Chp Chp-10 Electric fields in Matter 10 Electric fields in Matter Dielectrics Matter comes in many forms Notes mainly based on Griffiths Chp 4 These respond differently to an applied E-field Conductors & Insulators or Dielectrics Conductors & Insulators or Dielectrics We already considered conductors Conductors have charges that can freely move Conductors have charges that can freely move In metals it is typically one or two typically one or two electrons per atom That are not associated with a specific nucleus Electron density In a metal
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ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Dielectrics
Matter comes in many forms
Notes mainly based onGriffiths Chp 4
These respond differently to an applied E-field
Conductors & Insulators or DielectricsConductors & Insulators or Dielectrics
We already considered conductors
Conductors have charges that can freely moveConductors have charges that can freely move
In metals it is typically one or twotypically one or twoelectrons per atomThat are not associated with a specific nucleus Electron density
In a metal
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Insulators or Dielectrics
All charges are attached to a specific atom or molecule
They are not free to roam
The can move small distances within the confines of anThe can move small distances within the confines of an
atom or molecule
Yet the small displacements of charge in an E field accountYet the small displacements of charge in an E-field account
for the properties of dielectrics
Two types of displacements are possible…stretching and
rotating
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Induced Dipoles
What happens when a neutral atom is place in an E-field?
E-Field
+ Positively chargedNucleus
- Negatively chargedElectron cloud
Nucleus is pushed in the direction of the field
Electron cloud is attracted toward the field
If the E-field is strong enough it will ionize the atom (hence
forming ions and it becomes a conductor)
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
In the Limit of low magnitude E-field
Equilibrium is reached between the displacement induced
by the E-field and Coulombic attraction of nucleus and
electronselectrons
The atom is polarized in the E-field
r dr
+qq
d
+q+q-q
No E-field Applied E-field
+q-q
Get an induced dipole moment p points in the same
direction as the applied E-field
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
In the Limit of low magnitude E-field
Induced dipole moment is proportional to E
litypolarizabiatomic
Ep
H He Li Be C Ne Na Ar K Cs0.667 0.205 24.3 5.60 1.67 0.396 24.1 1.64 43.4 59.4
Atomic Polarizabilities /4o in units of 10-30 m3
Handbook of Chemistry & Physics 91st Edition CRC Pressy y
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Assuming that the electron cloud retains its spherical
symmetry under small displacement in an E-field, calculate
the atomic polarizability for the atom.
dr d
+q-q
No E-field Applied E-field
+q-q
eEE Internal field due to e- pulling nucleus leftIs balanced by the externalIs balanced by the external E-field pushing right
What is the field a distance d from the center of a uniformly charged Sphere?
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
The charge from outside shell does not contributed4V
3
sphere -q
Applied E-field
rd
d4density arg
3
3
sphere
echApplied E field
eEE 3d4q' so
soqE r4 3
2
334
2
2o
ddq'E'
so r4
E
densityarg3r4Vsphere
ech
32
r4q33
34
o2
o2
o
qddE'
3d4
d4E
3
3r4q so
yg3
What is the field a distance d from the center of a uniformly charged Sphere?
