EE 207
ELECTROMAGNETICS I
Electromagnetics (EM) - the study of electric and magnetic phenomena.
A knowledge of the fundamental behavior of electric and magnetic
fields is necessary to understand the operation of such devices as resistors,
capacitors, inductors, diodes, transistors, transformers, motors, relays,
transmission lines, antennas, waveguides, optical fibers and lasers.
All electromagnetic phenomena are governed by a set of equations
known as Maxwell’s equations.
Maxwell’s Equations
E - electric field intensity
H - magnetic field intensity
D - electric flux density
B - magnetic flux density
J - current density
Dv - volume charge density
EE 207 Electromagnetics I
Static fields and applications, Introduction to Maxwell’s equations.
EE 307 Electromagnetics II
Time-varying fields and applications.
Vector Algebra
The quantities of interest appearing in Maxwell’s equations along
with other quantities encountered in the study of EM can almost always be
classified as either a scalar or a vector (tensors are sometimes encountered
in EM but will not be covered in this class).
Scalar - a quantity defined by magnitude only.
[examples: distance (x), voltage (V ), charge density (Dv), etc.]
Vector - a quantity defined by magnitude and direction.
[examples: velocity (v), current (I ), electric field (E), etc.]
Note that vectors are denoted by boldface. The magnitude of a vector may
be a real-valued scalar or a complex-valued scalar (phasor).
Vector Addition (Parallelogram Law)
Vector Subtraction
Note:
(1)
(2)
The magnitude of the vector A!B is the separation distance d
between the points a and b located by the vectors A and B,
respectively [d = *A!B* = *B!A*].
The vector A!B is the vector pointing from b (origination
point) to a (termination point).
Multiplication and Division By a Scalar
Coordinate Systems
A coordinate system defines points of reference from which specific
vector directions may be defined. Depending on the geometry of the
application, one coordinate system may lead to more efficient vector
definitions than others. The three most commonly used coordinate systems
used in the study of electromagnetics are rectangular coordinates (or
cartesian coordinates), cylindrical coordinates, and spherical coordinates.
Rectangular Coordinates
The rectangular coordinate system is an orthogonal coordinate
system with coordinate axes defined by x, y, and z. The coordinate axes in
an orthogonal coordinate system are mutually perpendicular. By
convention, we choose to define rectangular coordinates as a right-handed
coordinate system. This convention ensures that the three coordinate axes
are always drawn with the same orientation no matter how the coordinate
system may be rotated. If we position a right-handed screw normal to the
plane containing the x and y axes, and rotate the screw in the direction of
the x axis rotated toward the y axis, the direction that the screw advances
defines the direction of the z axis in a right-handed coordinate system.
Component Scalars and Component Vectors
Given an arbitrary vector E in rectangular coordinates, the vector E
can be described (using vector addition) as the sum of three component
vectors that lie along the coordinate axes.
The component vectors can be further simplified by defining unit vectors
along the coordinate axes: ax, ay, and az. These unit vectors have
magnitudes of unity and directions parallel to the respective coordinate
axis. The component vectors can be written in terms of the unit vectors as
Thus, using component scalars, any rectangular coordinate vector can be
uniquely defined using three scalar quantities that represent the magnitudes
of the respective component vectors.
To define a unit vector in the direction of E, we simply divide the
vector by its magnitude.
where the magnitude of E is the diagonal of the rectangular volume formed
by the three component scalars.
Example (Unit vector)
Given
, determine the unit vector in the
direction of E at the rectangular coordinate location of (1,1,1).
Note that the component scalars are
functions of position (the direction of the
vector changes with position).
At the point (1,1,1) [x = 1, y = 1, z = 1],
Example (Vector addition)
An airplane with a ground speed of 350 km/hr heading due west flies
in a wind blowing to the northwest at 40 km/hr. Determine the true air
speed and heading of the airplane.
Dot Product
(Scalar Product)
The dot product of two vectors A and B (denoted by A @ B) is defined
as the product of the vector magnitudes and the cosine of the smaller angle
between them.
