Dr. Champak B. Das (BITS, Pilani) Electric Fields in Matter Polarization Electric displacement Field...
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Transcript of Dr. Champak B. Das (BITS, Pilani) Electric Fields in Matter Polarization Electric displacement Field...
Dr. Champak B. Das (BITS, Pilani)
Electric Fields in Matter
Polarization
Electric displacement
Field of a polarized object
Linear dielectrics
Dr. Champak B. Das (BITS, Pilani)
Matter
Insulators/Dielectrics
Conductors
All charges are attached to specific atoms/molecules and can only have a
restricted motion WITHIN the atom/molecule.
Dr. Champak B. Das (BITS, Pilani)
electron cloud
nucleus
• The positively charged nucleus is surrounded by a
spherical electron cloud with equal and opposite
charge.
A simplified model of a neutral atom
Dr. Champak B. Das (BITS, Pilani)
• The electron cloud gets displaced in a direction
(w.r.t. the nucleus) opposite to that of the applied electric field.
When the atom is placed in an external electric field (E)
E
Dr. Champak B. Das (BITS, Pilani)
• For less extreme fields
► an equilibrium is established
=> the atom gets POLARIZED
• If E is large enough
► the atom gets pulled apart completely
=> the atom gets IONIZED
Dr. Champak B. Das (BITS, Pilani)
-e +e
► The net effect is that each atom becomes a small charge dipole which
affects the total electric field both inside and outside the material.
Dr. Champak B. Das (BITS, Pilani)
Induced Dipole Moment:
Ep
α
Atomic Polarizability
(pointing along E)
Ep
Dr. Champak B. Das (BITS, Pilani)
To calculate : (in a simplified model)The model: an atom consists of a point
charge (+q) surrounded by a uniformly charged spherical cloud of charge (-q).
At equilibrium, eEE ( produced by the negative charge cloud)
+qa
-q -q
E
d
+q
Dr. Champak B. Das (BITS, Pilani)
304
1
a
qdEe πε
At distance d from centre,
304
1
a
qdE
πε
Eaqdp 304πε
va 03
0 34 επεα (where v is the volume of the atom)
Dr. Champak B. Das (BITS, Pilani)
Prob. 4.4:
A point charge q is situated a large distance r from a neutral atom of polarizability .
Find the force of attraction between them.
Force on q :
Attractiverr
qF
1
42
2
20
πεα
Dr. Champak B. Das (BITS, Pilani)
Alignment of Polar Molecules:
when put in a uniform external field:
0netF
Ep
τ
Polar molecules: molecules having permanent dipole moment
Dr. Champak B. Das (BITS, Pilani)
Alignment of Polar Molecules: when put in a non-uniform external field:
FFFnet
d
F+
F- -q
+q
Dr. Champak B. Das (BITS, Pilani)
F-
d
F+
-q
+qE+
E-
EEqFnet
EpFnet
Dr. Champak B. Das (BITS, Pilani)
For perfect dipole of infinitesimal length,
Ep
τ
the torque about the centre :
the torque about any other point:
FrEp
τ
Dr. Champak B. Das (BITS, Pilani)
Prob. 4.9:A dipole p is a distance r from a point
charge q, and oriented so that p makes an angle with the vector r from q to p.
(i) What is the force on p?
(ii) What is the force on q?
rrppr
qF pon
ˆˆ.34
13
0
πε
prrpr
qF qon
ˆˆ3
4
13
0πε
Dr. Champak B. Das (BITS, Pilani)
Polarization:When a dielectric material is
put in an external field:
A lot of tiny dipoles pointing along the direction of the field
Induced dipoles (for non-polar constituents)
Aligned dipoles (for polar constituents)
Dr. Champak B. Das (BITS, Pilani)
A measure of this effect is POLARIZATION
defined as:
P dipole moment
per unit volume
Material becomes POLARIZED
Dr. Champak B. Das (BITS, Pilani)
The Field of a Polarized Object= sum of the fields produced by infinitesimal dipoles
2
0
ˆ
4
1
s
s
r
rprV
πε
prs
r r
Dr. Champak B. Das (BITS, Pilani)
τd rs
r r
p
τ dPp
Total potential :
τ
πε τ
dr
rrPrV
s
s2
0
ˆ
4
1
Dr. Champak B. Das (BITS, Pilani)
21 sss rr̂r
Prove it !
