DIELECTRICS – Macroscopic Point of View
Y&F Chapter 24 Capacitance & Dielectrics
Dielectric are insulators – charges tend not to move easily in non-metallic solids
??? How does an object acquire a charge?
??? Why does a charged object attract a neutral object?
??? How do we investigate the electrical properties of an insulator or dielectric material?
Inserting a dielectric between the plates of a capacitor  capacitance increases
BACKGROUND CAPACITORS
Capacitors – two conducting plates separated by an dielectric
Capacitors – uses (basic component of most electronic circuits): timing circuits, filtering,
smoothing fluctuating voltages, transmission of ac signals, resonance circuits, flash lights in
cameras, pulsed lasers, air bag sensors, ac circuits, etc
Capacitor – stores charge on conducting plates, stores electric potential energy due to the
work done is separating the charges.
Capacitance – depends only on geometry of capacitor and dielectric – “ability” to store
charge
Q
C
V
??? What is meant by potential difference (voltage)?
??? What produces an electric field?
Dielectric in capacitors
1 prevent plates touching  small separation  higher C
2 reduction in electric field  same charge at a lower voltage
3 reduction in shorting by sparking  higher dielectric breakdown voltage
Charge on plates  electric field between plates
E  V
 Ex  
V
x
f
V  V f  Vi    E  dl
i
Parallel plate capacitor – no dielectric (vacuum or air)
E=V/d
Gauss’s Law
V=Ed
E 
 E  dA 
qenclosed
o

q free
o
physics/p2/emdielectric_1.doc
1
E A
V A q free

d
0
 C
E
q free

V
 free
0
0 A
d
V
d  free
0
q free   free A
Energy stored and energy density (energy stored by the field)
1 Q2 1
1
1
1
U
 CV 2  QV   0 E 2 Ad  u   0 E 2
2 C 2
2
2
2
Proof
Work done to charge capacitor
q dq
1 Q
1 Q2
dW  v dq 
W   q dq 
C
C 0
2 C
The operation of assembling upon a conductor a group of charges that mutually repel one
another requires work and therefore results in the production of potential energy - this
potential energy is possessed by the charged conductor itself but it may be more correct to
picture the energy stored in the field surrounding the conductor.
??? How is the charge located on the plates of a capacitor?
Electric displacement or electric flux density
D   0 E   free
E
D
0
Capacitors in series (charge on
each plate is the same)
1
Ctotal 
1
1

 ...
C1 C2
Capacitors in parallel (voltage across each
capacitor is the same)
Ctotal  C1  C2  ...
Parallel plate capacitor – isotropic dielectric inserted that fills space between plates
0   = r 0
Permittivity of the dielectric
  r 0
Capacitance increases when dielectric is present – redistribution of the charges in the
dielectric material (polarization).
Dielectric increases the breakdown voltage (can use a larger V)  dielectric strength – if the
voltage becomes too large a disruptive discharge occurs.
Fixed charge on plates of capacitor (Q = Qfree same value without or with dielectric inserted)
Dielectric fills space between the capacitor plates.
physics/p2/emdielectric_1.doc
2
Insertion of dielectric  reduction in V and E  increase in capacitance
Capacitance – no dielectric
C0
Capacitance – with dielectric Cd
Dielectric constant or relative permittivity  r or K
r  K 
Vd 
Cd
>1
C0
V0
Ed 
r
Cd   r Co 
r 0 A

A
d
d
capacitance increases with insertion of dielectric
E0
D  Dd  D0   free   f
r
??? How can we explain the reduction in the electric field between the capacitor plates?
Reduced electric field – net charges induced on the surface of the dielectric (dielectric still
neutral) – polarization. The polarization does not change the electric displacement
P   bound   b
E  Ed 
Ed 
1
0
1
0
D   free   f
( D  P) 
1
0
( free   bound ) 
( free   bound )
E0
r

