Transcript ohms law chart

Chapter 3. Static Electric Fields
3.1 Introduction
– Field : Spatial distribution of a physical quantity, function of position
What we will learn in this year.
- Static Electric Fields
Electric field intensity, Electric potential, Electric dipole, Electric properties of materials, Electric flux density
(Displacement Vector) Capacitors, Electrostatic Energy and Forces, Solutions of Electrostatic Problems
- Static Electric Currents
Ohm’s Law, Electromotive force, Kirchhoff’s Voltage and Current Law, Continuity equation, Power and Joule’s
law, Resistance calculation
- Static Magnetic Fields
Magnetic flux density, Magnetic potential, Magnetic dipole, Magnetic properties of materials, Magnetic field
intensity, Magnetic circuits, Inductors, Magnetostatic Energy and Forces and torques.
- Time-varying Electromagnetic Fields (2nd Semester)
Plane Waves, Reflection, Refraction, Transmission lines, Transients on Transmission lines, Smith Chart
Static Electric Fields
- In electrostatics, electric charges (the sources) are at rest, electric field do not change with time, there is no magnetic
fields.
- Lightning, corona, St. Elmo’s fire, grain explosion
- Oscilloscope, ink-jet printer, xerography, electret microphone, Digital Mirror Device
• Electrostatics
– Source charges are at rest (Test charges can move).
– Electric fields do not change with time : no magnetic fields
• Coulomb’s law
– Basic law about the electric force between two charged bodies
– Experimental law discovered by Charles de Coulomb in 1785.
F 12 
1 q1q2 ˆ
R12
2
40 R12
: Coulomb’s law
F12
F21
Coulomb’s torsion balance
• Electric field intensity : the force per unit charge that a very small stationary
test charge experiences when it is placed in a region where an electric field
exists.
E  lim
q 0
F
q
(N/C  V/m)
F  qE
Here, q is a test (probe) charge, which means that
it is small enough not to disturb the charge distribution
of the source.
(N)
If a charge q is placed in a region
where an electric field E exists,
it feels a force of F = qE
• Fundamental postulates of electrostatics in free space
– Differential forms :
Divergence theorem
– Integral forms :
E 

v

T
0
(   E )dv =
S
volume
charge
density
E ds 
1
ε0
 ρ dv
QT
0
: Gauss’s law
v
T
 E  0
 (  E )  ds  0
s
Stokes’ theorem
total charge
enclosed in the
surface S
 E d  0
C
: Energy Conservation
• Characteristics of electric fields
(1) Charge (density) forms an electric field that diverges (for static and dynamic cases).
E 
Gauss’s Law
T
0
Every point?

S
E ds 
QT
0
: The total outward flux of the electric
field intensity over any closed surface
in free space is equal to the total charge
enclosed in the surface divided by 0.
(2) The electrostatic field is conservative (only for static case).
 E  0
 E d  0
This law is an expression of
Kirchhoff’s voltage law in circuit theory.
: The algebraic sum of voltage drops
around any closed circuit is zero.
(Line integral of electric field is voltage.)
C
: The scalar line integral of the static
electric field intensity around any
closed path vanishes.
 E d  0   E d   E d
C

P2
C1
P1
E  d    E  d 
P1
Along C1
P2
Along C2
C2

P2
P2
E d   E d
P1
Along C1
P1
Along C2
• Electric field due to a single point charge q
(1) By Coulomb’s law
q
The force felt by a small test charge q at P is
F q' 
1
qq ' ˆ
R
2
40 R
By the definition of the electric field,
E at P 
F q'
1 q ˆ

R
2
q' 40 R
(2) By Gauss’s law

S
Eds 

S
(aˆR ER )  aˆ R ds  ER ds

S
2 
 ER  Rd R sin  d  ER (4 R 2 )  q
0
0 0
 E  aˆR ER  aˆ R
q
4 0 R
2
(V/m)
– If the charge q is not located at the origin, but at the specific position whose position
vector is represented as R´ ,
ˆ
q
EP 

, where   R  R'
3
2
40
40 R  R'
q( R  R')
• Electric field due to multiple point charges
: The principle of superposition applies here, and E is the vector sum of fields caused
by all the individual charges.
E
1
40
n

k 1
qk ( R  Rk ')
R  Rk '
3
ˆ '
qk 
k


40 k 1  k '2
1
n
Example 3-1. 3.2 p78
• Example 3-3 : Cathode-ray oscilloscope (electrostatic deflection)
The electrostatic deflection system of a cathode-ray oscilloscope is depicted in Fig. 3-4. Electrons from a heated
cathode are given an initial velocity u 0  zˆu0 by a positively charged anode (not shown). The electrons enter at z =
0 into a region of deflection plates where a uniform electric field E d   yˆEd is maintained over a width w.
Ignoring gravitational effects, find the vertical deflection of the electrons on the fluorescent screen at z = L.
Sol.
* Inkjet printers are devices based on the principle of electrostatic deflection of a stream of charged particles.
• Electric dipole
– Electric dipole : two point charges of equal magnitude and opposite polarity +q and –q,
separated by a small distance d (e.g. molecular scale polarization in dielectric materials)
– Is an important entity in the study of electric field in dielectric media.
– Determines the electrical and optical properties of the material.
– Electric field due to an electric dipole



d
d 
R
R 
q 
2
2 
E

3
3
40 
d
d 
 R

R


2
2


3
d
R
2

d  
d 

  R
 R 



2
2 

 
 Rd 
 R 3 1  2 
R 

3
R
d
2
3 / 2
 3 Rd 
 R 3 1 
2 
 2 R 
(d  R)
3 / 2
 2
d2 
 R  R  d  
4

 3 Rd 
 R 3 1 
2 
2
R


Taylor expansion
3 / 2
 Rd

E
3 2 R  d
3 
40 R  R

q
(Cont’d)
– Electric dipole moment p : the product of the charge q and the vector d
 Rd

