electric fields

[kowalskil@alpha.montclair.edu]

The consideration below was wrong because it ignored the fact that the
plates, in the question asked, were connected to a battery. It was not a
matter of inserting a dielectric material into the space between the 
initially charged and disconnected plates. The battery supplies free 
charge to compensete for the effect of bound charges. E (volts/meter)
is not changed by the introduction of mica.

                                              I goofed, Ludwik Kowalski

> My answer would be that Eb<Ea (field in the mica capacitor is weaker). The
> way of justifying this is as follows. Draw a parallel plate C so that the
> left plate is negative. Think about all those dipoles inside mica. How do
> they orient in the field? Their negative sides are now pointing to the
> right. The net effect is as a negative layer of charges next to the plus
> plate and positive layer of charges next to the minus plate. The field
> created by these layers weakens the field which would the plates create
> in a vacuum.