Transcript Chapter 3 : Electrostatics

UNIVERSITI MALAYSIA PERLIS
EKT 241/4:
ELECTROMAGNETIC
THEORY
CHAPTER 3 – ELECTROSTATICS
PREPARED BY: NORDIANA MOHAMAD SAAID
[email protected]
Chapter Outline
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Maxwell’s Equations
Charge and Current Distributions
Coulomb’s Law
Gauss’s Law
Electric Scalar Potential
Electrical Properties of Materials
Conductors & Dielectrics
Electric Boundary Conditions
Capacitance
Electrostatic Potential Energy
Image Method
2
Maxwell’s equations
Maxwell’s equations:
  D  v
B
E  
t
B  0
D
H  J 
t
Where;
E = electric field intensity
D = electric flux density
ρv = electric charge density
per unit volume
H = magnetic field intensity
B = magnetic flux density
3
Maxwell’s equations
Maxwell’s equations:
Relationship:
  D  v
B
E  
t
B  0
D
H  J 
t
D=εE
B=µH
ε =
µ =
electrical permittivity of
the material
magnetic permeability
of the material
4
Maxwell’s equations
• For static case, ∂/∂t = 0.
• Maxwell’s equations is reduced to:
Electrostatics
  D  v
 E  0
Magnetostatics
 B  0
 H  J
5
Charge and current distributions
 Charge may be distributed over a volume,
a surface or a line.
 Electric field due to continuous charge
distributions:
6
Charge and current distributions
Volume charge
density, ρv is defined
as:
q dq
 v  lim

C/m 3
v 0 v
dv

Total charge Q
contained in
a volume V is:

Q    v dV
C 
v
7
Charge and current distributions
• Surface charge
density

q dq
 s  lim

C/m 2
s 0 s
ds
• Total charge Q on
a surface:

Q    S dS C 
S
8
Charge and current distributions
• Line charge
density
q dq
C/m 
 l  lim

l 0 l
dl
• Total charge Q
along a line
Q    l dl C 
l
9
Example 1
Calculate the total charge Q contained in a
cylindrical tube of charge oriented along the zaxis. The line charge density is l  2 z ,
where z is the distance in meters from the
bottom end of the tube. The tube length is 10
cm.
10
Solution to Example 1
The total charge Q is:
  dz   2 zdz  z 
0.1
Q
0.1
l
0
2 0.1
0
 0.01 C
0
11
Example 2
Find the total charge
over the volume with
volume charge density:
V  5e
10 5 z
C
m3
12
Solution to Example 2
The total charge Q:
Q   V dV

V
0.01 2
 
0.04
  5e
10 5 z
dddz
  0   0 z  0.02
 7.854  10 14 C
13
Current densities
• Current density, J is defined as:

J   v u A/m 2

• Where:
u = mean velocity of moving charges
• For surface S, total current flowing through
is
I   J  ds A 
S
14
Current densities
• There are 2 types of current:
1) Convection current
 generated by actual movement of electrically
charged matter; does NOT obey Ohm’s law
 E.g movement of charged particles in cathode
ray tube
2) Conduction current
 atoms of conducting material do NOT move;
obeys Ohm’s law
 E.g movement of electrons in a metal wire
15
Coulomb’s Law
Coulomb’s law for
a point charge:
ˆ
ER
q
(V/m)
2
4R
Where;
R = distance between P and q
R̂ = unit vector from q to P
ε = electrical permittivity of the medium containing
the observation point P
16
Force acting on a charge
• In the presence of an electric field E at a given
point in space,
F  q'E (N)
• F is the force acting on a test charge q’ when
that charge is placed at that given point in space
• The electric field E is maybe due to a single
charge or a distribution of many charges
Units:
F in Newtons (N), q’ in Coulombs (C)
17
For acting on a charge
 For a material with electrical permittivity, ε:
D=εE
where:
ε = εR ε0
ε0 = 8.85 × 10−12 ≈ (1/36π) × 10−9 (F/m)
 For most material and under most condition, ε is
constant, independent of the magnitude and
direction of E
18
E-field due to multipoint charges
• At point P, the electric
field E1 due to q1 alone:
E1 
q1 R - R1 
4 R  R1
(V/m)
3
• At point P, the electric
field E1 due to q2 alone:
E2 
q 2 R - R 2 
4 R  R 2
3
(V/m)
19
E-field due to multipoint charges
• Total electric field E at
point P due to two
charges:
E  E1  E 2

