Electric displacement
Now letzs see how to investigate the ¤elds that are produced by a collection
of charges, both bound and free. (Free charges are those not bound up in
atoms and molecules of the material. This includes charges on the surface of
conductors.)
We have Gaussz law
½ + ½b
~ ¢E
~ = ½ = f
r
"0
"0
But we also discovered that
so
~ ¢ P~
½b = ¡r
~ ~
~ ¢E
~ = ½f ¡ r ¢ P
r
"0
Now it is convenient to gather the two divegences together:
~ ¢E
~ +r
~ ¢ P~ = ½ = r
~ ¢D
~
"0 r
f
(1)
~ is de¤ned to be
where the electric displacement D
~ = "0E
~ + P~
D
(2)
~ = "0 E
~ + "Âe E
~ = " 0 (1 + Âe ) E
~ = "E
~
D
(3)
For LIH materials
" is the permittivity of the material and the dimensionless ratio "=" 0 = · is the
~ here is the net E
~ at position ~r due to all
dielectric constant. Note that the E
~
~ which satis¤es
sources, including P itself. So it is often easier to work with D;
Poissonzs equation in the form (1).
~
Boundary conditions for D:
~ as we did for E;
~ by putting a tunaWe ¤nd the boundary conditions for D;
can across the boundary, and integrating equation (1) across the boundary.
1
Z
~ ¢D
~ dV
r
I
~ ¢ dA
~
D
6=
=
Z
Z
½f dV
½f dhdA = ¾ f dA
The contribution from the sides of the box is negligible, because h ¿ d: So
³
´
~1 ¡ D
~2 ¢ n
D
^ = ¾f
(4)
~ E
~ = 0 to derive the boundary
Previuosly we used the di¤erential equation r£
~
~ = "E
~ in an LIH
condition that E ta n is continuous across a boundary. If D
~
~
~
~
material, then r £ D is also zero. But r £ D is not zero at the boundary
~ £D
~ is in¤nite at the boundary.
because " changes abruptly there. In fact, r
~
~:
So we have one boundary condition (4) for D and one for E
To see how this works, letzs look at our slab again. With the slabzs sides
parallel to the y ¡ z¡plane, and normals in the §x direction, the normal com~ is Dx: Thus Dx is continuous as we cross the boundary since there
ponent of D
is no free charge there. (The charge layer is entirely bound charge.)
Dx;ou t = "0 E x;ap p = "E xin = Dx; in
~ is zero both inside and outside. So
The tangential component of E
~ in = "0 E
~ ap p
E
"
~ gives us the bound charge density. On the
The boundary condition on E
right hand side
³
¾
"0 ´
E x;ou t ¡ E x;in =
= E x;a pp 1 ¡
"0
"
Thus
¾ b = (" ¡ "0 ) E x;a pp
Using the same method, convince yourself that the charge on the left side is the
exact negative of this.
2
Boundary value problems
If we have a boundary between two uniform regions (two uniform dielectrics,
or a dielectric and a conductor, or a dielectric and vacuum) then we have the
following set of equations:
~ ¢D
~ =½
r
(5)
f
and
~ £E
~ =0
r
~ £E
~ = 0; we may write E
~ = ¡ rV;
~
within each region. Since r
and equation
(5) may be written
³
´
~ ¢ "rV
~
~ ¢ rV
~
¡r
= ½f = ¡"r 2V ¡ r"
~ = 0; and the equation for
Since we are assuming our medium is uniform, r"
the potential is
½f
r2 V = ¡
(6)
"
~ 6= 0 right at the boundary. We may use equation (6) within each
Be careful! r"
region, but not across the boundary.
The boundary conditions are:
³
´
~1 ¡ D
~2 ¢ n
D
^ = ¾f
~ tan is continuous
E
V is continuous
Dielectric sphere in a uniform ¤eld
To see how to use these relations, consider a dielectric sphere of radius
~ 0 . Because the boundary is spherical, we use
a placed in a uniform ¤eld E
spherical coordinates with origin at the center of the sphere and polar axis
~ 0: There is no free charge, so V satis¤es Laplacezs equation both
parallel to E
inside and outside the sphere (but not ON the boundary), and we know the
potential may be written in terms of Legendre polynomials.