1
  bound   free 1  
 r 
 D   r  0 Ed   Ed
Lines of D connect free
charges (positive to
negative)
Lines of P connect bound
charges (negative to
positive)
Lines of E connect net
charges (free & bound:
positive to negative)

 free  free

r 0

V
 free d

C
A
d
 bound   free
+q free on inner surface
P
+ + + + + + + + + + + +
-q bound
Symmetry
– fields must be uniform
– field lines perpendicular to plates
+q bound
- - - - - - - - - - - -
D
In isotropic materials: D ,
E and P all have the same
direction
-q free on inner surface
Interior points electric
field must be zero
physics/p2/emdielectric_1.doc
E
E
1
0
( D  P)
3
conductor
+
- dielectric
Gauss’s Law
+
E 0
-
+
 free  bound
+
-
E

free
  bound 
0
E
+
Isotropic materials – assume polarization proportional to the electric field
P  e  0 Ed
  r  1  e
Electric displacement D is a useful quantity, since its value only depends upon the density of
free charges and in not changed by the introduction of the dielectric.
D   E  Gauss’s Law
 D
dA    f dv  q f
 D  f
D represents the flux density
Gauss’s Law – the total electric flux emerging from any closed surface is equal to the net
positive charge enclosed within the surface.
Maxwell’s displacement current
Switched closed – current through conductors to charge
capacitor. Maxwell showed that it is necessary to
assume a current of the same value also flowed in the
space between the capacitor plates.
d f
dt

-q
+q
dD
dE dP
 0

dt
dt dt
dP/dt rate of change of polarization – associated with the actual motion of charges in the
dielectric: rotation of permanent dipoles or induced dipoles – displacement of charges
– posses a current character.
0
dE
current associated with change in electric field strength even when a vacuum is
dt
between the plates.
physics/p2/emdielectric_1.doc
4
Parallel plate capacitor with compound dielectric
Portion of the inter-plate region occupied by a dielectric slab of thickness t, the rest of the
region being air filled
C
0 A
d  t  
t
r
??? How do you prove this ?
Forces between parallel plates / energy in electric field
The charges on the plates are all on the inner
surfaces  unbalanced electrical attractive force F
+ + + + + + + + +
between the plates.
Suppose the top plate is raised a distance dy. The
work done is
dW = F dy
dy
+ + + + + + + + +
F
- - - - - - - - -
The energy thus expended can be regarded as stored in the electric field between the plates of
the capacitor
The additional volume is dv = A dy, therefore the increase in the energy density. Energy is
stored in an electric field if the field exists in a dielectric medium because the rotation or
creation of polar molecules represents the production of potential energy. However,
1
u   0 E 2 implies that even empty space when pervaded by an electric field contains energy.
2
dW 1
  E2
dv 2
1
1
dW   E 2 dv   E 2 Ady  F dy
2
2
1
V2 A
F   E2A 
2
2d2
Lord Kelvin measured the pull between the two charged plates using a beam balance 
measurement of potential V.
Dielectric constant and dielectric strength
Dielectric Dielectric
constant
strength
V.m-1
Air
1.00058
~ 3106
Water
80
Mica
7-8
Polystyrene
2.5
Titanium dioxide ceramics
15 - 500
~ 2107
Barium titanate
500 - 6000 ~ 2106
physics/p2/emdielectric_1.doc
5
DIELECTRIC BEHAVIOR
Insulator (dielectric) – will not permit a current (passage of charge) although local
microscopic displacements of charge may take place.
All atoms or molecules can be temporarily polarized under the action of an electric field.
Permanent polar molecules (H+Cl-) may exist in a dielectric.
Separation of charge  electric dipole
A dipole consists of two equal and opposite charges + and –q separated by a vector distance d
p  qd
dipole moment p = q d
d
points from negative to positive
-q
+q
Induced dipole moment – helium atom
E
-e
+2e
Zero electric field –
helium atom
symmetric  zero
dipole moment
-e
-e
+2e
A
-e
d
B
effectively charge +2e at A and -2e at B
dipole moment
p = 2ed
p
physics/p2/emdielectric_1.doc
6
Potential and electric field from an electric dipole
V ( P) 
1 q q
q  r2  r1 
  