3o
2o r4d4
E
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Very crude atomic model but accurate within a factor of ~4
qdpbutqdEE' -q
Applied E-field
rd
3o
3o
e
Epbut E'r4p
qdpbut r4
EE
Applied E fieldeEE
Estimate the atomic polarizability of an H atom
atomo3
o V3r4
241
312123o
Fm10x64.1
)m10x9.52)(F/m10x85.8(4r4
331
o
m10x48.14
330
o
m10x667.04
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
What about molecules? They are not isotropic in their
polarizability
CO2 /NC1054
m/NC 10x2240
240
If applied field is at an angle then resolve into components
m/NC10x5.4 240||
Applied E-field perpendicular or parallel to molecular axisIf applied field is at an angle then resolve into components
||||EEp
However, the induced dipole moment may not be pointing in
the same direction as the E-field
For an asymmetrical molecule the most general linear
relationship between E and p is via a tensor
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Tensor: magnitude and multiple directions
Polarizability tensor (tensor of rank 2 so for 3D space 32=9
elements)
zxzyxyxxxx EEEp
zzzyzyxzxz
zyzyyyxyxy
yy
EEEpEEEp
zzzyzyxzxzp
Tensorility Polarizabelements of ij MatrixThe values of ij depend on orientation of the axes you use
Orient so you have the “principal” axes so that all off
diagonal elements are zero leaving: zzyyxx
Characterize an asymmetric molecule with 3 polarizabilities
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Griffiths Prob. 4.1
A H atom (with Bohr radius of 0.5 angstrom) is situated between two metal plates 1 mm apart, which are connected to opposite terminals of a 500 V battery. What fraction of the atomic radius does the separation distance d amount pto, roughly? Estimate the voltage you would need with this apparatus to ionize the atom [Use the value of in Table 4 1]4.1]
Expectation…The displacements due to polarization will be quite small
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Griffiths Prob. 4.1
V/10500V5VE 5 V/m10x5m10x
E 53-
330106670
241330123o
330
o
F/m10x42.7)m10x667.0)(F/m10x85.8(4r4
m10x667.04
o
m 10x32.2C106021
)V/m 10x5)(F/m 10x42.7(Ed so edEp 1619
5241
C 10x602.1e 19
x10394m 2.32x10d 6--16
Ratio
%x1039.4Or
x1039.4m 52.9x10r
4-
12-Ratio
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Griffiths Prob. 4.1…To ionize let d=radius of atom
VEr
V10x14.1m) 10x1C)( 10.602x1m)( 102.9x5(rexV
exer
8241
31912
soF/m10x42.7 241
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Griffiths 4.1.3 Alignment of Polar Molecules
What about molecules with a permanent dipole moment
E.g. Water has a large electronegativity difference between
hydrogen and oxygen & OH bonds are 105o aparthydrogen and oxygen & OH bonds are 105 apart
Electrons are more tightly pulled toward oxygen (although it
is still a covalent compound) pis still a covalent compound) pCm 10x1.6p 30
What happens when anE-field is applied?
Large dipole moment so water is a good polar solvent
E field is applied?
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Place water molecule in a uniform E-field
The force on the positive end will be exactly balance the
force on the negative end but there will be a torque
EFEF qandq
dd
FrFrNEFEF
q and q
torque
p EdEdEdN
qq
2q
2p
q FE q
dO
qF
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
The dipole moment in a uniform E-field experiences a
torquedp q
Note N is in a direction to line up p parallel to E
EpN
Note N is in a direction to line up p parallel to E
A polar molecule that is rotationally free will rotate in an
applied E fieldapplied E-field
A non uniform E-field will mean a net force in addition to Nq F O h l l di iE q
dOOver the molecular dimensionsVariation in E is usually small
qF
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
A non uniform E-field will mean a net force in addition to N
There will be a difference in E-field from + to - end EEEFFF qq
dzTdyTdxTdT
Recall that
zyxzyx ˆdzˆdyˆdxˆzTˆ
yTˆ
xTdT
zyx
q FE IdTdT
zyx
So if the dipole is very short
dEΔE A d f &q
F
dOE
E+ d xx EΔE And for x & y
EdΔE dp qAnd since
qF
d EpF
E-
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Polarization*
What happens when we put slab of a dielectric material in
an E-field?
If the dielectric is made of neutral atoms or nonpolar moleculesIf the dielectric is made of neutral atoms or nonpolar moleculesThe E-field will induce a tiny dipole moment pointing in the field directionIn asymmetric molecules the induced p will not be parallel to the field but if they are Randomly oriented then the components perpendicular to the field will cancel. If it is a crystalline solid then the cancelation does not occur!If th di l t i i d f l l l h l lIf the dielectric is made of polar molecules, each molecular dipole will experience a torque and rotate if freeNote that temperature dependent random thermal motions will always compete withTh li t i th li d fi ld
* Note that this is NOT the same polarization as in light polarization…which refers to the time dependent orientation of the E-field vector during propagation of an EM Wave
The alignment in the applied field
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Dielectric material becomes polarized in the E-field
Cartoon representation of dipoles aligning in an E-field
Note that there will still be someStretching of a polar molecule inStretching of a polar molecule inThe fieldBut it is easier to rotate a molecule than stretch it(Energy scale for rotations vs. vibrations?)
