Transcript Document

EEE 498/598
Overview of Electrical
Engineering
Lecture 5:
Electrostatics: Dielectric Breakdown,
Electrostatic Boundary Conditions,
Electrostatic Potential Energy;
Conduction Current and Ohm’s Law
1
Lecture 5 Objectives
To continue our study of electrostatics
with dielectric breakdown, electrostatic
boundary conditions and electrostatic
potential energy.
 To study steady conduction current and
Ohm’s law.

2
Lecture 5
Dielectric Breakdown

If a dielectric material is placed in a very
strong electric field, electrons can be torn
from their corresponding nuclei causing
large currents to flow and damaging the
material. This phenomenon is called
dielectric breakdown.
3
Lecture 5
Dielectric Breakdown (Cont’d)
The value of the electric field at which
dielectric breakdown occurs is called the
dielectric strength of the material.
 The dielectric strength of a material is
denoted by the symbol EBR.

4
Lecture 5
Dielectric Breakdown (Cont’d)
The dielectric strength of a material may
vary by several orders of magnitude
depending on various factors including the
exact composition of the material.
 Usually dielectric breakdown does not
permanently damage gaseous or liquid
dielectrics, but does ruin solid dielectrics.

5
Lecture 5
Dielectric Breakdown (Cont’d)
Capacitors typically carry a maximum
voltage rating. Keeping the terminal
voltage below this value insures that the
field within the capacitor never exceeds
EBR for the dielectric.
 Usually a safety factor of 10 or so is used
in calculating the rating.

6
Lecture 5
Fundamental Laws of
Electrostatics in Integral Form
Conservative field
 E  dl  0
Gauss’s law
C
D

d
s

q
dv
ev


S
V
D E
Constitutive relation
7
Lecture 5
Fundamental Laws of Electrostatics
in Differential Form
Conservative field
 E  0
  D  qev
Gauss’s law
D E
Constitutive relation
8
Lecture 5
Fundamental Laws of
Electrostatics


The integral forms of the fundamental laws are
more general because they apply over regions of
space. The differential forms are only valid at a
point.
From the integral forms of the fundamental
laws both the differential equations governing
the field within a medium and the boundary
conditions at the interface between two media
can be derived.
9
Lecture 5
Boundary Conditions

Within a homogeneous medium, there are
no abrupt changes in E or D. However, at
the interface between two different media
(having two different values of ), it is
obvious that one or both of these must
change abruptly.
10
Lecture 5
Boundary Conditions (Cont’d)

To derive the boundary conditions on the
normal and tangential field conditions, we
shall apply the integral form of the two
fundamental laws to an infinitesimally
small region that lies partially in one
medium and partially in the other.
11
Lecture 5
Boundary Conditions (Cont’d)

Consider two semi-infinite media separated by a
boundary. A surface charge may exist at the
interface.
Medium 1
rs
xxx x
Medium 2
12
Lecture 5
Boundary Conditions (Cont’d)

Locally, the boundary will look planar
1
2
E1 , D1
xxxxxx
rs
E 2 , D2
13
Lecture 5
Boundary Condition on
Normal Component of D
• Consider an infinitesimal cylinder (pillbox) with
cross-sectional area Ds and height Dh lying half in
medium 1 and half in medium 2:
Ds
Dh/2
Dh/2
1
E1 , D1
aˆ n
r
xxxxxx s
2
14
E 2 , D2
Lecture 5
Boundary Condition on Normal
Component of D (Cont’d)

Applying Gauss’s law to the pillbox, we have
 Dds  q
S
ev
dv
0
V
LHS   D  d s 
top
 Dds   Dds
bottom
side
 D1n Ds  D2 n Ds
RHS  qes Ds
15
Lecture 5
Boundary Condition on Normal
Component of D (Cont’d)

The boundary condition is
D1n  D2 n  r s

If there is no surface charge
For non-conducting
materials, rs = 0 unless
an impressed source is
present.
D1n  D2 n
16
Lecture 5
Boundary Condition on
Tangential Component of E
• Consider an infinitesimal path abcd with width Dw
and height Dh lying half in medium 1 and half in
medium 2:
Dw
Dh/2
d
Dh/2
c
a
b
1
2
17
E1 , D1
aˆ n
E 2 , D2
Lecture 5
Boundary Condition on Tangential
Component of E (Cont’d)
aˆ s  unit vecto r perpendicu lar to path abcd
in the direction defined by the contour
aˆt  aˆ s  aˆ n  unit vecto r tangenti al to the
boundary along path
aˆt
aˆ n
d
aˆ s
a
b
c
18
Lecture 5
Boundary Condition on Tangential
Component of E (Cont’d)

