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1
Electrostatics in material
z ˆ
ηˆ 0
P = P z ˆ
G
R
Q
ε
r
ε
ε
above E
G
below E
G
A
Z
s
0
ε
q
b
σ
z ˆ^ ηˆ
FH
G
F
d G
x A^ − x x ˆ
R
z
ε E^0^^ z ˆ
σ (^) b > 0
σ f
− σ (^) f
d
V
D top
G
D bottom
G
E^ ε
G
σ (^) b < 0
I
Bound charges
Displacement
field
Dielectric boundary
conditions
Dielectric
Forces
Stored
energy in a
dielectric
Laplacean methods and dielectrics
Image methods Separation of variables
From Physics 212, one might get the impression that going from electrostatics in
vacuum to electrostatics in a material is equivalent to replacing epsilon_0 to epsilon
> epsilon_0. This is more-or-less true for some dielectric materials such as Class A
dielectrics but other types of materials exist. For example, there are permanent
electrets which are analogous to permanent magnets. Here the electric field is
produced by “bound” charges created by a permanently “frozen-in” polarizability.
The polarizability is the electric dipole moment per unit volume that is often induced
in the material by an external electric field. We will introduce the displacement field
(or D-field) which obeys a Gauss’s Law that only depends on free charges. Free
charges are the charges controllable by batteries, currents and the like. To a large
extent one cannot totally control bound charges. We discuss the boundary
conditions that E-fields and D-fields obey across a dielectric boundary. We turn next
to a discussion of Laplace’s Equation in the presence of dielectrics. We give a
“method-of-images” solution for a charge above a dielectric surface and a dielectric
sphere placed in a uniform electric field. These examples make extensive use of the
dielectric boundary conditions. We turn next to a discussion of energy storage in the
presence of dielectrics. There are some interesting new issues that arise
concerned with whether or not one includes or excludes the energy stored by the
bound charges. We next consider the forces that act on a dielectric that – for
example– tend to pull the dielectric into the plates of a capacitor. We will conclude
on a more detailed model for the dielectric constants of a material that can be used
for material composites called the Clausius-Mossotti Equation.
2
Matter response to E: Induced dipole moment
200 210
2
ψ (^2)
=
e-^ “falls” in direction of
applied force – against E
2 0
0 0 0
0 0 0
with z 3 2
3 3 tanh
as kT 3 , tanh
and and ||
a
ea E a E p ea kT
ea E ea E ea Ee kT kT
p E p E
G G
G G
ψ 2 1±
exp
ea E
kT
G
exp
ea E
kT
Stark effect G
probabilities
< p >
G
1/ T
Massively oversimplified
Here is an old slide from my Quantum mechanic course that shows how a hydrogen
atom would respond to an external electric field and create a net dipole moment.
The basic idea is initially degenerate states – say the initially degenerate 200 and
210 states in hydrogen have no dipole moment since they have no preferred
direction. However once an external field is applied they can form a linear
combination of these two states which allows the electron to lower its energy by
“falling” into the electric force and creating an asymmetric wave function. That has
a dipole moment. In this case the dipole moment is parallel to the external electric
field and is an example of paramagnetism. The other combination has the electron
cloud on the other side of the hydrogen proton which produces an anti-parallel
dipole moment and thus lies higher in energy since U = - p dot E. Of course we
won’t always find the electron in the lower state at finite temperatures because of
thermal fluctuations. We thus expect the polarization to vanish at low 1/T which is
high temperatures.