We measure this effect as the polarization P
vibrations?)
P Dipole moment per unit volumeThe polarized material will produce its own field…what will this do?
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Induced fields in nanoparticles
Shafiei et al. Nature Nanotechnology 8,95–99 (2013)
ChpChp--10 Electric fields in Matter10 Electric fields in MatterInduced fields in nanoparticles
Shafiei et al. Nature Nanotechnology 8,95–99 (2013)
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
We now have two E-fields
1. External applied E-field
E
Dielectric Material
Induce many dipole moments(each atom or
Materialp
2. ``Internal field`` due to the induced Polarization
(each atom or molecule)
These fields will combine by superposition
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
The field of a Polarized object
Concept of bound charges
What is the field produced by the induced dipoles?
Strategy…We know the field/potential (far field limit) for aStrategy…We know the field/potential (far field limit) for a
single dipole Field Lines &Potential surfaces f i l di l
Subdivide the dielectric into infinitesimal dipoles & integratefor a single dipole
It is easier to work with the potential
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Notational Warning to Physics 242 & 340 students
Purcell’s Notation for the potential (Phys 242) r Purcell s Notation for the potential (Phys 242)
Griffiths Notation for the potential (Phys 340)
r rV
Generalized Multipole Expansion
Rr
dv'P
Or'
R
OrExpansion for the potential for anyLocalized charge distribution( i l id d i l t d(previously we considered one isolated dipole)
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Generalized Multipole Expansion
rdv'
P'
R
dv'r'1 r
Or'
R4r
2
sourcesallo
Law of cosines
cos
rr'2
rr'1rcosrr'2r'rR
22222
Law of cosines
cos2
rr'
rr' 1rR
i.e. for point P far from r largefor 1
S t bi i l charge distributionSuggests a binomial expansion for1/R (see integral above)
1nn
...x
!21-nnnx1x1 2n
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Generalized Multipole Expansion
rdv'
Pr'
R 53111111 322
1
r
Or ...1682
1r
1rR
Substitute back in for
cos2r'r'
cos2rr
...cos2r'r'165cos2r'r'
83cos2r'r'
21111 3322
rr16rr8rr2rR
Collect terms in powers of (r’/r)33322 ...
2cos3cos5
rr'
21cos3
rr'cos
rr'1
r1
R1
33322
Pn are the Legendre polynomials…It is a cos
rr'
r1
R1
0n
n
n
P
Pn are the Legendre polynomials…It is a generating function for multipole expansionsNow substitute back into the integral
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Generalized Multipole Expansion
rdv'
P'
R
dv''1 rr r
Or'
dv''cosPr'11
R4
n rr
r
sourcesallo
...dv''1cos3r'1dv''cos'r1dv''11
dvcosPrr4
2232
n0
1n
rrrr
rr
no
...dv2
cos2
rr
dvcosrr
dvr4 32 rrrr
o
M lti l E i f th t ti l i t f 1/Monopole Dipole Quadrupole
Multipole Expansion for the potential in terms of 1/r(Exact expression if all terms included)
Integral depends on the angle between r & r’If you orient so r is on z’ axis (i.e. P on z’ axis) becomes the polar angle ’
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Multipole Expansion as an approximation to the potential dv'
P
R dQhQ1 r
Or' dvQ whererQ
41
o
monopole r
Monopole term dominates at large r
Dipole term
dv''cos'r11 rr 'r'r ˆ'
2
2
dv''ˆ14
1
dv''cos'rr4
rr'rr
rr
dipole
odipole
r'r ˆcosr' so r'r
cos
2
ˆ1
dv''
r4
rr'p
o
dipole
p =dipole moment of the charge distribution
2r41 rpr
o
dipole For a single dipole
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Where were we? Bound charges
