I am not sure
I am afraid you don't really understand the math and this may be the source of your problem.
Why don't you look up the math and try to understand that?
 
  • #11
nasu said:
I am afraid you don't really understand the math and this may be the source of your problem.
Why don't you look up the math and try to understand that?
Nasu, how should I take a look at it when that is the source of my problem? I need someone who can guide and explain to me so that can open up my brain
 
  • #12
nasu said:
I am afraid you don't really understand the math and this may be the source of your problem.
Why don't you look up the math and try to understand that?
There is always an explanation to everything, the source of everything starts with thoughts- a theoretical reason for why this formula exists. And I want that so I can start to think and analyze.
 
  • #13
naiasetvolo said:
Nasu, how should I take a look at it when that is the source of my problem? I need someone who can guide and explain to me so that can open up my brain
You started by saying that you understand it mathematically.
I thought you mean you can understand how that specific formula comes from more general definitions of energy and work.
But obviously you mean something else. Probably even your concept of "understanding" may be something else. Or maybe not even well defined in your mind.
It happens in the beginning,

If you want to apply that formula to a specific situation, just try to solve end of chapter problems. This is their purpose: working with the formula in specific situations you become more familiar with it. This is part of what we call "understanding".
 
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  • #14
nasu said:
You started by saying that you understand it mathematically.
I thought you mean you can understand how that specific formula comes from more general definitions of energy and work.
But obviously you mean something else. Probably even your concept of "understanding" may be something else. Or maybe not even well defined in your mind.
It happens in the beginning,

If you want to apply that formula to a specific situation, just try to solve end of chapter problems. This is their purpose: working with the formula in specific situations you become more familiar with it. This is part of what we call "understanding".
I understand your point, do you yourself have an explanation to this formula of why it is like that?
 
  • #17
naiasetvolo said:
I cannot open the site.
And I have a bunch of resources but non of the textbooks I have used note* in different languages have answered my question, if you understand why it is the way it is, I appreciate your help
 
  • #19
If there is anyone else who can explain this question of mine please do so. Thank you
 
  • #20
Initially the capacitor is uncharged. You begin by moving a charge from one plate the the other so the work done is very small since potential difference only begins when the charge actually gets to the other plate. As you move more charges the potential difference increases. So the potential difference depends on the net charge Q moved but that charge was moved under a linearly increasing potential difference. Thus the total charge Q was moved under an average potential difference of Vfinal/2 so the net work done is QVfinal/2. Work done = energy stored

Since Q=CV -------> W = E =CV2/2

resistance plays no part in this process.
 
  • #21
naiasetvolo said:
If there is anyone else who can explain this question of mine please do so. Thank you
Why do you refuse the ready made explanations that are in all the textbooks and also the hyperphysics site? The factor of 'a half' is there as a result of integration. If you don't want to do the maths then you are pretty well doomed not to understand this. You need to put in some effort on your own, I think because there is no adequate and simple arm waving reason.
The 'one half' factor is there for the same reason that the Kinetic Energy, SUVAT and Spring Energy formulae have it. It's very basic.
 
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  • #22
Post 20 & 21 are correct. A more interesting capacitor question is this:

Two identical capacitors, one fully discharged and the other charged to 2V. Thus their total energy is 0.5C(2V)^2 = 2CV^2 but after they are connected by a condutor each holds half the original charge and is at voltage V. Each has stored energy of 0.5CV^2 but together their total energy is CV^2, half of what it was prior to the mutual connection.
Where did energy CV^2 disappear to? To make it tougher, note that the conductor was a superconductor, with zero resistance.

PS if you are working with large high-voltage capacitors, as I have in a controlled fusion project, the stress in the dielectric, which is where the energy is stored, like in a spring, will not be completely releaxed by briefly shorting the poles together. - It will relax in the next few minutes, recharging the capacitor - making it quite dangerous again. Leave the terminals connected for several minutes to fully discharge capacitor.
 
Last edited:
  • #23
BillyT said:
Where did energy CV^2 disappare to?

Into charging the uncharged capacitor.
 