4
Bound and Free charges
0
d 4
We next use this theorem from Potential Chp
with and
S V V
S
V r P r r r
f r r da f r d r f d
P f r r r
P
P r da V r r r r
τ πε
τ τ
πε (^0)
∫ Α^ =^ ∫ ∇ Α^ +^ ∫Α^ ∇
∫ =^ +
G GG
i (^) G G
G G^ G G G G^ G^ G^ G G
i i i
G G G
G G
G G
G (^) G G G i G G i G
V
S V
d r
P
V r P r da d r r r r
τ
τ πε (^0)
⎢⎣ −^ − ⎥⎦
G
G G
G (^) G G i G G i G G
We can write this as
1
4
1
4
where and
( ) '
( ') ' '
ˆ
b
S
b
V
b b
V r da r r
r d r r
P P
σ
πε
ρ τ πε
σ η ρ
0
0
= (^) ∫ −
= = −∇
G G
G
G G
G G G i i
Bound charges are real charges
tot b
We show total bound charge
is zero using divergence thm
Q 0
ˆ
b V V S
b S S S
b b S V
d Pd P da
da P da P da
da d
ρ τ τ
σ η
σ ρ τ
∫ = − ∇∫ = −∫
∫ =^ ∫ =∫
⇒ = (^) ∫ + (^) ∫ =
G G G (^) G
i i
G G (^) G i i
We next rearrange our V(r) expression using an integration by parts expression first
introduced in the Potential chapter. Essentially the integration allow us to move the
del operator from the 1/r term to the Polarization density. We are left with an
expression for the V(r) consisting of a surface integral over P and a volume integral
over the divergence of P. In both cases we have the 1/r factor that usually multiples
charges in potentials. The surface integral suggests that P creates a surface
charge density of bound charge, and the volume integral suggests that the
divergence of P represents a bound charge density -- in much the same way as the
divergence of E is proportional to the free charge density. In the surface integral,
we assign the area vector to be in the direction eta which is “out” of the dielectric.
Although the bound surface charge and bound charge volume density are first
revealed through some slick mathematical manipulations – they definitely represent
real charges that are bound in the dielectric. As such they will contribute to the E-
field. Griffiths discusses some models for the “bound” charge. We can easily show
that the total bound charge is zero using the divergence theorem. The bound
charge volume density is negative of the divergence of the polarization density. The
divergence theorem says this is equal to the negative of the surface integral over
the surface that bounds the dielectric. The integral of the surface density is the area
integral constructed of the dot product of the polarization vector and area vector.
The surface bound charge integral exactly cancels the volume bound charge
integral so there is no net bound charge when one considers the surface density
and volume density over a full dielectric region.
5
Bound charge reduces field from free charges
Thus with 1 1
r b b b
b f b f
f f f b
f f f
P E P E E
E (^) x E x x E E
x x E E E E E
x x x E E
0 0 0
0
0 0 0
0 0
0 0 0
A
G G G
i G G G G
G G G G G
G G
− − − − −
− − − − −
P
G + + + + +
ηˆ η ˆ Eb = σ (^) b /ε (^0)
G
Ef = σ (^) f /ε 0
G
x ˆ
phys 212 Griffiths
0
1 1 In capacitors:
f r
air
Q
E
A
Qd Q A A
V E C C
A V d d
ε σ κ ε χ ε ε ε
ε ε κ ε
0
Δ = × = = = = >
A
E
G
E
G
E
G
E
G
E
G
− − − − −
Dielectric molecules will naturally align
themselves to create dipole fields which
oppose the applied E field due to free
charges.
The bound charge is a formal way of describing how the molecules in a dielectric try to cancel the
applied (external) electric field due to free charges. We model the dielectric as polar molecules with
a positive and negative charge. They orientate them selves to create a dipole moment in the
direction of the applied electric field. In our cartoon this applied electric field is due to the free
charges on two capacitor plates and the orientation is due to the fact that the relatively negative end
of the molecule is attracted to the sheet of positive free charges on the left and the relatively positive
end of the molecule is attracted to the sheet of negative free charges on the right. The dipole will
create electric field lines which originate from the + charge and end on the – charge and are in the
opposite direction as the applied field. Hence the molecular field opposes the external field and thus
reduces it from the field that would be present from the free charges if no molecules were present.
We follow this with a formal argument based on bound charges. The polarization density is
proportional to the total electric field with proportionality constant given by the product of the electric
susceptibility chi and epsilon_0. Since the electric field is constant, its divergence is zero and hence
there is no bound charge volume density – all the bound charge must be on the dielectric surface. In
this case it has a surface density of P dot eta-hat where eta-hat points outwards from the dielectric.