E
3
R

d


4 0 R 3  R 2

q
p  aˆ z p  p(aˆR cos  aˆ sin  )
R  p  Rp cos 
p  qd
E

p  qd
 R p

3
R

p


4 0 R 3  R 2

1
p
4 0 R
3
(aˆ R 2 cos  aˆ sin  ) (V/m) : in SCS
: The electric field intensity of an electric dipole is inversely proportional to the cube
of the distance R.  It is the proof of the faster decrease of the electric field rather
than a single charge.
• Electric field due to a continuous distribution of charge
– The electric field caused by a continuous distribution of charge can be obtained by
integrating (superposing) the contribution of an element of charge over the charge
distribution.
A differential element of charge in dv΄ behaves like a point charge and
its contribution to the electric field intensity at P is
d E  aˆ R
 dv '
4 0 R 2
Source coordinate!
Therefore, the total electric field E at the point P due to a volume
charge distribution (density) ρ is given by
E
For the surface charge distribution, E 
For the line charge distribution,
E
1
4 0
1
4 0
1
4 0


aˆ R

aˆ R
V'
S'

L'
aˆ R
R
2
s
R2

R
2
dv ' (V/m)
ds ' (V/m)
d ' (V/m)
• Example 3-4 : Electric field of an infinitely long, straight, line charge of a
uniform density   in air
 By symmetry, cylindrical coordinate system (CCS) is adequate.
 It is an accepted convention to use primed coordinates for source points
and unprimed coordinates for field points when there is a possibility of
confusion.
E

ˆ   d'  1
R
R
d' (V/m)
40 L ' R 2
40 L ' R 3
1
R  rˆ r  zˆ z ' , d'  dz ' , R  r 2  z '2

rˆ r  zˆ z '
E   2
dz '
2 3/ 2
L
'
40 (r  z ' )
   rˆ r  zˆ z '

dz '
40  (r 2  z '2 )3 / 2
The 2nd term (z-component) is an odd function of z, and thus vanishes
(Please see the left figure).
r
 E  rˆ 
40

dz '
ˆ

r
 (r 2  z '2 )3 / 2 20 r

(For integration, let z = r tan )
• Gauss's law and applications
– Gauss's law : The total outward flux of the E-field over any closed surface in free
space is equal to the total charge enclosed in the surface divided by 0.
 It provides a convenient method for determining the electric fields of charge
distributions with some symmetry conditions.
 E ds 
S
Qin
0
– How to use Gauss's law?
(1) Check the given charge distributions are symmetric.
(2) Choose the Gaussian surface S such that, from symmetry conditions, the
magnitude of the electric field is constant and its direction is normal (or tangential)
at every point of the surface S.
• Example 3-5 : Electric field of an infinite (uniform) line charge
Sol.
(Step 1) Check the symmetry : infinite symmetry.
 E field must be radial and perpendicular to the line
charge ( E  rˆ Er ).
Er (top)
(Step 2) Choose and draw the Gaussian surface.
ds
(Step 3) Evaluate the integral part of Gauss's law.
Er (side)

At the side surface : E  d s 
S
ds
L
2
0
0

Er rddz  2rLEr
At the top (bottom) surface : E  ds  no contribution
(Step 4) Calculate the total charge Q enclosed by the surface.
Q    d   L
L
(Step 5) Determine E by equating both sides of Gauss's law.
2rLE r 
 L
0
 E  rˆ

20 r
• Example 3-6 : Electric field of an infinite (uniform) planar charge
Sol.
(Step 1) Check the symmetry : infinite symmetry.
 E field must be normal to the sheet of charge
( E  zˆ Ez ).
(
(Step 2) Choose and draw the Gaussian surface.
(Step 3) Evaluate the integral part of Gauss's law.
Top surface
: E  d s  ( zˆEz )  ( zˆds)  Ez ds
Bottom surface : E  d s  ( zˆEz )  ( zˆds)  Ez ds
 E  d s  2E  ds  2E A
S
z
A
z
(Step 4) Calculate the total charge Q enclosed by the surface  Q   S A .
(Step 5) Determine E by equating
both sides of Gauss's law.
2Ez A 
S A
0
Ez 
s
,
2 0
s
ˆ
ˆ
E   z Ez   z
,
2 0
 E  zˆ E z  zˆ
S
2 0
z0
z0
• Example 3-7 : Electric field of a spherical cloud of electrons (volume charges)
   0 (0  R  b),
Sol.
  0 ( R  b)
Spherical symmetry condition  Draw spherical Gaussian surface.
(2) R  b
(1) 0  R  b
E  Rˆ ER , d s  Rˆ ds
 E  d s  E  ds  E
S
R
Si
R
4R 2
4 3
Q    dv    0  dv    0
R
V
V
3