The dot product is commonly used to determine the component of a vector
in a particular direction. The dot product of a vector with a unit vector
yields the component of the vector in the direction of the unit vector.
Given two vectors A and B with corresponding unit vectors aA and aB, the
component of A in the direction of B (the projection of A onto B) is found
evaluating the dot product of A with aB. Similarly, the component of B in
the direction of A (the projection of B onto A) is found evaluating the dot
product of B with aA.
The dot product can be expressed independent of angles through the
use of component vectors in an orthogonal coordinate system.
The dot product of like unit vectors yields one ( 2AB = 0o ) while the dot
product of unlike unit vectors ( 2AB = 90o ) yields zero. The dot product
results are
The resulting dot product expression is
Cross Product
(Vector Product)
The cross product of two vectors A and B (denoted by A× B ) is
defined as the product of the vector magnitudes and the sine of the smaller
angle between them with a vector direction defined by the right hand rule.
Note: (1) the unit vector an is normal to
the plane in which A and B lie.
(2) ABsin2AB = area of the parallelogram
formed by the vectors A and B.
Using component vectors, the cross product of A and B may be
written as
The cross product of like unit vectors yields zero ( 2AB = 0o ) while the cross
product of unlike unit vectors ( 2AB = 90o ) yields another unit vector which
is determined according to the right hand rule. The cross products results
are
The resulting cross product expression is
This cross product result can also be written compactly in the form of a
determinant as
Example (Dot product / Cross product)
Given E = 3ay + 4az and F = 4ax!10ay + 5az, determine
(a.) the vector component of E in the direction of F.
(b.) a unit vector perpendicular to both E and F.
(a.) To find the vector component of E in the direction of F, we must dot
the vector E with the unit vector in the direction of F.
The dot product of E and aF is
(Scalar component of E along F)
The vector component of E along F is
(b.) To find a unit vector normal to both E and F, we use the cross
product. The result of the cross product is a vector which is normal
to both E and F.
We then divide this vector by its magnitude to find the unit vector.
The negative of this unit vector is also normal to both E and F.
Coordinate and Unit Vector Definitions
Rectangular Coordinates (x,y,z)
Cylindrical Coordinates (D,N,z)
Spherical Coordinates (r,2,N)
Vector Definitions and Coordinate Transformations
Vector Definitions
Vector Magnitudes
Rectangular to Cylindrical Coordinate Transformation
(Ax, Ay, Az )
Y (AD, AN, Az )
The transformation of rectangular to cylindrical coordinates requires
that we find the components of the rectangular coordinate vector A in the
direction of the cylindrical coordinate unit vectors (using the dot product).
The required dot products are
where the az unit vector is identical in both orthogonal coordinate systems
such that
The four remaining unit vector dot products are determined according to
the geometry relationships between the two coordinate systems.
The resulting cylindrical coordinate vector is
In matrix form, the rectangular to cylindrical transformation is
Cylindrical to Rectangular Coordinate Transformation
(AD, AN, Az )
Y (Ax, Ay, Az )
The transformation from cylindrical to rectangular coordinates can
be determined as the inverse of the rectangular to cylindrical
transformation.
The cylindrical coordinate variables in the transformation matrix must be
expressed in terms of rectangular coordinates.
The resulting transformation is
The cylindrical to rectangular transformation can be written as
Rectangular to Spherical Coordinate Transformation
(Ax, Ay, Az )
Y (Ar, A2, AN )
The dot products necessary to determine the transformation from
rectangular coordinates to spherical coordinates are
The geometry relationships between the rectangular and spherical unit
vectors are illustrated below.
The dot products are then
and the rectangular to spherical transformation may be written as
Spherical to Rectangular Coordinate Transformation
(Ar, A2, AN )
Y (Ax, Ay, Az )
The spherical to rectangular coordinate transformation is the inverse
of the rectangular to spherical coordinate transformation so that
The spherical coordinate variables in terms of the rectangular coordinate
variables are
The complete spherical to rectangular coordinate transformation is
Coordinate Transformation Procedure
(1)
(2)
Transform the component scalars into the new coordinate
system.
