Here's a simplistic view that may be sufficient. Some energy (in the
form of charge redistribution, which includes current flow) has to come
from the capacitor, and some from the input signal, to do the work
needed to push the leaf against gravity. When the input signal is
removed, some of the energy (charge) stored on the leaf is returned to
the cap as gravity restores the initial position - roughly the same
amount of work, depending on leakage and mechanical loss, and heating of
the protective series resistor.
With a quick review of electrostatics, you could estimate up a simple
model and the field equations to get a more satisfying, detailed answer.
It may be more straightforward to look at it from a circuit perspective.
Picture it as as a very small, non-linear capacitor (the leaf structure)
in series with a much much larger regular capacitor charged up to a
constant DC voltage. Any actual series resistance is just resistance,
and the mechanical loss can be represented as more resistance added in
series. Presuming the leaf never actually touches or emits particles* or
arcs to the reference capacitor node, it's basically a capacitive
voltage divider, and the applied signal may be considered to be
transient, or even AC - it steps to the applied voltage, then returns to
zero (or open), then the cycle may be repeated. Each experiment is
adding or subtracting charge, then reversing the process. Ideally, the
cap would never lose its DC bias if there were no losses.
The problem is that figuring out all these details may not be trivial.
It may be more fun to just try some experiments and see how it goes -
you'll get some idea of how the real thing holds up, and figure what
amount of C is OK.
*You shouldn't have to worry too much about corona discharge if the
maximum voltage on anything is below 3 kV or so. Beyond that, it could
cause problems.
Ed
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