2. Electric field of a Polarized Object Question: Calculate E -field produced BY (not: IN) the polarized object. A dielectric object will become polarized. Available: An external E -field: E ex . E ex
3.
4. Analysis and Symmetry Assume: p // E ex ; n and p homogeneous Z Y X Coordinate axes: assume Z-axis // E ex E ex Result of polarization: Dipole distribution: n dipoles/m 3 ; each dipole moment p [Cm]
5. Approach to solution Approach: = calculate potential V ; = E from V by differentiation Distributed dipoles: dV - integration over volume elements dv filled with d p. E ex Z Y X v Question: calculate E -field in arbitrary point P outside v P
6. Calculations (1) E ex Z Y X P Z' r e r dp dz volume element dv with dp = np.dv = np.dS . . dz ; with dS Z-axis Dipole:
7. Calculations (2) E ex Z Y P Z' X r e r dp dz cross section through P and Z’-axis : P r dS r b r t z t : top z b : bottom
8. Calculations (3) dz -integration dr -integration : dr = -dz. cos dz = -dr / cos P r b r t r z t z b dS r dz dr P
9. Calculations (4) P r b r t r z t z b dS r dz dr P “Polarization” P = n p
10. Calculations (5) bound charges: d b = P.dS E ex Z Y P Z' X r e r dp dz P dS P.dS = P.dS
11. Conclusions the end bound charges: d b = P.dS P E ex Z Y X r bottom r top + - Conclusion: the field of a polarized volume is equivalent to that of bound surface charges (provided homogeneous polarization) + - + - - - + + + + + + + - - - -