# Charge on capacitor plates when seperated

1. Jun 26, 2014

### curiouschris

As I understand it if you charge two plates as in a capacitor and then disconnect the voltage source and pull the plates apart to an arbitrary distance the voltage at the terminal increases.

Is that correct?
Will that voltage continue to increase?
Is there a point that the voltage will stop increasing?
If so why?
Will the plates discharge into the surrounding environment if the plates are physically separated by a large distance?

Curious.

2. Jun 26, 2014

### analogdesign

If you're able to pull the plates apart without upsetting the charge stored then yes, the voltage will increase.

Q = CV where Q is charge, C is capacitance and V is voltage. If you pull the plates away you reduce C which increases V since Q remains constant.

There is a point at which the voltage will stop increasing because at some point the dielectric between the plates will cease to function. As the capacitance gets smaller and smaller other capacitances in the environment will play increasingly important roles in the system and eventually will dominate and share charge with your idealized capacitor.

The capacitor will only discharge to the environment if other conductors are present. Since in the real world this is always the case I will say as a practical answer yes, the plates will discharge.

Since you're curious, there is a fascinating type of device call a parametric amplifier used in very high speed RF systems. Basically it works by charging up a capacitor (usually a part of a specific type of diode) and then changing the capacitor to increase or decrease the voltage. Voila! An amplifier.

3. Jun 26, 2014

### ehild

The voltage increases between the capacitor plates with distance d till the distance is still much less then the size of the plates. Only in that case is approximately true that the electric field E is homogeneous and thus V=Ed.
See picture. One plate is grounded, the other having positive charge, is moved away from it. The edge effect appears and the field lines scatter out more and more from between the plates as the distance increases. At very high distance the plate behaves more and more as a point charge.

ehild

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4. Jun 26, 2014

### analogdesign

This is quite true but it doesn't give a clean break where we can declare "The Capacitor is no longer operating". In my experience you'll get parasitic capacitors dominating long before you reach a situation where the plates behave as point charges.

5. Jun 26, 2014

### ehild

Even a single charged piece of metal operates as a capacitor, as it has some voltage with respect to the ground, or infinity, and C=Q/U. Only the capacitance is no more C=εA/d when d is comparable to the size of the plates.

ehild

6. Jun 27, 2014

### curiouschris

Thanks all for your input. I am much clearer on it. That's an interesting 'proof' analogdesign using a cap as an amplifier. makes total sense though when you think about it.

My wondering was in relation to whether you could charge one plate up on a capacitor and then use it as a sort of battery. but the fact the voltage rises in relation to the charge and capacitance it would quickly become lethal even with a moderate charge.

Its very analogous with a collapsing field on an open cct inductor.