Insert the component scalars into the coordinate transformation
matrix and evaluate.
Steps (1) and (2) can be performed in either order.
Example (Coordinate Transformations)
Given the rectangular coordinate vector
(a.) transform the vector A into cylindrical and spherical
coordinates.
(b.) transform the rectangular coordinate point P (1,3,5) into
cylindrical and spherical coordinates.
(c.) evaluate the vector A at P in rectangular, cylindrical and
spherical coordinates.
(a.)
(b.) P (1, 3, 5)
P (1, 3, 5)
(c.)
Y x = 1, y = 3, z = 5
Y P (3.16, 71.6o, 5) Y P (5.92, 32.3o, 71.6 o)
Separation Distances
Given a vector r1 locating the point P1 and a vector r2 locating the
point P2, the distance d between the points is found by determining the
magnitude of the vector pointing from P1 to P2, or vice versa.
Volumes, Surfaces and Lines in Rectangular,
Cylindrical and Spherical Coordinates
We may define particular three-dimensional volumes in rectangular,
cylindrical and spherical coordinates by specifying ranges for each of the
three coordinate variables.
Rectangular volume (2×2×5 box)
(1 # x # 3)
(2 # y # 4)
(0 # z # 5)
Cylindrical volume (cylinder of length = 5, diameter = 2)
(0 # D #1)
(0 # N # 2B)
(0 # z # 5)
Spherical volume (sphere of diameter = 4)
(0 # r # 2)
(0 # 2 # B)
(0 # N # 2B)
Specific lines and surfaces can be generated in a given coordinate
system according to which coordinate variable(s) is(are) held constant. A
surface results when one of the coordinate variables is held constant while
a line results when two of the coordinate variables are held constant.
Surface on the Rectangular volume (front face of the box)
x=3
(2 # y # 4)
(0 # z # 5)
(x ! constant)
Surface on the Cylindrical volume (upper surface of the cylinder)
(0 # D #1)
(0 # N # 2B)
z=5
(z ! constant)
Surface on the Spherical volume (outer surface of the sphere)
r=2
(0 # 2 # B)
(0 # N # 2B)
(r ! constant)
Line on the Rectangular volume (upper edge of the front face)
x=3
(2 # y # 4)
z=5
(x ! constant)
(z ! constant)
Line on the Cylindrical volume (outer edge of the upper surface)
D =1
(0 # N # 2B)
z=5
(D ! constant)
(z ! constant)
Line on the Spherical volume (equator of the sphere)
r=2
2 = B/2
(0 # N # 2B)
(r ! constant)
(2 ! constant)
Differential Lengths, Surfaces and Volumes
When integrating along lines, over surfaces, or throughout volumes,
the ranges of the respective variables define the limits of the respective
integrations. In order to evaluate these integrals, we must properly define
the differential elements of length, surface and volume in the coordinate
system of interest. The definition of the proper differential elements of
length (dl for line integrals) and area (ds for surface integrals) can be
determined directly from the definition of the differential volume (dv for
volume integrals) in a particular coordinate system.
Rectangular Coordinates
Cylindrical Coordinates
Spherical Coordinates
Example (Line / surface / volume integration)
Using the appropriate differential elements, show that
(a.) the circumference of a circle of radius Do is 2BDo.
(b.) the surface area of a sphere of radius ro is 4B ro2.
(c.) the volume of a sphere of radius ro is (4/3) B ro3.
(a.)
(b.)
(c.)
Example (Surface / volume integration in spherical coordinates)
A three-dimensional solid is described in spherical coordinates
according to
0 # r #1
0 # 2 #B/4
0 # N #2B
(a.) Sketch the solid.
(b.) Determine the volume of the solid.
(c.) Determine the surface area of the solid
(a.)
(b.)
(c.)
Line Integrals of Vectors
Certain parameters in electromagnetics are defined in terms of the
line integral of a vector field component in the direction of a given path.