Subdivide the dielectric into infinitesimal dipoles & integrate
r 2r
ˆ4
1 rpr
odipole
P Dipole moment per unit volume
dv' Ppp
dv'ˆ1
dv
2
rr'Pr
Pp
dvr4 2
volumeo
r
But note ˆ1 ‘ notation indicatesBut note…2rr
1' r
notation indicates differentiation wrt the source coordinates (r’)
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Where were we? Bound charges
Subdivide the dielectric into infinitesimal dipoles & integrate
r
dv'r1'P
41
r
r4 volumeo
fff AAAIntegrate by parts using the product rulep
fff Integrate by parts using the product rule
dv''r1dv'
r'
41
ll
PPr
dv''14
1da'ˆ14
1
rr4
volumevolumeo
PnPr
r4r4 volumeoSurfaceo
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Resembles potentials of surface & volume charges
dv''14
1da'ˆ14
1 PnPr
r4r4 volumeoSurfaceo
nP ˆb Pb
dv'r4
1da'r4
1 l
b
S f
b
r
The potential of a polarized dielectric material is the same as is produced by:
r4r4 volumeoSurfaceo
P ˆ1. Surface charge density
2 Volume charge density
nP ˆb
Pb
EThis is true for the E-field too since the gradient is a linear operator
2. Volume charge density b E
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Net result we can avoid integrating all the dipoles
We find the bound charges then calculate their fields
E.g. we could apply Gauss’s Law as we have seen
d '1d '1 bb dv'
r4da'
r4 volume
b
oSurface
b
o
r
n̂P ˆb Pb
ChpChp--10 Electric fields in Matter10 Electric fields in MatterFind the electric field produced by a uniformly polarized sphere, radius R
p R
z Set z axis in the direction of polarization
Then determine bound charge densitiesn̂
nP ˆb Pb
p R Then determine bound charge densities
0 Pb Vol. bound charge density zero (since Polarization is uniform) cosPˆ nPb Surface bound charge density integrated over a sphere
d 'cosP1
da'r
cos4
Sphereo
r
RfP Inside
RP
Rrfor cosr3
r,
3
o
Solved in Ex 3.9 inGriffiths
cosrz Inside
Rrfor cosrR
3Pr, 2
o
Griffiths Outside
ChpChp--10 Electric fields in Matter10 Electric fields in MatterFind the electric field produced by a uniformly polarized sphere, radius R
p R
z n̂ Rrfor cosrPr, Insidep R
zPr
3,
o
cosrz
Potential varies linearly in z inside the sphere
z3
r,o
PotentialRrfor
31 ˆ
3P z
3P
PzE
ooo
ooo
E-field
http://homepage.ntu.edu.tw/~chernrl/
ChpChp--10 Electric fields in Matter10 Electric fields in MatterFind the electric field produced by a uniformly polarized sphere, radius R
p R
z n̂ Rrfor cosr
3Pr, Insidep R
RrforcosRPr
3,
3
o
Outside
Potential outside the sphere is identical to perfect dipole at the
Rrfor cosr3
r, 2
o
Outside
Potential outside the sphere is identical to perfect dipole at the origin!
rp2 rrforˆ1θ)(r,
Potential E-field
Pp 3
2
R4
rrfor r4
θ)(r,
o
Pp R3
http://homepage.ntu.edu.tw/~chernrl/
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Physical Interpretation of bound charges
E-field of a polarized dielectric is produced by the combined
distribution of bound volume and surface chargesnP ˆbPb
Was this just abstract mathematical wizardry/bookkeeping?No the bound charges represent physical accumulations
How do we understand this? Consider a line of dipoles in a
o t e bou d c a ges ep ese t p ys ca accu u at o sOf charge!
polarized material…We get effective cancelation except at
the end points Looks like a large charge displacementbut really many small displacements
but really many small displacements
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
We call the net charge at the ends a bound charge
It is still attached (bound) to a specific atom or molecule
But it is still a charge!
How do we calculate the amount of bound charge?
