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When the dielectric is inserted assuming no energy is lost I would expect the voltage to fall to 1/10-.5 KV ie 316.2 V so that CV^2/2 remains the same.
I am very puzzled by the opinion of JP that 90% of the energy in the system has disappeared by the voltage being reduced to 10% of what it was while the capacitance is increased 10 times.Where has it gone ? it cannot surely be the work done moving the dielectric sheet in.
Why? A dielectric with ε>ε0 polarizes under the action of the field, so it's attracted between the plates, so it gains kinetic energy. Neglecting the energy conveyed into heat because of Joule effect inside the dielectric and into radiated electromagnetic waves because of time variable currents, this kinetic energy is what you loose from the capacitor.
I was just about to say that, but Lightarrow beat me to it  (Yeah! Sure you were Geezer!)Should it not be a gain in potential energy rather than kinetic energy?
Quote from: Geezer on 21/09/2009 19:53:29I was just about to say that, but Lightarrow beat me to it  (Yeah! Sure you were Geezer!)Should it not be a gain in potential energy rather than kinetic energy? Of course it is potential energy at the beginning and kinetic energy with the dielectric exactly between the plates; in other positions it's both potential and kinetic, thanks for pointing it out .
The capacitor would have to be mounted on a track sufficiently far from the dielectric sheet along with its voltage measuring equipment.After charging it is moved slowly into the vicinity of the dielectric sheet which is drawn between the plates.by measuring the deflection of the sheet we can calculate how much energy has been transferred to it an see how it corelates to the voltage on the capacitor.It should be conducted in a very dry atmosphere or vacuum chamber as an moisture on the surface of the dielectric sheet would distort the results.