"Syd Rumpo" <
use...@nononono.co.uk> wrote in message
news:k6eee9$rjf$1...@dont-email.me...
> In (say) a parallel plate air-gap capacitor, the plates will bend
> towards each other slightly as charge is added. Is the electrical
> energy stored the same as the mechanical energy which would be required
> to bend the uncharged plates by the same amount?
Not the same as, but in addition to. If the plates did not flex (an
approximation you could base on the small-signal or high-frequency
capacitance, where plate deflection is negligible), you would expect that,
at some large static potential, the capacitor would store such-and-such,
projecting Q = V*C on a straight line (constant C). If you actually
perform the experiment, you'll discover the graph of Q vs. V bends up,
i.e., C goes up with voltage, storing much more charge at that "large
static potential". The difference is the excess energy consumed by
deflecting the plates that distance, and that difference is exactly the
amount of mechanical energy*.
*Average. The real world has rate losses (ESR, damping of the springy
plates, etc.) and static losses (hysteresis of the spring, dielectric,
etc.), so the actual zero-to-one integral may be higher than the
one-to-zero path. If you obtain a loop in the Q vs. V plot, draw a
straight line through the diagonal corners and see how that compares.
Tim
--
Deep Friar: a very philosophical monk.
Website:
http://webpages.charter.net/dawill/tmoranwms