3. ELECTROSTATICS Ruzelita Ngadiran
3. ELECTROSTATICSRuzelita Ngadiran
Chapter 4 Overview
Maxwell’s equations
Where;
E = electric field intensityD = electric flux densityρv = electric charge density
per unit volumeH = magnetic field intensityB = magnetic flux densityt
t
DJH
0B
BE
D v
Maxwell’s equations:3
Maxwell’s Equations
God said:
And there was light!
Maxwell’s equations
Relationship:
D = ε E
B = µ H
ε = electrical permittivity of the material
µ = magnetic permeability of the material
Maxwell’s equations:
t
t
DJH
0B
BE
D v
5
Maxwell’s equations
For static case, ∂/∂t = 0. Maxwell’s equations is reduced to:
Electrostatics Magnetostatics
0
E
D vJH
B
0
6
Charge and current distributions
Charge may be distributed over a volume, a surface or a line.
Electric field due to continuous charge distributions:
7
Charge and current distributions
3
0C/m lim
dv
dq
v
qv
v
C VdQv
v
Volume charge density, ρv is defined as:
Total charge Q contained in a volume V is:
8
Charge and current distributions
Surface charge density
2
0C/m lim
ds
dq
s
qs
s
• Total charge Q on a surface:
C SdQS
S
9
Charge and current distributions
Line charge density
C/m lim0 dl
dq
l
ql
l
• Total charge Q along a line
C ldQl
l
10
Charge Distributions
Volume charge density:
Total Charge in a Volume
Surface and Line Charge Densities
Example 1
Calculate the total charge Q contained in a cylindrical tube of charge oriented along the z-axis. 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.
zl 2
12
Solution to Example 1
The total charge Q is:
C 01.021.0
02
1.0
0
1.0
0
zzdzdzQ l
13
Example 2
Find the total charge
over the volume with
volume charge
density:
3105
5m
Ce zV
14
Solution to Example 2
The total charge Q:
C
dzdde
dVQ
z
z
V
V
10854.7
5
14
01.0
0
2
0
04.0
02.0
105
15
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
For a material with electrical permittivity, ε:
D = ε E
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
19
At point P, the electric field E1 due to q1 alone:
At point P, the electric field E1 due to q2 alone:
(V/m)
4
-3
1
111
RR
RRE
q
(V/m)
4
-3
2
222
RR
RRE
q
E-field due to multipoint charges
20
Electric Field Due to 2 Charges
Example 3
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.
C102 51
q C104 52
q
22
Solution to Example 3
The electric field E with ε = ε0 (free space) is given by:
The vectors are:
2ˆˆ3ˆ ,2ˆˆ3ˆ ,ˆ3ˆˆ 21 zyxRzyxRzyxR
32
223
1
11
21
--
4
1
RR
RR
RR
RR
EEE
23
Solution to Example 3
a) Hence,
b) We have
V/m 10108
2ˆ4ˆˆ 5
0
zyx
E
N 1027
4ˆ8ˆ2ˆ10
108
2ˆ4ˆˆ108 10
0
5
0
53
zyxzyxEF q
24
Electric Field Due to Charge Distributions
Field due to:
Cont.
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→∞.
28
Solution to Example 4
A ring of radius r and width dr has an area ds = 2πrdr
The charge is:
The field due to the ring is:
dqrdrds ss 2
rdr
hr
hdE s
2
4ˆ
2/3220
z
29
Solution to Example 4 The total electric field at P is
With plus sign corresponds to h>0, minus sign corresponds to h<0.
For an infinite sheet of charge with a =∞,
22
002/322
0
12
ˆ2
ˆha
h
hr
rdrh sa
s
zzE
charge ofsheet infinite 2
ˆ0
szE
30
Gauss’s Law
Application of the divergence theorem gives:
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.
32
Solution to Example 5
Construct a cylindrical Gaussian surface.The integral is:
Equating both equations, and re-arrange, we get:
(1) .... 2
ˆˆ0
2
0
rhDQ
dzrdDQ
r
h
z
r
rr
(2) .... But hρQ l
rD l
r 2
33
Solution to Example 5
Then, use , we get:
Note: unit vector is inserted for E due to the fact that E is a vector in direction.
charge of line infinite 2
ˆˆ000 r
D lr
rr
DE
space freefor 0ED
r̂r̂
34
Applying Gauss’s Law
Construct an imaginary Gaussian cylinder of radius r and height h:
Electric Scalar Potential
Minimum force needed to move charge against E field:
Electric Scalar Potential
Electric Potential Due to Charges
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 a point charge, V at range R is:
For continuous charge distributions:
Relating E to V
Cont.
(cont.)
Poisson’s & Laplace’s Equations
In the absence of charges:
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, in a conducting
material is in the opposite direction to the externally applied electric field E:
Hole drift velocity, is in the same direction as the applied electric field E:
where: µe = electron mobility (m2/V.s)
µh = hole mobility (m2/V.s)
(m/s) Eu ee
(m/s) Eu hh
eu
hu
44
Conductivity
Conductivity of a material, σ, is defined as:
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)
torsemiconduc S/m e μNμN
μρμ- ρσ
hhee
hvheve
conductor S/m eμNμ ρ eeeve
45
Conductivity
Conductivities of different materials:
46
Conductivity
ve = volume charge density of electronshe = volume charge density of holese = electron mobilityh = hole mobilityNe = number of electrons per unit volumeNh = number of holes per unit volume
Conduction Current
Conduction current density:
Note how wide the range is, over 24 orders of magnitude
Resistance
For any conductor:
Longitudinal Resistor
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:
Piezoresistivity
The Greek word piezein means to press
R0 = resistance when F = 0F = applied forceA0 = cross-section when F = 0 = piezoresistive coefficient of material
Piezoresistors
Wheatstone Bridge
Wheatstone bridge is a high sensitivity circuit for measuring small changes in resistance
Dielectric Materials
Polarization Field
P = electric flux density induced by E
Electric Breakdown
Electric Breakdown
Boundary Conditions
Summary of Boundary Conditions
Remember E = 0 in a good conductor
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
Capacitance
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
Tech Brief 8: Supercapacitors
For a traditional parallel-plate capacitor, what is the maximum attainable energy density?
= permittivity of insulation materialV = applied voltage = density of insulation materiald = separation between plates
Mica has one of the highest dielectric strengths ~2 x 10**8 V/m. If we select a voltage rating of 1 V and a breakdown voltage of 2 V (50% safety), this will require that d be no smaller than 10 nm. For mica, = 60 and = 3 x 10**3 kg/m3 .
Hence:
W = 90 J/kg = 2.5 x10**‒2 Wh/kg.
By comparison, a lithium-ion battery has W = 1.5 x 10**2 Wh/kg, almost 4 orders of magnitude greater
Energy density is given by:
A supercapacitor is a “hybrid” battery/capacitor
Users of Supercapacitors
Energy Comparison
Electrostatic Potential Energy
Electrostatic potential energy density (Joules/volume)
Total electrostatic energy stored in a volume
Energy stored in a capacitor
Image Method
Image method simplifies calculation for E and V due to charges near conducting planes.
1.For each charge Q, add an image charge –Q2.Remove conducting plane3.Calculate field due to all charges