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Energy in a capacitor

  1. Oct 28, 2015 #1
    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?
  2. jcsd
  3. Oct 28, 2015 #2
    How do you understand it "mathematically"?
  4. Oct 28, 2015 #3
    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
  5. Oct 28, 2015 #4
    To what formula of energy do you apply the formula of charge?
  6. Oct 28, 2015 #5
    To the potential electrical energy
  7. Oct 28, 2015 #6
    And that formula is...?
  8. Oct 28, 2015 #7
    I am not sure
  9. Oct 28, 2015 #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?
  10. Oct 28, 2015 #9
    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)?
  11. Oct 28, 2015 #10
    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?
  12. Oct 28, 2015 #11
    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
  13. Oct 28, 2015 #12
    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.
  14. Oct 28, 2015 #13
    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".
  15. Oct 28, 2015 #14
    I understand your point, do you yourself have an explanation to this formula of why it is like that?
  16. Oct 28, 2015 #15
  17. Oct 28, 2015 #16
  18. Oct 28, 2015 #17
    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
  19. Oct 28, 2015 #18
  20. Oct 28, 2015 #19
    If there is anyone else who can explain this question of mine please do so. Thank you
  21. Oct 28, 2015 #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.
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