4   r1 r2  4   r 1r2 
Er
E
r  d
P
V ( P) 

q d cos 


d2
4   r 2  cos2  
4


r2  r + (d/2)cos
q d cos  p cos 
p r


2
2
4  r
4  r
4  r 3
r1  r – (d/2)cos
r
r extends from the centre of the
dipole to the point P
(d/2)cos
The radial and tangential components
of the field at point P are
Er  
V 2 p cos 

r
4  r 3
E  
1 V p sin 

r  4  r 3

-q
along the axis of the dipole
along the right bisector of the dipole
+q
d
 = 0  E = 0
 = /2  Er = 0
Electric field approaches zero much more quickly than a point charge. ??? Why ?
See Matlab plot page 10
Electric polarization and electric susceptibility
The extent to which permanent or induced dipoles become aligned is described by the
electric polarization P
dp
dv
Consider, polarization of the dielectric between the plates of a charged parallel plate capacitor
electric dipole moment per unit volume
P
dA
+ + + + + + + + +
d

For a cylinder of the dielectric of cross-sectional area
dA extending from one plate to the other 
+f



Throughout the body of the dielectric, the charges on
adjacent ends of the polar molecules neutralize one
another. At both the top and bottom of the dielectric
the charges do not neutralize each other  bound
surface charges b .
-f
physics/p2/emdielectric_1.doc
-b
+b
- - - - - - - - -
7
electric dipole moment p = q d
dp  ( b dA)d
P
( dA)d
dp
dp

 b
dv dA d
dA d
P  b
P   b nˆ
Polarization equals the magnitude of the bound (induced) charge per unit area on the surface
of the dielectric.
For isotropic dielectrics, (electrical properties identical in all directions) the polarization that
occurs due to the applied electric field has the same direction as the field. Also, the
magnitude of the polarization is proportional to the field
P  e  E
where  e is the constant of proportionality, known as the electric susceptibility.
For anisotropic dielectric P and E are not in the same direction and  e is not a constant but
a tensor.
Electrets Isotropic dielectric material - susceptibility is not constant – e.g. electrets
microphones
Frequency response of the dielectric constant
The capacitance of any capacitor is directly proportional to the dielectric constant of the
material between the capacitor plates. Hence, the dielectric constants of two materials can be
readily compared by introducing the materials, in turn, into a given capacitor and determining
the resulting capacitances. For a given material, the change in dielectric constant as a function
of pressure, temperature, or some other variable can be measured with high precision by
employing the material-filled capacitor as the capacitive element in a tuned circuit. If the
circuit is sharply resonant, a small change in the capacitance of the capacitor results in a
significant change in the resonant frequency of the circuit. By this means, for example, even
the small changes in the dielectric constants of gases which occur when the temperature is
altered have been accurately studied.
When a DC voltage is applied to a capacitor, the polar molecules in the dielectric orient
themselves under the action of the electric field. When the applied voltage is an alternating
one, the polar molecules again attempt to line up with the field and are, in fact, equally
successful if the frequency of the AC voltage is low. As the polarity of the voltage changes,
the polar molecules obligingly change their direction. when the frequency of the applied field
is high, however, the polar molecules may not have time to orient themselves to the same
extent before the polarity changes. For this reason, in a material that possesses permanent
polar molecules, the dielectric constant decreases with increasing frequency. If, on the other
hand, the polar molecules in the dielectric are induced ones, resulting from a displacement of
physics/p2/emdielectric_1.doc
8
the planetary electron systems there is no observed decrease with increasing frequency,
because this displacement is practically
instantaneous.
In most materials, both permanent and induced
polar molecules contribute to the polarization.
The dielectric constant of water falls from its
low frequency value of 80 to less than 2 at
optical frequencies (~1014 Hz).
dielectric
constant
frequency
Refractive Index
Maxwell  prediction of electromagnetic waves
Electromagnetic waves  time-varying electric and magnetic fields whose directions are
mutually perpendicular. In unbounded dielectric media the waves are transverse.
Velocity of propagation of em waves depends upon the electric and magnetic properties of
the medium. For an unbounded medium
1
v

