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Vector Analysis
Chapter 2
Static Electric Fields (3 Weeks)
Chapter 3.3 Coulomb’s Law
Chapter 3.4 Gauss’s Law and Applications
Chapter 3.5 Electric Potential
Chapter 3.6 Material Media in Static Electric Field
Chapter 3.8 Boundary conditions for Electrostatic Fields
Chapter 3.9 Capacitance and Capacitors
Intro.1
III. VECTOR ANALYSIS
3.1 Vector algebra
• addition
• associative law
A+(B+C) = (A+B)+C
• commutative law A+B = B+A
• multiplication by scalar
• distributive law
aB
B
a( B+C) = aB + aC
Intro.2
A.B = A B cos θ AB
• Scalar (or Dot) Product:
A.B = B.A
A.(B+C) = A.B + A.C
• Vector (or Cross) Product:
A × B ≡ a n A B sin( θ AB )
AxB
A
an
θ AB
B
Intro.3
Note for cross products,
A × B = −B × A ≠ B × A
A × (B × C) ≠ ( A × B) × C
Cartesian Co-ordinate System (unit vectors ax ay az)
A.B = A x B x + A y B y + A z B z
ax
ay
az
A × B = Ax
Bx
Ay
By
Az
Bz
Intro.4
Differential displacement vector dl
dl = dx ax + dy ay + dz az
dl
Example: To integrate F along the path y=x2 from a (1,1) to
b (2,4), where
F = ya x + a y
y = x2
b
b
∫ F.dl = ∫ (ya
a
x
+ a y )(dx a x + dy a y )
a
2
=
∫
1
4
1
ydx + ∫ dy = 5
3
1
Intro.5
Differential surface vector dS:
+ve in the ‘outward’ normal direction to the
dS
surface element for an enclosed volume
Example:
Find F.dS for the surface ABCD for the given F in the
following unit cube.
A
B
az
ax
F
C
D
ay
Intro.6
F = ( 2 y + z )a x − 2a y
d S = − dydz a x
for surface ABCD
1 1
∫ F. dS = − ∫ ∫ ( 2 y + z ) dydz
0 0
1
⎡
z ⎤
3
= − ∫ ⎢ 2 yz +
⎥ dy = −
2 ⎥⎦
2
⎣
0⎢
0
1
2
Note the minus sign in dS because positive direction is
outward from the enclosed volume
Intro.7
Cylindrical coordinate system
ρ=
⎧y⎫
φ = tan ⎨ ⎬
⎩x⎭
z=z
−1
z
a
ϕ
x
x2 + y2
ρ
y
ρ
× aφ = a
z
aφ × a
z
= a
×a
ρ
= aφ
a
z
ρ
Intro.8
Spherical coordinate system
r=
⎧⎪
z
θ = cos ⎨
⎪⎩ x 2 + y 2 + z 2
−1 ⎧ y ⎫
φ = tan ⎨ ⎬
⎩x⎭
−1
z
θ
r
ϕ
x
x2 + y2 + z2
y
⎫⎪
⎬
⎪⎭
a r × aθ = aφ
aφ × a r = aθ
aθ × aφ = a r
Intro.9
3.2 Scalar and Vector Calculus
Integrals of scalars and vectors over volumes, surfaces and lines
often used in electromagnetics. Examples:
ρv is the charge density (per unit volume)
D is the electric flux density (or electric displacement)
E is the electric field intensity
∫ ρ dv
Total charge enclosed within volume V
∫ D. dS
Total electric flux passing through the surface S
∫ E. dl
Potential difference between two points on the
line
v
V
S
C
Intro.10
Gradient of scalar field f
∂f
∂f
∂f
grad f = ∇ f =
ax +
ay +
az
∂x
∂y
∂z
gradf is a vector in the direction of maximum increase of the
field f.