τπε τ
drPV s14
1
0
τπε
τπε
τ
τ
dPr
drPV
s
s
14
1
4
1
0
0
Dr. Champak B. Das (BITS, Pilani)
Using Divergence theorem;
τπε
πε
τ
dPr
adPr
V
s
S s
1
4
1
1
4
1
0
0
Dr. Champak B. Das (BITS, Pilani)
Defining:
nPbˆ
σ
Volume Bound Charge
Pb
ρ
Surface Bound Charge
Dr. Champak B. Das (BITS, Pilani)
τρ
πε
σ
πε
τ
dr
adr
V
s
b
S s
b
0
0
4
1
4
1
Potential due to a surface charge density b
& a volume charge density b
Dr. Champak B. Das (BITS, Pilani)
Field/Potential of a polarized object
Field/Potential produced by a
surface bound charge b
Field/Potential produced by a
volume bound charge b
+
=
Dr. Champak B. Das (BITS, Pilani)
Physical Interpretation of Bound Charges
…… are not only mathematical entities devised for calculation;
perfectly genuine accumulations of charge !
but represent
Dr. Champak B. Das (BITS, Pilani)
BOUND (POLARIZATION) CHARGE DENSITIES
►Accumulation of b and b
Consequence of an external applied field
τ
τρσ 0dda bS b
Dr. Champak B. Das (BITS, Pilani)
P
E
nqdnpP ( n : number of atoms per unit volume )
Dr. Champak B. Das (BITS, Pilani)
P
E
A A A
2
d2
d
nAqdQ Net transfer of charge across A :
Dr. Champak B. Das (BITS, Pilani)
PAQ Net charge transfer per unit area :
P is measure of the charge crossing unit area held normal to P when the dielectric gets
polarized.
Dr. Champak B. Das (BITS, Pilani)
P
E
N M
Q Q
When P is uniform :
… net charge entering the volume is ZERO
Dr. Champak B. Das (BITS, Pilani)
PA
Volume bound charge
Net transfer of charge across A :APPA
θcos
Pb
ρ
Dr. Champak B. Das (BITS, Pilani)
P
E
N
M G
n̂ n̂
n̂
n̂2
d2
d
Net accumulated charge between M & N :
APQ nPbˆ
σ
Surface bound charge
Dr. Champak B. Das (BITS, Pilani)
Field of a uniformly polarized sphere
Choose: z-axis || P
P is uniform
0 Pb
ρ
θσ cosˆ PnPb
z
P R
n̂
Dr. Champak B. Das (BITS, Pilani)
Potential of a uniformly polarized sphere: (Prob. 4.12)
Potential of a polarized sphere at a field point ( r ):
τ
πε τ
dr
rrPV
s
s2
0
ˆ
4
1
P is uniform
P is constant in each volume element
Dr. Champak B. Das (BITS, Pilani)
τ
τρ
περ ss
rr
dPV ˆ
4
112
0
Electric field of a uniformly charged
sphere Esphere
rEP,rV sphere
ρθ
1
Dr. Champak B. Das (BITS, Pilani)
rrEsphere
03ε
ρ
At a point inside the sphere ( r < R )
rPrV
03
1,
εθ
z
PE
03ε
PE
03
1
ε
Dr. Champak B. Das (BITS, Pilani)
Field lines inside the sphere :
►► ► ► ►
P
PE
03
1
ε
( Inside the sphere the field is uniform )
Dr. Champak B. Das (BITS, Pilani)
r̂r
RrEsphere 2
3
03ε
ρ
rPr
RrV ˆ
3
1, 2
3
0
εθ
At a point outside the sphere ( r > R )
Dr. Champak B. Das (BITS, Pilani)
20
ˆ
4
1
r
rpV
πε
(potential due to a dipole at the origin)
prrpr
rE
ˆˆ31
4
13
0πε
Total dipole moment of the sphere: PRp 3
3
4π
Dr. Champak B. Das (BITS, Pilani)
► ►
Field lines outside the sphere :
P
Dr. Champak B. Das (BITS, Pilani)
►► ► ► ►► ►
Field lines of a uniformly polarized sphere :
Dr. Champak B. Das (BITS, Pilani)
Uniformly polarized Uniformly polarized sphere – A physical sphere – A physical
analysisanalysis Without polarization:
Two spheres of opposite charge, superimposed and canceling each other
With polarization:The centers get separated, with the positive
sphere moving slightly upward and the negative sphere slightly downward
Dr. Champak B. Das (BITS, Pilani)
At the top a cap of LEFTOVER positive charge and at the bottom a cap of negative charge
Bound Surface
Charge b
+ ++ + + + + +
+ +
+
-d
+ +
- - - - - - - -
Dr. Champak B. Das (BITS, Pilani)
Recall: Pr. 2.18
Two spheres , each of radius R, overlap partially.
dE
03ε
ρ+
-
_
+d
_
+
r r
d
Dr. Champak B. Das (BITS, Pilani)
dE
03ε
ρ
Electric field in the region of overlap between the two spheres+ +
+ + + + + + + +
+
-d
+ +
- - - - - - - - PE
03
1
ε
For an outside point:
20
ˆ
4
1
r
rpV
πε
Dr. Champak B. Das (BITS, Pilani)
Prob. 4.10:A sphere of radius R carries a polarization
rkrP
where k is a constant and r is the vector from the center.