R - R2  
1  R - R 1 
q1


 q2
3
3
4  R  R 1
R  R 2 

20
E-field due to multipoint charges
• In general for case of N point of charges,
E
1
N

4
i 1
qi R  Ri 
R  Ri
3
V/m 
21
Example 3
q1  2 105 C
q2  4 105 C
Two point charges with
and
are located in free space at (1, 3,−1) and
(−3, 1,−2), respectively in a Cartesian coordinate
system.
Find:
(a) the electric field E at (3, 1,−2)
(b) the force on a 8 × 10−5 C charge located at that
point. All distances are in meters.
22
Solution to Example 3
• The electric field E with ε = ε0 (free space) is
given by:
E  E1  E 2

R - R2  
1  R - R 1 
q1


 q2
3
3
4  R  R 1
R  R 2 

• The vectors are:
R1  xˆ  yˆ 3  zˆ , R 2  xˆ 3  yˆ  zˆ 2, R  xˆ 3  yˆ  zˆ 2
23
Solution to Example 3
• a) Hence,
xˆ  yˆ 4  zˆ 2
E
 10 5
1080
V/m 
• b) We have
xˆ  yˆ 4  zˆ 2
xˆ 2  yˆ 8  zˆ 4
5
F  q3 E  8  10 
 10 
 10 10
1080
270
5
N 
24
E-field due to charge distribution
continuous distribution
• Total electric field due to 3 types of continuous
charge distribution:
 v dv'
ˆ
E   dE 
R
(volume distributi on)
2

4 v '
R'
v
1
 s ds '
ˆ
E   dE 
R
(surface distributi on)
2

4 S '
R'
S'
1
 l dl '
ˆ
E   dE 
R 2 (line distributi on)

4 l '
R'
l'
1
25
E-field of a ring of charge
• Electric field due to a ring of charge is:
dE  zˆ

h
40 b  h
2

2 3/ 2
Q
26
Example 4
Find the electric field at a
point P(0, 0, h) in free
space at a height h on
the z-axis due to a
circular disk of charge in
the x–y plane with
uniform charge density
ρs as shown.
Then evaluate E for the
infinite-sheet case by
letting a→∞.
27
Solution to Example 4
• A ring of radius r and width dr has an area
ds = 2πrdr
• The charge is:
 s ds  2s rdr  dq
• The field due to the ring is:
dE  zˆ

h
40 r 2  h

2 3/ 2
2s rdr 
28
Solution to Example 4
• The total electric field at P is
sh
s
rdr
E  zˆ
  zˆ
3
/
2

2 0 0 r 2  h 2 
2 0
a


h
1 

2
2
a h 

With plus sign corresponds to h>0, minus sign corresponds to h<0.
• For an infinite sheet of charge with a =∞,
s
infinite sheet of charge 
E  zˆ
2 0
29
Gauss’s law
• Electric flux density D through an enclosing
surface is proportional to enclosed charge Q.
• Differential and integral form of Gauss’s law:
  D  ρv
Gauss' s law 
 D  ds  Q Gauss' s law 
S
30
Example 5
• Use Gauss’s law to obtain an expression
for E in free space due to an infinitely
long line of charge with uniform charge
density ρl along the z-axis.
31
Solution to Example 5
Construct a cylindrical Gaussian surface.
The integral is:
h
Q
2
  rˆD
r
 rˆrddz
z 0 0
Q  2hDr r .... (1)
But Q  ρl h
.... (2)
Equating both equations, and re-arrange, we get:
l
Dr 
2r
32
Solution to Example 5
Then, use D   0 E for free space , we get:
l
infinite line of charge 
E
 rˆ
 rˆ
0
0
20 r
D
Dr
Note: unit vector r̂ is inserted for E due to
the fact that E is a vector in r̂ direction.
33
Electric scalar potential
• Electric potential energy is required to
move a unit charge between 2 points
• The presence of an electric field between
two points give rise to voltage difference
34
Electric Potential as a function
of electric field
• Integrating along any path between point P1 and
P2, we get:
Potential difference between P1 and P2 ,
regardless of path 1, 2 or 3 :
P2
V21  V2  V1    E  dl
P1
35
Electric Potential as a function
of electric field
• Kirchhoff’s voltage law states that the net
voltage drop around a closed loop is zero.
• Line integral E around closed contour C is:
 E  dl  0 Electrosta tics 
C
• The electric potential V at any point is given by:
P2
V    E  dl (V)
P1
• Integration path between point P1 and P2 is
arbitrary
36
Electric potential due to point
charges
• For a point charge located at the origin of a
spherical coordinate systems, the electric field at
a distance R:
ˆ
ER
q
(V/m)
2
4R
• The electric potential between two end points:
V  
R

q

ˆ q  ˆ
 RdR
R
2 
 4R 
4R
ˆ dR (arbitrary path)
dl  R
(V)
37
Electric potential due to point
charges
• For charge Q located other than origin, specified
by a source position vector R1, then V at the
observation vector R becomes:
q
V R  
4 R  R1
V
• For N discrete point charges, electric potential is:
V R  
1
N