∂f
∇f =
∂l
an
max
an is unit vector in the direction of maximum
increase of f
Intro.11
Divergence of a vector field A:
div A ≡ lim
∆v→ 0
∫ A. d S
S
∆v
If we consider A as a flux density (per unit surface area), the
closed surface integral represents the net flux leaving the
volume ∆v
In rectangular coordinates,
∂Ay
∂Ax
∂Az
+
+
div A = ∇ .A =
∂y
∂z
∂x
Intro.12
Divergence Theorem:
If A is a vector, then for a volume V surrounded by a closed
surface S,
∫
V
∇ .A dv =
∫ A. d S
S
The above integral represents the net flex leaving the closed
surface S if A is the flux density
V
S
Intro.13
Curl of a vector field: Curl of a vector A is a measure of the
tendency of A to “push or pull” around a closed path that
encircles a point. Component of curlA in direction ai
( curl A ) i ≡ ( ∇ × A ) i ≡ lim
∆Si → 0
∫ A. d l
Ci
∆Si
∆Si
Ci
ai
Intro.14
Since the maximum value of any component of a vector is
equal to the magnitude of the vector,
⎡
∇ × A = a n ⎢ lim
⎢ ∆S → 0
⎣
⎤
d
A.
l
∫C
⎥
∆S ⎥
⎦ max
In rectangular coordinates
ax
∂
∇×A =
∂x
Ax
ay
∂
∂y
Ay
az
∂
∂z
Az
Intro.15
Stokes’s theorem: For an open surface S bounded by a
contour C,
∫
S
(∇ × A ) .d S =
∫ A. d l
C
C
S
The line integrals from adjacent cells cancel leaving the
only the contribution along the contour C which bounds the
surface S.
Intro.16
IV. ELECTROSTATIC FIELDS (Time-Invariant)
The electric field intensity E is defined as the force on an unit
positive charge at a point.
D is the electric flux density (or electric displacement density). The
direction of D is that of the electric flux at the point, and its
magnitude is the no.of flux lines passing through an unit normal
surface area.
4.1 Gauss’s Law :
∇ .D = ρ v
Point form
∫
Integral form
S
D. d S = Q
Intro.17
ρv is the space volume charge density (coulomb per unit
volume) at a point. Q is the total charge enclosed within a
closed surface S. Unit of D is C/m2.
Relation between D and E:
D = ε oε r E = εE
εo is permittivity of free space (vacuum), εr is the relative
permittivity (or dielectric constant) of the medium material,
and ε is the permittivity of the medium material. (Unit of εo and
ε is Farad/metre)
In free space,
εr = 1
D = ε oE
Intro.18
• Gauss’s law can determine the electric field pattern due to any
distribution of charges.
• In the simplest case, consider the field at a point P of distance r
from a point charge of magnitude Q in free space.
Construct a spherical surface with
centre at Q and radius r. E is the same
everywhere on the surface. Applying
integral form of Gauss’s law gives
P
r
+Q
4π r E =
Q
2
E=
εo
Q
4πε o r
2
ar
Intro.19
The force acting a test charge q at point P is then (Coulomb’s law):
qQ
F=
ar
2
4πε o r
1
Example : Find the electric field due to a spherical charge
distribution of radius a with uniform volume charge density ρv in free
space.
Gaussian
surface of
radius r
a
Intro.20
Solution:
Case (i): For a Gaussian surface of radius r less than the radius
a of the charge distribution, the total charge enclosed by the
Gaussian surface is:
4
Q =
πr3ρv
3
Applying Gauss’s Law to the Gaussian surface,
Q
2
4π r E ( r ) =
εo
rρ
E (r ) =
ar
3ε o
Case (ii): For a Gaussian surface of radius r greater than a :
4
Q =
πa3ρ
3
Intro.21
Applying Gauss’s law,
4π r 2 E ( r ) =
E (r ) =
Q
εo
a3ρ
3ε o r
2
ar
4.2 Electric (electrostatic) potential V:
The electrostatic field is conservative, i.e.
∇×E = 0
∫ E. dl = 0
C
According to vector identity, for any scalar V
∇ × (∇ V ) ≡ 0
Intro.22
• So we can define a scalar electric potential V which is easier to
work with,
E = −∇ V
(Hence unit of E is volt/metre)
b
Integrating along a path
V b − V a = − E. d l
∫
a
• Note that the integral is independent of the path taken.