 E   Rˆ 0 R
3 0
4 3
b
3
3

b
0
 E   Rˆ
3 0 R 2
Q   0
 Outside the charge cloud, the E field is exactly the same
as though the total charge is concentrated on a single
point charge at the center.
 This is true, in general, for a spherically charged region
even though  is a function of R.
 special property of the spherical charge distribution.
• Electric potential
– Concept of electric potential
 In electric circuits, we work with voltages and currents, not electric or magnetic fields.
 Electric potential is the same as voltage : The voltage V between the two points in the
circuit represents the amount of work, or potential energy, required to move a unit charge
between the two points.
– Electric potential as a function of electric field
 If we attempt to move the charge along the positive y-direction, we will need to provide
an external force Fext to counteract Fe, which requires the expenditure of energy.
F ext  F e  q E
 The work done in moving any object by d  is
dW  F ext  d   q E  d   q yˆE  yˆdy  qEdy (J)
F ext  q E
E   yˆE
 The differential electric potential energy dW per
unit charge is called the differential electric potential
(or differential voltage) dV.
dV 
dW
  E  d  (J/C  V)
q
F e  qE
• Potential difference
– Potential difference : amount of work done against the electric field to move a unit
charge from one point to another point  The potential difference between any two
points is obtained by integrating differential electric potential along any path
between them.
P
P

2
P1
dV  V2  V1  V21    E  d  (V)
2
P1
 The result of the line integral on the right-hand side of the
above equation should be independent of the specific
integration path taken between points P1 and P2
(See lecture note 4-4)  This requirement is mandated by
the law of conservation of energy.
 In electrostatics, the potential difference between two points is uniquely defined.
• Electric potential (or voltage) at a point in space
 In electric circuits, the voltage at a point in a circuit is
P
the voltage difference with respect to a conveniently chosen
Vat P   E  d  (V)
point at which a reference voltage of zero is assigned (ground).

 The reference potential point is usually chosen to be at
reference point of V = 0 (infinity) infinity.

• Electric field from the electric potential
In lecture 5-5, we noticed that dV   E  d  . Also, from eq. (2-88), dV  V  d  .
By equating these two equations, we conclude that
E  V
(1) The inclusion of the negative sign is necessary
in order to conform with the convention that
in going against the E field the electric potential V
increases.
(2) When we defined the gradient of a scalar field,
the direction of V (i.e. E) is normal to the surfaces
Equipotential
surface
of constant V : field lines are everywhere perpendicular
to equipotential lines and equipotential surfaces.
Force lines
(flux lines)
• Electric potential due to point charges
– For a single point charge q at the origin,


q  ˆ
  R dR
V   E  d     Rˆ
2 


 40 R 
R
R

V
q
40 R
(V)
– The potential difference between any two points
q  1 1
  
V21  VP2  VP1 
40  R2 R1 
 The potential difference is determined by only
two endpoints.
 No work is done from P1 to P3.
– Electric potential due to the multiple point charges
V
1
40
n

k 1
qk
R  Rk '
Equipotential
lines
[HW]
See Example 3-8.
• Electric potential of an electric dipole
V
(d << R)
1  q q
 

40  R1 R2 
Since d << R, the lines R1 and R2 are
approximately parallel to each other.
R2  R1  d cos  ,
R1R2  R 2
Hence,
qd cos 
p  Rˆ
V

40 R 2 40 R 2
Electric field can be obtained by taking
negative gradient of the potential V.
V ˆ V
V
E  V   Rˆ

 ˆ
R
R
R sin 
p

Rˆ 2 cos   ˆ sin 
3
40 R


• Electric potential due to a charge distribution
– The approach is the same as the cases of electric field equations.
– The electric potential due to a continuous charge distribution confined in a given
region is obtained by integrating the contribution of an element of charge over the
charged region.
For the single point charge,
For the volume charge distribution,
V
V
For the surface charge distribution, V 
For the line charge distribution,
V
1
q
40 R
1
40
1
40
1
40

V'

S'

L'

R
s
R

R
dv' (V)
ds ' (V)
d' (V)
[HW]
Solve Example 3-10.
• Example 3-9 : Electric field on the axis of a circular disk
carrying a surface charge density
ds '  r ' dr ' d ' , R  z 2  r '2

V S
40
2
b
0
0
 
 E  V   zˆ


r'
(z  r' )
2
V
z
2
S
2 0
dr ' d ' 


1 / 2
S
1  z z 2  b 2 
,
2 0
1 / 2

 zˆ S 1  z z 2  b 2 
,
2 0
 zˆ
For very large z, z z  b
2
Q

ˆ
z
 4 z 2 ,
0
E
Q
 zˆ
,
2
 40 z

2 1 / 2
z0
z0
 b2 
 1  2 
 z 
1 / 2
b2
 1 2
2z
z
2
 b2  z
z0
z0
b  
E  zˆ
2
s
2
40 z
When the point of observation is very far away
from the charged disk (far field), The E field
approximately follows the inverse square law as if
the total charge were concentrated at a point.

