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Velocity of capacitor plates

  1. May 16, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    In the following RC circuit, the capacitor is in the steady state. The initial separation of the capacitor plates is x0. If at t = 0, the separation between the plates starts changing so that a constant current flows through R. Find the velocity of the moving plates as a function of time. The plate area is A.


    3. The attempt at a solution

    Applying Kirchoff's Law

    [itex]E=\dfrac{qvt}{\epsilon_0 A} + \dfrac{R dq}{dt} [/itex]

    But how do I solve this differential equation?
     

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  2. jcsd
  3. May 16, 2014 #2

    vela

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    The differential equation you derived isn't quite correct because the plate separation is initially ##x_0##, which doesn't appear anywhere in your equation.

    After you fix that, the quantities that potentially vary with time are q, v, and dq/dt, which makes the equation hard to solve. You want to eliminate q and dq/dt and get an equation that only depends on v and t. Use the information that the current through the resistor is constant.
     
  4. May 16, 2014 #3

    utkarshakash

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    The correct equation should be

    [itex]E=\dfrac{q(x_0-\int vdt )}{\epsilon_0 A} + \dfrac{R dq}{dt} [/itex]

    Here I've assumed that the separation is decreasing.
     
    Last edited: May 16, 2014
  5. May 17, 2014 #4

    rude man

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    ??
    That equation is dimensionally inconsistent in the 1st term on the right.

    Hint: what is the current thru the capacitor in terms of capacitance and capacitor voltage?
    Combine that with vela's hint, i.e. E = V_C + V_R. You wind up with a simple ODE in C(t), without either V or q.
     
  6. May 17, 2014 #5

    utkarshakash

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    [itex]q=CV_c \\
    \dfrac{dq}{dt} = i=V_c \dfrac{dC}{dt} \\
    V_r = iR \\
    E= \dfrac{q}{C} + V_c R \dfrac{dC}{dt} [/itex]

    Are the above equations correct?
    Why is my earlier equation dimensionally inconsistent?
     
  7. May 17, 2014 #6

    rude man

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    EDIT:
    They are correct but the fourth is unhelpful.
    What can you say about Vc in terms of E, i0 and R where i0 is the constant current? What is i0 in this problem?

    Also, get rid of q in your last equation.
     
    Last edited: May 17, 2014
  8. May 17, 2014 #7

    utkarshakash

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    Vc = E-i0 R
     
  9. May 17, 2014 #8

    rude man

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    Excellent idea! Now, get rid of q.
     
  10. May 17, 2014 #9

    utkarshakash

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    The last equation can be modified as

    [itex]E = (E-i_0 R)(1+R\dfrac{dC}{dt}) [/itex]
     
  11. May 17, 2014 #10

    rude man

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    Yes!
     
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