The eta-hat vector is along –x on the left and +x on the right which means we have a negative bound
charge surface density on the left and a positive bound surface charge density on the right. These
have the opposite polarity from the adjacent free surface charges. If we superimpose the fields from
the left and right bound charge sheets, we get E_bound = sigma_bound/epsilon_0 which given by the
negative of the susceptibility (chi) times the E-field. We also know that the E-field is given by the sum
of the E-field due to the free charges (sigma_free/epsilon_0) plus the field due to E_bound. We can
use this superposition formula to solve for the E-field due to sigma_free. We find that the usual E-
field for two sheets of opposite charge is reduced by a factor of (1 + chi). We can combine the (
+chi) factor with epsilon_0 to define a new permeability constant for materials (epsilon) which is
larger than that in vacuum (epsilon_0). In Physics 212 the ratio of epsilon/epsilon_0 was called
kappa, Griffiths calls it epsilon_r. We can calculate the capacitance for a parallel plane capacitor
using our electric field as a function of the free charge expression. The free charge surface density is
Q/A where A is the area of the plates and Q is the applied free charge. The voltage is just the E-field
times the plate separation d. We can then get the capacitance by dividing the charge by the voltage.
Basically the dielectric reduces the field and voltage by a factor of kappa and therefore increases the
capacitance by a factor of kappa over a air filled capacitor. Physics 212 may have lulled you into
thinking that you can account for dielectrics by simply changing epsilon_0 to epsilon in your formula
sheet but things can frequently be much more complicated.
7
The Spherical Electret
0 0
and
ˆ ˆ ˆ cos
b b
b
b b S
P P
P z
P P z P da
G G G
i i
G i G i i
3 1 2
3 0
1
2
We start with
From Laplace Chp
1 thus 3
cos and E= 3
( cos
V r R
R
V r r
P R
V r R V r
σ θ
ε
θ ε
σ θ σ θ θ
0
0
A
G
2 1 0 3
0 0 1 0
0 0
0 0
P (cos piece
where and B 3
and 3 3
Thus an
We also hav
d
a
e
( )
( ) cos
V r R A r
P
A B R R
P P
A V r R r
P P
V r R z E V z
− +
0
0
A A
A A A A
G
The E field lines
Here is a bound charge example
where ε and χ aren’t even defined
q = 0
Δ z
σ (^) b > 0
σ (^) b < 0
Offset charge picture
+ +^ +
−^ −
2 1 0 3
0 0 1 0
0 0
0 0
P (cos piece
where and B 3
and 3 3
Thus an
We also hav
d
a
e
( )
( ) cos
V r R A r
P
A B R R
P P
A V r R r
P P
V r R z E V z
− +
0
0
A A
A A A A
G
z^ ˆ ηˆ
0 P = P z ˆ
G
θ
R
We illustrate the concept of polarization density and bound charges with the unusual
example of spherical electret Here the polarization density is frozen in and P is
independent of E and hence a susceptibility cannot be defined. Since P a constant
within the sphere, it has no divergence and hence there is no bound state volume
density. There will be a bound surface charge on the sphere surface. The form of
this will be eta dotted into P where eta points out of the dielectric and is in the r
direction. This means the bound surface charge is proportional to cos (theta) We
can think of the bound surface charge as “glued” on to the sphere and we can
analyze the potentials using the same technique as glued charge in the Laplace
Chp. Here we match the Legendre Polynomial to the cos(theta) dependence of the
glued charge which means the potential is constructed from P1. The potential in the
r>R will go as 1/r^3 since the r term will blow up at infinity. Following the glued
charge example we can write the potential both inside and outside the sphere. In
particular in the r 8
The Displacement Field
Because one has no real control of bound
charges since matter will respond as it
will, it is desirable to cast Gauss’s law in
a form that only depends on free charges.
