Week 2 Parts of Chapter 2 of Text
2.6 : Electric Flux Density
2.7: Gauss Law and its applications
2.8 : Divergence and the point form of Gauss’ Law
2.9: Electric Potential
2.10 Conductors and Ohm’s Law
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
(a) Metallic sphere of radius a with charge +Q is enclosed by a pair of hemispherical
shells of radius b. (b) The outer shell is briefly grounded, (c) allowing a charge –Q to
accumulate on the outer shell.
Figure 2-28 (p. 44
Electric Flux density D is measured in
Coulombs/meter2
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Electric Flux (Ψ) Density (Coulombs/m2)
D = Ψar/4πr2
Electric Field Intensity
E = Qar/4πεor2 (volts/m)
The two expressions are very similar if
D = εoE
When we deal with D, we think of Electric flux
lines starting from a + charge and terminating at –
charge independent of the medium.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Electric Flux Ψ = |D| |S| cosθ
Ψ = |D| |S| cosθ or the dot product = D.S
If D varies over the surface over which Ψ is desired then,
Ψ = ∫D.dS
Figure 2-29 (p. 45)
The flux through a surface that is at an angle to the direction of flux (a) is less than
the flux through an equivalent surface normal to the direction of flux (b).
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Dot product of A and B Vectors
Definition:
Cartesian coordinates
Spherical Polar Coordinates:
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Gauss Law
“ The net Electric flux through any closed surface is
equal to the total charge enclosed by that surface”
Gauss law in an integral form would then be
Used to obtain E-fields or E-Flux density D for
different charge distributions. Have to construct a
Gaussian surface consistent with charge
distribution symmetry.
MAXWELL’S 1st EQUATION IN AN INTEGRAL FORM
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 1 :Using Gauss’ law , calculate Electric Field due to a point charge Q
Consider a spherical Gaussian surface
surrounding a point charge at the origin.
See Figure 2-30 (p. 48). We calculate the
flux of D through the Gaussian surface
and equate it to the total charge enclosed
in that surface.
See Slide 35 week 1
Er = Qar/4πεor2
Er is spherically symmetric and decreases as 1/r2
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 2. :Calculate E at point P due to line charge ρL everywhere in spaceWe construct a cylindrical Gaussian surface
containing the point P placed around a
section of an infinite length line of charge
density ρL occupying the z-axis. We calculate
the flux of D through the Gaussian surface.
Then equate this flux to the charge inside this
surface- Gauss law. Why use cylindrical
coordinates?
E =ρLaρ/2πεoρ
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 3 Electric Field due to an
infinite plane of charge ρs every where
in spaceOur Gaussian surface is a rectangular
pill box, of dimensions 2x,2y and 2h.
The charge enclosed in the pill box is
Q = ρs (2x) (2y).
Flux of D from the sides is zero. We
need to consider flux through the top
and bottom surfaces.
8Dzxy = ρs 4xy or Dz = ρs/2 or Ez = ρsaz/2εo
The Ez points up at z > 0, and it points down at z < 0.
Figure 2-32 (p. 50)
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 4 :Infinitely long Coax cable.Obtain Efield at all points. We consider (a) Coaxial cable with
a solid inner conductor of radius a and charge density
ρv surrounded by a grounded thin conductive shell at
radius b. (b) Cross section of the inner conductor
with field components for a pair of differential charge
elements
Figure 2-33ab (p. 51)
.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Figure 2-33c (p. 52). Consider cross section of the
coaxial cable showing the three Gaussian surfaces, GS1,
GS2, and GS3.
There are three regions of interest, ρ < a,
a < ρ < b, and finally ρ> b.
Flux of D through GS1 is
Charge enclosed in Gaussian
surface can be obtained
The D-field in region 1, ρ < a can be calculated using Gauss’ law
The D-field at the 2nd GS , ρ= a, once the total charge Qencl = πρvha2
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
In region 3 , since the outer conductor is grounded, the total net charge
emclosed by the 3rd GS is therefore zero. The figure 2.34 gives the Dfield in the three regions, for the example Matlab 2.6.
Region 1 , D-field increases
linearly with ρ.
Region 1
Region 2
Region 3
Region 2, 3 < ρ < 6, the D-field
decreases as 1/ ρ, and finally
In region 3, ρ > 6, the D-field is
zero.
Figure 2-34 (p. 54)
Dρ versus radius for the coaxial cable in MATLAB 2.6.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Divergence of a vector D at some point
P is the flux of that vector emerging
from a small volume ∆x∆y∆z placed at
that point, as the volume gets
infinitesimally smaller.
By calculating the flux of D from the
cube, we show in equa. 2.58 on p.56
div D = lim∆v→0 [∫D.dS]/∆v (1)
div D = ∂Dx/∂x + ∂Dy/∂y + ∂Dz/∂z . We can
now cast Gauss’ law in a point form. In
equation (1) above, we now replace, [∫D.dS]
by Qencl (The integral form of Gauss’ law),
and obtain,
Figure 2-35 (p. 55)
A differential volume element
used to derive divergence.
divD = ρv
This is Gauss’ law in a differential
form.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Divergence of D
Consider a vector field DP =Dxoax +Dyoay+Dzoaz
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Divergence of a vector in different coordinate
systems
Cartesian
Cylindrical
Spherical Polar
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
The divergence Theorem
The divergence of a vector at a point is related to the Flux of
the vector out of a small volume placed at that point.
