Wed. (C14) 4.3 Electric Displacement Washington 3-2 Rep 7pm AHoN 116 Thurs. HW5 Fri. (C14) 4.4.1 Linear Dielectrics (read rest at your discretion) From last Time: Polarization Observer   dp P   r  d r     1  rˆ   1  b b V (r)    Prd   da  d   dips   2  4   4 0 volume r  0 surface r volume r  Polarized object      where r  b  P aˆ and b   P Q   d   da b  b  b volume surface Polarization & “Bound Charge” Exercise     b  P aˆ and b   P

A thick spherical shell (inner radius a and outer radius b) is made of dielectric k material with a “frozen-in” polarization Pr  rˆ r b Locate all of the bound charge and use Gauss’s law to calculate the electric field in a the three regions.

Cross-sectional view The Electric “Displacement” Quite Generally, Gauss’s law says    E  1   o (all)

Now we also relate “bound”  charge (due to variation in density of dipoles)   P  b So, if you have a region with dipoles and free charges,

(all)   free  bound or,  free  (all)  bound      free   o E    P     free   o E  P     o E  P  D Electric Displacement So Gauss’s Law for free charge and Electric Displacement  free   D and   Q   d  D da free  free  Polarization & Electric Displacement Exercise – take 2      Q   d  Dda  E  P  D free  free  o

A thick spherical shell (inner radius a and outer radius b) is made of dielectric k material with a “frozen-in” polarization P  r   r ˆ . There are no free charges. r Find D from our new Gauss’s Law in all three regions, and then find E from it. Polarization & Electric Displacement Example      Q   d  Dda  E  P  D free  free  o  Consider a huge slab of dielectric material initially with uniform field, E  o and corresponding uniform polarization and electric displacement D o   o E o  P o.

You cut a small spherical hole out of it. What is the field in  its center in terms of and P o ? E o For illustrative purposes only, take the polarization to be anti- Po parallel to the field, and imagine both to be in the z direction. By Superposition Principle, cutting out a sphere is the same as inserting a sphere of opposite polarization. Quoting Example 4.2 (which in turn builds on 3.9), the field inside a uniformly polarized sphere is  1  Esphere   Psphere  3 o So, we ‘add in ‘ a sphere of polarization  Po  1  Adding field E  P added 3 o   o E  E  1 P in.sphere o 3 o 0 Polarization & Electric Displacement Example      Q   d  Dda  E  P  D free  free  o  Consider a huge slab of dielectric material initially with uniform field, E  o and corresponding uniform polarization and electric displacement D o   o E o  P o.

You cut a small spherical hole out of it. What is the field in    its center in terms of and P ? 1  o Esphere   Psphere E    3 o o E  E  1 P in.sphere o 3 o 0 Po  What is the electric displacement in its center in terms of D o and ?    Dsphere   o Eshere  Psphere There is no material in the sphere, so       D   E  1 P  where E  1 D  P  sphere o o 3 o 0 o  o o o so       1 1 2 D    D  P  P   Do  P0  sphere o  o o o 3 o 0 3 Polarization & Electric Displacement Exercise         Q   d  Dda  E  P  D  b  Paˆ and b   P free  free  o  Consider a huge slab of dielectric material initially with uniform field, E  o and corresponding uniform polarization and electric displacement D o   o E o  P o.  You cut out a wafer-shaped cavity perpendicular to P o .  What is the field in its center in terms of and ? E o Hint: Think of inserting the appropriate waver-sized capacitor.  Po What is the electric displacement in its center in terms of D o and ? The Electric “Displacement” So Gauss’s Law for free charge and Electric Displacement           D and Q   d  D da where D   E  P free free  free  o

Of practical use – with polarizable materials, you might directly

control free, but bound unavoidably changes in response (reminiscent of the free energies in thermo)     Warning: while   E  0 in electrostatics, and so E  1 rˆd  4o  r 2       For electric displacement  D   o E   P   and so D  1 free rˆd  Not necessarily 0 4o  r 2 Which means it can’t necessarily be expressed as gradient of a scalar field Boundary Conditions Electric field, across charged surface   datop Etop   encl  E     o daBottom   Q E da  encl    o Ebottom       Q dasurface E da  E da  E da  encl    top top  bottom bottom  sides sides   o o Send side height / area to 0     dasurface E da  E da    top top  bottom bottom  A o Etop A  Ebottom A(1)   o  Etop  Ebottom   o Boundary Conditions Electric Displacement, across charged surface   datop Dtop    D   free.encl  free  daBottom   D da  Q   free.encl Dbottom       D da  D da  D da  Q   freedasurface  top top  bottom bottom  sides sides free.encl  Send side height / area to 0     D da  D da   da  top top  bottom bottom  free surface

Dtop A  Dbottom A(1)   free A

Dtop  Dbottom   free Boundary Conditions (static) Electric field, along charged surface   E dltop top    E  0     dlbottom  E dl  0  Ebottom       E dl  E dl  E dl  0  top top  bottom bottom  sides sides Send side height to 0     E dl  E dl  0  top top  bottom bottom

E||top L  E||bottom L(1)  0

E||top  E||bottom  0 Boundary Conditions (static) Electric displacement, along charged surface   D dltop top      D   P    D     dlbottom bottom  D dl   P dl             D dl  D dl  D dl  P dl  P dl  P dl  top top  bottom bottom  sides sides  top top  bottom bottom  sides sides Send side height to 0         D dl  D dl  P dl  P dl  top top  bottom bottom  top top  bottom bottom

D||top L  D||bottom L(1)  P||top L  P||bottom L(1)

D||top  D||bottom  P||top  P||bottom Boundary Conditions  D Electric and Displacement fields top    datop dl Etop top aˆ Along E  E  0   ||top ||bottom da  Bottom dl bottom  D||top  D||bottom  P||top  P||bottom  Ebottom    Dbottom (could have guessed as much from D   o E  P .) Across     Etop  Ebottom  E aˆ  E aˆ   top bottom  o or   o D aˆ  D aˆ   Dtop  Dbottom  free top bottom free     ˆ (could have guessed as much from D   o E  P and    free   b  P  a .) Boundary Conditions Electric and Displacement fields    D   E  P Along o  Across   free   b  P aˆ  E||top  E||bottom  0 Etop  Ebottom   o D||top  D||bottom  P||top  P||bottom Dtop  Dbottom  free Exercise Bar Electret (like an electric bar magnet): uniform P along axis Sketch 푃, 퐸, and 퐷 as to obey the boundary conditions. 퐸 퐷 푃

Only surface charge is the bound There is no free surface charge, so surface charge on two faces no discontinuity in D. D   o E  P Mon. (C14) 4.3 Electric Displacement Mon. (C14) 4.4.1 Linear Dielectrics (read rest at your discretion) Wed. 6.1 Magnetization HW10