Inside the sphere, the potential must not blow up at r = 0; so we have only
positive powers of r :
V (r < a) =
1
X
Al r l Pl (cos µ)
l=0
~ 0 : That
Outside, we must have a potential that gives us the uniform ¤eld E
potential is V u n if = ¡E 0 z = ¡E 0 r cos µ = ¡E 0 rP 1 (cos µ) : Other than that, all
the other terms must ! 0 as r ! 1; so
V (r > a) = ¡E0 rP1 (cos µ) +
3
1
X
Bl
P l (cos µ)
rl+1
l=0
Now we apply all our conditions, starting with the easiest one. V is continuous.
1
X
l=0
Al al P l (cos µ) = ¡E 0 aP1 (cos µ) +
1
X
Bl
Pl (cos µ)
al+1
l=0
Because the P l are orthogonal, the coe¢cient of each P l must separately equal
zero. Thus
B1
l = 1 : A1 a = ¡E 0 a + 2
(7)
a
Bl
l 6= 1 : Al al = l+1
(8)
a
~ The normal component is the radial
Now we use the boundary condition for D:
component, and there is no free charge on the surface, so we have
¯
¯
@V (r < a) ¯¯
@V (r > a) ¯¯
"
=
¯
¯
@r
@r
r=a
r=a
1
1
X
X
Bl
"
lAl al¡1 P l (cos µ) = ¡E 0 P1 (cos µ) +
¡ (l + 1) l+2 P l (cos µ)
a
l=0
l=0
Again the coe¢cient of each Pl must separately equal zero.
l=1:
"A1 = ¡E 0 ¡ 2
Bl
a3
(9)
Bl
(10)
al+2
For l =
6 1; we use equation (8) to eliminate Al from equation (10), and get
l 6= 1 :
l 6= 1 :
"lAl al¡1 = ¡ (l + 1)
"l
Bl
Bl
= ¡ (l + 1) l+ 2
al+2
a
which could only be true if " = ¡ (l + 1) =l: This is impossible because " is not
negative for ordinary materials, and cannot equal ¡ (l + 1) =l for more than one
l; if any, in any case. Thus we must conclude that Bl = Al = 0:
It is a di¤erent story for l = 1: We use equation (7) to eliminate A1 from
equation (9).
µ
¶
B1
Bl
" ¡E0 + 3
= ¡E 0 ¡ 2 3
a
a
B1
(" + 2) = E0 (" ¡ 1)
a3
So
B1 = E 0 a3
and then
A1 = ¡E 0 + E 0
"¡ 1
"+ 2
" ¡1
E0
= ¡3
" +2
"+ 2
4
The potentials are
3
E 0 r cos µ
"+2
V (r < a) =
¡
V (r > a) =
¡E 0 r cos µ + E 0
" ¡ 1 a3
cos µ
" + 2 r2
The ¤eld inside is uniform:
~ (r < a) =
E
3 ~
E0
"+2
and outside we have the applied, uniform ¤eld, plus a dipole ¤eld with dipole
moment
~ 0 a3 " ¡ 1
~p = 4¼" 0 E
"+2
The dipole moment is zero if " = 1 (the sphere is vacuum, ie there is no sphere!).
The ¤eld inside the sphere is less than the applied ¤eld for " > 1; as expected.
Since " measures the degree to which the material can reduce an applied ¤eld
in its interior, we might imagine that a conductor, which reduces an applied
¤eld to zero, coresponds to a very large value of ": In fact some properties of
conductors may be retrieved in the limit " ! 1: This limit applied to our
~ 0 a3 : These are
sphere gives us zero ¤eld inside and a dipole moment ~p = 4¼" 0 E
the results we found previously for a conducting sphere.
Energy
We have learned several things about electric energy that are relevant to our
present discussion.
² The energy density stored in the electric ¤eld in vacuum is u = 12 "0 E 2
2
² The energy stored in a capacitor is U = 12 C (¢V )
~
² A dipole in an electric ¤eld has potential energy U = ¡~p ¢ E
Letzs look at the second of these ¤rst. If we ¤ll a capacitor with dieletric we
increase the capacitance by ": To see why, compute the capacitance by putting
a charge Q on the capacitor. The ¤eld inside the dielectric is a factor · = "=" 0
smaller than with no dielectric, so the potential di¤erence is also a factor of ·
smaller, and thus the capacitance C = Q=¢V is a factor · larger. Thus the
stored energy also increases by a factor of · (if the potential di¤erence is held
¤xed).
But we know that the energy stored in a capacitor is stored in the electric
¤eld, and, without dielectric
Z
1
1
2
U = C0 (¢V ) =
"0 E 2 dV
2
2
5
where the integral is over the volume of the capacitor. With dielectric:
Z
1
1
1
2
2
U =
C (¢V ) = ·C0 (¢V ) =
·"0 E 2 dV
2
2
2
Z
Z
1 2
1~ ~
=
"E dV =
E ¢ D dV
2
2
So it looks as if the energy density is
u=
1~ ~
E ¢D
2
which is greater than 12 "0 E 2 :(for the same E :) Why is that? Well, as we increase
the ¤eld by bringing up free charge, that ¤eld acts to create and/or align the
atomic dipoles in the material. The work done by the ¤eld is negative (3rd point
above) and thus the stored energy in the ¤eld is increased by the work done.
We can con¤rm these conjectures by imagining a process to put the system
together. So we bring up the free charges, bit by bit. We donzt have control
over the bound chargesw they just do what they must do. So when the potential
has the value V and we bring up the next piece of free charge, we have to do
work
Z
¡
¢
¢W =
¢½f V d¿
Z ³
´
~ ¢ ¢D
~ V d¿
=
r
Now we do the usual "integration by parts" trick.
³
´ ³
´
~ ¢ ¢ DV
~
~ ¢ ¢D
~ V + ¢D
~ ¢ rV
~
r
= r
So
¢W
=
=
Z h
³
´
i
~ ¢ ¢DV
~
~ ¢ rV
~
r
¡ ¢D
d¿
Z
Z ³
´
~ ¢n
~ ¢E
~ d¿
¢ DV
^ dA +
¢D
S1
The ¤rst term is zero, for the usual reasons. Now if our material is LIH; then
~ = "¢E
~ and
¢D
³
´
³
´
³
´
~ ¢E
~ = " ¢E
~ ¢E
~ = 1 "¢ E
~ ¢E
~
¢D
2
1 ³~ ~´
=
¢ E ¢D
2
Thus
¢W = ¢
Z
6
1~ ~
E ¢ D d¿
2
and hence
u=
1~ ~
E ¢D
2
as we conjectured above.
Note: I found Gri¢thzs discussion somewhat misleading. All the energy is
electric, including his so-called "spring" energy. For a more complete discussion,
see Jackson §4.7.
Forces:
We have already noted from energy arguments that dielectrics are sucked
into higher ¤eld regions. Here wezll investigate the force that does the sucking.
Wezll use as our example system a parallel plate capacitor. The plates measure
w £ ` and the plate separation is d: Letzs charge up the capacitor, and then
disconnect it from the battery. This gives us a nice, clean isolated system with
the charge on each plate ¤xed at Q = Q 0 :. Now we insert a dielectric slab that
¤lls the space between the capacitor plates. The initial energy is
Ui =
1 Q 20
1
2
= C i (¢V 0 )
2 Ci
2
Uf =
1 Q20
1 Q 20
Ui
=
=
2 Cf
2 ·Ci
·
The ¤nal energy is
The energy decreases because the capacitor does work on the slab as it sucks it
in.
To ¤nd the force, we look at the system when the slab is part way in, as
shown:
We may model this system as two capacitors in parallel: one with plate area
A1 = w(` ¡ x) and one with plate area A2 = wx: The two capacitances are
C1 =
"0 A1
"A2
and C 2 =
d
d
The total capacitance for the two in parallel is
C (x) = C1 + C2 = "0
w
w
[` ¡ x + ·x] = "0 [` + (· ¡ 1) x]
d
d
and the stored energy is
U (x) =
1 Q20
1
Q20 d
=
2 C (x)
2 "0 w [` + (· ¡ 1) x]
7
The potential di¤erence between the plates is ¢V (x) = Q=C (x) and may be
measured along any path between any two points on the plates. Thus the
~ = ¡ rV
~ ) is the same in the part with the
electric ¤eld between the plates (E
~ = "E
~ is greater in the part with the
dielectric and the part without. But D
dielectric, and thus the free charge density on the conducting plates is greater
where the dielectric is. As the slab moves in, charge migrates across the plates.
