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Subject: Re: What's E-field in a capacitor?
Date: Wed, 06 Nov 2002 22:56:50 +0000
NNTP-Posting-Date: 6 Nov 2002 22:56:44 GMT
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Christopher R. Carlen wrote:
> 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
> dielectric medium.
> 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.
The result is half your conclusion.
To reply to me directly:
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