The component of a vector along a given path is found using the dot
product. The resulting scalar function is integrated along the path to obtain
the desired result. The line integral of the vector A along a the path L is
then defined as
where
dl = al dl
al ! unit vector in the direction of the path L
dl ! differential element of length along the path L
Al ! component of A along the path L
Whenever the path L is a closed path, the resulting line integral of A is
defined as the circulation of A around L and written as
Differential Lengths on Arbitrary Paths
Line integrals on paths in arbitrary directions may be defined using
general differential lengths or differential displacements. The general
differential displacements in rectangular, cylindrical and spherical
coordinates are
These differential lengths are valid for integration in any general direction
but the resulting integrands must be parameterized in terms of only one
variable (the variable of integration).
Example (Line integral)
Given
, evaluate the line
integral of H along the path L made up of the three straight line paths
shown below.
Surface Integrals of Vectors
Certain parameters in electromagnetics are defined in terms of the
surface integral of a vector field component normal to the surface. The
component of a vector normal to the surface is found using the dot product.
The resulting scalar function is integrated over the surface to obtain the
desired result. The surface integral of the vector A over the surface S (also
called as the flux of A through S) is then defined as
where
ds = an ds
an ! unit vector normal to the surface S
ds ! differential surface element on S
An ! component of A normal to the surface S
For a closed surface S, the resulting surface integral of A is defined
as the net outward flux of A through S assuming that the unit normal an is
an the outward pointing normal to S.
Example (Surface integral / net outward flux)
Given
, determine the net outward flux through
the closed hemispherical surface defined by (0 # r # 5), (0 # 2 #B/2), and (0
# N #2B).
S = S1 + S2
S1 ! hemispherical surface (ro = 5)
S2 ! circular surface (2o = 90o)
Line, Surface and Volume Integration using MATLAB
The evaluation of line integrals (1D integrals), surface integrals (2D
integrals) and volume integrals (3D integrals) can all be performed using
the simple 1D integration tools within MATLAB assuming that the 2D or
3D integrand of interest is separable into a product of functions which are
each dependent on only one variable. For example, given the general
volume integration in rectangular coordinates of a function F(x,y,z), this
3D integration reduces to three 1D integrals if the function F can be
written as
MATLAB provides three distinct 1D integration tools which employ
different types of integration schemes.
The 1D integration defined by the “trapz” command employs the
trapezoidal rule. In order to evaluate the integral
the MATLAB command syntax for this technique is
result = trapz(xvec,yvec)
xvec
yvec
vector containing the evaluation points for the integration
from xmin to xmax used in the integration.
vector containing the values of the integrand f (x) at the
evaluation points defined in xvec.
The xvec and yvec vectors are easily generated using the linspace
command in MATLAB:
vec = linspace(xmin,xmax,N)
which creates a vector vec containing N equally spaced points from xmin
to xmax.
Example (Numerical integration using MATLAB)
Evaluate the integral
using the trapezoidal rule in MATLAB.
» theta=linspace(pi/4,pi/2,100);
» f=cos(theta).*(sin(theta)).^3;
» result=trapz(theta,f)
result =
0.1875
theta
cos(theta)
sin(theta)
f
=
=
=
=
[0.7854
[0.7071
[0.7071
[0.2500
0.7933
0.7015
0.7127
0.2539
» plot(theta,f)
» xlabel('\theta')
» ylabel('f(\theta)')
0.8013
0.6958
0.7182
0.2578
...
...
...
...
1.5708]
0.0000]
1.0000]
0.0000]
In addition to the trapz command, MATLAB provides two other
1D integration tools: quad and quad8. These quadrature-based
integration schemes use higher-order approximations than the trapezoidal
rule. The formats of these MATLAB commands are
quad(‘f’,xmin,xmax)
quad8(‘f’,xmin,xmax)
where the function f to be integrated must be defined within MATLAB.
The quad8 integration scheme is simply a higher-order version of the
quad scheme.
