Current through parallel plate capacitor vibration sensor

[HappMatt]

Homework Statement

A parallel plate capacitor is used as a vibration sensor. The plates have an area of 100cm^2, the dieelectric is air, and the dist between the plates is given by

d(t)=1+.01 sin(200 t) mm

A constant voltage of 200 V is applied to the sensor. Determine the current through tthe sensor as a function of time by using the aproximation1/(1+x) = 1-x for x<<1.

Homework Equations

C=(1*10000mm^2)/(d(t)
i(t)= C* dv/dt

The Attempt at a Solution

this is about as far as i have gotten because I am not sure how to approach this exactly because I am not sure what the dv/dt is since i know it has a constant voltage across it there is 200 v across the capacitor but that leaves me with dv/dt =0.

 

[buffordboy23]

What is the equation for electric potential (voltage) of a parallel plate capacitor? Look at your variables in the equation; something must be changing with time to keep a the potential constant.

 

[HappMatt]

What is the equation for electric potential (voltage) of a parallel plate capacitor? Look at your variables in the equation; something must be changing with time to keep a the potential constant.

The only thing that seems to be changin with time is the capacitance. And the equation for voltage aross a parallel plate cap is

v(t)=(1/c)(INTEGRAL(i(t)) + V0

and so far that has done me no good when I use that to substitute into the equation for i(t). I have written the question down exactly as it is in the book and I am not sure where to go with this. I'm thinking about it but nothing has seemed to work yet.

 

[buffordboy23]

Sure, capacitance depends inversely on the distance between the plates. If the capacitance varies, the charge q must vary as well.

So that we are on the same page, these were the equations that came to my mind.

$$V = \frac{Qd}{A\epsilon_{0}}$$

and

$$C = \frac{A\epsilon_{0}}{d}$$

 

[HappMatt]

yeah that makes sense i realize that something is varying and I had not seen this form of the equation yet
[tex] V = \frac{Qd}{A\epsilon_{0}} <br /> $$and i think it might make a little more sense seeing that but I am not sure on how i would get the charge then.$$[/tex]

 

[HappMatt]

SO here is what i ended up with. I scanned my paper and hopefully this works. If anyone has any better ideas that i can implement before class tomorrow i would be more than happy for the help along with the help i have recied already.
thanks Matt

 

[buffordboy23]

Yes, your dV/dt = 0, if V is constant. You said that the capacitance is changing though, so this must mean that the amount of charge Q is changing:

$$C = \frac{Q}{V}$$

Since Q is changing, there must be a current:

$$i\left(t\right) = \frac{dQ\left(t\right) }{dt}$$

This is a general algorithm for how I would solve the problem, which is different from your method. First, since the dielectric is air, I would assume that $$\kappa = 1$$ (but of course you could leave it in there). Rearrange the following equation for $$Q\left(t\right)$$

$$V = \frac{Qd}{A\kappa\epsilon_{0}}$$

and then substitute the time-dependent distance into the equation . Next apply your small x approximation and then take the time-derivative of $$Q\left(t\right)$$ to get your current $$i\left(t\right)$$. The amplitude of the current is really small (on the order of femto-amps). Does this amplitude make sense to you? I am not an EE major, so I briefly looked at some commercial sites that sold sensors and found ones with this level of sensitivity.