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PH504
Dielectrics
Introduction
In an insulator, an electron is attached to a particular atom.
When certain non-conducting materials are used to fill the space
between the two conductors of a capacitor the capacitance is
found to increase. Such materials are known as dielectrics.
Relative permittivity: If the capacitance of a capacitor is Co
when the region between the conductors is a vacuum and Cm
when it is filled with a given dielectric then the ratio
Cm/Co= r
defines the relative permittivity or dielectric constant of the
dielectric.
Dielectric strength: Dielectrics normally act as insulators but
above a given electric field (known as their dielectric strength)
their insulating property breaks down and they start to conduct.
This limits the maximum potential allowed between the two
conductors and hence from Q=CV the maximum charge and
energy that can be stored.
Polarization of Dielectric
If a material contains polar molecules, they will generally be in random
orientations when no electric field is applied. An applied electric field
will polarize the material by orienting the dipole moments of polar
molecules:
This decreases the effective electric field between the plates and will
increase the capacitance of the parallel plate structure. The dielectric
must be a good electric insulator so as to minimize any DC leakage
current through a capacitor
Parallel Plate with Dielectric
The capacitance of a set of charged parallel plates is increased by
the insertion of a dielectric material. The capacitance is inversely
proportional to the electric field between the plates, and the presence
of the dielectric reduces the effective electric field. The dielectric is
characterized by a dielectric constant k, and the capacitance is
multiplied by that factor.
Dielectrics represent a class of materials which, although
insulators, exhibit a number of effects when placed in an E-field. A
good example is their effect on capacitors.
A capacitor has capacitance C0 when the space between its two
conductors is a vacuum, filling this space with a dielectric
increases the capacitance to a new value Cm. The ratio Cm/C0=r is
known as the relative permittivity of the dielectric.
The plates of a capacitor are charged to some value Q and then
isolated. Because of the relationship Q = CV, if C increases when
a dielectric is placed between the conductors, as Q must remain
constant V must decrease. Because V is related to the E-field by
E=-V, the introduction of the dielectric must decrease E by r.
The aim of this lecture is to understand why this occurs.
LIH dielectrics
In order to simplify our treatment we will consider a sub-class of
dielectrics known as LIH ones. Such dielectrics exhibit the
following properties.

Linearity (LIH): All properties are proportional to the magnitude
of any applied E-field. This implies that r is a constant and is
independent of applied voltage or E-field.

Isotropy (LIH): r is independent of the orientation of the
material i.e. it is independent of the direction of the E-field.

Homogeneity (LIH): r has the same value at all points in the
dielectric.
Static relative permittivity: For the majority of dielectrics,
r is found to vary with temperature and the frequency (if any) at
which the applied E-field is modulated. The value of r in a steady
E-field is known as the static relative permittivity. It is this value
that is relevant to electrostatic situations.
The effect of an E-field on a dielectric (qualitative
treatment)
Because dielectrics are insulators they contain no free charges
(charges which are able to move around freely within the material
when an E-field is applied). However although unable to move far,
the positive and negative charges may be able to move slightly
(but in opposite directions) when an E-field is applied.
Consider an atom which consists of negatively charged electrons
orbiting the positively charged nucleus. In general the centroids of
the positive and negative charges coincide and the atom produces
no external E-field.
Neither the electrons or the nucleus is free to move over large
distances when an E-field is applied.
However the application of an E-field will cause the electrons to be
moved, on average, a small distance in one direction and
(possibly) the nucleus to move in the opposite direction.
The centroids of the positive and negative charges now no longer
coincide and we have an electric dipole which we know produces
an external E-field. In this case the atoms are said to be
polarised.
Referring to the above diagram. For no external field the atoms are
unpolarised (centroids of q coincide) and hence they produce no
external E-field. When an external field (Eext) is applied the atoms
are polarised and form small electric dipoles.
In the present case the field moves the positive charges upwards
and the negative ones downwards. As their centroids no longer
coincide they produce an external field E*. Because E-fields point
from positive to negative charges, in the present case E* will point
downwards and hence will oppose Eext. The total E-field within the
dielectric will hence be less than the external field (Eext-E*)
explaining our previous observation that the E-field between the
conductors of a capacitor decreases when a dielectric is
introduced.
The effect of an E-field on a dielectric (quantitative treatment)
Polarisation. An applied E-field causes the atoms or molecules in
a dielectric to become small electric dipoles. The polarisation P at
any point in the dielectric is defined as the electric dipole
moment per unit volume at that point.
Hence if we have a small volume d it will contain a total dipole
moment dp of Pd. This is the definition of the polarisation P (units
Cm-2).
Consider now
a small volume element of length dL and cross-sectional area dS.
The application of an E-field causes the displacement of positive
charge (+dq) out of one end of the element and an equal but
negative charge (–dq) out of the other end.
This results in a dipole moment (p=sQ) of dp=dq.dL.
If instead of dq we use a surface charge density  dq=.dS
and hence dp=.dS.dL
But dS.dL is simply d, the volume of the element and from the
definition of P
P = dp/d  P = 
So polarisation charge (dq) crossing area dS is PdS.
More generally if the area dS is not normal to the polarisation
vector P then
Polarisation charge crossing dS: = PdS
(A)
Surface polarisation charges
Consider a slab of dielectric in an applied E-field. Along the field
direction we can consider the dielectric in terms of a sequence of
small elements. If the dielectric is homogeneous (P is the same at
all points) then the positive polarisation charge at the end of an
element is balanced by an equal but opposite negative charge at
the end of the neighbouring element.
Only at the surface of the dielectric is there an unbalanced charge.
From above if Pn is the normal component of P at the surface then
a surface polarisation charge density (P) exists given by
P = Pn.
Volume Effects
If E is not uniform, then P(x,y,z).Or, if the material is not
homogeneous then there will also be a volume polarisation
charge. The total charge density accumulated per unit volume is:
b = - P
since the excess charge leaving a small volume is
(P)(dx dy dz).
Potential
Therefore, the potential produced by a polarized object can be split
into the potentials produced a surface charge density and a
volume charge density. Previously we had for a point charge:
So we need to replace Q with surface and volume terms,
and integrate over the surface and volume.
Gauss’s Law in dielectrics, D-field
Consider a general situation in which we have both free charges
(labelled Qf) and charges due to the polarisation of one or more
dielectrics (labelled QP). Gauss’s law can still be applied but we
must now take account of both the free and polarisation charges.
Applying Gauss’s law to the surface S

