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Fri. 10/23
(C14) 4.4.1 Linear Dielectrics (read rest at your discretion)
Mon. Wed. Thurs. Fri.
(C 17) 12.1.1-.1.2, 12.3.1 E to B; 5.1.1-.1.2 Lorentz Force Law: fields and forces (C 17) 5.1.3 Lorentz Force Law: currents (C 17) 5.2 Biot-Savart Law
HW6
Polarization & Electric Displacement Example
Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .
DPEo
adDdQ freefree
oE
oooo PED
oE
You cut a small spherical hole out of it. What is the field in its center in terms of and ? oP
oE
For illustrative purposes only, take the polarization to be anti-parallel to the field, and imagine both to be in the z direction.
Quoting Example 4.2 (which in turn builds on 3.9), the field inside a uniformly polarized sphere is
By Superposition Principle, cutting out a sphere is the same as inserting a sphere of opposite polarization.
oP
sphere
o
sphere PE
3
1
So, we ‘add in ‘ a sphere of polarization oP
Adding field o
o
added PE
3
1
031
. PEEoospherein
Polarization & Electric Displacement Example
Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .
DPEo
adDdQ freefree
oE
oooo PED
oE
You cut a small spherical hole out of it. What is the field in its center in terms of and ? oP
oE
oP
sphere
o
sphere PE
3
1
There is no material in the sphere, so
031
. PEEoospherein
What is the electric displacement in its center in terms of and ?
oD
oP
sphereshereosphere PED
031 PED
ooosphere
where ooo PDEo
1
so
0311 PPDD
oo ooosphere
032 PDo
You cut out a wafer-shaped cavity perpendicular to .
Polarization & Electric Displacement
Exercise: Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .
DPEo
adDdQ freefree
oE
oooo PED
oE What is the field in its center in terms of and ?
oP
oE
oP
What is the electric displacement in its center in terms of and ?
oD
oP
PaP bb
andˆ
Hint: Think of inserting the appropriate waver-sized capacitor. oP
d
pdP
Boundary Conditions Electric field, across charged surface
o
enclsidessidesbottombottomtoptop
QadEadEadE
Send side height / area to 0
o
bottomtop EE
topad
Bottomad
bottomE
topE
o
enclQadE
o
surfaceda
o
surface
bottombottomtoptop
daadEadE
o
bottomtop
AAEAE
)1(
o
enclE
Boundary Conditions Electric Displacement, across charged surface
enclfreesidessidesbottombottomtoptop QadDadDadD .
Send side height / area to 0
freebottomtop DD
topad
Bottomad
bottomD
topD
free
enclfreeQadD .
surfacefreeda
surfacefreebottombottomtoptop daadDadD
AADAD freebottomtop )1(
enclfreeD .
Boundary Conditions (static) Electric field, along charged surface
0 sidessidesbottombottomtoptop ldEldEldE
Send side height to 0
0|||| bottomtop EE
topld
bottomE
topE
bottomld
0 ldE
0 bottombottomtoptop ldEldE
0)1(|||| LELE bottomtop
0 E
Boundary Conditions (static) Electric displacement, along charged surface
sidessidesbottombottomtoptopsidessidesbottombottomtoptop ldPldPldPldDldDldD
Send side height to 0
topld
bottomD
topD
bottomldPD
)1()1( |||||||| LPLPLDLD bottomtopbottomtop
ldPldD
bottombottomtoptopbottombottomtoptop ldPldPldDldD
bottomtopbottomtop PPDD ||||||||
Boundary Conditions Electric and Displacement fields
(could have guessed as much from .)
0|||| bottomtop EE
topld
bottomE
topE
bottomld
topad
Bottomad
o
bottomtop EE
aAlong
freebottomtop DD
bottomtopbottomtop PPDD ||||||||
PED o
(could have guessed as much from and .) PED o
aPbfree
ˆ
or o
bottomtop aEaE
ˆˆ
freebottomtop aDaD ˆˆ
bottomD
topD
Across
Boundary Conditions Electric and Displacement fields
0|||| bottomtop EEo
bottomtop EE
Along
freebottomtop DD bottomtopbottomtop PPDD ||||||||
PED o
aPbfreeˆ
Across
Exercise Bar Electret (like an electric bar magnet): uniform P along axis
Sketch 𝑃, 𝐸, and 𝐷 as to obey the boundary conditions.
𝑃
𝐸
Only surface charge is the bound surface charge on two faces
PED o
𝐷
There is no free surface charge, so no discontinuity in D.
Atom on a stick
Recall: Atom’s Response to Electric Field
𝑬𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍
𝑭𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 = 𝑞𝑬𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍
𝑭𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 = −𝑭𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍
...0
intint
ss
FF
ss
For small stretch, first term in Taylor Series (Hook’s Law)
...