Applying conservative law to the path, we have
 E  dl  0
C
b
c
d
a
a
b
c
d
LHS   E  d l   E  d l   E  d l   E  d l
Dh
Dh
Dh
Dh
  E1n
 E2 n
 E2t Dw  E1n
 E2 n
 E1t Dw
2
2
2
2
 E1t  E2t )Dw
19
Lecture 5
Boundary Condition on Tangential
Component of E (Cont’d)

The boundary condition is
E1t  E2t
20
Lecture 5
Electrostatic Boundary
Conditions - Summary

At any point on the boundary,
components of E1 and E2 tangential to
the boundary are equal
 the components of D1 and D2 normal to the
boundary are discontinuous by an amount
equal to any surface charge existing at that
point
 the
21
Lecture 5
Electrostatic Boundary
Conditions - Special Cases

Special Case 1: the interface between two
perfect (non-conducting) dielectrics:
 Physical
principle: “there can be no free
surface charge associated with the surface of
a perfect dielectric.”
 In practice: unless an impressed surface
charge is explicitly stated, assume it is zero.
22
Lecture 5
Electrostatic Boundary
Conditions - Special Cases

Special Case 2: the interface between a
conductor and a perfect dielectric:
 Physical
principle: “there can be no
electrostatic field inside of a conductor.”
 In practice: a surface charge always exists
at the boundary.
D1n  r s
E1t  0
23
Lecture 5
Potential Energy


When one lifts a bowling ball and places it on a
table, the work done is stored in the form of
potential energy. Allowing the ball to drop back to
the floor releases that energy.
Bringing two charges together from infinite
separation against their electrostatic repulsion also
requires work. Electrostatic energy is stored in a
configuration of charges, and it is released when the
charges are allowed to recede away from each other.
24
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution

Q1

25
Consider a point
charge Q1 in an
otherwise empty
universe.
The system stores no
potential energy since
no work has been
done in creating it.
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)

Q1
Q2

R12
Now bring in from
infinity another point
charge Q2.
The energy required to
bring Q2 into the
system is
W2  Q2V12
• V12 is the electrostatic potential due to Q1
at the location of Q2.
26
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)

Q3 R13
R23
Q2
R12
Q1

Now bring in from
infinity another
point charge Q3.
The energy required
to bring Q3 into the
system is
W3  Q3 V13  V23 )
27
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)

The total energy required to assemble the
system of three charges is
We  W2  W3
 Q2V12  Q3 V13  V23 )
28
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)



Now bring in from infinity a fourth point
charge Q4.
The energy required to bring Q4 into the
system is
The total energy required to assemble the
system of four charges is
We  W2  W3  W4
 Q2V12  Q3 V13  V23 )  Q4 V14  V24  V34 )
29
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)


Bring in from infinity an ith point charge Qi
into a system of i-1 point charges.
The energy required to bring Qi into the
i 1
system is
Wi  Qi V ji
j 1

The total energy required to assemble the
system of N charges is
N
N
i 1
i 2
i 2
j 1
N
i 1
We  Wi   Qi V ji   QiV ji
30
i  2 j 1
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)

Note that
QiV ji  Qi
Qj
40 R ji
 Qj
Qi
40 Rij
 Q jVij
 Physically, the above means that the partial
energy associated with two point charges is equal no
matter in what order the charges are assembled.
31
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)
i 1
1 N i 1
We   QiV ji   Q jVij  QiV ji )
2 i  2 j 1
i  2 j 1
N
1
 Q1V21  Q2V12
2
 Q1V31  Q2V32  Q3V13  Q3V23
 Q1V41  Q2V42  Q3V43
 Q4V14  Q4V24  Q4V34  ...
32
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)
1
We  Q1 V21  V31  V41  ...)
2
 Q2 V12  V32  V42  ...)
 Q3 V13  V23  V43  ...)
 Q4 V14  V24  V34  ...)  ...
1
1 N
 Q1V1  Q2V2  Q3V3  ...)   QiVi
2
2 i 1
33
Lecture 5
Electrostatic Energy in a Discrete
Charge Distribution (Cont’d)
where
N
Vi   V ji
j 1
j i
 Physically, Vi is the potential at the location of the ith
point charge due to the other (N-1) charges.
34
Lecture 5
Electrostatic Energy in a
Continuous Charge Distribution
Q  qev dv
n
QV
i 1
i i
  qev r ) V r ) dv
V
1
We   qev r ) V r ) dv
2V
35
Lecture 5
Electrostatic Energy in a Continuous
Charge Distribution (Cont’d)
qev    D
1
We   V   D ) dv
2V
vector identity :
V   D )    V D )  D  V
1
1
We     V D ) dv   D  V dv
2V
2V
36
Lecture 5
Electrostatic Energy in a Continuous
Charge Distribution (Cont’d)
Divergence theorem and
E  V
1
1
We   V D  d s   D  E dv
2S
2V
37
Lecture 5
Electrostatic Energy in a Continuous
Charge Distribution (Cont’d)