Note that any dielectric contained entirely within S (e.g. A) does
not contribute to the surface flux as it contributes equal amounts of
negative and positive flux.
Dielectric B however, which is cut by the surface S, will give a
contribution to the flux through S as in general there will be a net
polarisation charge crossing S.
From (A) above the polarisation charge crossing a surface element
dS is PdS. Hence the total polarisation charge which leaves the
volume enclosed by S is
(the integration can be taken over the whole of S as P=0 for all
points on S which are outside of the dielectric).
The above gives the polarisation charge which has moved
outwards through S. Therefore the polarisation charge which
remains within S (and is hence relevant for Gauss’s law) is the
negative of this value.
Hence Gauss ’s law is altered to:
/
Rearranging:
o
because the term 0E + P occurs very frequently it is given a
special name, the electric displacement or D-field, symbol D
D=0E+P .
The units of D are C m-2.
Gauss’s law in terms of D is now
(integral form) .
In words, this states ‘the outward flux of D over any closed
surface S equals the algebraic sum of the free charges enclosed
by S’.
The differential form of Gauss’s law for D is
where f is the volume density of free charges.
From the form of the above equations we see that while both
free and polarisation charges are sources of E, only free charges
are sources of D.
Because D is given by a modified form of Gauss’s law techniques
used to find E for a given charge distribution can also be used to
find D.
In free space (no dielectric) P=0 and hence D=0E.
Example. The D-field a distance r from a point charge Q is
Relationships between E, D and P
Electric susceptibility.
If the E-field at a point within a dielectric is E and the
polarisation at that point is P then the electric susceptibility (e)
is defined via the equation
P=0eE
For an LIH dielectric e is independent of the size and direction
of E and position within the dielectric.
Because D=0E+P D=0(1+e)E
(B)
Relationship between e and r
Consider a system of free charges which produce an E-field and
hence a D-field. If we now introduce a dielectric of relative
permittivity r we know that D can not change and hence (B)
above tells us that within the dielectric E must decrease to
1/(1+e) of its value in vacuum.
However from our definition of r in terms of capacitance we
know that introducing a dielectric results in a decrease in E by a
factor r. Hence we must have the general result
r = 1+e
and hence we can also write
D=r0E and P=0(r-1)E
Sometimes  = r0 is used where  is the permittivity of the
dielectric.
Summary of E, D and P
The above diagram summarises the results. A parallel plate
capacitor carries charge densities of f. A dielectric partially
fills the volume between the plates.
 D has the same value in the dielectric and in the vacuum.
 The polarisation P is only non-zero in the dielectric.
 The polarisation P results in surface polarisation charges P
on the top and bottom surfaces of the dielectric.
 The polarisation surface charges produce an additional Efield, Ep which opposes the vacuum field Evac to give a reduced
total field Em within the dielectric.
 Lines of D pass through the dielectric
 Some lines of E terminate at the polarisation surface charges.
Effect of dielectrics on the E-field circuital law
in vacuum. Charges make no contribution to
the integral of E around a closed path and this is true for both
free charges and polarisation charges. Hence this result is still
valid in a dielectric
or
in a vacuum or dielectric (but not when there is a time varying
magnetic field!!).
General electrostatic equations in the presence of
dielectrics
In general all the previously derived equations for V and E can be
used if 0 is replaced by 0r. (This is only strictly true for LIH
dielectrics).
For example if a point charge Q is placed in a dielectric then the
electric potential V, at a distance r is given by
The only equation which can not be modified in this manner is
Coulomb's equation for electrostatic forces. This is because to
place a charge in a dielectric we would have to make a small
hole in which to place the charge. Hence the charge would not
strictly be in the dielectric. The solution of this problem is not
trivial but fortunately we are generally interested in E-fields not
forces.
．
Boundary conditions for D and E
These are required for problems where we are concerned with
what happens at the interface between two different media (e.g.
vacuum and a dielectric, two different dielectrics etc).
