PH 316 Bound Charge in Dielectrics 10/6/05 MJM (worksheet on 2nd page)
This is intended as a brief summary of some of what we have been up to.
Vdipole = kp cos /r2, or V dipole = k (pr)/r3 .
In a dielectric material, we integrate to get the effect of all the dipoles. We use the polarization P (the
dipole moment per unit volume, so that a small batch of dipoles is represented by P .
The integral starts out as
V(r) = k d P(r-r')/|r-r'|r-r'|r-r'|3 .
We showed that ' 1/|r-r'|r-r'|r-r'| = (r-r')/|r-r'|r-r'|r-r'|3 . ( ' is the gradient with respect to r'.).
This gave
V(r) = k d P[ ' 1/|r-r'|r-r'|r-r'|] => => k da bound /|r-r'|r-r'|r-r'| + k d bound/|r-r'|r-r'|r-r'|.
where b = Pn^, and b = - 'P . (some vector identities were used)
From gauss's law, we had o E = , and now we recognize both free charge and bound charge. Free
charge could exist on conductors, or it might exist as net charge within a dielectric, where teeny charges
got put into molten dielctric, then hardened. Bound charge is the ends of dipoles, either sticking out the
surface of a dielectric, or as a net charge (a batch of dipole ends where P is spreading out (diverging). So
o E = f + b = f + ( -P) .
Then (oE + P) = f and then D = f, where D = oE + P .
In a linear dielectric P = oE, and then
D = o (1+)E = E = r o E ,
where is the dielectric constant, is the electric susceptibility (I'm omitting the subscript for now)
and r is the 'relative' dielectric constant, the ration of to o .
Applying the divergence theorem to a tiny pillbox, we showed that D is continuous across a boundary
between dielectrics (there must be no free charge right at the boundary), and from curl E = 0, and Stokes
theorem, that E|r-r'||r-r'| is continuous. (It turns out that the potential V is also continuous across a boundary.
Why isn't E continuous? From o E = f + b and a tiny pillbox, we find that E would jump at the
boundary if there was bound charge present at the boundary: E 2 - E 1 = b/o, even if no free charge
was present.
(Next page for a short worksheet).
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