Let the volume V be all of space. Then the
closed surface S is sphere of radius infinity.
All sources of finite extent look like point
charges. Hence,
1
1
2
V
D  2 ds  R
R
R
lim  V D  d s  0
R 
S
38
Lecture 5
Electrostatic Energy in a Continuous
Charge Distribution (Cont’d)
1
We   D  E dv
2V
1
1 2
We   we dv  we  D  E  E
2
2
V
Electrostatic energy
density in J/m3.
39
Lecture 5
Electrostatic Energy in a Continuous
Charge Distribution (Cont’d)
D  0 E  P
1
1
1
2
We   D  E dv    0 E dv   P  E dv
2V
2V
2V
energy required to
set the field up in
free space
40
energy required to
polarize the dielectric
Lecture 5
Electrostatic Energy in a Capacitor
V2
+Q
+ V
12
-
V1
-Q
1
We   qev r ) V r ) dv
2V
1
1
 V1  qes1 r ) ds  V2  qes 2 r ) ds
2 S c1
2 Sc 2
1
1
 V2  V1 ) Q  QV12
2
2
41
Lecture 5
Electrostatic Energy in a Capacitor

Letting V = V12 = V2 – V1
1
1
2
We  QV  CV
2
2
42
Lecture 5
Electrostatic Forces: The
Principle of Virtual Work


Electrostatic forces acting on bodies can be
computed using the principle of virtual
work.
The force on any conductor or dielectric
body within a system can be obtained by
assuming a differential displacement of the
body and computing the resulting change in
the electrostatic energy of the system.
43
Lecture 5
Electrostatic Forces: The Principle
of Virtual Work (Cont’d)

The electrostatic force can be evaluated as
the gradient of the electrostatic energy of
the system, provided that the energy is
expressed in terms of the coordinate
location of the body being displaced.
44
Lecture 5
Electrostatic Forces: The Principle
of Virtual Work (Cont’d)

When using the principle of virtual work,
we can assume either that the conductors
maintain a constant charge or that they
maintain a constant voltage (i.e, they are
connected to a battery).
45
Lecture 5
Electrostatic Forces: The Principle
of Virtual Work (Cont’d)

For a system of bodies with fixed charges,
the total electrostatic force acting on the
body is given by
F Q  We
46
Lecture 5
Electrostatic Forces: The Principle
of Virtual Work (Cont’d)

For a system of bodies with fixed
potentials, the total electrostatic force
acting on the body is given by
F V  We
47
Lecture 5
Force on a Capacitor Plate

Compute the force on one plate of a charged
parallel plate capacitor. Neglect fringing of the
field.
• The force on the
upper plate can be
found assuming a
system of fixed
charge.
y
+Q
-Q
48
Lecture 5
Force on a Capacitor Plate
(Cont’d)

The capacitance can be written as a function of the
location of the upper plate:
C

0 A
d
 C y) 
0 A
y
The electrostatic energy stored in the capacitor may
be evaluated as a function of the charge on the
upper plate and its location:
Q2
Q2
We  y ) 

y
2C  y ) 2 0 A
49
Lecture 5
Force on a Capacitor Plate
(Cont’d)

The force on the upper plate is given by
We  y )
Q
F Q  We  aˆ y
 aˆ y
y
2 0 A
2

Using Q = CV,
CV
F Q  aˆ y
2d
50
2
Lecture 5
Force on a Capacitor Plate
(Cont’d)

Compute the force on one plate of a charged
parallel plate capacitor. Neglect fringing of the
field.
• The force on the
upper plate can be
found assuming a
system of fixed
potential.
y
V = V12
V=0
51
Lecture 5
Force on a Capacitor Plate
(Cont’d)

The capacitance can be written as a function of the
location of the upper plate:
C

0 A
d
 C y) 
0 A
y
The electrostatic energy stored in the capacitor may
be written as a function of the voltage across the
plates and the location of the upper plate:
2

