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An explanation of bound charges in dielectric materials, integrating the effect of all the dipoles using polarization p. Gauss's law is used to recognize free and bound charges, and the relationship between the electric field, polarization, and dielectric constant is discussed.
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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 /r^2 , or V (^) dipole = k ( p r )/r^3. 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 = P n^ , 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 (o E + P ) = f and then D = f, where D = o E + 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). Name___________________________________ Box __________ 1
A sphere of radius R carries a polarization. P = C r , where C is a constant. [ This is basically Prob. 4.10, p. 169 ] a) Calculate the bound charges b and b. b) Find the electric field inside the sphere (think Gauss's law) c) Find the electric field outside the sphere 2