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Homework Help: Capacitors - long Q - full procedure, very grateful if someone will check

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Please see attachment.
    I really need someone to check this properly please- I know this is meant to help people but checking this would be a huge help to me since its worth marks....(Ive done all the work)

    THANKS A LOT to anyone who helps..

    2. Relevant equations

    3. The attempt at a solution
    a) C=[tex]\epsilon[/tex]0*A/d

    C[tex]\propto[/tex] A
    C[tex]\propto[/tex] 1/d



    t=1.2ms (just used solver for this)

    [tex]\Sigma[/tex] C=6uF

    Attached Files:

    • caps.png
      File size:
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  2. jcsd
  3. Jul 19, 2010 #2


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    (a) OK
    (b) The number is correct but you should learn to do it without solver. You may not have solver available if a question like this shows up on a test.
    (c) Incorrect. What is the "difference" that you are asked to explain all about? Is potential energy stored in the capacitor(s) conserved or is there something else that is conserved?
  4. Jul 19, 2010 #3
    thanks for the responce
    b) i can solve this - just take ln both sides - but i dont have my calc with me so i used internet solver.
    c) is the p.d i found correct??? Yeah I just realised i forgot to do half the question. No potential energy cant be stored in the capacitor and that makes me think that my numerical answer is wrong to but I dont know how else to do it?????
    thanks again for responding.
  5. Jul 19, 2010 #4


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    Try conserving charge instead of energy.
  6. Jul 19, 2010 #5
    so qi=qf which I can see would be true as long as the system is closed which it is.
    My problem is now that I cant find an appropriate formula anywhere in my textbook - online I found Q=CV which would give me
    Qi=1.6*10^-3 C
    therefore 1.6*10^-3= (2+4uf)V
    so V=266.66V ????????????

    for the energy can I use U=0.5CV^2 ???
    and get
    Ubefore = .5*4*10^-6*400^2 =0.32J ???
    Uafter =.5*6*10^-6*266.66^2 =0.21333J

    If that is correct (I doubt) than that proves that energy is not conserved, do you think that that is what they want explained???
  7. Jul 19, 2010 #6


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    If your textbook does not have Q = CV (the definition of capacitance) you should toss it and get a better textbook.
    I agree with these numbers.
    That's what they want you to explain, the apparent non-conservation of energy.
  8. Jul 19, 2010 #7
    thankyou!!!!!!!!!!! just looking textbook does have it im just an idiot -it was in the form C=Q/V (actually knew this of the top of my head just didn't realize its use.) so i flicked past it thinking it was useless to me....... no harm
  9. Jul 19, 2010 #8
    Hey kuruman, just one more question if you have time or see this.. with the explanation - I dont understand where the energy could have gone? How can energy not be conserved in a closed system???
  10. Jul 20, 2010 #9


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    Not all closed systems necessarily conserve mechanical energy which is what we are talking about here. Consider the totally inelastic collision where mass m1 is moving with initial speed v0 and collides with m2 initially at rest. The two masses move together after the collision and their common speed is the speed of the center of mass, namely


    If you calculate the final energy of the "closed" system, you get (after a bit of algebra)

    [tex]K_f=\frac{1}{2}m_1v^{2}_{0} \left( \frac{m_1}{m_1+m_2} \right)[/tex]

    This is clearly less than the initial energy


    Where did the energy go? The stock answer is "Heat generated by friction."

    This capacitor problem is exactly analogous to the totally inelastic collision. The symbols are different but the math is the same. Here is why

    Momentum (p) is conserved becomes charge (q) is conserved.
    Definition p = mv is replaced by definition q = CV, i.e. capacitance (C) is the analogue of mass (m) and voltage (V) is the analogue of velocity (v).
    Kinetic energy=(1/2)mv2 becomes Energy stored=(1/2)CV2.

    So the prediction is that the energy remaining in the system is the analogue of Kf, namely

    [tex]E_f=\frac{1}{2}C_1V^{2}_{0} \left( \frac{C_1}{C_1+C_2} \right)[/tex]

    Put in the numbers and see what you get.

    As for "where did the energy go?", complete the analogy. If heat production by friction accounts for loss of mechanical energy in a collision, what accounts for loss of mechanical energy in a circuit and how is it generated?
  11. Jul 20, 2010 #10
    I know from empirical evidence that heat is a product of some reaction in most if not all circuits but i dont know why. building a TC in the holidays my caps where literally bursting from heat and although I had read a lot on the net about it happening to other people, I never understood where the heat was coming from???? Having said that I guess that there is energy loss in other areas too??? I cant imagine that a few resistors would produce much heat if they are rated for the use?
  12. Jul 20, 2010 #11


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