1. Electric field of a Polarized Object © Frits F.M. de Mul

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

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) + - + - - - + + + + + + + - - - -