Energy in a Capacitor: Explained

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
45 replies · 7K views
naiasetvolo
Messages
22
Reaction score
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?
 
Physics news on Phys.org
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
 
nasu said:
To what formula of energy do you apply the formula of charge?
To the potential electrical energy
 
nasu said:
And that formula is...?
I am not sure
 
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?
 
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)?
 
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?
 
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
 
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.
 
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".
 
  • Like
Likes   Reactions: sophiecentaur
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?
 
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
 
If there is anyone else who can explain this question of mine please do so. Thank you
 
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.
 
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.
 
  • Like
Likes   Reactions: Mark Harder and nasu
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:
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.)
 
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?
 
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.
 
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).