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Capacitors with dielectric in series

  1. Dec 13, 2014 #1
    I would like to know if everything is good in my reasoning.
    I've never done exercises like that and I don't have the results.

    1. The problem statement, all variables and given/known data
    A generator which mantains a constant potential difference of 400V is connected to two capacitors in series C1 = 500pF and C2=1000pF. A dielectric is inserted into C1 (##\epsilon_r = 4##).
    Calculate the charge variation on C2, the polarization charge on both faces of the dielectric and the energy supplied by the generator.

    3. The attempt at a solution
    I calculated the ##C_eq## of the system of the capacitors without the dielectric.

    ## \frac{1}{C_eq} = \frac{1}{C1} + \frac{1}{C2} ##

    ## C_eq = 333 pF ##

    Since the capacitors are connected in series, q is the same on both capacitors.

    ##q = CV = 1.33x10^{-7}C ##

    Now I calculate C1 with the dielectric.
    ## C1 = \epsilon_r C1 = 2x10^-9 F ##

    The new ##C_eq## is ##6.66x10^-10F##.

    ##q = CV = 2.66x10^{-7} C##

    I can calculate easily the variation of the charge.

    The potential difference is constant, the capacity (as I expected) increased. q could only increase.

    Now I find the polarization charge on both faces.
    I find it with the vector ##\vec{P} = \epsilon_0 \chi \vec{E}##

    ## P = \sigma_p = \frac{q_p}{S} = \epsilon_0 (\epsilon_r -1) \frac{q}{S} ##

    ##q_p## is 1.99x10^-7. It is +qp near the negative face of the capacitor, and it is -qp near the positive one.

    The energy is

    ##U = \frac{1}{2} CV^{2} ##

    Could you please tell me if that's right?
    If not, please tell me something to think about!

    Thank you
     
  2. jcsd
  3. Dec 14, 2014 #2

    rude man

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    Good for the first two parts (delta Q of C2 and polarization charge of C1).
    Careful with the last part.
    The generator does not supply just the energy represented by the new capacitor configuration.
    The new total energy storage is indeed U = 1/2 CV^2 where C is the new equivalent capacitance of 6.667e-10 F and V = 400V.
    However, that energy is not all that the generator needs to supply. Moving the dielectric into C1 constitutes a temporary linear motor, with work provided by the pull-in force. The generator needs to supply this work as well!
     
  4. Dec 14, 2014 #3
    How can I consider that last part into the energy supplied by the generator?
     
  5. Dec 14, 2014 #4

    ehild

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    After inserting the dielectric Δq additional charge appears on the capacitor plates. That charge is supplied by the generator, which makes Δq charge move though 400 V potential difference. Part of the work increases the energy of the capacitors, other part is dissipated as heat on the connecting wires, the internal resistance of the generator, and electromagnetic radiation.
     
  6. Dec 14, 2014 #5

    rude man

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    The total energy supplied by the generator is more than that. Even if there is zero wire heat and zero generator impedance, and e-m radiation can safely be ignored, the generator must also supply the energy provided by the work realized as the dielectric is pulled into the plates of C1.
     
    Last edited: Dec 14, 2014
  7. Dec 14, 2014 #6

    rude man

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    Do it in two parts:
    1. Compute energy U1 used to set up the circuit before the dielectric is inserted. You know that U1 is 100% supplied by the generator.
    EDIT: actually, the generator supplies more energy than 1/2 CeqV^2 even for this first part. No matter how small the wire resistance, half the power furnished by the generator will change to heat. The other half charges up the two capacitors.

    2. Compute the extra charge ΔQ supplied by the generator after the dielectric was put into C1's plates. Then the generator supplied extra energy U2 = VΔQ.
    Then the total energy supplied by the generator = U1 + U2.

    Note that (U1 + U2) > 1/2 C'eqV2 where C'eq is the equivalent series capacitance after the dielectric is inserted. I explained why in post 2.
     
    Last edited: Dec 14, 2014
  8. Dec 16, 2014 #7
    That was very clear, thank you.
    I know that half of the energy furnished by the generator during the process of charge changes to heat, but I don't think this problem wants me to consider that.
     
  9. Dec 16, 2014 #8

    rude man

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    Why not? The problem asks for " ... the energy supplied by the generator."
     
  10. Dec 16, 2014 #9
    Shouldn't I know something else? It doesn't provide me any data..
     
  11. Dec 16, 2014 #10

    rude man

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    N
    No. As I said, the generator furnishes twice the energy stored in phase 1 (before the dielectric is inserted) irrespective of the wiring reistance so long as it's not zero. So if the capacitor energy is 1/2 CeqV2 you know the generator supplied twice that energy to set up C1 and C2 before the dielectric is inserted.

    Now, to make your life even more confusing: I initially assumed that for the second phase, when the dielectric is inserted, there would be no further energy dissipation in the wiring. That unfortunately I now believe is incorrect.

    Here's the picture: after C1 and C2 are set up, assume the dielectric is instantaneously inserted into C1. Also assume finite wire resistance R. C1 rises from 500 pF to 2000 pF instantaneously. And during that zero time no charge has had time to flow from the generator. So now the problem is you have C1 at 2000 pF, a lower C1 voltage than before, same C2 voltage as before, and current starting to flow until C1 and C2 voltages are redistributed per the final configuration. Then you can compute V∫i(t)dt which is the energy the generator supplies during phase 2. As a check you can add ∫i2Rdt which is the energy dissipated by the wiring in phase 2, and add to that the change in stored energy (between phase 1 and 2).

    Sorry, this is a big problem, and quite possibly your instructor didn't mean you to include wiring dissipation. But then the problem should have said so.
     
    Last edited: Dec 17, 2014
  12. Dec 18, 2014 #11

    rude man

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    Afternote: I tried to find the energy dissipated in the wiring for the second phase. It's straight-forward but the math is so messy that a mistake or two is likely. So again I suspect you were not really supposed to include the wiring energy dissipated in either phase.
     
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