Start with plates of a charged capacitor (P=positive charge,
N=negative charge)
P N
P N
P N
P N
Then insert a single dipole (n = negative end, p = positive end ).
P N
P np N
P N
P N
The potential difference between the row of P's and the row of N's
is reduced because the potential created by the np dipole is negative
on the left and positive on the right (the negative end of the dipole
(n) is closer to the row of Ps than the positive end (p) of the same
dipole). This argument extends to filling the whole space with
dipoles.
Mike
===============================================================
Mike Moloney
moloney@nextwork.rose-hulman.edu
Dept of Physics & Applied Optics (812) 877 8302
Rose-Hulman Institute of Technology Terre Haute, IN 47803 http://www.rose-hulman.edu/~moloney