Bound Charge
How do we calculate the amount of bound charge?d
A P
n̂
A= q- q
AendAConsider a cylinder
Of dielectric parallel to P qdAdPp p PAq
qdAdPp
Bound charge accumulated on right face
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
How do we calculate the amount of bound charge?d
A P
n̂
A= q- q
AendAP
Consider a cylinderOf dielectric parallel to P dd Of dielectric parallel to P
PAqqdAdPp
Bound charge accumulated on right facef
PAqb
If the slab is cut perpendicular to the cylinder
If th l b i t t l t th li d h i thIf the slab is cut at angle to the cylinder, charge is the same
nP ˆcosPA
q
end
bnP ˆend
Physically, Polarization “paints” bound charge over the surfacenP b
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
If the polarization is nonuniform bound charge will also
accumulate inside the dielectric as well as at the surface
d
Uniformly polarized sphere
E+ charges all move up
charges all move down
Without polarizationCh l
With polarizationCh t i t
E- charges all move down(slightly!)
Diverging P Charges cancel(neutral)
Charges separate intoCaps (surface bound charge)
The Integrated bound volume charge will be equal and opposite to the charge pushed out to the surface
d 'dˆd ' PnP ConcludePb
dv' dadv' Surface
PnP VolVol
bConclude
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
The field inside a dielectric
We started the derivation with formula for the potential for
an ideal or perfect dipole
Atoms or molecules are real or physical dipoles (but tiny!)Atoms or molecules are real or physical dipoles (but tiny!)
Also the Polarization P is a continuous vector function
Does this raise flags of concern?Does this raise flags of concern?
Outside the dielectric we are in the far field (R>>d) so the
dipole term dominates the expansion and we do not notice
the discrete graininess of the real dipole
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
What about inside the dielectric?
At the microscopic level of atoms/subatomic particles it
would be impossible to integrate
Electric field would vary wildly in space and time at the
But we are interested in the macroscopic field
Electric field would vary wildly in space and time at the microscopic level
But we are interested in the macroscopic field
i.e. the average over regions containing thousands of atoms
But not too large!
So we average (integrate) over regions smaller than object
itself dv' Vol
b Physicists and Mathematicians willDiverge here!
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Assume we want to measure the macroscopic field near a
point r 2a1000R oIntegrate over a sphere of radius R radiusatomic ao
o
RrChargesO t id & I idOutside &
sphereInside
Macroscopic Field at r will be the sum of E-fields due to charges Inside & Outside the sphere
inoutr EEE
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
RCh
Macroscopic Field at r will be the sum of E-fields due to RrCharges
Outside &sphere
Insidecharges Inside & Outside the sphere
inoutr EEE The average E-field over a sphere produced by charges outside, is equal to the field they produce at the center
Eout is the is the field at r due to dipoles outside the sphere
dv'ˆ1
rr'Pr dv
r4 2outsideo
r
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
RCh
Macroscopic Field at r will be the sum of E-fields due to RrCharges
Outside &sphere
Insidecharges Inside & Outside the sphere
inoutr EEE
Ein dipoles are too close…we can’t integrate safely!
But wait…we just need the average fieldFall of Icarus byM ti
The average field within a sphere of radius R due to all the charge contained in the volume is
1 p
Matisse
3Ave R41 pE
o (This is Eqn 3.105 in Griffiths)
(p is the total dipole moment)
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
RCh
Macroscopic Field at r will be the sum of E-fields due to RrCharges
Outside &sphere
Insidecharges Inside & Outside the sphere
EEE inoutr EEE
PppE
R34 where
R41 3
3in
PE
o
31
3R4
in Regardless of the charge distributionInside the sphere
o3 Inside the sphere We implicitly assumed our sphere is small enough that P does not vary much over its volume…Wait what was the field of a uniformly polarized sphere?