(i) Calculate the bound charges b and b.
(ii) Find the field inside and outside the sphere.
kRb σ kb 3ρ
rkE inside
0ε 0outsideE
Dr. Champak B. Das (BITS, Pilani)
The Electric Displacement
Polarization
Accumulation of Bound charges
Total field = Field due to bound charges + field due to free charges
Dr. Champak B. Das (BITS, Pilani)
Gauss’ Law in the presence of dielectricsWithin the dielectric the total charge density:
fb ρρρ
bound charge free charge
caused by polarization
NOT a result of polarization
Dr. Champak B. Das (BITS, Pilani)
Gauss’ Law
fD ρ
enclfQadD
Defining Electric Displacement ( D ) :
PED
0ε
( Differential form )
( Integral form )
Dr. Champak B. Das (BITS, Pilani)
D & E :
τρ drr
rKrE
s
s
2
ˆ
τρ drr
rKrD f
s
s
2
ˆ
… “looks similar” apart from the factor of 0 ( ! )
…….but :
Dr. Champak B. Das (BITS, Pilani)
D & E :
0 PD
0 E
Field = - Gradient of a Scalar Potential
No Potential for Displacement
Dr. Champak B. Das (BITS, Pilani)
Boundary Conditions:
fbelowabove DD σ
||||||||belowabovebelowabove PPDD
On normal components:
On tangential components:
Dr. Champak B. Das (BITS, Pilani)
Prob. 4.15:
A thick spherical shell is made of dielectric material with a “frozen-in” polarization
a
b
rr
krP ˆ
where k is a constant and r is the distance from the center. There is no free charge.
Find E in three regions by two methods:
Dr. Champak B. Das (BITS, Pilani)
(a) Locate all the bound charges and use Gauss’ law.
a
b
Prob. 4.15: (contd.)
For r < a : 0E
For r > b:
For a < r < b: rr
kE ˆ
0
ε
0E
Answer:
Dr. Champak B. Das (BITS, Pilani)
(b) Find D and then get E from it.
a
b
Prob. 4.15: (contd.)
0encfreeQ 0 D
)&(0 brarforE
)(ˆ0
braforrr
kE
ε
Answer:
Dr. Champak B. Das (BITS, Pilani)
The Equations of Electrostatics Inside Dielectrics
00
EandE
ε
ρ
0 EandD f
ρ
or
with
PEDandVE
0ε
Dr. Champak B. Das (BITS, Pilani)
For some material (if E is not TOO strong)
EP e
χε0
Electric susceptibility of the medium
Linear DielectricsRecall: Cause of polarization is an Electric field
Total field due to (bound + free) charges
Dr. Champak B. Das (BITS, Pilani)
In such dielectrics;
EED e
χεε 00
)1(0 ewithED χεεε
Permittivity of the material
The dimensionless quantity:
0
1ε
εχε er
Relative permittivity or Dielectric constant of the material
Dr. Champak B. Das (BITS, Pilani)
EP e
χε0 ED
εand / or
Electric Constitutive Relations
Represent the behavior of materials
Dr. Champak B. Das (BITS, Pilani)
Location ► Homogeneous
Magnitude of E
► Linear
Direction of E ► Isotropic
In a dielectric material, if e is independent of :
Most liquids and gases are homogeneous, isotropic and linear dielectrics at least at low electric fields.
Dr. Champak B. Das (BITS, Pilani)
But in a homogeneous linear dielectric :
00 DP
00 DP
Generally, even in linear(& isotropic) dielectrics :
Dr. Champak B. Das (BITS, Pilani)
DE
ε
1 vacEE
ε
ε0
vacr
EE
ε
1
When the medium is filled with a homogeneous linear dielectric, the field is reduced by a factor of 1/r .
vacED 0ε
0 DandD f
ρ
Free charges D , as:
In LD :
Dr. Champak B. Das (BITS, Pilani)
Capacitor filled with insulating material of dielectric constant r :
vacr
EE
ε
1
vacr
VVε
1
vacrCC ε
Dr. Champak B. Das (BITS, Pilani)
So far…………source charge distribution at
restELECTROSTATICS
ρε0
1 E
0 E
1st/4 Maxwell’s Equations
Dr. Champak B. Das (BITS, Pilani)
Coming Up…..
MAGNETOSTATICS
ELECTROMAGNETISM
…source charge distribution at motion
A New Instructor