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The field inside the conducting plate is zero. For that reason
my expectation would be:
MMMMMMMMMMMMMMMMMMMMMM__________
MMMMMMMMMMMMMMMMMMMMMM
UPPER METALLIC PLATE (no field inside)
++++++++++++++++++++
| | | | | | | | | contributes +sigma/eps-o
v v v v v v v v v
| | | | | | | | | contributes -sigma/eps_o
v v v v v v v v v
- - - - - - - - - - - - - - - - - - - -
LOWER METALLIC PLATE (no field inside)
MMMMMMMMMMMMMMMMMMMMMM__________
MMMMMMMMMMMMMMMMMMMMMM
What is wrong with this reasoning?
Ludwik Kowalski
John Mallinckrodt wrote:
On Sun, 3 Feb 2002, Ludwik Kowalski wrote:
... two layers of sigma in a capacitor reside on metallic
plates and each plate produces E"=sigma/eps_o.
No. Gauss' law shows that each layer produces sigma/(2*eps_o) on
each side of the layer. The result from the + layer looks like
this:
^ ^ ^ ^ ^ ^ ^ ^ ^
| | | | | | | | |
++++++++++++++++++
| | | | | | | | |
v v v v v v v v v
When the effect of the - layer is included we find that the field
between the plates is sigma/eps_o and elsewhere it is 0.
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm