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LC circuit

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    In an oscillating circuit consisting of of a parallel-plate capacitor and an inductance coil with negligible active resistance the oscillations with energy ##W## are sustained. The capacitor plates were slowly drawn apart to increase the scillation frequency ##\eta##-fold. What work was done in the process?

    2. Relevant equations
    ##C = \frac{\epsilon_0 \cdot S}{d}##
    ##C'=\frac{\epsilon_0 \cdot S}{\eta^2 d} ##


    3. The attempt at a solution
    So ohm's law is written like this: (before the stretching of the capacitor)

    ##L \cdot \ddot{q} = -\frac{q}{C}##
    This implies:
    ##\omega^2=\frac{1}{LC}##
    ##q_{(t)}=q_m \cdot sin(\omega t)##

    The electrostatic force exerted by one plate on the other is:

    ## F = E_{(t)} \cdot q_{(t)} ##
    ## E_{(t)} = \frac{\sigma}{2\epsilon_0} = \frac{q_{(t)}}{2\epsilon_0S} ## ##\Bigg\} ## ##F = \frac{q_{(t)}^2}{2\epsilon_0S}=\frac{q_m^2 \cdot sin(\omega t)^2}{2\epsilon_0S}##

    Because the streching of the capacitor happened slowly it means that we can consider the average force over one period:

    ##\bar{F}=\frac{q_m^2}{4\epsilon_0S}##
    So , the work done is then:
    ## \Delta W= \bar{F}\cdot d\cdot (\eta^2-1) = \frac{q_m^2d}{4\epsilon_0S} \cdot (\eta^2-1)##

    Now,
    ##W=\frac{q_m^2}{2C}=\frac{q_m^2d}{2\epsilon_0S}##

    From here results that :
    ##\Delta W= \frac{W}{2} \cdot (\eta^2-1)##

    This formula implies that the initial charge ##q_m## is not conserved. Is it ok? It looked odd to me...
     
  2. jcsd
  3. Oct 31, 2016 #2

    TSny

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    I believe this is correct.

    This assumes that ##\bar{F}## remains constant while the plates are separated. But as you found, ##q_m## is not conserved. So, the average force is not constant.

    Yes

    For notational convenience, I suggest letting ##x## stand for the distance between the plates. Consider the work done by the average applied force for an infinitesimal increase in separation, ##dx##.
     
  4. Oct 31, 2016 #3
    Thank you!

    I solved it. Obtaining that :

    ##\Delta W = W \cdot (\eta-1)##
     
  5. Oct 31, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    Looks good!
     
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