MATLAB will also perform a two-dimensional integration using the
dblquad function. To evaluate an integral of the form
the function F(x,y) must be defined in a separate m-file. For example, to
evaluate the integral
an m-file named fxy.m is generated which contains the following.
function z=fxy(x,y)
z=cos((x.^2+1).*y);
The following command line determines the integral result
» dblquad('fxy',-1,1,0,2)
ans =
0.7875
The 2-D integrand in the previous example can be plotted (in 3-D) using
the following commands.
»
»
»
»
»
»
x=linspace(-1,1,50);
y=linspace(0,2,50);
[xx,yy]=meshgrid(x,y);
zz=fxy(xx,yy);
mesh(xx,yy,zz)
xlabel ('x'),ylabel('y'),zlabel('f(x,y)')
Electrostatic Fields
Electrostatic fields are static (time-invariant) electric fields produced
by static (stationary) charge distributions. The mathematical definition of
the electrostatic field is derived from Coulomb’s law which defines the
vector force between two point charges.
Coulomb’s Law
Q1, Q2
point charges (C)
F12
vector force (N) on Q2 due to Q1
r1, r2
R12 = r2
vectors locating Q1 and Q2
r1
R = R12 = r2
aR12 = R12 /R
o
vector pointing from Q1 to Q2
r1
separation distance (m) between Q1 and Q2
unit vector pointing from Q1 to Q2
free space (vacuum) permittivity
[8.854×10 12 F/m]
Coulomb’s law can also be written as
Note that the unit vector direction is defined according to which
charge is exerting the force and which charge is experiencing the force.
This convention assures that the resulting vector force always points in the
appropriate direction (opposite charges attract, like charges repel).
The point charge is a mathematical approximation to a very small
volume charge. The definition of a point charge assumes a finite charge
located at a point (zero volume). The point charge model is applicable to
small charged particles or when two charged bodies are separated by such
a large distance that these bodies appear as point charges to each other.
Given multiple point charges in a region, the principle of
superposition is applied to determine the overall vector force on a
particular charge. The total vector force acting on the charge equals the
vector sum of the individual forces.
Force Due to Multiple Point Charges
Given a point charge Q in the vicinity of a set of N point charges (Q1,
Q2,..., QN), the total vector force on Q is the vector sum of the individual
forces due to the N point charges.
F
total vector force on Q
due to Q1, Q2,..., QN
Electric Field
According to Coulomb’s law, the vector force between two point
charges is directly proportional to the product of the two charges.
Alternatively, we may view each point charge as producing a force field
around it (electric field) which is proportional to the point charge
magnitude. When a positive test charge Q is placed at the point P (the
field point) in the force field of a point charge Q located at the point P ,
the force per unit charge experienced by the test charge Q is defined as the
electric field at the point P. Given our convention of using a positive test
charge, the direction of the vector electric field is the direction of the
force on positive charge. A convention has been chosen where the source
coordinates (location of the source charge) are defined by primed
coordinates while the field coordinates (location of the field point) are
defined by unprimed coordinates.
Q - point charge producing
the electric field
Q - positive test charge used
to measure the electric field
r - locates the source point
(location of source charge Q )
r - locates the field point
(location of test charge Q)
From Coulomb’s law, the force on the test charge Q at r due to the charge
Q at r is
The vector electric field intensity E at r (force per unit charge) is found by
dividing the Coulomb force equation by the test charge Q.
Note that the electric field produced by Q is independent of the magnitude
of the test charge Q. The electric field units [Newtons per Coulomb (N/C)]
are normally expressed as Volts per meter (V/m) according to the following
equivalent relationship:
For the special case of a point charge at the origin (r = 0), the electric field
reduces to the following spherical coordinate expression:
Note that the electric field points radially outward given a positive point
charge at the origin and radially inward given a negative point charge at
the origin. In either case, the electric field of the a point charge at the
origin is spherically symmetric and easily defined using spherical
coordinates. The magnitude of the point charge electric field varies as r 2.
