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The Electrical energy of a conducting cylinder

  1. Dec 1, 2016 #1
    1. The problem stsatemesnt, all variables and given/known data
    A cylinder conductor of length ##l##, inner radius ##a## and outer radius ##b## is half-filled with a liquid dielectric as shown in the picture. The cylinder is attached to a battery of constant voltage ##U## and then separated from it. Imagine that the dielectric is leaking and calculate the change of electrical energy when there is no dielectric present. ( meaning the difference between the beginning state and the one were there is no dielectric)
    There is a picture below.
    IMG_2019.JPG
    2. Relevant equations
    3. The attempt at a solution
    I'm gonna guide you through my work and thinking so you can correct me and help me finish the problem.

    ##ΔW_e=W_{e2}-W_{e1}## ( The state 2 is when its empty)
    ##W_{e1}=\frac{1}{2}\frac{Q_1^2}{C_1}## ( ##C_1,Q_1## are the total charge and capacitance of the system 1)
    Because there is a dielectric in the first system i have two electric fields, the one above the ##h## and the one below going radial.
    ##E_1=\frac{Q'_1}{2 \pi ε_0 r}##
    ##U_1=\frac{Q'_1}{2 \pi ε_0}\ln (b/a)##
    ##E_2=\frac{Q'_2}{2 \pi ε_0 ε_r r}##
    ##U_2=\frac{Q'_2}{2 \pi ε_0 ε_r} \ln (b/a)##
    Then i express the linear charge densities ##Q'_1,Q'_2##
    ##Q'_1=\frac{2 \pi ε_0 U_1}{\ln (b/a)}##
    ##Q'_2=\frac{2 \pi ε_0 ε_r U_2}{\ln (b/a)}##
    ##Q_{total}=Q'_1*(l/2) + Q'_2*(l/2)##
    Something does not seem right here, can you help me with what i got wrong?
     
    Last edited: Dec 1, 2016
  2. jcsd
  3. Dec 1, 2016 #2

    BvU

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    Hi,

    I'm missing the relevant equations in section 2 :smile: like ##Q_{\rm total}\ ## conserved, ##U_1 = U_2 ## and such...
    And what about ##\ C_1, C_2\ ## ?
     
  4. Dec 1, 2016 #3
    It's not stated but im guessing that ##Q## must be conserved and ##C## and ##U## are changing. :) The whole work is my guess so i dont know the answer if anything is conserved xD I was thinking someone can provide an insight.
     
  5. Dec 1, 2016 #4

    rude man

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    I would forget about E fields. Just compute C1 and C2, then take the difference in 1/2 CV^2. Yes, Q is conserved; removing the dielectric does not remove any charge. So you know that V1 and V2 will be different also.
     
  6. Dec 1, 2016 #5
    But to compute C i need to have V and to have V i need E
     
  7. Dec 1, 2016 #6

    rude man

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    No you don't. C is a property of the geometry etc etc and is independent of V, E or Q. (We are not yet dealing with nonlinear capacitors at your stage of the game).
     
  8. Dec 1, 2016 #7
    ##C=\frac{Q}{V}## by definition. What do you mean its not dependant on ##Q## or ##V##?
     
  9. Dec 1, 2016 #8

    rude man

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    OK take two large parallel plates. Area = A, separation = d. What's the capacitance?
     
  10. Dec 1, 2016 #9
    ##C=\epsilon_0\frac{A}{d}##
    I see your point, it is the conductors geometric property.
    For the cylindrical system i then have two capacitances ##C_1=\frac{2\pi\epsilon_0L/2}{ln(b/a)}##, ##C_2=\frac{2\pi\epsilon_0\epsilon_rL/2}{ln(b/a)}## Now i need to know the ##V_1,V_2## for the system in case it's half filled and for that i need ##E##. When i find that what is my ##C## for the case where its half filled? I have two C's?
     
  11. Dec 1, 2016 #10

    rude man

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    You don't need E.
    You have two capacitors in "parallel", do you see that? Call them CA and CB. So C = CA + CB. Say CA has the dielectric, initially. You can compute CA1; also CB1 = CA2 = CB2.
    You know V1, = same for CA and CB. You then can find Q = V1(CA1 + CB1).
    For V2 you invoke the fact that Q does not change. This enables you to compute V2.
    Finally, as I said, energy = 1/2 Δ{CV2}, or = 1/2 Q ΔV.
    I can say no more.
     
  12. Dec 1, 2016 #11

    BvU

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    All a matter of writing down the relevant equations, including the conservation relationships ...
    (not to rub it in, but to make clear the ingredients of a solid problem solving approach ... )
    No guessing. Physics. Where would Q go ?
     
  13. Dec 2, 2016 #12
    Dielectric polarization can make Q change at conductor surfaces. The net change in Q is 0 globally.
     
  14. Dec 2, 2016 #13

    rude man

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    Don't think so. The dielectric does not assume net charge. All charge stays on the metal surfaces.
     
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