• Electrical properties of materials
– Electromagnetic constitutive parameters of a material medium
 Electrical permittivity, 
 Magnetic permeability, 
 Conductivity,  : a measure of how easily electrons can travel through the material under
the influence of an external electric field.
– A material is said to be :
 homogeneous : if its constitutive parameters do not vary from point to point
 isotropic : if its constitutive parameters are independent of direction
• Classification of materials based on their electrical property
– Conductors (metals) : having a large number of loosely attached electrons in the
outermost shells of the atoms  migrating easily with or without an electric field
– Dielectrics (insulators) : having electrons tightly held to the atoms  difficult to
detach them even under the influence of an electric field
– Semiconductors : their conductivities fall between those of conductors and
insulators
Band Theory
(a) Probability density
function of an isolated
hydrogen atom
(b) Overlapping probability density
functions of two adjacent hydrogen
atoms
(c) The splitting of the n = 1 state.
As atoms are brought together from
infinity, the atomic orbits overlap and
give rise to bands. Outer orbits overlap
first. The 3s orbits give rise to the 3s
band, 2p orbits to the 2p band, and so
on.
Schematic of an isolated silicon atom.
The splitting of the 3s and 3p states of
silicon into the allowed and forbidden energy
bands.
electro
n
Insulator
Conduction band
hole
Valence band
Semiconductor
Hole: bubble in water, unoccupied chair in crowded
theater
unoccupied state in valence band
• Conductors
– Conductors : materials having a lot of free electrons, resulting in high conductivities
 Most metals :  ~ 106 to 107 S/m (c.f. good insulators :  ~ 10-10 to 10-17 S/m)
 Perfect conductor :  =  (c.f. perfect dielectric :  = 0)
 Superconductor : having a practically infinite conductivity and Meissner effect at very
low temperature (Meissner effect: the effect of expelling magnetic field )
– Ohm's law for conductors
The drift velocity ue of electrons in a conducting material is given by
u e  e E (m/s)
(e : electron mobility)
The current density in a conducting medium containing a volume charge density e
(electrons) moving with a velocity u is J   u . Therefore,
e
J   e e E   E (A/m 2 ) : point form of Ohm's law (for conductors)
 In a perfect dielectric ( = 0) : J = 0 regardless of E
(However, there is no perfect dielectrics.)
 In a perfect conductor ( =  ) : E = J/  = 0 regardless of J
(However, there is no perfect conductor, but superconductor.)
Example: Resistance and speed of electrons in conductor.
Find the resistance of a 1 mile (1.609 km) length of #16 copper wire, which has of
e (copper)=0.0032 m 2 / Vs
 (copper)=5.8
107 (S / m)
1.291
mm.
R
L
1609

 21.2
7
 S 5.8 10 ( S / m)   (1.291103 / 2) 2
This #16 wire can safely carry about 10 Adc. Speed of electron?
J
10
 7.65 106 (A/m2 )
3
2
 (1.29110 / 2)
V  IR  212 (V)
E  V / l  212 /1609  0.312 (V/m)
J   e e E   E (A/m 2 )
    e e
ue  e E
ue  0.0032(m 2 /Vs)  0.312(V/m)  0.0009(m / s)  0.9(mm / s)
 (copper)=-5.8 107 / 0.0032  1.8 1010 C/m3
1.8 1010 (C/m3 ) /1.6 1019 (C)  1029 (개/m3 )  1011 (개/ m3 )
Very slow!!!
• Properties of conductors
(1) Inside a (perfect) conductor (under static conditions):
  0,
E 0
 Inside a conductor, no volume charge density can exist.
 All charges move to the surface of a conductor and redistribute themselves in
such a way that both the charge densities and the electric field inside vanish.
(2) Under static condition, the electric field E on the conductor surface is everywhere
normal to the surface.
 Otherwise, the charges would move tangential to the surface.
(3) A conductor under static condition is an equipotential medium
 Potential is constant inside a conductor and on the conductor surface.
(4) Potential is continuous across any surfaces, but the electric field is not.
 The electric field may be discontinuous across the boundary that has a surface
charge density.
• Boundary condition at an interface between a conductor and free space
(1) Normal component : Use Gauss's law.
 E  d s  En S 
h  0
S
 s S
0

(2) Tangential component : Use
En 
s
0
 E  d   0.
C
 E  d   E w  0
abcda
t
B. C. at a conductor/free space interface : Et  0,

Et  0
s
En 
0
• Uncharged conductor in an external static electric field
-
Induced field
E1
E =0
conductor
Applied external field E0 moves electrons inside a
conductor to the left, and the right end becomes charged
positively.
 Induced field E1 is formed inside the
conductor and it makes the field inside the
External field E0 conductor zero.
+
+
+
+
+

•
Example 3-11 : A point charge surrounded by a spherical conducting shell (Ro, Ri).
Find E and V everywhere.
E  aˆR ER , Gauss's Law!!
P
(a) R  Ro (Start from infinity!!)
+
1
+
E2  0
+

-
- +Q ---
Einduced
+
+
+
+
S
E ds 
P1
V   E  d
0
ER1 4 R 2 

(b) Ri  R  Ro
Q 
0
Q
R
(V), V1    aˆ R

+
Q
4 0 R
ER1  aˆ R
Q
4 0 R 2
R
ˆR dR) V1    aˆ R
2 (a
Q
4 0 R 2
(aˆR dR)
ER 2  0
Ri
Q
Ro
4 0 Ro
V2  V1 R  R  (  ER 2 dl ) 
o
(c) R  Ri
ER 3 
Q
4 0 R 2
V3  V2
V3 
R
R  Ri
  (ER 3 )dR
Ri
Q 1 1 1 
 
 
4 0  R Ro Ri 
• Dielectrics in static electric field (1)
– Polarization (分極) mechanism in dielectrics
 The electrons in the outermost shells of a dielectric are strongly bound to the atom.
 In the absence of an external electric field, the electrons
in any material form a symmetrical cloud around the nucleus,
with the center of the cloud being at the same location as the
center of the nucleus.
 When an external electric field is applied to dielectrics,
it can polarize the atoms or molecules : distorting the center
of the electron cloud and the location of nucleus.
 The polarized atom or molecule may be represented by an
electric dipole. Such induced dipoles set up a small electric field.
This induced field (polarization field) modifies the electric field
inside and outside of the dielectric material.
• Dielectrics in static electric field (2)
– Polarized medium
 Within the dielectric material, the induced
dipoles align themselves in a linear arrangement,
into the direction of the applied external field.
 Along the upper and lower edges of the material,
the dipole arrangement exhibits a positive and a
negative surface charge density, respectively.
 If the polarization not uniform, a polariztion volume charge
density can be created inside the dielectric material.
– Nonpolar and polar materials
 Nonpolar material
p
 Molecules do not have permanent dipole moments.
 Polarized only when an external electric field is applied
 Polar material
 Molecular structure shows built-in permanent dipole moments
in the absence of an external electric field (e.g. water)
H+
105o
O-
H+
• Equivalent charge distributions of polarized dielectrics
To analyze the macroscopic effect of induced dipoles, we define a polarization vector P.
nv
P  lim
p
k 1
v 0
k
(C/m 2 )
v
: P is the volume density of the induced dipole moments.
(n is the number of molecules per unit volume.)
The dipole moment dp of an elemental volume dv' is d p  Pdv' . This moment produces a potential
P  Rˆ
dV 
dv ' 
4 0 R 2
V
V
1
4 0
1
4 0
P  Rˆ
dv ' : potential produced by polarized material of
4 0 V ' R 2
volume V'.
R  xˆ ( x  x)  yˆ ( y  y ) zˆ( z  z )