The vector force on a test charge Q at r due to a system of point
charges (Q1 , Q2 ,..., QN ) at (r1 , r2 ,..., rN ) is, by superposition,
The resulting electric field is
Example (Electric field due to point charges)
Determine the vector electric field at (1, 3,7) m due to point charges
Q1 = 5 nC at (2,0,4) m and Q2 = 2 nC at ( 3,0,5) m.
Charge Distributions
Charges encountered in many electromagnetic applications (e.g.,
charged plates, wires, spheres, etc.) can be modeled as line, surface or
volume charges. The electric field equation for a point charge can be
extended to these charge distributions by viewing these distributions as
simply a grouping of point charges.
Charge
Distribution
Charge
Density
Units
Total Charge
In general, the various charge densities vary with position over the line,
surface or volume and require an integration to determine the total charge
associated with the charge distribution. Uniform charge densities do not
vary with position and the total charge is easily determined as the product
of the charge density and the total length, area or volume.
Uniform Charge Distributions
Uniform line charge (
L
= constant)
(Lo = total line length)
Uniform surface charge (
S
= constant)
(Ao = total surface area)
Uniform volume charge (
V
= constant)
(Vo = total volume)
Electric Fields Due to Charge Distributions
Each differential element of charge on a line charge (dl ), a surface
charge (ds ) or a volume charge (dv ) can be viewed as a point charge. By
superposition, the total electric field produced by the overall charge
distribution is the vector summation (integration) of the individual
contributions due to each differential element. Using the equation for the
electric field of a point charge, we can formulate an expression for dE (the
incremental vector electric field produced by the given differential element
of charge). We then integrate dE over the appropriate line, surface or
volume over which the charge is distributed to determine the total electric
field E at the field point P.
Point Charge
Line Charge (
L
dl
Q)
Surface Charge (
S
Volume Charge (
V
ds
Q)
dv
Q)
Example (E due to a line charge)
Evaluate E at P = (x,y,z) due to a uniform line charge lying along the
z-axis between (0,0,zA) and (0,0,zB) with zB > zA.
The integrals in the electric field expression may be evaluated analytically
using the following variable transformation:
For the special case of a line charge centered at the coordinate origin
(zA = a, zB = a) with the field point P lying in the x-y plane [P = (x,y,0)],
the electric field expression reduces to
(E-field in the x-y plane due to a
uniform line charge of length 2a
centered at the origin)
To determine the electric field of an infinite length line charge, we take the
limit of the previous result as a approaches .
(E-field due to a uniform line
charge of infinite length lying
along the z-axis.)
Note that the electric field of the infinite-length uniform line charge
is cylindrically symmetric (line source). That is, the electric field is
independent of due to the symmetry of the source. The electric field of
the infinite-length uniform line charge is also independent of z due to the
infinite length of the uniform source. In comparison to the electric field of
a point charge (which varies as r 2), the electric field of the infinite-length
uniform line charge varies as 1. If L is positive, the electric field points
outward radially while a negative L produces an electric field which points
inward radially.
Infinite-length uniform
line charge ( L >0)
Infinite-length uniform
line charge ( L <0)
Example (E due to a surface charge)
Evaluate E at a point on the z-axis P = (0,0, h) due to a uniformly
charged disk of radius a lying in the x-y plane and centered at the
coordinate origin.
The unit vector a , which is a function of the integration variable , can
be transformed into rectangular coordinate unit vectors to simplify the
integration.
The first two integrals in the electric field expression are zero given the
sine and cosine integrals with respect to
over one period.
The electric field expression reduces to
(E-field on the z-axis due to a uniformly
charged disk of radius a in the x-y plane
centered at the origin, h = height above disk)
The electric field produced by an infinite charged sheet can be determined
by taking the limit of the charged disk E as the disk radius approaches .
(E-field due to a uniformly
charged infinite sheet)
Note that the electric field of the uniformly charged infinite sheet is
uniform (independent of the height h of the field point above the sheet).