AV
1
We  y )  CV 2  0
2
2y
52
Lecture 5
Force on a Capacitor Plate
(Cont’d)

The force on the upper plate is given by
We  y )
 0 AV 2
F V  We  aˆ y
 aˆ y
y
2 y2

Manipulating, we obtain
CV
F Q  aˆ y
2d
53
2
Lecture 5
Steady Electric Current
Electrostatics is the study of charges at
rest.
 Now, we shall allow the charges to move,
but with a constant velocity (no time
variation).
 “steady electric current” = “direct current
(DC)”

54
Lecture 5
Conductors and Conductivity




A conductor is a material in which electrons
are free to migrate over macroscopic
distances within the material.
Metals are good conductors because they
have many free electrons per unit volume.
Other materials with a smaller number of
free electrons per unit volume are also
conductors.
Conductivity is a measure of the ability of
the material to conduct electricity.
55
Lecture 5
Semiconductor
A semiconductor is a material in which
electrons in the outermost shell are able to
migrate over macroscopic distances when a
modest energy barrier is overcome.
 Semiconductors support the flow of both
negative charges (electrons) and positive
charges (holes).

56
Lecture 5
Conduction Current


When subjected to a field, an electron in a
conductor migrates through the material
constantly colliding with the lattice and losing
momentum.
The net effect is that the electron moves
(drifts) with an average drift velocity that is
proportional to the electric field.
v d   e E
57
electron
mobility
Lecture 5
Conduction Current (Cont’d)

Consider a conducting wire in which charges
subject to an electric field are moving with
drift velocity vd.
current
electron
E
vd
Ds
aˆ n
cross-section
58
Lecture 5
Conduction Current (Cont’d)

If there are nc free electrons per cubic meter of
material, then the charge density within the wire
is
qev  enc

Consider an infinitesimal volume associated with
Ds:
Dv  DsDl
Ds
Dl
59
Lecture 5
Conduction Current (Cont’d)

The total charge contained within Dv is
DQ  qev Dv  enc DsDl

This charge packet moves through the
surface Ds with speed
aˆn  v d  e E  aˆn

The amount of time it takes for the charge
packet to move through Ds is
Dl
Dt 
aˆ n  v d
60
Lecture 5
Conduction Current (Cont’d)


Current is the rate at which charges passes
through a specified surface area (such as the
cross-section of a wire).
The incremental current through Ds is given by
DQ
DI 
 enc e DsE  aˆ n )
Dt
61
Lecture 5
Current Density


The component of the current density in the
direction normal to Ds is
DI
J  aˆ n 
 enc e E  aˆ n )
Ds
In general, the current density is given by
J  enc e E
62
Lecture 5
Current Density (Cont’d)

The constant of proportionality between the
electric field and the conduction current
density is called the conductivity of the
material:
  enc e

Ohm’s law at a point:
J  E
63
Lecture 5
Current Density (Cont’d)


The conductivity of the medium is the
macroscopic quantity which allows us to treat
conduction current without worrying about the
microscopic behavior of conductors.
In semiconductors, we have both holes and
electrons
hole
  eN e e  N p  p )
mobility
hole density
64
Lecture 5
Current Density (Cont’d)

The total current flowing through a crosssectional area S may be found as
I   J ds
S

If the current density is uniform throughout
the cross-section, we have
I  J  aˆn )A
65
cross-sectional
area
Lecture 5
Current Flow

Consider a wire of non-uniform crosssection:
E
A1
A2
66
Lecture 5
Current Flow (Cont’d)
To maintain a constant electric field and a
steady current flow, both E and J must be
parallel to the conductor boundaries.
 The total current passing through the
cross-section A1 must be the same as
through the cross-section A2. So the
current density must be greater in A2.

67
Lecture 5
Ohm’s Law and Resistors

Consider a conductor of uniform cross-section:
V2
I

A E
• Let the wires and the two
exposed faces of the
“resistor” be perfect
conductor.
A F1
• In a perfect conductor:
J is finite
 is infinite
E must be zero.
l
+V68
Lecture 5
Ohm’s Law and Resistors
(Cont’d)



To derive Ohm’s law for resistors from Ohm’s law at
a point, we need to relate the circuit quantities (V
and I) to the field quantities (E and J)
The electric field within the material is given by
V12 V2  V1 V
E


l
l
l
The current density in the wire is
I
J
A
69
Lecture 5
Ohm’s Law and Resistors
(Cont’d)



Plugging into J = E, we have
l
V
I
A
Define the resistance of the device as
l
R
A
Thus,
Ohm’s law for
V  RI
resistors
70
Lecture 5