This is done by inventing a new field: D
D ε E P 0
= +
G G G
The divergence of D only depends on
free charges.
f b ; b
f
D E P D E P
P
D
E
ε ε
ε ρ ρ ρ
ρ
0 0
0
G G G G G G G G G
i i i
G G G
G G
i
G
i i
We can also cast this into an integral
Gauss’s law form.
d d ;
inc f f S
∫ ∇^ D^^ τ^ =^ ∫ ρ^ τ ∫ D da^ = Q
G G G G
i i
Q
r
Consider a class A dielectric
surrounding a charged metal sphere
2
2 2
; E=
S
D da r D Q
Q D Q
D r r
r r
π
π ε πε
∫ =^ =
G G
i
G
G G
Class-A
D = ε E
G G
Q
r
A very convenient concept is the D-field (displacement field) which is a combination
of the E-field and the polarization density P. In Class –A dielectrics D is
proportional to E with a proportionality constant epsilon where epsilon = episilon_o(
- chi) but as in the case of the electret other dielectrics are possible. The power of
D is apparent once one takes its divergence. The divergence of E will be the total
bound charge volume density which is the sum of the free charge density and the
bound charge density. The divergence of P is (-) the bound charge density. This
means that the diverge of D gives just the free charge volume density. One reason
that this is a worthwhile concept is that one has control over the free charges, the
bound charges are created as the material responds to the electric field and in
some sense “come along for the ride”. We can use the divergence theorem to
create a Gauss’s Law for D where the surface integral of D over a Gaussian surface
is equal to the enclosed free charge. If the situation has enough symmetry, one can
compute D using Gauss’s Law and then use (for example) D=epsilon E to compute
E. Here is a particularly simple example of a Class A dielectric surrounding a
charged metal sphere carrying a free charge of Q. We use the Physics 212
technique to compute E and D for this situation in a region where we are in the
dielectric. It is best to think of D as a mathematical convenience. The electrical
force, for example is tied to E not D and E responds to both free and bound
charges.
10
Two approaches to the class-A capacitor
( )
( )
0 0
(^0 )
Alt view is sum of bound & free cap fields
σ
ε
ε ε ε
ε ε ε ε
ε ε ε ε
ε ε ε ε
ε
ε ε
ε (^0 0 0 )
0 0 0 0
E b^ Q^ P D E a
D E P E P E
Q E Q
E E
a a
Q Qd Q a E V Ed C
Q
a V
a
a d
Hence bound charge cancels part of E, lowers V and raises C by factor of ε/ε 0
σ b = − P
σ b = + P
σ f = + Q a /
σ f = − Q a /
b
Eb
σ
ε 0
f
Q
E
a^ ε 0
d
V
P
G
σ b > 0
σ f
− σ f
d
V
D top
G
D bottom
G
E^ ε
G
σ b < 0
f
f f
f
f
2D
D & D
D D ˆ
∫
inc f top S
top bottom
top bottom
D da Q a a
D z
z Qz
D E E
a
G G
i
G G G
G G G
This slide compares two ways of computing the electrical field between the
capacitor plates. They are actually infinite planes. The easiest approach is on the
left. Here we apply Gauss’s law for the D-field in the Physics 212 way. We can
ignore the bound charge on the top and bottom of the dielectric since Gauss’s law
for the D-field includes free charge only. The D-fields superimpose just like E-fields.
At the end we use D=epsilon E to compute E.
The right slide explicitly computes the field contributions of the free and bound
charges by essentially using Gauss’s law for E field. We can think of this as the
field from free charge capacitor, subtracted from the field of the bound state
capacitor. We have ignored any bulk charge volume density since the P vector is
constant and has no divergence. We then use the D expression to compute the P
vector which allows us to compute the field from the bound state capacitor. We get
the same answer as before from E. We continue to compute the capacitance which
(as in Physics 212) is just the usual formula with epsilon substituted for epsilon_0.
11
Lets calculate force on dielectrics
f 0 0
0 f
0 f
f
Force where
is due to free charge + bottom
bound charge. Find
Force on the Top
Dielectric
bot b
bot b
bo
top b b f
b f
b
b
f b f
t b
E
F a
E
E
E
P E
P
E
E
E
σ
σ ε ε ε ε ε
ε σ σ ε
σ ε σ
ε ε ε
σ
ε
0 0
0
G G
G
0 f
f f f
E b f
ε σ
ε ε
σ σ σ (^) ε
ε ε ε ε
0
0 0 0
[ ]{ }
f 0
top
Top
2 2 0
0 f
Top
f 0
and no net force
b f
f
bottom t
b b f
op
E z
F a E
F a
a
z
F z
F F
0
0
0
⎣ ⎝^ ⎠ ⎦
⎣⎝^ ⎠
⎢⎣ ⎝^ ⎠⎥⎦
⎨ −^ + ⎬
⎦⎩ ⎝^ ⎠ ⎭
G
G
G
G
G
G
G
− σ b
bot
Eb
G
E f
G
E other
G
F top
G
Top
Bot
F bot
G
+ σ b
z^ ˆ
This problem is tricky since when computing force on top of the dielectric we cannot
just use the E-field calculated on last slide but rather must subtract the field due to
the top of the dielectric itself.