Electric Flux Density
The electric flux density D in free space is defined as the product of
the free space permittivity ( o) and the electric field (E):
Given that the electric field is inversely proportional to the permittivity of
the medium, the electric flux density is independent of the medium
properties.
The units on electric flux density are
so that the units on electric flux density are equivalent to surface charge
density.
The total electric flux ( ) passing through a surface S is defined as
the integral of the normal component of D through the surface.
where an is the unit normal to the surface S and Dn is the component of D
normal to S. The direction chosen for the unit normal (one of two possible)
defines the direction of the total flux.
For a closed surface, the total electric flux is
Gauss’s Law
Gauss’s law is one of the set of four Maxwell’s equations that govern
the behavior of electromagnetic fields.
Gauss’s Law - The total outward electric flux through any closed
surface is equal to the total charge enclosed by the surface.
Gauss’s law is written in equation form as
where ds = an ds and an is the outward pointing unit normal to S.
Example (Gauss’s law, point charge at origin)
Given a point charge at the origin, show that Gauss’s law is valid on
a spherical surface (S) of radius ro.
Gauss’s law applied to the spherical surface S surrounding the point charge
Q at the origin should yield
The electric flux produced by Q is
On the spherical surface S of radius ro, we have
Note the outward pointing normal requirement in Gauss’s law is a direct
result of our electric field (flux) convention.
By using an outward pointing normal, we obtain the correct sign on the
enclosed charge.
Gauss’s law can also be used to determine the electric fields produced
by simple charge distributions that exhibit special symmetry. Examples of
such charge distributions include uniformly charged spherical surfaces and
volumes.
Example (Using Gauss’s law to determine E )
Use Gauss’s law to determine the vector electric field inside and
outside a uniformly charged spherical volume of radius a.
k = constant
S
spherical surface of
radius r = a
S+
spherical surface of
radius r > a
S
spherical surface of
radius r < a
Gauss’s law can be applied on S to determine the electric field inside the
charged sphere [E(r <a)].
By symmetry, on S (and S+), Dr is uniform and has only an ar component.
or
Gauss’s law can be applied on S+ to determine the electric field outside the
charged sphere [E(r >a)].
or
Electric Field for the uniformly charged
spherical volume of radius a
Divergence Operator / Gauss’s Law (Differential Form)
The differential form of Gauss’s law is determined by applying the
integral form of Gauss’s law to a differential volume ( v). The differential
form of Gauss’s law is defined in terms of the divergence operator. The
divergence operator is obtained by taking the limit as v shrinks to zero (to
the point P) of the flux out of v divided by v.
Gradient operator
The divergence operator in rectangular coordinates can be determined
by performing the required integrations. The electric flux density within
the differential volume is defined by
while the electric flux density evaluated at the point P is defined as
The total flux out of the differential volume v is
The electric flux density components can be written in terms of a Taylor
series about the point P.
For points close to P (such as the faces on the differential volume), the
higher order terms in the Taylor series expansions become negligible such
that
The flux densities on the six faces of the differential volume are
front face
back face
right face
left face
top face
bottom face
The integrations over the six sides of the differential volume yield
The divergence operator in rectangular coordinates is then
Note that the divergence operator can be expressed as the dot product of
the gradient operator with the vector
The same process can be applied to the differential volume element
in cylindrical and spherical coordinates. The results are shown below.
Cylindrical
Spherical
Example (Divergence)
Given
, determine
.
V
Divergence Theorem
The divergence theorem (Gauss’s theorem) is a vector theorem that
allows a volume integral of the divergence of a vector to be transformed
into a surface integral of the normal component of the vector, or vice verse.
Given a volume V enclosed by a surface S and a vector F defined
throughout V, the divergence theorem states
Gauss’s law can be used to illustrate the validity of the divergence theorem.
Example (Divergence theorem and Gauss’s law)
Using the divergence theorem, calculate the total charge within the
volume V defined by 2 r 3, 0
/2, 0
2 given an electric flux
density defined by
by evaluating
(a.)
(b.)