Physically the potential difference is equal to the work done
in moving an unit charge from point A to point B.
• Potential due to N point charges:
V =
1
4πε o
N
∑
k =1
Qk
r − rk
Intro.23
4.3 Capacitance
Capacitance between two conductors is defined as:
Q
C=
V
where Q is the charge on the conductor and V is the
potential difference between the conductors.
Intro.24
Example: Find the potential difference between two coaxial
cyclinders of radius a, b in the following diagram. Assume the
surface charge density (per unit area) on the inner conductor is ρs.
Hence find the capacitance per unit length of the cylinder.
Solution:
ρs
Construct a Gaussian cylindrical
surface of unit length and radius r.
The E-field at radius r is:
r
b
ρ s 2π a
2π r E ( r ) =
ε
ρsa
E(r ) =
ar
ε r
Intro.25
The potential difference Vab between outer and inner conductor:
a
V ab = V a − Vb = − ∫ E ( r ).d r
b
ρ s adr
ρ sa ⎛ b ⎞
ln ⎜ ⎟
= −∫
=
ε r
ε
⎝a⎠
b
a
Capacitance per unit length
2πa ρ s
C=
Vba
2πε
=
⎛b⎞
ln ⎜ ⎟
⎝a⎠
Intro.26
4.5 Dielectric material
• Dielectrics are insulating materials, and contain bound charges
which cannot move freely to generate currents.
• Bound charges can move short distances under an electric field to
form electric dipoles. (A dipole is a pair of equal and opposite
charges (+Q,-Q) separated at a small distance r. The dipole
moment p is equal to Qrar. )
p
t
+ve
E
-ve
Zero field
Under a field
Intro.27
• From a macroscopic view point, all the minute dipole moments
pk add up to result in a net dipole moment per unit volume called
the polarization vector P (C/m2):
volum e
∆v
N ∆v
P = lim
∆v → 0
∑p
k =1
k
pk
∆v
where N is the number of dipoles per unit volume.
• The polarization (or bound) charges can also contribute to the
electric field E as the free charges. (Note ρv defined before is
free charge density.) It can be shown that D is related to E and
P according to the following equation:
Intro.28
D = ε oE + P
• In most materials, P is in the same direction and
proportional to E (P = εoχE), so that
D = ε o E + ε o χE
= ε oε r E = εE
where εo is the permittivity of vacuum, εr is the relative
permittivity (or dielectric constant) of the material, ε is
the permittivity of the material, and χ is the susceptibility
of the material.
Intro.29
4.6 Presence of a conductor
Electric current will flow in a conductor when there is an electric
field. The current density J (A/m2) is proportional to the electric field
intensity E.
J = σE
where σ is the conductivity of the material. This is another form of
Ohm’s Law. The total current I flowing through a surface S is
given by:
I = ∫ J. d S
S
In the interior of a perfect conductor, σ is very large so that
E can be assumed to be zero.
Intro.30
END
Intro.31
dS
D
θ
D. d S = D d S cos θ
If D is the flux density (per
unit surface area), D.dS
represents the flux leaving the
surface element dS in the
normal direction. So
∫ D. dS
S
is the total flux leaving the entire
surface S.
Intro.32
The physical meaning of the integral form of Gauss Law: the
total electric flux leaving a closed surface
∫ D. dS is equal to the net charge enclosed by the
S
surface. The net charge is just equal to the summation of all the
positive or negative charges inside. Note that the variation of
the positions of the charges does NOT alter the total flux
leaving the surface; the variation only affects the distribution
pattern of the electric lines of force.
Intro.33
∫ E. dl = 0
An electrostatic field is conservative, i.e.
C
It means the energy required to move any charge around a
closed loop is zero! The reason is that in moving along the
path, sometimes work is done by the electric force on the
charge, and sometimes work is done against the electric force.
If we regard ∫ E.d l as the potential
difference between two points on the path, ∫ E. dl must
C
be zero because it is the potential difference between the same
points.
Note that
∫ E. dl
is not necessarily zero in a time-
C
varying magnetic field because of the induced e.m.f. due to the
magnetic flux.
Intro.34