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3. ELECTROSTATICS
Ruzelita Ngadiran
Chapter 4 Overview
Maxwell’s equations
3
Maxwell’s equations:
  D  v
Where;
B
E  
t
B  0
D
H  J 
t
E = electric field intensity
D = electric flux density
ρv = electric charge density
per unit volume
H = magnetic field intensity
B = magnetic flux density
Maxwell’s Equations
Maxwell’s equations
5


For static case, ∂/∂t = 0.
Maxwell’s equations is reduced to:
Electrostatics
Magnetostatics
  D  v
 B  0
 H  J
 E  0
Maxwell’s equations
6
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
Charge and current distributions
7


Charge may be distributed over a volume, a surface
or a line.
Electric field due to continuous charge distributions:
Charge and current distributions
8
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
v
C 
Charge and current distributions
9

• Total charge Q on
a surface:
Surface charge density

q dq
 s  lim

C/m 2
s 0 s
ds

Q    S dS C 
S
Charge and current distributions
10

Line charge density
q dq
C/m 
 l  lim

l 0 l
dl
• Total charge Q
along a line
Q    l dl C 
l
Charge Distributions
Volume charge density:
Total Charge in a Volume
Surface and Line Charge Densities
Example 1
12
Calculate the total charge Q contained in a
cylindrical tube of charge oriented along the z-axis.
, l  2 z
The line charge density is
where z is the distance in meters from the bottom
end of the tube. The tube length is 10 cm.
Solution to Example 1
13
The total charge Q is:
  dz   2 zdz  z 
0.1
Q
0.1
l
0
0
2 0.1
0
 0.01 C
Example 2
14
Find the total charge over
the volume with volume
charge density:
V  5e
10 5 z
C
m3
Solution to Example 2
15
The total charge Q:
Q   V dV
V
0.01 2

0.04
    5e
105 z
 0  0 z  0.02
 7.854 10 14 C
dddz
Current Density
For a surface with any orientation:
J is called the current density
Convection vs. Conduction
Coulomb’s Law
Electric field at point P due to single charge
Electric force on a test charge placed at P
Electric flux density D
For acting on a charge
19

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
E-field due to multipoint charges
20

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)
Electric Field Due to 2 Charges
Example 3
22
5
q

2

10
C
with 1
q2  4 105 Care
Two point charges
and
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.
Solution to Example 3
23

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
Solution to Example 3
24


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 
Electric Field Due to Charge Distributions
Field due to:
Cont.
Example 4
28
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→∞.
Solution to Example 4
29

A ring of radius r and width dr has an area
ds = 2πrdr
 s ds  2s rdr  dq

The charge is:

The field due to the ring is:
dE  zˆ

h
40 r 2  h

2 3/ 2
2s rdr 
Solution to Example 4
30

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
Gauss’s Law
Application of the divergence theorem gives:
Example 5
32

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.
Solution to Example 5
33
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
Solution to Example 5
34
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.
Electric Scalar Potential
Minimum force needed to move charge
against E field:
Electric Scalar Potential
Electric Potential Due to Charges
For a point charge, V at range R is:
In electric circuits, we usually select a
convenient node that we call ground and
assign it zero reference voltage. In free
space and material media, we choose infinity
as reference with V = 0. Hence, at a point P
For continuous charge distributions:
Relating E to V
Cont.
(cont.)
Poisson’s & Laplace’s Equations
In the absence of charges:
Conductivity
42



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
Conductivity
43

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 is
h 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)
Conductivity
44

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  Ne 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)
Conductivity
45

Conductivities of different materials:
Conductivity
ve = volume charge density of
electrons
he = volume charge density of
holes
e = electron mobility
h = hole mobility
Ne = number of electrons per unit
volume
Nh = number of holes per unit
volume
Conduction Current
Conduction current density:
Note how wide the range is, over 24 orders
of magnitude
Resistance
Longitudinal Resistor
For any conductor:
G’=0 if the insulating material is air or a
perfect dielectric with zero conductivity.
Joule’s Law
The power dissipated in a
volume containing electric field E
and current density J is:
For a resistor, Joule’s law reduces to:
For a coaxial cable:
Dielectrics
53





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.
Dielectric Materials
Polarization Field
P = electric flux density induced by E
Electric Breakdown
Electric Breakdown
Electric Boundary Conditions
57




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
Boundary Conditions
Dielectric-conductor boundary
59


Assume medium 1 is a dielectric
Medium 2 is a perfect conductor
Dielectric-conductor boundary
60





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
Conductors
Net electric field inside a conductor is zero
Field Lines at Conductor Boundary
At conductor boundary, E field direction is always
perpendicular to conductor surface
Summary of Boundary Conditions
Remember E = 0 in a good conductor
Example 1

Find E1
two different
Dielectrics
 i)

ii)
Conductor- conductor boundary
65


Boundary between two conducting media:
Using the 1st and 2nd boundary conditions:
E1t  E2t V/m  and  1 E1n   2 E2 n   S
Conductor- conductor boundary
66


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.
Capacitance
Capacitor – two conducting bodies
separated by a dielectric medium
Capacitance
For any two-conductor configuration:
For any resistor:
Application of Gauss’s law gives:
Q is total charge on inside of outer
cylinder, and –Q is on outside surface of
inner cylinder
Previous coaxial
conductance Slide 51
Electrostatic potential energy
71





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
Electrostatic potential energy
72

Electrostatic potential energy,
1
We  CV 2
2
Q
Q
A
C 

V Ed
d

The capacitance:

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)
Electrostatic Potential Energy
Electrostatic potential energy density (Joules/volume)
Energy stored in a capacitor
Total electrostatic energy stored in a volume
example

Given that radius of inner and outer conductor in
coaxial cable are 2 cm and 6 cm respectively,
permittivity for the dielectric media are 4. The
charge density is 10^-4 C/m. Calculate the energy
stored in 20 cm length cable. Given :
l
infinite line of charge 
E  rˆ
2r
Image Method
Image method simplifies calculation for E and V due Image theory states that a
to charges near conducting planes.
charge Q above a grounded
perfectly conducting plane is
1. For each charge Q, add an image charge –Q
equal to Q and its image –Q with
2. Remove conducting plane
ground plane removed.
3. Calculate field due to all charges