Charge on capacitor plates when separated

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Discussion Overview

The discussion revolves around the behavior of a charged capacitor when its plates are separated after disconnecting from a voltage source. Participants explore the implications of distance on voltage, capacitance, and potential discharge into the environment, touching on theoretical and practical aspects of capacitors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that when the plates of a charged capacitor are pulled apart, the voltage at the terminals increases due to a decrease in capacitance while charge remains constant.
  • Others argue that there is a limit to how much the voltage can increase, suggesting that the dielectric may cease to function effectively at larger separations, and environmental factors may influence the system.
  • A participant mentions that the voltage increases until the distance between the plates is significantly smaller than their size, beyond which edge effects and the behavior of the plates as point charges come into play.
  • Some participants note that the capacitor may discharge into the surrounding environment if other conductive elements are present, indicating that practical scenarios often involve additional capacitances that can affect the system.
  • One participant introduces the concept of using capacitors in parametric amplifiers, highlighting a practical application of the principles discussed.
  • Another participant reflects on the potential dangers of using a charged capacitor as a battery, noting that the rising voltage could become lethal with moderate charge levels.

Areas of Agreement / Disagreement

Participants express varying views on the behavior of voltage and capacitance as the plates are separated, with no consensus reached on the exact limits of capacitor functionality or the conditions under which discharge occurs.

Contextual Notes

Participants acknowledge that the behavior of capacitors can be influenced by factors such as plate size, distance, and the presence of other conductive materials, but these aspects remain unresolved in the discussion.

Who May Find This Useful

This discussion may be of interest to those studying capacitor behavior in electrical engineering, physics, or related fields, particularly in contexts involving high voltage applications and theoretical explorations of capacitance.

curiouschris
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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.
 
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curiouschris said:
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.

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.
 
curiouschris said:
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?

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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ehild said:
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

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.
 
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
 
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.
 

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