Energy in a capacitor

Main Question or Discussion Point

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?
 

Answers and Replies

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How do you understand it "mathematically"?
 
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
 
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To what formula of energy do you apply the formula of charge?
 
To what formula of energy do you apply the formula of charge?
To the potential electrical energy
 
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And that formula is...?
 
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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?
 
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)?
 
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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?
 
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
 
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.
 
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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".
 
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?
 
I cannot open the site.
And I have a bunch of resources but non of the text books 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
 
gleem
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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.
 
sophiecentaur
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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 text books 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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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
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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.)
 
gleem
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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?
 

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