One way tp dp this is to add the fields from the top and bottom metal plates (free
charge) to the bound charge from the bottom of the dielectric. This approach is
slightly different from that used for the pressure on metal discussed in the Laplace
chapter. There we used a self-field subtraction method that consisted of
subtracting the field due to a patch of induced charge from the total electrical field
close to the conducting plate. Here we are explicitly adding all fields but the field
due to the upper plate. We can think of upper bound charge as being attracted to
upper plate and repelled from lower plate. You will use the self-field, patch
subtraction method in homework. We find that the force on the top of the dielectric
causes the dielectric to be attracted to the top plate. The force depends on the how
different epsilon is from epsilon. This makes sense since if epsilon were equal to
epsilon_0 there would be no bound charge.
13
Boundary conditions for D and E
below above || ||
below above || ||
In electrostatics
Thus at boundary
∇ × = → =
∫
∫
E E d
E d E E
E E
G G^ G^ G
i A G G i A A A
v
v
Some components of D and E are continuous across a dielectric boundary and others are not. We begin with E.
Df = σ f / 2
G
a^ ˆ
Df = σ f / 2
G
σ f
a^ ˆ
enc enc f free f V below above
If σ ,
⊥ ⊥ ⊥ ⊥
∫
above below
D da Q Q
D a D a D D
G G
i
f above below
If is present it creates D =
⊥ ⊥ ⊥ ⊥
⊥ ⊥
f f
f f above below f
D D D D
D D
||
In electrostatics
E is always continuous
D ⊥is often continuous
a^ ˆ
a^ ˆ
ε
ε
above D
G
below D
G
0
ε
ε
above
E
G
below
E
G
A
0
ε
ε
above
E
G
below
E
G
A
In Laplace’s equation problems (which we will do shortly), we need to know what
quantities are continuous across a dielectric boundary and which are not. We will
assume that potentials are continuous. On the left we argue that the parallel (or
tangential) component of the E-field is continuous across a dielectric boundary in
electrostatic problems. Here the curl of the electric field vanishes. Stokes’ theorem
implies that the line integral of the electric field around a complete path is zero. To
establish the continuity of the parallel or tangential component of the E-field, we use
a small loop of length L that extends on just below and just above the dielectric
boundary. The line integral involves the parallel component of the electrical fields
above and below the boundary. For it to vanish the parallel components of E must
be the same on either side of the dielectric boundary. We will argue (on the right
side) that the perpendicular (or normal) component of the D-field is often continuous
across the boundary, and hence for a Class A dielectric, the normal components of
E will be discontinuous across the boundary as shown.
The argument for the displacement field uses Gauss’s law. We use a pillbox
surface. In the absence of a free surface charge density , the enclosed charge free
charge is zero and surface integral will vanish. The surface charge will be
proportional to the difference of the normal D-fields because the area vector
changes sign going from the above to below the boundary. In the absence of a free
surface charge the normal component of D will be continuous. Nearly always there
will be a bound surface charge on the boundary and hence the normal component
of the electrical field will be discontinuous across a dielectric boundary. Usually
there will not be a free charge surface density on the dielectric boundary and hence
normal component of the D field will be continuous. If there happens to be a free
charge density the discontinuity in the normal component will be proportional to the
free charge surface density.