Rˆ
1
 1  R̂ 
1



V ' P  '  R dv '   '  R   R 2 cf)   R    R 2
  x̂
 ŷ
 ẑ
x
y 
z 

P
 ' P 
dv '
 ' f A  f  ' A  A  ' f
 V '  '  dv ' V '
R
R
 


V
1

divergence theorem
V
V
P  nˆ
1
ds
'

40 S ' R
40
1
1
40

S'
s
R
ds '
 

A polarized dielectric can be replaced by an
('P)
equivalent
dv
'
:
V ' R
surface bound charge density (ps) and an
equivalent
1

 p  (
 ps bound
 P  nˆcharge density
V
dv'
volume
p). P

V
'
40 R
• Electric flux density and dielectric constant
Because a polarized material gives rise to an equivalent volume bound charge density p, the
Gauss's law should be modified to include the effect of polarization.
 E 
1
0
(  free   p )     0 E   free    P    ( 0 E  P)   free
Where is  ps ?
D   0 E  P (C/m 2 ) : Electric flux density (Electric displacement)
* The use of the vector D enables us to write a divergence relation between
the electric field and the distribution of free charges in any medium without the
necessity of dealing explicitly
with the polarization vector P or the polarization charge density p .
  D   free
(C/m 3 )
– Properties of medium
Divergence
theorem
 Linear, isotropic dielectrics :
 Dds  Q
S
free
* The D-field is
(C)
medium-independent .
electric susceptibility
P   0 e E
D   0 (1   e ) E   0 r E   E
r  1 e 

0
dielectric constant
(relative permittivity)
absolute permittivity
(permittivity)
- Free space : 1.0
- Air : 1.00059
 Simple medium
linear, homogeneous, isotropic
medium
P
E
nonlinear, homogeneous, isotropic
medium
P
éDx ù
éEx ù
ê ú
r ê ú
êDy ú= e ( E ) êE y ú
ê ú
ê ú
êD ú
êE ú
zû
ë
ë zû
E
éDx ù éEx ù
ê ú ê ú
êDy ú= e êE y ú
ê ú ê ú
êD ú êE ú
ë zû ë zû
Anisotropic medium
éDx ù
ê ú
êDy ú=
ê ú
êD ú
ë zû
Ferroelectric medium
P
ée11 e12
ê
êe21 e22
ê
êe
ë 31 e32
Tensor
Biaxial medium
E
éDx ù
ê ú
êDy ú=
ê ú
êD ú
ë zû
ée11 0
ê
ê0 e22
ê
ê0 0
ë
e13 ùéEx ù
úê ú
e23 úêE y ú
úê ú
ú
e33 úê
ûëEz û
0 ùéEx ù
úê ú
0 úêE y ú
úê ú
ú
e33 úê
ûëEz û
Uniaxial medium
éDx ù
ê ú
êDy ú=
ê ú
êD ú
ë zû
ée11 0 0 ùéEx ù
ê
úê ú
ê0 e11 0 úêE y ú
ê
úê ú
ê0 0 e úêE ú
33 ûë z û
ë
• Example 3-12 : A point charge at the center of a spherical dielectric shell
Dielectric problem : 1) find D instead of E, 2) find E, 3) find P fr
a) R  Ro (Start from infinity)
DR1 
Q
,
4 R 2
ER1 
DR1
Q
dR 
2
Q
V1   
R

4 0 R
0
Q

4 0 R 2
PR1  0
,
4 0 R
b) Ri  R  Ro
DR 2 
Q
,
4R 2
ER 2 
DR 2
 0 r

Q
40 r R 2

Q
4R 2
PR 2   0  e ER 2   0 ( r  1) ER 2

1
  1 
 r
V2  V1 R  R  
o
c) R  Ri
DR 3 
Q
,
4R 2
V3  V2
ER 3 
R
R  Ri
 Q

1



1

2
 
r
 4R

Q
40 R 2
  ER 3 dR 
Ri
,
R
Ro
 Q

2
 4R
Q
Q 
1

1 
2
4R
40   r
PR 3  0
Q 
1
1 
40   r
 1 
1
  1 
 Ro   r
 1 1
  
 Ri R 
 1
1 
 

 Ro  r R 
(Cont'd)
– Surface and volume bound charges

1 Q
  PR 2 R  R  1  
(1) On the inner shell surface :  ps R  R  P  ( Rˆ )
i
i
R  Ri
 r  4Ri 2

 ps  P  nˆ

1 Q
 PR 2 R  R  1  
(2) On the outer shell surface :  ps R  R  P  Rˆ
2
RR
  r  4Ro
o
o
o
(3) Inside the dielectric volume :  p    P  
 p    P
 Induced surface bound charges reduce the
1 
( R 2 PR 2 )  0
2
R R
Field by Q
Field by
+
polarization
electric field in the dielectric made from the charge Q. +
ER 2 
• Dielectric strength
Q
4 0 r R 2