Consider the interface between two media 1 and 2 (these can be
vacuum, conductors or insulators). For generality we assume
there is a free charge density f at the interface. E- and D-fields
are as shown in the diagram.
For a Gaussian surface we take the cylinder shown. If its height is
made infinitesimally small then we need only consider the flux
through the two ends.
Applying Gauss’s law for D:
D1dS-D2dS=f dS
In the limit dS0 D1n- D2n = Dn = f
where Dn is the component of D normal to the surface.
This result shows that the normal component of the D-field is
discontinuous by f across any surface. If the free charge is zero,
f=0, then the normal component of D is continuous.
For the E-field take the rectangular path as the closed loop. If its
height is made infinitesimally small then the only contribution to
the line integral is along the top and bottom edges
where Et is the tangential component of E.
Hence E1t = E2t; the tangential component of E is continuous
across any interface.
Now consider a general case where an E-field (and hence D-field)
passes from a first dielectric to a second.
If there is no surface charge at the interface then we must have
the following
(E-tangential)
(C)
(D-normal)
(D)
But D1=0rE1 and D2=0rE2
Hence we can write
(E)
Finally dividing equation (E) by (C)
Note: Refraction of lines of D and E at a surface.
Note: V is continuous across the interface even in the presence
of free charge.
Stored energy in terms of E and D
In the absence of dielectrics, we have shown that electrical
potential energy (U) could be written in terms of the E-field (E)
as
This equation is modified in the presence of dielectrics.
For a parallel plate capacitor filled with a dielectric having a
relative permittivity r we have the following
Hence
But
Hence
This can be shown to be a general result.
Hence for situations where E (and D) are not uniform
The dot product of D and E is required because for nonLIH dielectrics D and E may not be parallel to each other.
Values:
r = 1.00059 air
= 2–3
ice
= 5 -10
glass
= 80
water
= 4.69 (perp) or 5.06 (parallel) quartz
Stress on a Conductor (Force on dielectric)
Suppose we try to separate two conductors. We have seen that
Therefore, at constant Q (open circuit)
U = Q2x / ( 2A) where x is the plate spacing.
Hence the force is:
Fx = - dU/dx = - Q2 / ( 2A)
Hence the force per unit area (stress) is:
Fx = - Q2 / ( 2A2) = - 2/ 2
- exerted by the conductor.
At constant V (closed circuit), energy supplied by
battery = VdQ = V2 dC = (Q2/C2) dC.
However, change in U = CV2/2 is dU = (Q2/2C2) dC.
Battery supplies equal amounts of energy to increase
internal energy and do external work.
Examples:
1. Two long coaxial cylindrical tubes, length L,
inner radius a, outer radius b, stand vertically in
a tank of water. The inner one is maintained at
potential V and the outer one is grounded. To
what height h does the water rise in the (narrow)
space between the tubes? Take r = 80, V= 300v,
d = a-b = 1mm.
For parallel capacitors:
C = Cair + C water
= (2r (L-h)) o / d + (2r h)  / d
U = CV2/2 = (2r (L + h(r - 1)) o V2 / 2d
The force on the liquid is:
Fh = (U/h )V
= 2r (r - 1) o V2 / 2d
WHY??? Because of the ‘Fringing Field’:
As the dielectric is introduced into the fringe field, the
dielectric molecules start to align with the external field. This
creates an attractive force as the plus charges of the
dielectrics are aligned so that they are closest to the negative
plate of the capacitor (and of course the same process occurs
for the negative dielectric charges and positive plate of the
capacitor).
The horizontal components of the fringe field yield a
horizontal, attractive force which pulls the dielectric into the
space between the capacitor plates.
The work done by this force in accelerating the dielectric into
the capacitor volume decreases the energy stored in the
capacitor, thereby lowering the energy stored in the capacitor.
Some energy is also converted into electromagnetic radiation
and into heat.
This force balances the weight of liquid: 2 r d gh
Finally:
h = (r - 1) o V2 / 2d2g
(independent of L). Values yield 3.2 mm.
Conclusions










Effects of a dielectric on E, V and C
LIH dielectrics
Qualitative explanation of dielectrics
Quantitative explanation of dielectrics, definitions of P
Surface polarisation charge
Electric displacement (D-field)
Gauss's law for D-fields
Relationships between E, D and P (r, e)
Electrostatics in the presence of dielectrics
Boundary conditions for D and E
 Electrical potential energy in terms of D and E