14
Images in a Class-A Dielectric
2 (^4 ) 4
We begin by computing (^) ˆ
. Can we find , from
is true but kills
spherical symmetry as shown:
b
S
b S
z P
D E P D P
qr D da r D q D r
D da q
σ
ε
π π
σ
0
∫ =^ =^ →^ =
G
i G G G G G
G G G
i
G (^) G i
E-field lines for q
above & below
dielectric
( )
( )
0
2 2
2 2
0 0
0 0
2 23 2 0
/
Use & superposition
(just below boundary
cos
cos ; ˆ
since ( ) ;
q z z
b z
b z
z z z b
b z
b b
b
b z
D E E P
E E
q E Z s
Z
P P
Z s
P E E
E
qZ
Z s
E
ε ε
α σ
πε ε
α σ η
ε ε χε σ
σ ε ε χ χε
σ σ
χε πε ε
χ σ χ
0 0
0 0
G G G G
G G
G
G
G
i
( )
2 23 2 2
/
qZ
πε 0 Z s
Z
s
ε
ε
q
b
σ
α
z ˆ
ηˆ
This is factor is typo
Here is an elegant problem (Griffith’s Example 4.8) consisting of a charge placed a
distance Z above a dielectric slab. We will have electrical fields due to free charge
and bound charges on the surface of the dielectric. Prior to this example, we have
been able to compute fields using the D version of Gauss’s law and are tempted to
try this again under the false assumption that D only “feels free charges and is
thus blind to the bound charges. It is true that only free charges are included in the
D Gauss’s law but bound charges will affect D and destroy the symmetry since D =
epsilon_0 E above the dielectric and E is affected by sigma_b. The field line figure
is a nice illustration of the boundary conditions illustrated on slide11 since field lines
bend away from normal in the dielectric.
16
1 0 1 2
0 1 2^1
0 0
in , ,
out , ,
in
We write ( ) P (cos )
and ( ) P (
To avoid ( ) , All.
cos )
Unlike the conducting sphere case,
both and have non-zero V
V
B V r A r r
E
V r r r
r B
r R r R
θ
β α θ
=
= +
⎛ ⎞ = (^) ∑ (^) ⎜ + ⎟ ⎝ ⎠
→ = ∞
⎛
=
⎞ = (^) ∑ (^) ⎜ + ⎟ ⎝ ⎠
A (^) A A A A A
A (^) A A A
A
A A
G
G
G
G ( ) 0 ( ) 0
1 0 1 0
ˆ V out cos
Hence and
r E z r E r
E
θ
α α (^) ≠
→ ∞ = ⇒ → ∞ = −
= − (^) A =
G
Example of Dielectric BC : Sphere in a Uniform E-field
in out
1 1 1 0 2 1 0 3
2 1 1
in out
0
We next apply B.C.
R= ; =
R = ;
We next apply D D B.C. or
or
For =
r r r
1
For 1
( ) ( )
( ) ( )
in out in ou
R R R
V R V R
A E R A E R R
A A R R
R R
V V V V
β β
β β
ε ε κ
⊥ ⊥
=
− + − +
=
⎡∂ ⎤ ⎡ ∂^ ⎤ ⎡ ∂^ ⎤ ∂ ⎢ ⎥ =^ ⎢ ⎥ ⎢ ⎥ = ⎣ ∂^ ∂^ ∂ ⎦
≠
⎦ ⎣ ⎦ ⎣
A A A A A A A
A
A
1 (^1 0 )
1 2 1 2
For =
For
r
2 = where
1 ; 1
1 -
t
R
A E R
A A R R R
β ε κ κ ε
β κ κ β
0
− + ≠ +
⎡ ⎤ ⎢ (^) ∂ ⎥ ⎣ ⎦
− − =
= = −
A A A A A (^) A A
A (^) A A A
A
A
in 0 1 2
out 0 1 0 1 2
Hence:
P
P
, ,
, ,
( ) (cos )
( ) cos (cos )
V r A r
V r E r r
=
=
A A A A
A A A A
G
G
0
We have a dielectric
sphere in a uniform
electric field E=E z ˆ.
G
R
z
ε 0
E z ˆ
We apply the D-normal continuity BC to the familiar problem of a sphere in a
uniform E-field. This differs from the grounded conducting sphere in a uniform E-
field since there are fields inside the sphere as well. I write separation of variable
solution in terms of A and B coefficients inside the sphere and alpha and beta
coefficients outside the sphere. Since the inside solution includes r=0 and we need
a finite V at the origin, we know that all B coefficients which go as 1/ powers of r
must vanish. As r becomes large V must approach r cos(theta) because of the
external E field, but we want no higher powers of r but 1. This means that alpha_
can be non-zero but all other alpha’s must vanish. Hence we are limited to the form
at the top of the right side with unknown A and beta which we must find using BC.