For non-magnetic materials
1
1
v

 0
 r  0 0
For a vacuum
c
1
 0 0
The change in its velocity as it passes from one medium to another is responsible to
refraction. Refractive index n (non-magnetic materials)
n
  
c
 r 0 0
v
 0 0
n  r
This prediction of Maxwell’s electromagnetic theory originally served as a basis for
criticizing the theory, for example DC values for air and water
air
n = 1.000294
water n = 1.3
 r = 1.000295
 n  r
 r = 8.9
 n  r
It was not known at the time that water contained permanent polar molecules and as a result
the value of  r decreases with increasing frequency. The polarization of the air molecules is
entirely due to the displacement under the action of the applied electric field of the electron
physics/p2/emdielectric_1.doc
9
clouds of their constituent atoms – since this displacement occurs with great rapidity, r
displays no frequency dependence.
When the frequency is comparable to the orbital frequency of the electrons in the material,
absorption and emission can take place – the index of refraction can display appreciable
frequency dependence, e.g., dispersion of visible light in passage through a glass prism.
Ferroelectricity
crystalline dielectric materials – permanent electric polarization (analogy with ferro-magnetic
materials – it has nothing to do with iron)
Rochelle salt, potassium dihydrogen phosphate and barium titanate
Piezoelectricity
crystalline dielectric materials – mechanical pressure exerted upon the crystal results in
appearance of electric charge along its surface (polarization effect)
mechanical stress  emf
emf  mechanical stress – change in dimensions of crystal
Quartz crystal – mechanically vibrating  ac voltage across its face – frequency of
mechanical vibration depends upon the dimensions and other parameters of the crystal  ac
voltage generated possesses a very constant frequency.
ac voltage applied to the faces of crystal – if frequency of the voltage is identical to natural
frequency of crystal’s mechanical vibrations  crystal vibrate “violently”  ultrasonic
waves
transducer: electrical energy  mechanical energy.
Pyroelectricity
crystal heated or cooled  changes in charge at the surface – these are piezoelectric changes
from the strain associated with thermal expansion or contraction.
Electric Dipole PLot– Matlab
??? Why is difficult to plot the potential in a plane passing through the axis of the diopole?
Potential: Electric Dipole
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
physics/p2/emdielectric_1.doc
10
% electric_dipole.m
% Ian Cooper School of Physics,University of Sydney
close all
clear all
clc
% emconstants ------------------------------------------------------c = 3.00e-8;
% speed of light
e = 1.602e-19;
% elementary charge
eps0 = 8.85e-12;
% permittivity of free space
NA = 6.02e23;
% Avogadro constant
me = 9.11e-31;
% electron rest mass
mp = 1.673e-27;
% proton rest mass
mn = 1.675e-27;
% neutron rest mass
h = 6.626e-34;
% Planck's constant
kB = 1.38e-23;
% Boltzmann's constant
kC = 8.988e9;
% Coulomb constant
mu0 = 4*pi*1e-7;
% permeability of free space
amu = 1.66e-27;
% atomic mass unit
% Setup ------------------------------------------------------------q = e;
% dipole charge
d = 1.6795e-018;
% dipole separation distance
q1 = q; q2 = -q;
% separated charges
kc = 1/(4*pi*eps0);
% constant in Coulomb's Law
x1 = d/2; x2 = -d/2;
% position of dipole
y1 = 0; y2 = 0;
scale = 1.25;
% plotting region
xmax = scale * d;
ymax = xmax; xmin = -xmax; ymin = -ymax;
% plane above diople
num = 100;
x = linspace(xmin,xmax,num);
y = x;
[xx yy] = meshgrid(x,y);
r1 = sqrt((xx-x1).^2 + (yy-y1).^2);
% distance from charges
% to test point to calc. potential
r2 = sqrt((xx-x2).^2 + (yy-y2).^2);
V1 = kc .* q1 ./ (r1);
% potential from each charge
V2 = kc .* q2 ./ (r2);
Vtot = V1 + V2;
Vmax = max(max(Vtot));
sat = 0.5;
% saturate the potential
Vtot(Vtot > sat*Vmax) = sat * Vmax;
% potential near a charge
%
is extremely large
Vtot(Vtot < -0.5*Vmax) = -sat * Vmax;
Vtot = Vtot/(max(max(Vtot)));
figure(2);
% [3D] plot
surf(xx/d,yy/d,Vtot,'FaceColor','interp',...
'EdgeColor','none',...
'FaceLighting','phong')
daspect([1 1 1])
axis tight; view(-45,20)
camlight left; colormap(jet)
grid off; axis off
colorbar
title('Potential: Electric Dipole')
physics/p2/emdielectric_1.doc
11