Re: E field and capacitance question

[Herb Schulz <herbs@interaccess.com>]

> Good evening everyone,
>
> My fellow physics teacher and I ran into a problem that has us perplexed
> so I thought I'd run it by any of you who would be interested. My own
> background in electricity is rather weak so I wasn't able to offer any
> insight. Here's the problem stated as clearly as possible in the absence
> of a diagram.
>
> Two capacitors, A and B, are connected in parallel with a 45 V voltage
> source. They are identical in construction except that A uses air as a
> dielectric and B uses mica as a dielectric (K=6.0). If the electric
> field for A is Ea and the electric field for B is Eb, then
>
> (a) Ea = Eb
> (b) Ea > Eb
> (c) Eb > Ea
>
> Your best answer and an explanation would be greatly appreciated. I
> can't recall my colleague's reasoning exactly, but he disagrees with the
> textbook's stock answer. (I can't imagine that the textbook could
> contain an incorrect answer. Can you? Yeah, right...)
>
> Later,
> Tim
>
> --
> Timothy D. Wilson                    "A little song, a little dance,
> Henry Sibley High School              a little seltzer down your pants."
> 1897 Delaware Ave, W St. Paul, MN          -Chuckles the Clown as
> wilson@chemsun.chem.umn.edu                 eulogized by Ted Baxter

Howdy,

Since the capacitors are connected in parallel they must have the same
potential difference across their plates: therefore they will have equal
electric fields between the plates (i.e., answer (a) above).  The capacitor
with the dielectric between the plates will have the larger amount of
charge on the plates to compensate for the surface charge on the dielectric
material.

If the capacitors were connected in series the charges would be equal and
the electric field (and therefore the potential difference across the
plates of that capacitor) would be smaller in the dielectric filled
capacitor.

Good Luck,

Herb Schulz
(herbs@interaccess.com)