Re: capacitor questions
- From: "Tom Bruhns" <k7itm@xxxxxxx>
- Date: 4 Apr 2006 16:58:50 -0700
Charge would be pumped until you reached the open-circuit voltage of
the solar cell. Why would you think it would go beyond that? (Unless
of course the capacitor or something else breaks down before that
point.)
If on the other hand you were to shoot electrons at one plate and
positive ions at the other, you'd eventually get to the point where the
electrons were repelling each other with a force great enough that they
would fly off (toward the positive plate), and/or the ions would do
similarly. With an air dielectric, you reach a point where the field
strength is great enough that the stray electron (there are pretty much
always some around) gets accelerated enough by the field that when it
hits an air molecule, it knocks off one or more electrons, and pretty
soon you have an arc. The lower the air pressure, the longer the mean
free path before the electron strikes something, so the lower the
electric field required to start an arc, at least up to the point where
the mean free path approaches the plate separation.
Capacitors with a vacuum as dielectric don't store energy in
polarization of the dielectric, I suppose. The energy is stored very
similarly to the energy stored when you lift a weight and place it on a
shelf. There is a potential for doing work on the stored charge; the
potential is the voltage. If you wish, you can say that the energy is
stored in an electric field; but an equally valid way to think of it is
that the energy is stored in the fact that two charges (positive and
negative) are attracting each other, with a force that can be allowed
to act over a distance, just as the gravitational force can be allowed
to act on the weight over a distance.
Imagine a capacitor formed of two large plates with some separation
which is small compared with the size of the plates, with a vacuum
between the plates. Charge the capacitor with charge Q. The plates
will see a force pulling them together. Allow the plates to come
together half the initial separation. The charge remains the same, but
the voltage drops by half, because you have removed half the available
energy. At the same time, the capacitance has doubled. Note that this
is all consistent with C=Q*V and stored energy = (C*V^2)/2. Also note
that if you pull the plates apart, you put energy into the system, and
the voltage goes up. (It's how some high voltage generators work...)
Cheers,
Tom
===================================
Hi,
if you hooked up a solar cell in space to a capacitor, and let the sun
power the solar cell and charge up the capacitor indefinitely, more and
more electrons will be pumped to the negative plate of the capacitor,
until what limit? Assuming the dielectric can handle the voltage, will
all of the free electrons eventually be pumped to the negative side of
the capacitor, or is there no limit to the number of electrons that can
be pumped leading to a voltage only limited by the dielectric breakdown
voltage? If there is no limit to the number of electrons that can be
pumped to the negative plate, where do these electrons come from?
I am a bit confused about where the capacitor stores its energy, is it
stored as an electron differential in the two plates or is it stored as
a polarization in the dielectric? (similar to an inductors core
magnetization)
cheers,
Jamie
.
- References:
- capacitor questions
- From: Jamie Morken
- capacitor questions
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