0
intint
qsqs
FF
s
...0
intint
pp
FF
s
Electric Dipole moment
...0
intint
pp
FFF
s
ext
...0
intint
pp
FFqE
s
ext
So, for small enough stretch, weak enough field
extqEp or extEp
≡polarizability
Linear Dielectrics Chunk of induced dipoles
𝐸
Ep
Everyone’s field but its own
Point along field Linearly proportional
d
pdP
Polarization = Dipole density
so E
d
d
d
EdP
Define “electric susceptibility” to be the proportionality constant (and provide convenient factor of o.)
P 0eE
d
d
o
e
1
PED o
Always linear dielectric
EE eoo
Eeo
1
ED
0 1 e Permittivity (of not-so-free space)
Dielectric Constant
er o
1
ED ro
Or, in terms of polarization
PDe
11
or PD
e
e
1
for individual induced dipole
For chunk of induced dipoles
Linear Dielectrics Chunk of induced dipoles
𝐸
Example: consider a simplified version of problem 4.18 . Say we have only one dielectric material, of constant r between two capacitor plates distance a apart. a. Electric Displacement, D. enclfreeQadD .
Gaussian box
free
free
free
bottom
enclfree
inside
enclfreebottominside
enclfreebottominsidetopoutside
A
QD
QAD
QADAD
.
.
.
0
Expect only perpendicular to surface and only inside capacitor
outside 0
inside zD
free
b. Electric Field, E.
zor
freeˆ
𝐷
DE
z
freeˆ
c. Polarization, P.
zfreerˆ1 1
EDP o
zz
r
free
freeˆˆ
𝑃
d. Potential Difference across plates, DV.
aor
free
D
top
bottom
ldEV
top
bottom or
freezdz
ˆ
a DV
e. Bound charge, b and b.
bottombottom
toptop
b
nP
nPnP
ˆ
ˆˆ
0 Pb
zntopˆˆ
znbottomˆˆ
zzfreerˆˆ1 1
freer
11
zzfreerˆˆ1 1
freer
11
f. E from charge distribution.
b
or
f
o
enclQadE
o
enclbenclf
bottominside
QQAE
..
o
bf
insideE
o
ff r
11
Linear Dielectrics
𝐸o
Example: Alternate / iterative perspective on field in dielectric. Consider again a simple capacitor with dielectric. We’ll find the electric field in terms of what it would have been without the dielectric. We’ll do this iteratively and build a series solutions.
free
free
𝑃o
b.o
0. Say we start with no dielectric. Initially there’s the field simply due to the free charge; Eo.
oEE
and the associated surface charges contribute a field of their own, where so in the opposite direction.
E1 P0 0 eE0
oeoo EP
zEo
ob ˆ.1
nPobˆ
0.
oe E
𝐸1
1. This field induces a little counter polarization,
P1 0eE1 0e2E0
E2 P1 0 e 2E0
oe E
2
2. See a pattern?
...
𝑃1 𝐸2
o
n
n
e EE
0
As long as e< 1, this converges to
Einside 1
1 e
E0
1
rE0
Same result as we got previously
We insert the dielectric and that field induces a polarization,
Which is means a surface charge and resulting field contribution of its own
Linear Dielectrics
𝐸o
Exercise: Try it for your self. A sphere made of linear dielectric
material is placed in an otherwise uniform electric field 𝐸o. Find the electric field inside the sphere in terms of the material’s dielectric constant, r.
free
free oEE
You can take it as a given that a sphere of uniform polarization contributes field E P 30
Example: A coaxial cable consists of a copper wire of radius a surrounded by a concentric copper tube of inner radius c. The space between is partially filled (from b to c) with material of dielectric constant r as shown below. Find the capacitance per length of the cable.
a b
c
V
QC
D
L
For the sake of reasoning this out, say there’s charge Q uniformly distributed along the surface of the central wire.
r
enclfQadD .
D
c
a
ldEV
QsLD 2
Gaussian cylinder of some radius a<s<c.
sL
QD
2
csbssL
Q
bsassL
Q
E
ro
o
ˆ2
ˆ2
D
c
a
ldEV
D
c
b ro
b
a o
dssL
Qds
sL
QV
22
bc
ab
orL
QV lnln
21
D
csbD
bsaDE
ro
o
1
1
c
b
b
a
ldEldE
bc
ab
o
rL
C
lnln
21
Exercise: There are two metal spherical shells with radii R and 3R. There is material with a dielectric constant r = 3/2 between radii R and 2R. What is the capacitance? R
2R
3R
r
Fri. 10/23
(C14) 4.4.1 Linear Dielectrics (read rest at your discretion)
Mon. Wed. Thurs. Fri.
(C 17) 12.1.1-.1.2, 12.3.1 E to B; 5.1.1-.1.2 Lorentz Force Law: fields and forces (C 17) 5.1.3 Lorentz Force Law: currents (C 17) 5.2 Biot-Savart Law
HW6