Sadiku Elements of EM
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Curl of a vector
Curl of a vector H at a point provides a measure of
the line integral or circulation of the vector on a
small loop placed at that point. Stoke’s Theorem.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Consider a piston arrangement enclosing a finite volume
of air. Let us focus on the spherical volume in blue. At
equilibrium air molecules come and leave the spherical
volume in equal amounts. There is no divergence.
Now let us move the piston up as air expands. With an increased volume more air
molecules will leave the spherical volume than go in. We view that to be the case
of a positive divergence. Moving the piston down will compress air, and will cause
more molecules to flow in the blue spherical region than come out- corresponding
to a negative divergence. Figure 2-36 (p. 57).
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 2.16 p. 58
Suppose D = ρ2aρ. Find the flux through the surface of a
cylinder with 0 < z < h and ρ = a by evaluating the (a) left hand
side of the Divergence Theorem and (b) the right hand side of
the Divergence Theorem.
●D = ρ2aρ implies a D-field
that is increasing radially out
in a cylindrical geometry. So
let us calculate the LHS of the
Divergence Theorem
.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
What can we say about the surface area elements dS on the
side and the top and the bottom?
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
So we just calculated the LHS of the Divergence Theorem for the
problem, and found
Ψ = 2πha3
Next let us calculate the RHS of the Divergence Theorem
How do we write the Divergence of a vector in cylindrical
coordinates? See back page of the book.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Electric Potential
Work done by a force F in displacing an object by a small length dL is defined as,
dW = F.dL
For a finite displacement, we would then write,
Work done by an E-field in moving a charge Q from point a to b
FE = QE
o→E o→E
b
a
If an external force moves a charge against the field, then the work done is
negative. So in the presence of an E-field, an external force must be applied to
overcome the field repulsion,
Definition of Electric Potential difference
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
The Electric potential at a point P distant r from a positive charge Q
at the origin.
Definition of Pot. Difference between points a and b,
Let “a” be a point at infinity and “b” a point r, then
This is an important relation. Electric field due to a charge Q goes as ~ 1/r2 but V ~ 1/r.
For a collection of charges , we would sum contributions, i.e.,
V is a scalar and so one adds numbers
For a continuous distribution of charges, we would write an integral,
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Figure 2-38 (p. 60)
Three different paths to calculate work moving from the origin to a point P against
an electric field.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Electric Potential and Electric Field due to a charge Q
Consider a charge Q in air. Surrounding the charge there is an electric field that
is directed radially out as shown by the blue arrows. What about the Electric
potential? All points having the same potential (equipotential) form concentric
spheres, or circles in 2D with the charge Q at the center.
Equipotential lines are shown
orthogonal to field lines for a point
charge. Figure 2-39 (p. 64)
.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Given a charge distribution we can calculate the E-field in
one of three ways.
(i) If there is sufficient symmetry we can use Gauss’ law,
div D=ρv
(ii) Use Coulomb’s law directly , E = Σ[Qi/4πεori2]ai
(iii) Calculate V, and take its gradient,
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Consider a disk of radius “a” possessing a charge density ρs.
Obtain the E-field at a point h on the z-axis. We want to
calculate V, then the - gradient of V to get E. We visited the
problem earlier on p,38, Matlab 2.3.
p.65, Fig.2.40
h2
+ρ2=
α2
Hint: use
to
solve first integral for V
.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Matlab 2.8 p. 65
ML0208
Plots the Potential V and the Electric Field E due to
a 1 nC charge as a function of radial distance.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Figure 2-41 (p. 66)
E and V for a point charge in MATLAB 2.8
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Field due to an Electric dipole
●A pair of charges of
opposite sign
separated by a
distance d is a dipole.
●Dipole moment
Qd
●What is the Potential V at some
point P(r,θ), where r>>d ?
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
The gradient of a scalar function Vp in spherical
polar coordinate system
See back
page of text
Vp(r,θ) no φ dependence
E = [Qd/(4πεor3)][ 2cosθ ar + sinθ aθ]
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Notice that Vp varies as cosθ (broken line curves)
E = [Qd/(4πεor3)][ 2cosθ ar + sinθ aθ]
continuous lines
Fig. taken from
Shen/Khong
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Electrical conductivity of
several materials at RT,
Figure 2-43 (p. 68)
I = ∫J.dS
Current density J
J = ρv u
J = σE Ohm’s Law
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Consider a metallic bar of length L , cross-section area S,
and conductivity σ. We apply a voltage Vab and obtain the
current flow I.
J = σE
Ohms Law
I/S = σ Vab/L
R = Vab/I = L/(σS)
R = Vab/I = - ∫E.dL/ ∫σE.dS
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Example 2.21p.71 Consider a coax cable of length L
with an inner conductive shell of radius a and an outer
conductive shell of radius b filled with material of
conductivity σ. Assume a charge Q on the inner shell. The
E-field between the conductors
●Recall E-field decreases as
1/ρ between the two conductive
shells
● Voltage would appear across the electrodes, and a
current should flow Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
The voltage drop will give rise to a current flow through
the conductive medium,
This will permit us to define the Resistance or conductance
of the medium, since we know Vab and I
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
If we divide Vab/I we can obtain a resistance R
The resistance between the conductors of a coaxial
cable can be used to define a conductance G ( = 1/R)
The conductance per unit length , G’, can then be
written as
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
In Transmission line problems we will
introduce distributed parameters in Chapter 6,
when we will use G and G’
Joule’s Law
Electric Force on a charge dQ is given as
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Work done to move charge dQ through the E-field
Power is rate of doing work, and so
Where u = dL/dt. Since J = ρv u , we can rewrite
dP as
Upon integrating over the volume,
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Watt= Joule/sec = Volt x Ampere
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.