For any given L value, we have two unknowns A_L and beta_L. There will be no
“coupled” equations which relate two different L values such as A_1 and beta_
since such an equation will can never be satisfied at all values of cos(theta). We
thus need two equations for each L value. The first of these is continuity of the
potential itself. We assume that Vin and Vout join continuously at r=R. We write this
a separate BC for L=1 and one for all other L. The other BC is based on continuity
of D_perp or the radial component of D. We assume that D_perp which is D_r is
continuous at r=R which means that epsilon E_r is continuous E_r is the derivative
of V with respect to r. This gives us the necessary conditions to solve for A_L and
beta_L.
17
Completing the dielectric sphere
2 1 2 1
1 1
Now put together the 1 conditions
A
A R R
A
A A
A
κ β β
κ κ
β
≠ ≠
A A A A A A
A A A
A A
A
A
A
A A
A A
0
3
(^0 )
Putting it all together we have
a remarkably simple form:
cos
cos
in
out
E
V r
R
V E r
r
θ κ
κ θ κ
⎣ ⎝^ + ⎠ ⎦
( )
1 3 1 0 3 1 1 0
1 1 0 3 1 0 3 1
1 0 0 1 0 1
0 3 3 1 0 1 0
Put together the 1 conditions
A E A E R
R
A E
A E R
R
A E E
A E A
E
E R E R
β β
β κ κ β
κ
κ
κ β β κ κ
⎛ −^ ⎞ ⎛ −1⎞
⎝ +^ ⎠ ⎝ + ⎠
A
0
0
3 0 0
Hence for there is a constant
field:
For we have the E field
added to ideal dipole with a moment of:
in
r R
E z
E r R V
r R z
p R E z
κ
κ πε κ
= ⎡^ ⎤
G G
G
G
Putting the two L=1 BC together we can solve for A_1 and beta_1 in terms of E_
and kappa (or epsilon/epsilon_0). When we put together the L ne 1 terms we find
that all of these must vanish. The reason is that the L ne 1 equations are
“homogenous” – they are linear in A_L and beta_L but with no additional constants
(source terms) to set the scale for the A, beta coefficients. This means one could
double all A and beta for L ne 1 and get an equally valid solution. For the case of
L=1 E_0 breaks the homogeneous force and thus sets the scale for A_1 and
beta_1. We finally end up with a fairly simple form for the potential inside and
outside of the sphere. Inside the sphere we have a constant electrical field. Perhaps
this isn’t that surprising since we found a constant field for the uniformly polarized
sphere which also had a cos(theta) dependence of bound charge on the surface.
Outside of the sphere we have a pure dipole contribution added to our constant field
of strength E_0.
19
Energy of free charges in dielectrics
( ) ( ) ( )
( ) ( ) ( )
( )
The stored energy from potential chapter
was obtained from
V where
Here =. But usually
we don't control since it is due to
respon
V
f b
b
U E E d
U r r d r E
r r r
r
ε τ
ρ τ ρ ε
ρ ρ ρ
ρ
0
0
= (^) ∫ 2
1 = (^) ∫ = ∇ 2
G G i
G G G G^ G i
G G G
G
V
( )
' ( ) ( ) ( )
1 ' ( )
se of dielectric. Can we find an
energy density that only depends on free
charge Start with
V where
V
We get to the field energy from integrat
f
f f V
V
r
U r r d r D
U r D d
ρ
ρ τ ρ
τ
1 = (^) ∫ = ∇ 2
→ = (^) ∫ ∇ 2
G
G G G G^ G i
G G^ G i
( ) ( ) ( ) ( )
ion
by parts expression.