Q
Q

4 R 2 4 0 R 2
+
+
+
+
-
+
-
+
-
+
+
+
+
 Dielectric breakdown : permanent damage under strong electric field
 Dielectric strength : maximum E field that a dielectric material can withstand without
breakdown (See Table 3-1)
• Boundary conditions for electrostatic fields
(1) Tangential component
 E d  E
abcda
1
  w  E 2  ( w)  E1t w  E2t w  0
 E1t  E2t
D1t

:
Tangential component of an E field is
continuous across an interface.
D2t
1
2
(2) Normal component

S


D  d s  D1  nˆ2  D 2  nˆ1 S  nˆ2  ( D1  D 2 )S   fs S
nˆ2  ( D1  D 2 )   fs
or
 D1n  D2 n   fs
:
Normal component of a D field is
discontinuous across an interface
where ‘free’ surface charge exists
: the amount of discontinuity is equal to
the free surface charge density.
Where the reference unit normal is outward from 2. (p117)
D2 : 0  D1n  1E1n  s
 If medium 2 is a conductor
space,
)
 Two dielectric are in contact with no free charges at the
interface :
s  0  D1n  D2n , 1E1n  2E2n
s
(If medium 1 Eisn free
0
•
Example 3-14 : An interface between the dielectric sheet and free space
A lucite sheet (r = 3.2) is introduced perpendicularly in a uniform electric
field E 0  xˆE0 in free space. Determine Ei, Di, and Pi inside the lucite.
Sol) We assume that the introduction of the sheet does not disturb the
original uniform electric field. (Why?)
Only normal field components need to be considered.
nˆ2  ( D1  D2 )   fs ,
D1n  D2 n   fs
Where the reference unit normal is outward from 2.
(p117)
No free surface charge
 0
fs
exists !
①
②
(1) nˆ2  ( D1  D2 )  0
aˆx  ( D1  D2 )  0, D1  D2  Di
(2) D1n  D2n
D1n  aˆx  D1  aˆx  (aˆx 0 Ex )   0 Ex
D2n  aˆx  D2   0 Ex
D2  Di  aˆx 0 Ex
Ei 
1

Di 
1
 0 r
Di  aˆ x
E0
 aˆ x E0
3.2
1 

Pi  Di   0 E i  aˆ x 1 
  0 E0
 3.2 
 aˆ x 0.6875 0 E0 (C/m 2 )
•
Example 3-16. Coaxial cable to carry electric power. The radius of inner conductor is
determined by the load current, and the overall sized by the voltage and type of
insulating material used. ri=0.4 cm, εrr (rubber)=3.2, εrp (polystyrene)=2.6. Design a
cable that is to work at a voltage rating of 20 kV. In order to avoid breakdown due to
voltage surges, the maximum electric field intensities in the insulating materials are
not exceed 25% of their Sol
dielectric
) Er max strength.
0.25  25 106 (V / m), E p max  0.25  20 106 (V / m)


E  aˆr Er (r )  ?
 D  ds  Q, Dr  2lr , E  aˆr 2l r
l
l
E

,
E

rp  1.54ri
r
max
p
max
ri
2


3.2

r
2


2.6

r
0
i
0
p
rp
l
l
or E p max 
, Er max 
rr  0.65ri
 
εrr
2


2.6

r
2 0  3.2  rr
0
i
ro
εrp
a voltage rating of 20 kV
rp
ri
ro
rp
l
 8 104 , rp  1.54ri  0.616(cm)
2 0
 E p dr   E p dr  20,000(V )
rp 
l  1
ro
1
ln

ln

  20, 000
2 0  2.6 rp 3.2 ri 
ro  2.08ri  0.832(cm)
• Capacitance and capacitors
– Capacitance of an isolated conductor
Assume that a single conductor has charge Q and its potential is V.
If we increase the charge to kQ, the potential is also increased to kV (Why?).
 The charge of a conductor and its potential is proportional to each other.
Q  CV
: Capacitance C is the constant of proportionality.
– Capacitor : two conductors separated by free space or a dielectric medium
C
Q
V12
(C/V  F)
–
The capacitance of a capacitor depends on :
–
(1) Geometry of the capacitor (size, shape, relative
–
–
positions of two conductors)
(2) Permittivity of the medium between conductors
•
How to calculate the capacitance?
(1) Choose an appropriate coordinate system  (2) Assume charge +Q and –Q on the
conductors  (3) Find E from Q  (4) Find V12 by evaluating the line integral of E from –Q
to +Q  (5) Find C by taking the ration Q/V12.
•
Example 3-17 : Capacitance of a parallel-plate capacitor
+Q
(neglect fringing field.)
(1) Coordinate system : RCS (left figure)
(2) Put charges +Q(upper) and –Q(lower)
: The charges are assumed to be uniformly
distributed over the conducting plate with
surface charge density +s and –s.
–Q
②
(3) D1n  D2 n   s  Q / S

①
In conductor, D2 n  0 (
E2 n  0, P  0)
nˆ2  ( D1  D2 )  aˆ y  ( D1 )  D1n  s

Q
 D1  aˆ y  s  E   aˆ y s   aˆ y

S
(4) V12   
y d
y 0
E  d  
d
0
Q
Qd
( )dy 
  Ed 
S
S
C
Q
S

V12
d
Important!
Solve Example 3-19.
• Example 3-18 : Capacitance of a cylindrical capacitor
(1) CS : CCS (left figure)
(2) Put charges +Q(inner) and –Q(outer)
 Find the E field by Gauss’s law.
Assuming D  rˆD
r
Dds  Q
S
 Dr  2rL  Q
Q
D
Q
 Er  r 
2rL

