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I-V Relation for Capacitors

  1. Jul 20, 2014 #1
    I'm having a brain fart so this is just another silly question but...

    when deriving the I-V relation for the capacitor:


    $$\frac{d}{dt}C=\frac{d}{dt} (\frac{dq}{dV})=\frac{d}{dt}C=\frac{di}{dV}$$

    from here, normally we're supposed to do the following




    but even before integrating, where i have quantity: $$\frac{dC}{dt}$$ isn't this just zero?
    in which case, if we integrate both sides with V i just get 0 on the LHS so i know it's not valid..
    but why is it not valid?
    Last edited: Jul 20, 2014
  2. jcsd
  3. Jul 21, 2014 #2


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    Your initial equation is wrong. C isn't the derivative of charge with respect to voltage. C is the charge divided by voltage (Q = CV). Big difference.

    Remember that current is defined as charge per unit time and you'll get it right.
  4. Jul 21, 2014 #3
    the quantity:


    is a density. but all densities can be expressed in terms of differentials no?

    qualitatively, all i have done was reduce "the total amount of charge per the given amount of voltage"
    "the unit charge per unit voltage"

    why is doing this invalid? we do this in physics all the time! we take bulk quantities and reduce them down to their "fundamental" unts. and because capacitance is a density (a ratio), the value should be the same, why shouldn't it be?
  5. Jul 21, 2014 #4


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    C=dq/dV holds if C is constant. Because q(t)=CV(t) hence differentiating wrt time and because C is constant in time we get dq/dt=CdV/dt or dq=CdV. However the mistake is in the 2nd line because using the chain rule of differentiation we get

    ...Well i have to say you ve blocked my mind as well. But somewhere along the 2nd line is the mistake cause we know dC/dt=0 while dI/dV isnt zero


    Well something very strange , if we take the definition of [tex]d^2q[/tex] as second order differential we get two different things:

    1) if we consider [tex]q(V)=CV[/tex] then [tex]d^2q=q''(V)dV^2=0 [/tex] since C is constant
    2) if we consider [tex]I(t)=\frac{dq}{dt}[/tex] then [tex]\frac{dI}{dt}=\frac{d^2q}{dt^2}, d^2q=I'(t)dt^2[/tex] which obviously isnt identical zero...
    Last edited: Jul 21, 2014
  6. Jul 21, 2014 #5
    aha! got it

    indeed it was silly..

    order of integration:


    $$\int CdV=\int dq$$

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