S
∫ f r^ Α^ r^ da^ =^ ∫ f r^ ∇ Α^ d^^ τ+ Α∫^ r^ ∇ f d^ τ
G G^ G G G G^ G^ G^ G G i i i V V
Here and
V
Since - we have:
Usually we let volume
so
V
S
S
S
D f r V r
r D d
V r D r da D r V d
U V D da D Vd
V E
U V r D da D E d
V r
τ
τ
τ
τ
∫ ∇^ =
∫ −^ ∫ ∇
G G G G
G G^ G
i
G G^ G G G^ G G
i i
G G G G
i i
G G
G G^ G G^ G
i i
G
V
V
V
all space
on S &
S
V r D da
U D E d τ
G G G
i
G G
i
To answer this question, we review the derivation of the U stored energy
expression. We obtained this from the voltage times the charge density and wrote
the charge density as the gradient of the E-field. We then did an integration by parts
manipulation to get our final form for U. All of this is true in the presence of a
dielectric but when we write that the charge density as the gradient of the E-field we
are including both the free charge and the bound charge. In our W_bat calculation,
we were only including the free charge that is pushed from the battery to the
capacitor plates. Usually the free charge work is the most relevant since we can
control the free charge and the bound charge is due to the response of the dielectric
and “comes along for free”. The clue to writing a general expression for a “free
charge” stored energy starts from the divergence of D (only free charges contribute)
rather than divergence of E (both bound and free contribute). We then do the same
integration by parts trick, throw way the surface term by putting the surface at
infinity where the voltage vanishes to obtain a “free charge” stored energy involving
the product of D and E.
20
U versus U’
all space
all space
Griffith's discusses incremental change
in U' since D depend history of how
E was applied:
Usually:
can
U D E d
U D E d
τ
τ
G
G G
i
G G
i
( )
2 1
0
1
' 2 2
' '
Now lets compute
and as before
U differs form U' since U' doesn't include
(negative) work on the bound charges.
U E E d
Q Q Q A U Ad U A A Ad
U U U U
ε τ
ε ε ε ε
ε ε ε ε
ε
ε
0
− 0 0 0
= (^) ∫ 2
⎛ ⎞ ⎛ ⎞ = (^) ⎜ ⎟ = (^) ⎜ ⎟ = ⎝ ⎠ ⎝ ⎠
⎛ ⎞ → = (^) ⎜ ⎟ > ⎝ ⎠
G G i
( )
2 1 2 2
1 '
1 ' 2 2 2 2
Lets compute U D E d
Q Q Q a Q CV U ad A A Ad C
τ
ε
ε
−
= (^) ∫ 2
⎛ ⎞ ⎛ ⎞ = (^) ⎜ ⎟ = (^) ⎜ ⎟ = = ⎝ ⎠ ⎝ ⎠
G G i
Using Gauss's law:
Q Q
D E
A ε A
σ (^) b > 0
σ f
f
d
V
D top
G
D bottom
G
E^ ε
G
σ (^) b < 0
I
Here we apply the U’ expression to the usual dielectric capacitor but this time we
concentrate on expressions in terms of the free charge on the capacitor plates
rather than the work supplied by the battery to give us more practice in dealing with
D. Our first step is to compute D_top using Gauss’s Law for the free charge on the
top plate and D_bottom using Gauss’s law for free charge on the bottom plate. The
total D is the sum of D_top + D_bottom and is just Q/A where again Q is just the
free charge. We are assuming a linear medium so D= epsilon E which allows us to
calculate E. We then can compute U’ from our integral expression involving D and
E which is simple since the D dot E integrand is constant and the volume is just A d.
Putting it altogether we get the Physics 212 expression for the stored energy in a
capacitor. If we next compute the stored energy of the free plus bound charge we
get a smaller stored energy. This is somewhat surprising that it takes less work to
supply the capacitor with bound plus free charge than free charge alone. Evidently it
takes negative work to create the bound charge. The reason is that bound charge
on the top plate is negative and the bound charge on the bottom plate is positive
which is the opposite of the free charge. This must be true since the E field in the
dielectric is smaller than you get with no dielectric for the same Q so the bound
charge field opposes the free charge field. Hence takes negative work to put a
negative bound charge on the top plate and a positive bound charge on the bottom
plate since this is the direction bound charges would like to move in a downward E-
field. Is it always true that U’ > U?