1 31 PE
o The term omitted for the outside
integration gets put back in by Ein! See last lecture
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
Thus the macroscopic E-field can be obtained from the potentialWhich is obtained by integrating over the entire dielectric!
dv'r
ˆ4
12
rr'Pr
r4 Volumeo
We were properly calculating the average macroscopic field for insideaverage macroscopic field for inside the dielectric
All due to the fact that…Th fi ld h d h h i idThe average field over any sphere due to the charge inside Is the same asThe field at the center of a uniformly polarized sphere with y p pthe same total dipole moment
*It would hold for other averaging geometries…but harder to prove
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
The Electric Displacement
Gauss’s Law in the Presence of Dielectrics
E-field due to Polarization is the field due to bound charges
Bound Bound Surface Charge
nP ˆb Pb
Volume Charge
Let’s put it all together!
The E-field due to all other charges (e.g. electrons in a
conductor, ions in the material, etc.) is called the free charge
Free Charge f Any charge that is not due to Polarization of the dielectric material
ChpChp--10 Electric fields in Matter10 Electric fields in Matter
In the Dielectric total charge density
Total Charge Densityf b
Gauss’s law for a dielectric material is
ff PE b ff PE bo
E is the total field not just that due to Polarization P
Combine the two divergence terms and re-write defining the Electric Displacement D
PE f
PED
PE
o
o
define D is the Electric Displacement
f D Gauss’s Law for Dielectrics in Differential Form
ChpChp--10 Electric fields in Matter10 Electric fields in MatterGauss’s Law in terms of the electric displacement
f D enclosed fQda ˆ nDSurfaceDifferential Form Integral Form
Griffith’s Eqn 4.22 Griffith’s Eqn 4.23
Note this only refers to the free charge
But we can only control the free chargeTada
“free of charge”But we can only control the free chargeBound charge “comes along for the ride” once we put the free charge in place (creating an E-field…which induces the Polorization P)
free of charge
Polorization P)Initially we start with knowing the free chargeBut we do not initially know the bound charge y gGiven the correct symmetry, we can use Eqn 4.23 to calculate D using the usual Gauss’s Law methods
ChpChp--10 Electric fields in Matter10 Electric fields in MatterEx. 4.4 A long straight wire of charge line density is
surrounded by rubber insulation out to radius a. Find D
aGaussian surface
Flux through the cylinder endss Dn̂
L
Flux through the cylinder endsWill be zero by symmetry
Apply Integral form of Gauss’s Law (Eqn 4.23) Charge enclosed
enclosed fQ daDda ˆ SurfaceSurface
nD
pp y teg a o o Gauss s a ( q 3)
LQ enc f
gBy surface
sL2A
sD ˆ s 2
so LsLD 2 s This equation holds both inside & outside
The insulator…Outside P=0
asfor ˆs2
so sDEPEDoo
o
Outside
(we would need toKnow P to calculateE inside)
ChpChp--10 Electric fields in Matter10 Electric fields in MatterGauss’s Law in terms of the electric displacement
f D enclosed fQda ˆ nDSurfaceDifferential Form Integral Form
Griffith’s Eqn 4.22 Griffith’s Eqn 4.23
Wait a minute…what about the surface bound charge b ?We cannot apply the differential form of Gauss’s law
f f fprecisely at the surface of a dielectric surface as b blows up along with the DivEThe integral form of Gauss’s law is free of this issuegWe can also treat the edge as having a finite thickness so thePolarization decays to zero…in which case there is no b
* The polarization drops precipitously to zero outside the dielectric material…so its derivative is a delta function…The surface bound charge is this term (see Griffith’s Prob. 1.46)
ChpChp--10 Electric fields in Matter10 Electric fields in MatterA Deception Parallel…beware!
f D enclosed fQda ˆ nDSurfaceDifferential Form Integral Form
Griffith’s Eqn 4.22 Griffith’s Eqn 4.23
Eq. 4.22 Looks exactly like Gauss’s law except with the free charge density and DCan we treat D and E the same way? No!Can we treat D and E the same way?There is no Coulomb’s law for D
No!
'dv'ˆ1 f r'rD
2'4
1rfrD
Moreover the curl of D is not always zero PPED o
Research Aside Linear & Nonlinear OpticsResearch Aside Linear & Nonlinear Optics