Download Electric Fields in Dielectrics

Electric Fields in Dielectrics
Any kind of matter is full of positive and negative electric charges. In a dielectric, these
charges cannot move separately from each other through any macroscopic distance, so when
an electric field is applied there is no net electric current. However, the field does push the
positive charges just a tiny bit in the direction of E while the negative charges are pushed in
the opposite directions. Consequently, the atoms and the molecules comprising the dielectric
acquire tiny electric dipole moments in the direction of E.
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To see the the net effect of all these dipole moments on the macroscopic scale, imagine
smearing all the positive charges into a large uniform charge density +ρ and likewise all the
negative charges into uniform charge density −ρ. Without the electric field, these densities
overlap each other over the whole dielectric, so the net charge density cancels out. But when
we turn on the field, the positive density moves a tiny bit in the direction of E while the
negative density moves in the opposite direction:
−ρ only
positive charges
positive charges
move left
move right
overlap
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+ρ only
As the result of this move, the bulk of the dielectric — where the positive and negative
charges continue to overlap each other — remains electrically neutral. But in a thin surface
layer on the left side of the dielectric there are only negative charges while in a similar layer
on the right side there are only positive charges. Altogether, the dielectric develops surface
charges as shown on the picture below:
neutral
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Note the signs of the surface charges induced by the electric field: On the side of the
positive external electrode, the induced surface charge is negative, while on the side of the
negative external electrode, the induced surface charge is positive. Consequently, the induced
surface charges reduce the electric field in the dielectric from what it would have been without
them:
The black lines here are the electric field lines. There are fewer of them inside the dielectric
then outside it, which indicates a weaker electric field inside the dielectric.
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The surface charges on the dielectric are proportional to the electric field, which is in
turn proportional to the external charges on the capacitor plates. In the Electrodynamics
class (PHY 352 K) you should take in your junior year, you will learn how this works in
some detail. For the moment, let me simply state that for a dielectric slab between parallel
capacitor plates
Qsurface
κ−1
= −
Qplate
κ
(1)
for some constant κ > 1, and therefore the overall charge on one side of the dielectric is
1
Qplate + Qsurface = + × Qplate
κ
(2)
— same sign as the charge on the plate, but smaller. Consequently, the electric field in the
dielectric is κ times weaker than the electric field in the air gaps between the plates and the
dielectric. Specifically, for the plates of area A,
Qplate
,
ǫ0 A
Qplate + Qsurface
1 Qplate
=
×
,
E[in the dielectric] =
ǫ0 A
κ
ǫ0 A
1
× E[in the gaps].
=
κ
E[in the gaps] =
(3)
(4)
(5)
The constant κ — which depends on a dielectric in question but not on the charges — is called
the relative permittivity or the dielectric constant. Table 26.1 on textbook page 841 gives the
values of κ for some common dielectrics; a similar table is available on a nyperphysics web
page http://hyperphysics.phy-astr.gsu.edu/hbase/tables/diel.html .
Eq. (4) has immediate application to capacitors. Suppose almost all the space between
the plates is filled with a dielectric with relative permittivity κ. If this space has width D,
then the potential difference between the capacitor plates is
∆V = D × E[in the dielectric] = D ×
1 Qplate
D
×
= Qplate ×
.
κ
ǫ0 A
κǫ0 A
(6)
As far as the capacitor is concerned, the charge Q it stores is
right
Q = +Qleft
plate = −Qplate
(7)
since its the plates which are connected by wires to the outside world, while the induced
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charges on the dielectric’s surface are not connected to anything. Consequently, the capacitance is
C =
κǫ0 A
Q
=
,
∆V
D
(8)
κ times more than one would get without the dielectric.
For example, consider a ceramic capacitor with plates of area A = 1.03 cm2 separated
by 25.0 µm gap filled with barium titanate, Ba Ti O3 . This particular ceramics has very
high permittivity κ ranging from 1,000 to 10,000; for our example let κ = 2500. Then the
capacitor in question has
C =
κǫ0 A
1.03 · 10−4 m2
= 2500 × 8.85 · 10−12 F/m) ×
≈ 91 nF.
D
25.0 · 10−6 m
(9)
But without the high-permittivity dielectric, this capacitor would have a much smaller capacitance of only a few picoFarads.
⋆
⋆
⋆
Now let’s go beyond the capacitors and consider a point charge +Q stuck in the middle
of a dielectric medium. The electric field of the point charge induces microscopic dipole
moments throughout the dielectric, although the dipole moment density is no longer uniform
in space but decreases with the distance from the charge. Nevertheless, the collective effect of
all the microscopic dipole moments is to move the negative charges towards the point charge
while the positive charges move outward; these charges cancel in the bulk of the dielectric,
but there is a net negative screening charge
Qscreening = −
κ−1
×Q
κ
(10)
immediately surrounding the point charge Q, and the opposite positive charge
Qsurface = +
on the outer surface of the dielectric:
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κ−1
×Q
κ
(11)
Let’s assume the dielectric piece has large size and focus on the neighborhood of the point
charge Q, where we are much closer to Q than to the dielectric’s surface. In this case, the
electric field is dominated by the Q and by the screening charge immediately surrounding it,
thus
E =
Q + Qscreening ~ır
Q
~ır
× 2 =
× 2,
4πǫ0
r
4πκǫ0 r
(12)
1/κ of the Coulomb field charge Q would create in the vacuum.
A point of terminology: In dielectric context, a free charge is a charge outside the
dielectric (as on a capacitor plate) or put inside the dielectric (like an ion into water), but
not not a charged induced in the dielectric itself by the electric field. Thus, the Coulomb
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field in a dielectric is
E =
Qfree
~ır
× 2,
4πκǫ0 r
(13)
and the Coulomb force between two free charges is
free
Qfree
1 × Q2
F =
.
4πκǫ0 × r 2
(14)
In particular, an ion dissolved in water has free charge
Qfree = e × valence,
(15)
for example Qfree = +2e for the calcium Ca++ ion or Qfree = −e for the chlorine Cl− ion.
Thus, the Coulomb force between two such ions is
e2
valence1 × valence2
F =
×
,
4πκǫ0
r2
(16)
for κ(water) ≈ 80.
In terms of free charges, the Gauss Law for a uniform dielectric becomes
ZZ
1
× Qfree [enclosed within S].
ΦE [S] =
E · d2 Area =
κǫ0
(17)
S
This formula helps to find electric fields in a symmetric setups. For example, consider a
coaxial cable where the inner conductor has charge +Q, the outer conductor has charge −Q,
and the space between the conductors is filled with a dielectric with relative permittivity κ,
then the electric field in the dielectric is
E =
Q/L
1
×
2πκǫ0
r
(18)
where L is the cable’s length. Consequently, the voltage between the conductors is
Q/L
∆V =
×
2πκǫ0
ZR2
dr
R2
Q/L
× ln
=
r
2πκǫ0
R1
(19)
R1
where R1 is the outer radius of the inner conductor while R2 is the inner radius of the outer
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conductor. Consequently, the cable has capacitance
C =
Q
2πκǫ0
=
× L.
∆V
ln(R2 /R1 )
(20)
For example, for R2 = 4.0 × R1 and polystyrene dielectric with κ ≈ 2.5, we get
C
≈ 1.0 pF/cm.
L
(21)
Note: in MKSA units, the relative permittivity κ of a dielectric always appears in combination κ × ǫ0 . This product is called the absolute permittivity of the dielectric. In this
context, the vacuum permittivity ǫ0 is simply the absolute permittivity of vacuum viewed as
just another dielectric!
A point of notation: In many books and articles, the relative permittivity of a dielectric
is denoted ǫ rather than κ. But sometimes ǫ is used for the absolute permittivity κ × ǫ0 ,
although this is quite uncommon in modern literature.
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Let me conclude these notes by defining the electric displacement field
D = κǫ0 E,
(22)
D = ǫ0 E + P
(23)
or more generally
where P is the density of the induced dipole moment,
P =
dipole moment vector
.
volume
In terms of the displacement field D, the Gauss Law in a dielectric becomes
Z
D · d2 Area = Qfree [enclosed within S].
(24)
(25)
S
This formulation of the Gauss Law is quite general and works for all closed surfaces S, even
if part of it is inside a dielectric an another part is outside. Or if several dielectrics with
different κ’s are involved, we may have different parts of S be inside different dielectrics. In
particular, letting S be a squat cylinder spanning a border between two dielectrics
D1
S
D2
and applying the Gauss Law (25) tells us that for the electric field ⊥ to the border,
D1 = D 2
=⇒
κ1 E1 = κ2 E2 .
(26)
Finally, for non-uniform dielectrics where κ varies from place to place, it is convenient
to use the differential form of the Gauss Law for D(r) = κ(r)ǫ0 E(r):
∇ · D = ρfree .
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(27)