W9_parallel_resonance.eps

152 Week 4: Capacitance

a function of coordinates. This latter derivation has the advantage in that the first two lines hold for

anysource of the free-space fieldE 0 , not just a presumed external parallel plate capacitor with its

uniform field. For example, if we surround a bare point charge with a dielectric shell as portrayed

in figure 49:

+Q

b

+Qb εr

−Q

a

b

Figure 49: A bare charge +Qsurrounded by a dielectric shell with relative permittivityǫr.

Hopefully we all know quite well at this point that the “bare” field of the free charge +Qin the

center is just

Er=

Q

4 πǫ 0 r^2

From the reasoning above:

σb = −χe

Q

(4πa^2 )ǫr

=−

Q(ǫr−1)

(4πa^2 )ǫr

(268)

σb = +χe Q

(4πb^2 )ǫr

= +Q(ǫr−1)

(4πb^2 )ǫr

(269)

Note well that thetotalbound charge oneithersurface has magnitude:

Qb=Qǫr−^1

ǫr

(270)

The charge on the inner surface reduces the field produced by Gauss’s Law “just right” to produce

a field ofE/ǫrin the dielectric; the charge on the outer surface puts it back so that the usual field

obtains outside of the dielectric sphere!

Advanced: This can safely be skipped to the next separator line if you are not a physics major.

Before we go on to energy density, we should at least put down the more advanced relations that

you will derive and learn in a more advanced course in Electrodynamicsand hint at how such a

derivation would proceed. Supposenˆis a normal unit vector perpendicular to a dielectric surface,

where the polarization density is e.g.P~=ǫ 0 χeE~forE~justinsidethe material. Thenσb, which is

ascalar, is given by:

σb=P~·nˆ

Our treatment above was valid for the special case thatnˆ||P~, but note that the dot product gets

the sign ofσbrightandcorrects for the “tilt” of the surface relative to the field! We had to put

the former in “by hand” above, and had no clue about the latter (although you can show it easily

enough if you recapitulate the original argument connectingσbtoPabove for a tilted surface).