Now in the ideal models we have been using, the ¤eld lines, both in the vacuum and in the dielectric, are everywhere perpendicular to the plates, and the
charge density, both bound and free, forms layers parallel to the plates. But
thatzs not quite right. At the very end of the slab there is some bound charge
density and some ¤eld lines connect from the plates to the end of the slab. The
volume occupied by these curved ¤eld lines is very small, and consequently our
calculation of the capacitance is extremely accurate, as is the energy. This is
just as well, because it is very hard to calulate this "fringing ¤eld" accurately.
But it is precisely this ¤eld that gives rise to the sucking force! Fortunately we
can calculate the force using energy arguments, without having to ¤nd the ¤eld.
dW
Fx dx
Work done by the ¤elds reduces the stored energy
= F~ ¢ d~s = U (x) ¡ U (x + dx)
dU
= ¡
dx
dx
and so
Fx
dU
(constant Q)
dx
d 1 Q20
= ¡
dx 2 C (x)
d 1
Q 20 d
= ¡
dx 2 "0 w [` + (· ¡ 1) x]
!
µ
¶Ã
1 Q20 d
¡ (· ¡ 1)
=
¡
2
2 "0w
[` + (· ¡ 1) x]
= ¡
=
F x (x)
=
(11)
(12)
1 Q 20 d
(· ¡ 1)
2
2 "0 wl [1 + (· ¡ 1) x=`]2
1 Q 20
(· ¡ 1)
2 C0 l [1 + (· ¡ 1) x=`]2
(13)
Dimensionally this is correct, since Q=C 0 = ¢V 0 and Q¢V 0 =l is dimensionally
8
charge£electric ¤eld = force. Fx is positive, so the force does pull the slab in,
as we already concluded. The force decreases as x increases, and is maximum
just as the slab enters the capacitor.
We can aslo write the force (13) as
Fx =
1 C0 Q 20
(· ¡ 1)
1
C 0 Q20
1
C0
= (· ¡ 1)
= (· ¡ 1)
[¢V (x)]2
2
2
2 ` C 0 [1 + (· ¡ 1) x=`]
2
` C (x)2
2
`
(14)
Letting F0 =
2
1 Q0
2 C 0l ;
we plot F =F 0 versus x=`:
1
0.9
0.8
0.7
Fx/F0
0.6
0.5
0.4
0.3
0
0.2
0.4
0.6
x/l
0.8
1
Now what happens if we leave the capacitor hooked up to the battery? The
capacitor is no longer isolated, so the discussion is a bit more complicated. This
time the potential di¤erence stays ¤xed at ¢V = ¢V 0 , and the battery pumps
extra charge onto the plates as the slab goes in. The ¤nal energy is then
Uf =
1
Cf (¢V 0 )2 = ·Ui
2
The energy increases! So the battery adds energy as well as charge. Yet the
slab still gets sucked in. How can this be? Energy balance looks like this:
Work done by battery - work done by ¤elds = change in stored energy.
The work done by the battery to add charge dQ is
dWb at te ry
=
dQ (¢V0 )
=
dC (¢V0 )
dC
2
dx (¢V 0 )
dx
2
=
The work done by the ¤elds on the slab is thus
dW¤ e ld s = dWb att ery ¡ dU =
9
dC
dU
2
(¢V 0 ) dx ¡
dx
dx
dx
But the energy now is
U (x) =
and
1
1 w
C (x) (¢V0 ) 2 = "0 [` + (· ¡ 1) x] (¢V0 ) 2
2
2 d
dU
1 dC
2
=
(¢V0 )
dx
2 dx
So
dW ¤ e ld s
Fx
Fx
·
¸
dC
1
1 dC
2
2
= F x dx =
(¢V 0 ) 1 ¡
dx =
(¢V 0 ) dx
dx
2
2 dx
1 dC
dU
=
(¢V 0 )2 =
(constant ¢V )
2 dx
dx
1
C0
2
=
(· ¡ 1)
(¢V0 )
2
l
(15)
(16)
If we compare (11) and (15) it appears that the force has changed sign. But
because the energy U depends on C; Q and ¢V ; it is not that simple. When
we actually calculate the force in the two cases we ¤nd that it is in the same
direction, although with the battery connected, the force remains constant as
the slab is inserted. Comparing (14) and (16), we see that the force in the
second case equals the initial force in the ¤rst case.
10