2 rL
(Fringing field is also neglected.)
 Dr 
Vab   
r a
r b
Q
 Q

b
ˆ 
E  d   
rˆ   rdr
ln  
b
2 L  a 
 2 Lr 
a
C 
Q
2 L

Vab
b
ln  
a
• Series and parallel connections of capacitors
Series connection
1
1
1
1


  
Csr C1 C2
Cn
Parallel connection
C||  C1  C2      Cn
• Capacitances in multi-conductor systems
– In the case of more than two conducting bodies in an isolated system,
Q1  c11V1  c12V2      c1NVN ,
Q2  c21V1  c22V2      c2 NVN ,

QN  c N 1V1  cN 2V2      cNN VN
Q1  Q2  Q3      QN  0
cii : coefficients of capacitance = ratio of charge Qi on
and the potential Vi of the i-th conductor with all
other conductors grounded.
cij : coefficients of induction
cii > 0, cij< 0 and cij = cji
• Four conductor systems
Q1  c11V1  c12V2  c13V3 ,
Q2  c12V1  c22V2  c23V3 ,
Q3  c13V1  c23V2  c33V3
Q1  C10V1  C12 (V1  V2 )  C13 (V1  V3 ),
Q2  C20V2  C12 (V2  V1 )  C23 (V2  V3 ),
Q3  C30V3  C13 (V3  V1 )  C23 (V3  V2 )
C10 , C20 , C30 : self partial capacitanc es,
Cij (i  j ) : mutual partial capacitanc es
• Electrostatic shielding (Faraday cage)
Q1  C10V1  C12 (V1  V2 )  C13 (V1  V3 )
Since conductor 2 is grounded, V2  0.
Q1  (C10  C12  C13 )V1  C13 (V3 )
When Q1  0, no electric field inside shell 2  V1  V2  0
 C13  0
 V3 will not affect Q1, and vice versa : shielding
• Electrostatic energy
– Work needed to locate three point charges q1, q2, q3 at locations P1, P2, P3
Work needed to place a charge q at the point P is W = qV(P) .
P3
q3
(2) Locating q2 at P2 : Work must be done against the E-field
produced by q1.
R23
R13
P2
R12
P1
q2
q1
(1) Locating q1 at P1 : No work is needed because of no field.
W2  q2Vat P2  q2
1  q1 


4 0  R12 
(3) Locating q3 at P3 : Work must be done against the E-field
produced by q1 and q2.
W3  q3Vat P3  q3
q2 
1  q1



4 0  R13 R23 
The total energy required to locate the three charges is
1  q1q2 q1q3 q2 q3 




4 0  R12
R13
R23 
 q1
 q1
q3 
q3 
q2  
1   q2
  q1 


q


q

 2
 3

2   4 0 R12 4 0 R13 
 4 0 R12 4 0 R23 
 4 0 R13 4 0 R23  
1
  q1V1  q2V2  q3V3 
Note : Vi is the potential caused by charges except qi
2
W
• General expression for potential energy of a group of N discrete point charges
1
W   q1V1  q2V2  q3V3 
2
N
1
 qkVk (J)
2 k 1
– Two remarks
We 
Vk 
1
40
N
qj
R
j 1
( j k )
jk
: electric potential at qk position caused by all the other charges
(1) We can be negative.
(2) We represents only the interaction energy (mutual energy) and does not include
the work required to assemble the individual point charge themselves.
– Units of energy : Joule (J)
cf) electron-volt (eV) : work required to move an electron against a potential
difference of 1 V
1 (eV) = 1.610-19 (J)
• Electrostatic potential energy of a continuous charge distribution
We 
1
V dv (J)

V
'
2
• Electrostatic energy in terms of field quantities
1
1

V
dv

(  D)V dv


V
'
V
'
2
2
1
1
    (V D) dv   D  V dv (   (V D)  V  D  D  V )
2 V'
2 V'
1
1
  V D  nˆ ds   D  E dv
2 S'
2 V'
We 
= 0 for infinitely large volume, why?
V' : any volume including all the charges  very large sphere with radius R.
As we let R, (1) V and D fall off at least as fast as 1/ R and 1/ R2, respectively.
(2) The area of the bounding surface S' increases as R2.
 The first surface integral term decreases as fast as 1/ R and will vanish as R.
Electrostatic energy density, we:
1
We  entire D  E dv (J)
2 space
2
1
1
D
 entire  E 2 dv  entire
dv (J)
2 space
2 space 
We   we dv
V'
1
1 2 D2
we  D  E   E 
2
2
2
(J/m3 )
•Three Methods to find electrostatic energy for assembling a uniform spherical
volume charge densities  of radius b.
(1) Find V at Rx , then use dWe =dqV , We =  Vdq. (J) (Ex.3.22)
V ( Rx ) 
b
Rx
Q( Rx )
4 o Rx 2
4 Rx 3
Q( Rx )  
3
dQ   4 Rx 2 dR
V(Rx)
4 2b5
We =  VdQ 
R 0
15 0
R b
3Q 2

20 0b
(2) Find V everywhere, then use We 
1
V dv. (Ex3-23)
2 V '
(3) Find E everywhere, then use We 
1
2
entire  E dv.