4
i 1
qi
R  Ri
V 
38
Electric potential due to
continuous distributions
• For a continuous charge distribution:
V (R ) 
v
1
dv' (volume

4 R '
distributi on)
v'
V (R ) 
s
1
ds ' (surface distributi on)

4 R '
S'
V (R ) 
l
1
dl ' (line distributi on)

4 R '
l'
39
Electric field as a function of
electric potential
• To find E for any charge distribution easily,
E  V
where:
V = gradient of V
40
Poisson’s & Laplace’s equations
• Differential form of Gauss’s law:
  D  ρv
• This may be written as:
ρV
E 

• Then, using E  V , we get:
ρV
  V   

41
Poisson’s & Laplace’s equations
• Hence:  2V   V

 2V  0
Poisson' s equation 
Laplace' s equation 
• Poisson’s and Laplace’s equations are used
to find V where boundaries are known:
• Example: the region between the plates of a
capacitor with a specified voltage difference
across it. (we will see in capacitance topic)
42
Conductivity
• Conductivity – characterizes the ease with
which charges can move freely in a material.
• Perfect dielectric, σ = 0. Charges do not move
inside the material
• Perfect conductor, σ = ∞. Charges move freely
throughout the material
43
•
Conductivity
• Drift velocity of electrons, u e in a conducting
material is in the opposite direction to the
externally applied electric field E:
u e    e E (m/s)
• Hole drift velocity, u h is in the same direction as
the applied electric field E:
u h   h E (m/s)
where:
µe = electron mobility (m2/V.s)
µh = hole mobility (m2/V.s)
44
Conductivity
• Conductivity of a material, σ, is defined as:
σ  - ρve μe  ρvh μ h
 N e μe  N h μ h  e S/m 
semiconduc tor 
   ρve μe  N e μee S/m  conductor 
where ρve = volume charge density of free electrons
ρvh = volume charge density of free holes
Ne = number of free electrons per unit volume
Nh = number of free holes per unit volume
e = absolute charge = 1.6 × 10−19 (C)
45
Conductivity
• Conductivities of different materials:
46
Ohm’s Law
• Point form of Ohm’s law states that:
J  E
A/m  Ohm' s law 
2
Where: J = current density
σ = conductivity
E = electric field intensity
• Properties for perfect dielectric and conductor:
Perfectdielectric with   0 : J  0, regardless of E
Perfectconductor with    : E  0, regardless of J
47
Example 6
A 2-mm-diameter copper wire with conductivity
of 5.8 × 107 S/m and electron mobility of 0.0032
(m2/V·s) is subjected to an electric field of 20
(mV/m).
Find (a) the volume charge density of free
electrons, (b) the current density, (c) the current
flowing in the wire, (d) the electron drift velocity,
and (e) the volume density of free electrons.
48
Solution to Example 6

5.8 107
 1.811010 C/m 3 
a) ve    
e
0.0032
b) J  E  5.8 107  20 103  1.16 106 A/m 2 
6
c) I  JA  1.16 106    4 10   3.64 A



4

d) u    E  0.0032  20  10 3  6.4  10 5 m/s
e
e
ve
10
1
.
81

10
29
3
e) N e  


1
.
13

10
electrons/
m
e
1.6 1019
49
Resistance
• The resistance R of a conductor of length l and
uniform cross section A (linear resistor) as
shown in the figure below:
l
R
( )
A
50
Resistance
• For a resistor of arbitrary shape, the resistance
R is:
V
R 
I
  E  dl
l