  • #24
gleem said:
Into charging the uncharged capacitor.
No. that is not correct. I'll wait a little more before answering my own question., but give a hint to part of the answer: All capacitors have some inductance. Thus, as the charged capacitor dumps energy into the uncharged one, its inductance is storing energy too. When that magnetically stored energy is again zero, the initially charged capacitor will have LESS than V across its terminals and the other more than V.
I. e. the systems "rings" like a bell initially, but does end up with both charged to V.

To make the problem even tougher, assume the internal resistance of the capacitors is, like that of their interconnection, also zero. (Often nearly true. I.e. not much of the missing energy heated the capacitors.)
 
  • #25
Inductance would only be a factor when the charge is flowing into the uncharged capacitor. What happens to the energy when the charge stops flowing?
 
  • #26
This problem was discussed on the forum. More than one time, I believe.
It keeps coming back. :smile:
 
  • #27
nasu said:
This problem was discussed on the forum. More than one time, I believe.
It keeps coming back. :smile:
Do you have link to earlier one, giving the correct answer?
 
  • #28
Pf has a search facility. You could find it useful.
 
  • #29
sophiecentaur said:
Pf has a search facility. You could find it useful.
Yes. Here is link with relatively complete and correct dicussion:
http://www.smpstech.com/charge.htm
Although it does not mention that when all resistance is assumed absent, the circuit "rings" like damped bell. The oscillating current flowing in the supper conducting interconnection wire, radiates half the originally stored energy into space.
 
  • #30
That's a good link.
However 'ideal you make your circuit, the Radiation loss will always be there because of the finite dimensions of the set up. The inherent Inductance will 'Ring' with the Capacitances and, despite notionally zero resistance in the conductors, the Power Loss due to radiation will appear in the form of a series resistance in the loop. This is referred to as Radiation Resistance and this also what a Transmitter 'sees' when it is feeding a transmitting antenna.
It is interesting that the factor of one half comes in, whatever the Inductance and Resistance happens to be (however many cycles of ringing and whatever frequency).
 
  • #31
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).
 
  • #32
naiasetvolo said:
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.
cape15.gif

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

Cheers
 
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  • #33
litup said:
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.
 
  • #34
Mark Harder said:
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.

Mark Harder said:
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.
 
  • #35
sophiecentaur said:
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.
 
<h2>What is a capacitor?</h2><p>A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.</p><h2>How does a capacitor store energy?</h2><p>When a voltage is applied to a capacitor, the electric field between the plates becomes charged. This causes one plate to accumulate positive charge, while the other accumulates negative charge. This separation of charges creates the stored energy in the capacitor.</p><h2>What is the relationship between voltage and energy in a capacitor?</h2><p>The amount of energy stored in a capacitor is directly proportional to the voltage applied to it. This means that the higher the voltage, the more energy the capacitor can store.</p><h2>How does the size of a capacitor affect its energy storage capacity?</h2><p>The size of a capacitor, specifically the surface area of its plates and the distance between them, directly affects its energy storage capacity. A larger surface area and smaller distance between plates will result in a higher energy storage capacity.</p><h2>What are some common uses of capacitors?</h2><p>Capacitors have a wide range of applications, including storing energy in electronic circuits, filtering out unwanted frequencies in audio equipment, and power factor correction in power grids. They are also commonly used in electronic devices such as cameras, computers, and televisions.</p>

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.

How does a capacitor store energy?

When a voltage is applied to a capacitor, the electric field between the plates becomes charged. This causes one plate to accumulate positive charge, while the other accumulates negative charge. This separation of charges creates the stored energy in the capacitor.

What is the relationship between voltage and energy in a capacitor?

The amount of energy stored in a capacitor is directly proportional to the voltage applied to it. This means that the higher the voltage, the more energy the capacitor can store.

How does the size of a capacitor affect its energy storage capacity?

The size of a capacitor, specifically the surface area of its plates and the distance between them, directly affects its energy storage capacity. A larger surface area and smaller distance between plates will result in a higher energy storage capacity.

What are some common uses of capacitors?

Capacitors have a wide range of applications, including storing energy in electronic circuits, filtering out unwanted frequencies in audio equipment, and power factor correction in power grids. They are also commonly used in electronic devices such as cameras, computers, and televisions.

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