From: "Christopher R. Carlen"
Subject: What's E-field in a capacitor?
Date: Wed, 06 Nov 2002 11:21:56 -0800
Organization: Sandia National Laboratories, Albuquerque, NM USA
NNTP-Posting-Date: Wed, 6 Nov 2002 18:20:53 +0000 (UTC)
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Here is an interesting question:
At the boundary between a conductor and a dielectric, we know that:
E_vec = n_hat Q/(A p)
where E_vec is obviously the electric field, n_hat is the normal vector
to the surface pointing from the conductor into the dielectric medium, Q
is the charge, A is the area, and p is the permittivity of the
For a thin parallel plate capacitor in which the thickness d is much
less than the area dimensions of the plates, the "top" plate connected
to source V has charge Q given by Q=CV, and the "bottom" plate has
charge -Q. Let's consider that the z-axis points upward, against the
E-field that is developed in the spatial arrangement described.
Thus the electric field should be given by the superposition of the
fields contributed by the two individual charge densities. With the top
(+) plate's field being labeled E_vec1 and the bottom (-) plate's field
being labeled E_vec2:
E_vec=(-z_hat)Q/(A p) + z_hat (-Q)/(A p)
E_vec=(-z_hat) 2 Q/(A p)
the important part of the result here being the 2.
Is this the correct result, or is the field actually only 1/2 of my
conclusion? If so, why?
Thanks for comments.
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA