Chapter 4 Electric Fields in Matter
Dielectrics
4.1 Polarization: 4.1.1 Dielectrics
Dielectrics : Microscopic displacements are not as
dramatics as the wholesale rearrangement of charge in
conductor, but their cumulative effects account for the
characteristic behavior of dielectric materials.
Most everyday objects belong to one of two large classes:
conductors and insulators (or dielectrics)
Conductors : Substances contains an “unlimited” supply
of charges that are free to move about through the
material.
Dielectrics : all charges are attached to specific atoms or
molecules. All they can do is move a bit within the atom or
molecule.
There are actually two principal mechanisms by which
electric fields can distort the charge distribution of a
dielectric atom or molecule: stretching and rotating.
1
4.1.2 Induced Dipoles
2
4.1.2 Induced Dipoles
What happens to a neutral atom when it is placed in an
electric field E ?
Although the atom as a while is electrically neutral, there is
a positively charged core (the nucleus) and a negatively
charged electron cloud surrounding it.
The atom or molecule now has a tiny dipole moment p,
which points in the same direction as E and is proportional
to the field.
p = αE,
α= atomic polarizability
Thus, the nucleus is pushed in the direction of the field, and
the electron the opposite way.
The electric fields pull the electron cloud and the nuclear
apart, their mutual attraction drawing them together - reach
balance, leaving the atom polarized.
3
4
Example 4.1 A primitive model for an atom consists of a point
nuclear (+q) surrounded by a uniformly charged spherical
cloud (-q) of radius a. Calculate the atomic polarizability of
such an atom.
Sol. The actual displacements involved are extremely small.
It is reason to assume that the electron cloud retains
its spherical shape.
The equilibrium occurs when the nucleus is displaced a
distance d from the center of the sphere.
The external field pushing the nucleus to the right exactly
balances the internal field pulling it to the left. How?
qd
4πε 0 a 3
α = 4πε 0 a 3 = 3ε 0 v
Ee =
1
p = qd = (4πε 0 a 3 ) E = α E
the atomic polarizability5
Sol.
The electric field inside a uniform charged sphere of radius a
1 43 πρ r 3
1
Ee (r ) =
rˆ =
ρ rrˆ
2
4π r
3ε 0
ε0
1 q
∴ Ee (a ) =
rrˆ , where q = 43 πρ a 3
3
4πε 0 a
The electric field produces by two uniform charged spheres
separated by d
1 q
E(r )=Eq + (r+ ) + Eq − (r− ) =
(r+ − r− )
4πε 0 a 3
1 q
1
1
1 qd
=
((r − d) − (r + d))= −
3
4πε 0 a
2
2
4πε 0 a 3
1
=−
p ∴ α = 4πε 0 a 3
3
6
4πε 0 a
See Problem 2.18
Polarizability of Molecules
Polarizability Tensor
CO2 is relatively simple, as molecules go, since the atoms
at least arrange themselves in a straight line.
For molecules the situation is not quite so simple, because
frequently they polarize more readily in some directions
than others.
For instance, carbon dioxide CO2
For a complete asymmetrical molecule, a more general
linear relation between E and p.
When the field is at some angle to the axis, you must resolve
it into parallel and perpendicular components, and multiply
each by the pertinent polarizability:
p = α⊥E ⊥ + α∥E∥
The set of nine constants αij constitute the polarizability
tensor for the molecular.
It is always possible to choose “principal” axies such that
the off-diagonal terms vanish, learning just three nonzero
polarizabilities.
In this case the induced dipple moment may not even be in
the same direction as E.
7
8
4.1.3. Alignment of Polar Molecules
Torque for a Permanent Dipole in Uniform Field
In a uniform field, the force on the positive end, F =qE,
exactly cancels the force on the negative end. However,
there will be a torque:
The neutral atom has no dipole moment to start with--p was induced by the applied field E. However, some
molecules have built-in, permanent dipole moment.
N = (r+ x F+) + (r- x F-)
= [(d/2) x (qE) + (-d/2) x (-qE)]
= qd x E
This dipole p = qd in a uniform field experiences a torque
N=pxE
The dipole moment of water is usually large : 6.1*10-30
C*m, which accounts for its effectiveness as solvent.
N is in such a direction as to line p up parallel to E, a polar
molecule that is free to rotate will swing around until it
points in the direction of the applied field.
What happens when polar molecules are placed in an
electric field ? Rotating
9
10
4.1.4. Polarization
Net Force due to Field Nonuniformity
What happens to a piece of dielectric material when it is
placed in an electric field?
If the field is nonuniform, so that F+ does not exactly
balance F-; There will be a net force on the dipole.
•Neutral atoms : Inducing tiny dipole moment, pointing in
the same direction as the field (stretching)
Of course, E must change rather abruptly for there to be
significant in the space of one molecule, so this is not
ordinarily a major consideration in discussing the behavior
of dielectrics.
•Polar molecules : experiencing a torque, tending to line it
up along the field direction (rotating).
The formula for the force on a dipole in a uniform field is of
some interest
@ different position
Results : A lot of little dipoles points along the direction of
the field---the material becomes polarized.
A convenient measure of this of this effect is
P ≡ dipole moment per unit volume, which is called the
polarization.
F = F+ + F- = q(E+ -E-) = q(∆E) = q((d．∇)E)
F = (p．∇)E
11
12
Prob.4.2 According to quantum mechanics, the electron
cloud for a hydrogen atom in ground state has a charge
density
ρ (r ) =
4.2 The Field of a Polarized Object
4.2.1 Bound Charges
q −2r / a
e
πa 3
Where q is the charge of the electron and a is the Bohr
radius. Find the atomic polarizability of such an atom. [Hint:
First calculate the electric field of the electron cloud, Ee(r);
then expand the exponential, assume r<<a.
Suppose we have a piece of polarized material with
polarization P. What is the field produced by this object?
(It is easier to work with potential)
1 rˆ ⋅ p
For a single dipole p, the potential is V (r ) = 4πε r 2
0
where r is the vector form the dipole to the point at which
we are evaluating the potential.
Sol. For a more sophisticated approach, see W. A. Bowers,
Am. J. Phys. 54, 347 (1986).
13
Example 3.10 A electric dipole consists of
two equal and opposite charges separated
by a distance d. Find the approximate
potential at points far from the dipole.
4.2.1 Bound Charges
For an infinitesimal dipole moment dp = Pdτ, the total
potential is
Sol:
1
1
q
−
)=
((1 − ε ) −1/ 2 − (1 + ε ) −1/ 2 )
d ˆ
d ˆ
4πε 0 r − 2 z r + 2 z
4πε 0 r
r′ r′
d
r′
where ε = ( − 2 cos θ ′) ≅ cos θ (if << 1, so θ ′ ≅ θ )
r r
r
r
q
q
d
1 qd cos θ
V (r ) =
((1 − ε ) −1/ 2 − (1 + ε ) −1/ 2 ) =
( cos θ ) =
4πε 0 r
4πε 0 r r
4πε 0
r2
V (r ) =
V (r ) =
q
14
(
V (r ) =
Note that
rˆ ⋅ dp
1
rˆ ⋅ P(r′)
=
dτ ′
4πε 0 υ∫ r 2
4πε 0 υ∫ r 2
1
rˆ
1
∇′( ) = 2
r
r
V =
qd cos θ
1 rˆ ⋅ p
p cos θ
=
=
2
2
4πε 0
r
4πε 0 r
4πε 0 r 2
1
where P = qd ↑ pointing form negative charge to the positive charge.
15
with respect to
the source coordinate.
1
1
′
⋅∇
(
) dτ ′
P
4πε 0 ∫v
r
commutative
16
Bound Charges
Bound Surface and Volume Charges
Integrating by parts, using product rule, gives
σ b = P ⋅ n̂
{ρ
A
g (∇ ⋅ A) − A ⋅ ∇g
∇⋅( ) =
g
g2
1
[ ∫ ∇′ ⋅ (
σ b = P ⋅ n̂
ρ b = - ∇′ ⋅ P
surface charge
volume charge
= - ∇′ ⋅ P
1
4πε 0
v∫
S
σb
r
da′ +
1
4πε 0
∫
ρb
v
r
dτ ′
This means that the potential of a polarized object is the
same as that produced by a surface charge plus a volume
charge density.
P
1
) dτ ′ - ∫ (∇′ ⋅ P ) dτ ′ ]
vr
r
4πε 0 v
1
P
1
1
=
⋅ da′ +
(-∇′ ⋅ P ) dτ ′
v
∫
∫
v
4πε 0 S r
4πε 0 r
V =
b
V =
Ex. 4.2 Find the electric field produced
by a uniformly polarized sphere of
radius R.
17
σ b = P ⋅ nˆ = Pcosθ ′
 ρ = -∇′ ⋅ P = 0
 b
V (r ) =
1
4πε 0
∫
π
0
da′ = (Rdθ ′)( Rsinθ ′dφ ′)
= R 2sinθ ′dθ ′dφ ′
1 ∞ R n
∑ ( ) Pn ( cos θ ′)Pcosθ ′2π R 2sinθ ′dθ ′ r ≥ R
4πε 0 ∫0 r n =0 r
11 ∞
1
R
=
( ) n Pn ( cos θ ′)Pcosθ ′2π R 2 d cos θ ′
∑
∫
-1
4πε 0
r n =0 r
1 1 PR 3
1 PR 3
 orthogonality 
2
′
′
=
=
cos
d
cos
θ
θ
 only n =1 survive 
2
2
∫
-1
2ε 0
3ε 0 r
r


V (r , 0, 0) =
Pcosθ ′
2π R 2sinθ ′ ⋅ dθ ′
r
1 1 ∞ R n
= ∑ ( ) Pn ( cos ϑ )
r r n =0 r
1 ∞ r
= ∑ ( ) n Pn (cos ϑ )
R n =0 R
r≥R
r≤R
ϑ : angle between
r and r′ ⇒ cos ϑ = rˆ ⋅ rˆ ′
r≥R
r≤R
19
1
π
1 ∞ r n
( ) Pn (cos θ ′)Pcosθ ′2π R 2sinθ ′dθ ′
∑
∫
0
4πε 0
R n =0 R
1 1 r
1
( ) P1 (cos θ ′) Pcosθ ′2π R 2 d cos θ ′
=
4πε 0 ∫-1 R R
P
 orthogonality 
=
r
 only n =1 survive 
3ε 0


V (r , 0, 0) =
It will be easier if we let r lie on the z axis, so that the angle
between them changes from ϑ to θ ′
1 1 ∞ R n
= ∑ ( ) Pn ( cos θ ′)
r r n =0 r
1 ∞ r
= ∑ ( ) n Pn (cos θ ′)
R n =0 R
18
1
π
 1 PR 3
cosθ

 3ε 0 r 2
V
(
r
,
0,
0)
=

Allow r a θ-dependence.
 P rcosθ
 3ε 0
r≤R
(r > R )
(r < R)
20
Electric field of a Uniformly Polarized Sphere
4.2.2 Physical Interpretation of Bound Charges
3
V(r) =
P R
cos θ
3ε0 r 2
What is the physical meaning of the bound charge?
(outside)
Consider a long string of dipoles.
4
( πP ⋅ R 3 )
1 3
cosθ
V (r ) =
4πε0
r2
1 p ⋅ rˆ
4
=
where p = πR 3 P
2
4πε0 r
3
p
p
r cos θ =
z
V(r) =
3ε0
3ε0
E = -∇V = -
(inside)
The net charge at the ends is called the bound charge. The
bound charge is no different from any other kind.
Consider a “tube” of dielectric with a given polarization P.
Why are the field
lines not continuous?
P
zˆ uniformly
3ε0
The net bound charge in a given volume is equal and
opposite to the amount that has been pushed out through
the surface.
q
= P cos θ = P ⋅ nˆ
Aend
22
Example 4.3 Two spheres of charge: a positive
sphere and a negative sphere. Without
polarization the two are superimposed and
cancel completely.
But when the material is uniformly polarized, all the plus
charges move slightly upward and all the minus charges
move slightly downward.
∫ ρ dτ = - ∫ P ⋅ da = ∫ (-∇ ⋅ P) dτ
S
σb =
Another Way of Analyzing
the Uniformly Polarized Sphere
If the polarization is nonuniform, we get accumulations of
bound charge within the material as well as on the surface.
b
q
=P
A
21
Nonuniform Polarization
Æ The Bound Volume Charge
v
σb =
v
gauss’s laws
The two spheres no longer overlap perfectly: at the top
there’s a “cap” of leftover positive charge and at the
bottom a cap of negative charge.
This is true for any volume
bound charge.
ρb = −∇ ⋅ P
23
This “leftover” charge is precisely the bound surface charge
σb.
24
4.2.3 The Field Inside a Dielectric
4.2.3 The Field Inside a dielectric
What kind of dipole are we actually dealing with, “pure”
dipole or “physical” dipole?
The electric field inside matter must be very complicated,
on the microscopic level, which would be utterly
impossible to calculate, nor would it be of much interest.
Outside the dielectric there is no real problem, since we
are far away from the molecules.
Inside the dielectric, however, we can hardly pretend to be
far from all the dipoles.
The macroscopic field is defined as the average field over
regions large enough to contain many thousand of atoms.
The macroscopic field smoothes over the uninteresting
microscopic fluctuation and is what people mean when
they speak of “the” field inside matter.
25
26
The Macroscopic Field
The macroscopic field at r, consists the average field over
the sphere due to all charge outside, plus the average due
to all charge inside.
E = E out + Ein
Problems: 2, 9, 10, 16, 31
rˆ ⋅ P(r′)
dτ ′
outside
r2
Vout =
1
4πε0
Ein = −
1 p
1 43 πR 3 P
1
P
=
−
=−
4πε0 R 3
4πε0 R 3
3ε0
V (r ) =
Homework #7
∫
1
4πε0
rˆ ⋅ P(r′)
dτ ′
R →0
r2
∫
where the integral runs over the entire volume of the
dielectric.
27
28
4.3 The Electric Displacement
4.3.1 Gauss’s Law in the Presence of Dielectric
Gauss’s Law in the Presence of Dielectric
The effect of polarization is to produce accumulation of bound
charge, ρb = −∇ ⋅ P within the dielectric and σ b = P ⋅ n̂ on the
surface.
Now we are going to treat the field caused by both bound
charge and free charge. ρ = ρ + ρ
f
In a typical problem , we know ρf , but not ρb. So this
equation allows us to deal with the information at hand.
What is the contribution of the bound surface charge?
where E is now the total field, not just that portion generated
by polarization .
ε 0∇ ⋅ E + ∇ ⋅ P = ρ f
∇ ⋅ (ε 0 E + P) = ρ f
Let D = ε 0 E + P the electric displacement
Gauss’s law reads
∇⋅D = ρ
f
29
30
4.3.2 A Deceptive Parallel
“To solve problems involving dielectrics, you just forget
all about the bound charge ─ calculate the field as you
ordinarily would, only call the answer D instead of E”
Sol : Drawing a cylindrical Gaussian surface, of radius s and
length L, and applying the new Gauss’s law , we find
⇑ This conclusion is false.
For the divergence along is insufficient to determine a
vector field; you need to know the curl as well.
∇×D=ε0(∇×E) +∇×P = ∇×P ←not always zero
λ
λ
sˆ ∴ E =
sˆ
2π s
2π sε r ε 0
Outside D(2πsL ) = λL ⇒ D =
The bound surface charge σb can be considered as ρb
varies rapidly but smoothly within the “skin”.
So Gauss’s law can be applied elsewhere .
Example 4.4 A long straight wire, carrying
uniform line charge λ, is surrounded by
rubber insulation out to a radius a. Find
the electric displacement .
⇒ D=
The total free charge
enclosed in the volume
b
= ρ f − ∇ ⋅ P = ε 0∇ ⋅ E
Inside D ( 2π sL ) = λ L
∇ ⋅ D = ρ f ⇒ ∫ D ⋅ da = Q f enc
Since the curl of D is not always zero, D cannot be
expressed as the gradient of a scalar.
λ
λ
sˆ ∴ E =
sˆ
2πs
2πsε 0
31
Advice : If the problem exhibits spherical, cylindrical, or
plane symmetry, then you can get D directly from the
generalized Gauss’s law.
32
4.4 Linear Dielectric
4.4.1 Susceptibility and Permittivity
4.3.3 Boundary Conditions
The electrostatic boundary condition in terms of E
σ
ε0
⊥
⊥
E above
− E below
=
∇⋅E =
//
//
Eabove
− Ebelow
=0
For many substances, the polarization is proportional to
the field, provided E is not too strong.
ρ
ε0
∇×E = 0
P = ε0χeE
The electrostatic boundary condition in terms of D
dimensionless
∇⋅D = ρf
⊥
⊥
Dabovt
− Dbelow
=σ f
χe :the electric susceptibility of the medium
materials that obey above equation is called linear dielectrics .
//
above
D
−D
//
below
=P
//
above
−P
//
below
∇×D = ∇×P
The total field E may be due in part to free charges and in
part to the polarization itself .
33
Permittivity and Dielectric Constant
Linear Media & Dielectric Constant
In linear media ,
D = ε 0 E + P = ε 0 E + ε 0 χ e E = ε 0 (1 + χ e )E = εE
We cannot compute P directly from this equation:
the external field
will polarize the
this polarization will
material
produce its own field and
contributes to the total field.
where, ε = ε 0 (1 + χ )
↑
Permittivity of the material
E 0 → P0
P0 → E 0 + ∆ E ′P
34
εr =
ε
= 1 + χe
ε0
Relative permittivity
or dielectric constant
The new total
field will polarize
the material.
E 0 + ∆ E ′P → P0 + ∆ P0′
Will this series converge?
Depends.
35
36
Example 4.5 A metal shpere of radius a carries a charge Q.
It is surrounded, out to radius b , by linear dielectric material
of permittivity ε. Find the potential at the center (relative to
infinity).
Sol: Use the generalized Gauss’s law
Q
D=
rˆ
for all points r > a
4πr 2
 Q ˆ
 4πε r 2 r
E=
 Q 2 rˆ
 4πε 0 r
for a < r < b
a
b
∞
∞
V = − ∫ E ⋅ dl = − ∫
P = ε 0 χ eE =
Q
4πε 0 r
2
dr − ∫
a
b
Q
4πε 0ε r r
2
dr =
1 
Q 1 1
 +

−
4πε 0  b ε r a ε r b 
ε 0 χ eQ
Q  χe 
rˆ

rˆ =
2
4πεr
4πr 2  1 + χ e 
∇×D = 0
ε
D=
at the outer surface
at the inner surface
The surface bound charge at inner surface is negative. It is
this layer of negative charge that reduces the field, within
the dielectric by a factor of ε r .
38
Shielding Effect & Susceptibility Tensor
The polarization of the medium partially
“shields” the charge, by surrounding it
with bound charge of the opposite sign.
However, if the space is entirely filled with a homogenous
linear dielectric, then this objection is void.
1
 ε 0 χ eQ
2

σ b = P ⋅ nˆ =  4πεb
−ε χ Q
 0 e2
 4πεa
surface bound charge
37
In general, linear dielectrics
∇×P ≠ 0
cannot escape the defect that
E=
ρ b = −∇ ⋅ P = 0
In this respect a dielectric is rather like an imperfect conductor.
Stokes’ Theorem for the Polarization
∇⋅D = ρf
volume bound charge
Note that n̂ always points outward with respect to the dielectric ,
which is + r̂ at b but − r̂ at a .
for r > b
The metal sphere is equalpotential
Cont’: Bound Charges in the Dielectric
1
εr
For some material, it is generally easier to polarize in some
directions than in others .
E vac
P = ε 0 χ eE
Remark : When all the space is filled with a homogenous
linear dielectric, the field everywhere is simply reduced by
a factor of one over the dielectric constant .
39
linear dielectric
general case
the susceptibility tensor
40
Prob. 4.18 The space between the planes of a parallel-plate
capacitor is filled with two slabs of linear dielectric material. Each
slab has thickness a, so the total distance between the plates is 2a.
Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric
constant of 1.5 the free charge density on the top plate is σ and on
the bottom plate -σ.
(a) Find the electric displacement D in each slab.
(b) Find the electric field E in each slab.
(c) Find the polarization P in each slab.
(d) Find the potential difference between the plates.
(e) Find the location and amount of all bound charge.
(f) Now that you know all the charge (free and bound), recalculate
the field in each slab, and confirm your answer to (d).
4.4.2 Boundary Value Problems with Linear Dielectrics
Relation between bound charge and free charge
χ
D

ρb = −∇ ⋅ P = −∇ ⋅  ε 0 χ e  = − e ρ f ← in a homogenous
ε
1+ χe

linear dielectric
shielding effect
The boundary conditions that makes reference only to the
free charge .
⊥
⊥
⊥
⊥
Dabovt
− Dbelow
= σ f ⇒ ε above Eabove
− ε below Ebelow
=σ f
(ε above∇Vabove − ε below ∇Vbelow ) = −σ f nˆ
or (ε above
∂Vabove
∂V
∂Vabove
− ε below below ) = −σ f where
= ∇V ⋅ nˆ .
∂n
∂n
∂n
41
42
∞
V (r , θ ) = ∑ ( AA r A + BA r −( A +1) ) PA (cosθ )
Homogeneous Linear Dielectric Sphere
A =0
Example 4.7 A sphere of homogeneous linear dielectric
material is placed in a uniform electric field E. Find the
resulant electric field.
∞

(
,
)
V
r
θ
AA r A PA (cosθ )
r≤R
=
∑
in

A =0

∞
Vout (r ,θ ) = − E0 r cos θ + ∑ BA r −( A +1) PA (cosθ ) r ≥ R

A =0
B.C. (iii)
Sol: Look at Ex. 3.8 an uncharged conducting sphere. In
that case the field of the induced charge completely
canceled E within the sphere; However, in a dielectric the
cancellation is only partial.
B.C. (i) : AA R A PA = − E0 R cos θ + BA R − ( A +1) PA
 A R = − E0 R + B1 R − 2 A = 1
⇒ 1
A
− ( A +1)
A ≠1
 AA R = BA R
A −1
B.C. (ii) : ε r AAA R PA = − E0 cos θ − (A + 1) BA R −( A + 2) PA
The boundary conditions
no free charge
at the surface
43

A =1
ε r A1 = − E0 − 2 B1 R −3
⇒
A −1
−( A + 2)
A ≠1
ε r AAA R = −(A + 1) BA R
44

A =1
ε r A1 = − E0 − 2 B1 R −3

A −1
−( A + 2 )
A ≠1
ε r AAA R = −(A + 1) BA R
 A1 R = − E0 R + B1 R −2 A = 1

A
− ( A +1)
A ≠1
 AA R = BA R
3E0
ε −1 3

; B1 = r
R
 A1 = −
⇒
εr + 2
εr + 2
 AA = BA = 0
A =1
A ≠1
3E0

Vin (r , θ ) = −
r cos θ

εr + 2

ε − 1 3 −2
Vout (r , θ ) = − E0 r cos θ + ( r
) R E0 r cos θ

εr + 2
3E0
Ein = −∇Vin = −
zˆ ← uniform
εr + 2
z>0
V=
z<o
V=
∞
qb = 2π ∫ −
r =0
45
How to express the energy for a dielectric filled capacitor?
Suppose we bring in the free charge, a bit at a time. As ρ f is
increased by an amount ∆ρ f , the polarization will charge and
with it the bound charge distribution.
The work done on the incremental free charge is :
∆W = ∫ (∆ρ f )Vdτ
qb
q
1 

+
2
2 
2
2
4πε 0  x2 + y2 + ( z − d )
x + y +(z +d) 


( q + qb )
1 
 2
 ← why ?
2
4πε 0  x + y 2 + ( z − d ) 
 χ 
1  χe 
qd
1 ∞ χe
qd
rdr = − ∫
dr2 = − e  q


3
3
0
2π  χe + 2  r2 + d 2 2
2 χe + 2 r2 + d 2 2
 χe + 246
(
)
(
)
Which Formula is Correct?
1
(ε 0 E ⋅ E)dτ derived in Chap. 2
speak to somewhat
2∫
different question
1
derived in Chap. 4
W = ∫ (E ⋅ D)dτ
2
W=
What do we mean by “the energy of a system“?
It is the work required to assemble the system.
the resulting change in D
0
∆W = ∫ (∇ ⋅ ∆D)Vdτ = ∫ (∇ ⋅ ∆DV − ∇V ⋅ ∆D)dτ
surface integral vanish if we
integral will over all of pace.
∆W = ∫ E ⋅ ∆Ddτ =
Example 4.8 Suppose the entire region below the plane z=0 is
filled with uniform linear dielectric material of susceptibility χ e .
Calculate the force on a point charge q situated at distant d
above the origin.
Sol: The surface bound charge on the xy plane is of opposite
sign to q , so the force will be attractive.

 ∂V
∂V 
1  χe 
qd
−
= σb
B.C. −ε0 

 =− 
3
 ∂z z=0+ ∂z z=0−  2π  χe + 2  ( x2 + y2 + d2 ) 2
4.4.3 Energy in Dielectric systems
∇ ⋅ D = ρ f ⇒ ∆ρ f = ∇ ⋅ (∆D)
Partial Image Charge
1
∆(εE 2 )dτ
∫
2
∴W =
1
(E ⋅ D)dτ
2∫
47
(1) Bring in all the charges (free and bound ), one by one,
with tweezers, and glue each one down in its proper final
position (Chap. 2).
(2) Bring in the free charges, with the unpolarized dielectric in
place, one by one, allowing the dielectric to respond as it see
fit (Chap. 4).
48
The Fringing Field Effect
4.4.4 Forces on Dielectric
The dielectric is attracted into an electric field, just like
conductor: the bound charge tends to accumulate near the
free charge of the opposite sign.
How to calculate the forces on dielectrics?
Fringing field are difficult to calculate, so we
adapt the following ingenious method.
Consider the case of a slab of linear
dielectric material, partially inserted
between the plates of a parallel-plate
capacitor.
The electric force on the slab is: F = −
C = C1 + C 2 =
2d
F =−
V 2 < 0 indicates that the force is in the negative
x direction; the dielectric is pulled into the capacitor.
1
2
Fixed voltage F = + V 2
dC
dx
ε 0ω x
d
+
ε 0ε r ω (A − x )
d
=
dW
dx
ε 0ω
d
Q2
2C
(ε r A − χ e x )
Fixed charge
49
ε 0 χ eω
1
2
2
The energy stored in the capacitor is: W = CV =
If the field is perpendicular to the
plates, no force would exert on the
dielectric. Is that true?
F =−
In reality a fringing field around the edges
is responsible for the whole effect.
It is this nonuniform fringing field that pulls
the dielectric into the capacitor.
1 Q 2 dC
1
dW
=
= V
dx
2 C 2 dx
2
2
ε χ ω
dC
=− 0 e V
dx
2d
2
50
Homework #8
Problems: 21, 27, 28, 33, 36.
pushed out. why?
To maintain a constant voltage, the battery must do work.
why?
work done by the battery
dW = Fme dx + VdQ
Fme : the force I must exert. (Fme = − F )
F =−
dW
dQ
1
dC
dC 1 2 dC
+V
=− V2
+V 2
= V
2
2
dx
dx
dx
dx
dx
51
52