2 space
Example 3-23 : Electrostatic energy for assembling a uniform spherical volume charge densities 
of radius b.
•
- Sol. We 
1


Vdv
'

2 V '
2
b

2
0
0
0
 
V ( R) R sin  dRd d 
2

2

b
0
V ( R)4 R 2 dR
Note that the potential V cannot move outside the integral because it is a function of R.
The electric field of a spherical electron cloud is already calculated in Example 3-7.
(1) R  b
E  Rˆ
ρ
(2) 0  R  b
E  Rˆ ER , d s  Rˆ ds

S
E  d s  ER  ds  ER 4 R
2
Si
Q    dv    dv   
V

 E   Rˆ
R
3 0
V
4 3
R
3
4 3
Q   0
b
3
 b3
 E  Rˆ
3 0 R 2
Q
4 0 R 2
R
ˆ 
V    E  dl    E  RdR

We 
E R1  Rˆ
b 3
ˆ
R
( R  b)
40 R 2
3 0 R 2
E R 2  Rˆ
QR
R
 Rˆ
2
40 R
3 0
Q
4 o R 2
1
Vdv '  0
2 V '
Q
(0  R  b )
R
b
R
V    E  d R     ER1dR   ER 2 dR 
 


b
R R
 b b 3
   2 b 2 R 2    3b 2 R 2 
 b  
 


 
dR  
dR  

 3 R 2
b 3
3

2
2
3

2
2


0
0
0 
0 


We 

2

b
0
  3b 2 R 2 
4 2b 5
2

4R dR 

3 0  2
2 
15 0
(3) Find E everywhere, then use We 
(1) R  b
1
2

E
dv.
entire
2 space
 b3
1
ˆ
ER
,
W
=
eout
3 0 R 2
2
2

  b3  2
Rb  0  3 0 R 2  R sin  dRd d

(2) 0  R  b

1
E   Rˆ
R, Wein = 
3 0
2
ρ

2
   2

R0 0  3 0 R  R sin  dRd d
b
We  Wein  Weout
It is very strange to find
b
b
Weout
We
Method 1, 2
Method 3
Wein
Which one is right?
Which has gravitation potential energy? Maybe not important?
• Example 3-24 : Stored electrostatic energy in a parallel-plate capacitor filled
with dielectric material
[HW] Important!
Solve Example 3-25.
E
2
V
(neglecting fringing field)
d
2
1
1 V 
1 S 2
V 
We      dv     Sd     V
2 V'  d 
2 d
2 d 
Capacitance of the parallel plate capacitor
1
We  CV 2 (J)
2
Q  CV
1
Q2
We  QV 
(J)
2
2C
 Although the above equations for electrostatic energy are derived for the parallelplate capacitor case, they hold true for any two-conductor capacitor.
• Electrostatic forces
– Calculating electrostatic forces of charged bodies
 We know that the force between two point charges are given by Coulomb's law.
 Then, how about the charged bodies (for example, two charged conductors) ?
For conductors, charges are distributed on the surface (surface charge densities 1 and 2).
 The force that the surface element dS2 of conductor 2 feels is given by F = 2 dS2 E1 .
The total force that conductor 2 feels (due to the electric field produced by conductor 1) is
F 12 

S2
 2 dS2 E1 , where E1 
1  1dS1
 S1 4 0 r122 uˆr12
 Evaluating this integral is a very tedious job  Is there a better idea ?
• Calculating electrostatic force from the energy consideration
– Principle of virtual displacement : A method to calculate the electrostatic force on
an object from the electrostatic energy of the system
(1) System of bodies with fixed charges
 Consider an isolated system of charged conducting, as well as dielectric, bodies separated
from one another with no connection to the outside world.
 The charges on the bodies are therefore constant.
Assume that the object 3 (carrying charge Q3) moves by d
due to the electric force F Q .
Q1
d  Q3
Q2
Then the mechanical work done by the system is given by
dWm  F Q  d
Since the system is isolated, mechanical work must be done at the expense of the stored
electrostatic energy.
dWm  dWe  F Q  d
By the way,
dWe  (We )  d 
F Q  We (N)
(Cont'd)
(2) System of conducting bodies with fixed potentials
 Consider a system in which conducting bodies are held at fixed potentials through
connections to such external sources as batteries.
 Uncharged dielectric bodies may be present.
V1
Q1
d Q
3
Q2
A displacement d by a conducting body would result in a
change in total electrostatic energy and would require the sources
to transfer charges to the conductors in order to keep them at
their fixed potentials.
V3
dWs  Vk dQk
V2
k
Total energy supplied by the sources
Charge transfers also change the
electrostatic energy of the system by
dWe 
dWm  F V  d
Mechanical work done by the system
From the energy conservation law
dWm  dWe  dWs
1
1
Vk dQk  dWs

2 k
2
dWm  F V  d  dWe  (We )  d
F V  We (N)
• Example 3-26 : Electrostatic force on the parallel-plate capacitor plates
Express electrostatic energy in terms of space coordinate in the direction of virtual
displacement (for example, x).  The capacitance C is then depending on x (C(x)).
 Energy is described in terms of C(x), and Q or V.  The force is then evaluated
by differentiate the energy with the coordinate variable.
(a) By assuming fixed charges
Q2
Q2 1
Q2
We 


x
2C ( x) 2  S 2 0 S
0
x
energy
We
Q2 
Q2
Fe  We   xˆ
  xˆ
 x    xˆ
x
2 0 S x
2 0 S
(b) By assuming fixed voltages
1
V 2 S  0 SV 2 1
2
We  C ( x)V 
0 
2
2
x
2 x
virtual displacement (increasing x)
+Q
x
x=0
S
-Q
virtual displacement (increasing x)
2
C 2V 2
V2  S 

 0 
2 0 S 2 0 S  x  V
coenergy
We
 0 SV 2   1 
 0 SV 2
Fe  We  xˆ
 xˆ
    xˆ
x
2 x  x 
2x2
x
S
x=0