  E  dl
l
 J  ds  E  ds
S
S
• The conductance G of a linear resistor:
1 A
G 
R
l
S or siemens 
51
Joule’s Law
• Joule’s law states that for a volume v, the total
dissipated power P is:
P   E  Jdv
W  Joule' s law 
v
• For a linear resistor, the dissipated power P:
P  VI
 I 2R
52
Dielectrics
• Conductor has free electrons.
• Dielectric electrons are strongly bounded to the
atom.
• In a dielectric, an externally applied electric field,
Eext cannot cause mass migration of charges
since none are able to move freely.
• But, Eext can polarize the atoms or molecules in
the material.
• The polarization is represented by an electric
dipole.
53
Dielectrics
• Fig (a) - Eext is absent: the center of the electron
cloud is co-located with the center of the nucleus
• Fig (b) - Eext is present: the two centers are
separated by a distance d
54
• Fig (c) – an electric dipole caused by Eext
Electric Boundary Conditions
• Electric field maybe continuous in each of two
dissimilar media
• But, the E-field maybe discontinuous at the
boundary between them
• Boundary conditions specify how the tangential
and normal components of the field in one
medium are related to the components in other
medium across the boundary
• Two dissimilar media could be: two different
dielectrics, or a conductor and a dielectric, or
two conductors
55
Dielectric- dielectric boundary
• Interface between two dielectric media
56
Dielectric- dielectric boundary
• Based on the figure in previous slide:
• First boundary condition related to the tangential
components of the electric field E is:
E1t  E2t
V/m 
• Second boundary condition related to the normal
components of the electric field E is:
D1n  D2 n   S
• OR
 1 E1n   2 E2n   S
57
Perfect conductor
• When a conducting slab is placed in an external
electric field, E0
• Charges that accumulate on the conductor
surfaces induces an internal electric field Ei  E0
• Hence, total field inside conductor is zero.
58
Dielectric-conductor boundary
• Assume medium 1 is a dielectric
• Medium 2 is a perfect conductor
59
Dielectric-conductor boundary
• Based on the figure in previous slide:
• In a perfect conductor,
E  D  0 everywhere in the conductor
• Hence, E2  D2  0
• This requires the tangential and normal
components of E2 and D2 to be zero.
60
Dielectric-conductor boundary
• The fields in the dielectric medium, at the
boundary with the conductor is E1t  E2t .
• Since E2t  0 , it follows that E1t  D1t  0 .
• Using the equation, D1n   s ,
• we get: D1n  1E1n   s
• Hence, boundary condition at conductor surface:
D1   1 E1  nˆ s
at conductor surface 
where n̂ = normal vector pointing outward
61
Conductor- conductor boundary
• Boundary between two conducting media:
• Using the 1st and 2nd boundary conditions:
E1t  E2t V/m  and  1 E1n   2 E2 n   S
62
Conductor- conductor boundary
• In conducting media, electric fields give rise to
current densities.
• From J  E, we have:
J 1t
1

J 2t
2
and
1
J 1n
1
2
J 2n
2
 S
• The normal component of J has be continuous
across the boundary between two different
media under electrostatic conditions.
63
Conductor- conductor boundary
• Hence, upon setting J 1n  J 2 n , we found the
boundary condition for conductor- conductor
boundary:
 1  2 
J1n     ρs
 1  2 
electrosta tics 
64
Capacitance
• Capacitor – two conducting bodies
separated by a dielectric medium
65
Capacitance
• Capacitance is defined as:
Q
C
V
C/V
or F
where: V = potential difference (V)
Q = charge (C)
C = capacitance (F)
66
Example 7
Obtain an expression for the capacitance C of a
parallel-plate capacitor comprised of two parallel
plates each of surface area A and separated by
a distance d. The capacitor is filled with a
dielectric material with permittivity ε.
67
Solution to Example 7
• The charge density on the upper plate is ρs =
Q/A. Hence,
E   zˆE
• magnitude of E at the dielectric-conductor
boundary:
E   s   Q / A
• The voltage difference is
d
d
0
0
V   E  dl    zˆE   zˆdz  Ed
• Hence, the capacitance is:
C
Q
Q
A


V Ed
d
68
Electrostatic potential energy
• Assume a capacitor with plates of good
conductors – zero resistance,
• Dielectric between two conductors has negligible
conductivity, σ ≈ 0 – no current can flow through
dielectric
• No ohmic losses occur anywhere in capacitor
• When a source is connected to a capacitor,
energy is stored in capacitor
• Charging-up energy is stored in the form of
electrostatic potential energy in the dielectric
medium
69
Electrostatic potential energy
1
2
W

CV
• Electrostatic potential energy, e
2
Q
Q
A

• The capacitance: C  
V Ed
d
• Hence, We for a parallel plate capacitor:
1 A
1 2
1 2
2
Ed   E ( Ad )  E v
We 
2 d
2
2
where V  Ed (voltage across capacitor)
v  Ad (volume of the capacitor)
70
Image Method
• Image theory states that a charge Q above a
grounded perfectly conducting plane is equal to
Q and its image –Q with ground plane removed.
71
Example 8
Use image theory to determine E at an arbitrary
point P (x, y, z) in the region z > 0 due to a
charge Q in free space at a distance d above a
grounded conducting plane.
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Solution to Example 8
• Charge Q is at (0, 0, d) and its image −Q is at
(0,0,−d) in Cartesian coordinates. Using
Coulomb’s law, E at point P(x,y,z) due to two
point charges:
 x̂x  ŷy  ẑz  d 

 2

2 3/ 2
2
1  QR1  QR 2 
Q  x  y  z  d 

 3 

E

x̂x  ŷy  ẑz  d  
40  R1
R23  40 


3
/
2
2
 x 2  y 2  z  d 





73