How is integration order determined for the I-V relation of capacitors?

[iScience]

I'm having a brain fart so this is just another silly question but...

when deriving the I-V relation for the capacitor:

$$C=\frac{dq}{dV}$$

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

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

$$\int\frac{dC}{dt}dV=i$$

$$C\frac{d}{dt}V=i$$

$$C\frac{dV}{dt}=i$$

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?

 

[analogdesign]

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.

 

[iScience]

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

the quantity:

$$\frac{q}{V}$$

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"
to..
"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?

 

[Delta2]

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
\frac{d}{dt}(\frac{dq}{dV})=\frac{d^2q}{dV^2}\frac{dV}{dt}=\frac{d^2q}{dV}\frac{1}{dt}

...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 isn't zero

Update:

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

1) if we consider q(V)=CV then d^2q=q''(V)dV^2=0 since C is constant
2) if we consider I(t)=\frac{dq}{dt} then \frac{dI}{dt}=\frac{d^2q}{dt^2}, d^2q=I'(t)dt^2 which obviously isn't identical zero...

 

[iScience]

aha! got it

indeed it was silly..

order of integration:

$$C=\frac{dq}{dV}$$

$$\int CdV=\int dq$$

$$\frac{d}{dt}CV=\frac{dq}{dt}$$