any current in an AC circuit if you moved the plates of a capacitor apart?

say you have an ac power source (120 v) connected to a circuit which just has one parallel plate open air capacitor, with the plates seperated by, say, 1 mm.

you measure the current flowing through the circuit and you get some rms value.

now say you move the plates of the open air capacitor apart - say to 1 meter. would you still measure an rms current in the ciruit? why/why not?
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 How is the distance between the plates related to the capacitance?
 Also consider that the dielectric constant of practical capacitors is greater than of air.

any current in an AC circuit if you moved the plates of a capacitor apart?

ok, so if you double the distance between the capacitor plates the capacitance halves. so I_rms=V/X_c

where X_C=1/(omega*C) where I is rms current, X_c is capacitive impedance, C is capacitance, and omega is 2pi*frequency.

So I_rms=v*omega*C.

So double the distance of the capacitor plates, C halves and so does the RMS current.

Now for my real question:

Why is it that everyone says a circuit needs a return path? By taking an RC circuit and greatly increasing the distance between the plates of a parallel plate capacitor you are effectively creating an open circuit -- there is no return path! Yet as demonstrated from the equation above there will still be an rms current. It will be smaller than if the plates were close together, but it will still be there, and it doesn't fall off exponentially, but is proportional to 1/d where d is the distance between capacitor plates.

Now this is only true for AC circuits, but based on the above it seems like you don't need a return path in order for an AC current to exist: it will exist in an RC circuit whose capacitors plates are separated by a great distance (say 1 m), which is effectively an open circuit, thus a circuit without a return path.

 Quote by mrmojorising By taking an RC circuit and greatly increasing the distance between the plates of a parallel plate capacitor you are effectively creating an open circuit -- there is no return path! Yet as demonstrated from the equation above there will still be an rms current.
Where is this (theoretical) current flowing then, if there's no return path?

 Quote by gnurf Where is this (theoretical) current flowing then, if there's no return path?

I get it from the equation I_rms=v*omega*C.

As D (the distance between the cacitor plates increases) to something like 1 m the circuit effectively becomes an open circuit, which means there's no return path.

However as D increases to something like 1 meter C and thus I_rms decrease, but from the equation above do not go to zero - hence there's a current with no return path.

This would only work for AC, with the charged particles in the wire jiggling about in periodic motion. So the current is not flowing anywhere, it's composed of charged particles jiggling in periodic motion.

 Quote by mrmojorising As D (the distance between the cacitor plates increases) to something like 1 m the circuit effectively becomes an open circuit [...]
Then, effectively, no current will flow. You can't have it both ways.

 Quote by gnurf Then, effectively, no current will flow. You can't have it both ways.
why current would not flow?The capacitor plate which is given a -ve potential will induce a +ve charge on the other plate and hence electron will flow from +ve plate to +ve terminal of the battery resulting in current.
 Since your capacitance will be extremely small after you separate the plates, you will need a very high frequency to get any (measurable) current. And now we are talking about radio frequency circuits, which as we all know *can* transmit a small (AC) current through air.

 Quote by amaresh92 why current would not flow?The capacitor plate which is given a -ve potential will induce a +ve charge on the other plate and hence electron will flow from +ve plate to +ve terminal of the battery resulting in current.
If capacitance is negligible then current is negligible. If current is not negligible, then there isn't an open circuit, and the capacitance of the gap must be taken into account...it can not be treated as an open circuit. Whether a gap can be considered an open circuit or not depends on what you're doing...if an AC voltage is applied, there will always be current as parts of the circuit charge and discharge, however miniscule.
 [QUOTE=cjameshuff;3006087]If capacitance is negligible then current is negligible. how you can say the current will be negligible if the capacitance is negligible without knowing the input current.

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 Quote by mrmojorising Why is it that everyone says a circuit needs a return path?
For the same sort of reason that 'they' say current takes the path of least resistance. They are both over-simple statements which on on the right side of wrong.
If there is some value of Capacitance then there will be some value of current.

Quote by amaresh92
 Quote by cjameshuff If capacitance is negligible then current is negligible.
how you can say the current will be negligible if the capacitance is negligible without knowing the input current.
If current through such a circuit is non-negligible, then neither is the capacitance. Read the full post, rather than fixating on one small portion of it.
 Does current flow AS the capacitor pulled apart? I think the answer is yes, but I'm trying to think of an expression for this current.
 There is a return path; it's through your very, very, very small capacitor which has a crapton of reactance. Because the reactance is so high, the current will be very, very, very small, so small that it can be ignored for the vast majority of applications.
 Capacitance decreases as the plates are separated. Let me know when you figure out the distance that reaches zero farads.

 Quote by Skaperen Capacitance decreases as the plates are separated. Let me know when you figure out the distance that reaches zero farads.
Infinity.

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