S1 - outer hemispherical surface
(r =3, 0
/2, 0
2 )
S2 - inner hemispherical surface
(r =2, 0
/2, 0
2 )
S3 - flat ring
(2 r 3, = /2, 0
(a.)
0
0
2 )
(b.)
0
Electric Scalar Potential
Given that the electric field defines the force per unit charge acting
on a positive test charge, any attempt to move the test charge against the
electric field requires that work be performed. The potential difference
between two points in an electric field is defined as the work per unit
charge performed when moving a positive test charge from one point to the
other.
From Coulomb’s law, the vector force on a positive point charge in an
electric field is given by
The amount of work performed in moving this point charge in the electric
field is product of the force and the distance moved. When the positive
point charge is moved against the force (against the electric field), the work
done is positive. When the point charge is moved in the direction of the
force, the work done is negative. If the point charge is moved in a
direction perpendicular to the force, the amount of work done is zero. For
a differential element of length (dl), the small amount of work done (dW)
is defined as
The minus sign in the previous equation is necessary to obtain the proper
sign on the work done (positive when moving the test charge against the
electric field). When the point charge is moved along a path from point A
to B, the total amount of work performed (W) is found by integrating dW
along the path.
The potential difference between A and B is then
The potential difference equation may be written as
where VA and VB are the absolute potentials at points A and B, respectively.
The absolute potential at a point is defined as the potential difference
between the point and a reference point an infinite distance away. The
definition of the potential difference in terms of the absolute potentials at
the starting and ending points of the path shows that the potential
difference between any two points is independent of the path taken
between the points.
For a closed path (point A = point B), the line integral of the electric
field yields the potential difference between a point and itself yielding a
value of zero.
Vector fields which have zero-valued closed path line integrals are
designated as conservative fields. All electrostatic fields are conservative
fields.
Example (Potential difference)
Determine the absolute potential in the electric field of a point charge
Q located at the coordinate origin.
The electric field of a point
charge at the origin is
The potential difference between two
points A and B in the electric field of
the point charge is
If we choose an inward radial path from r = rA to r = rB, the vector
differential length is
which yields
The absolute potential at point B is found by taking the limit as rA
approaches infinity.
Potentials of Charge Distributions
The previous formula can be generalized as the absolute potential of
a point charge at the origin (let rB = r).
(Absolute potential for a
point charge at the origin)
Note that the potential distribution of the point charge exhibits spherical
symmetry just like the electric field. The potential of the point charge
varies as r 1 in comparison to the electric field of a point charge which
varies as r 2. Surfaces on which the potential is constant are designated
as equipotential surfaces. Equipotential surfaces are always perpendicular
to the electric field (since no work is performed to move a charge
perpendicular to the electric field). For the point charge, the equipotential
surfaces are concentric spherical surfaces about the point charge.
The absolute potential of a point charge at an arbitrary location is
(Absolute potential for a point
charge at an arbitrary location)
The principle of superposition can be applied to the determine the potential
due to a set of point charges which yields
(Absolute potential of a
set of point charges)
The potentials due to line, surface and volume distributions of charge
are found by integrating the incremental potential contribution due to each
differential element of charge in the distribution.
Point Charge
Line Charge (
L
dl
Q)
Surface Charge (
S
Volume Charge (
V
ds
Q)
dv
Q)
Example (Potential due to a line charge)
Determine the potential in the x-y plane due to a uniform line charge
of length 2a lying along the z-axis and centered at the coordinate origin
Even integrand
Symmetric limits
(Absolute potential in the x-y plane due to a
uniform line charge of length 2a lying along
the z-axis centered at the coordinate origin)
Example (Potential due to a square loop)
Determine the potential at the center of a square loop of side length
l which is uniformly charged.
The uniformly charged square loop
can be viewed as four line charges. The
total potential at the center of the loop is
the scalar sum of the contributions from
the four sides
(identical scalar
contributions). Thus, the potential at P
due to one side of the loop is
Electric Field as the Gradient of the Potential
The potential difference between two points in an electric field can
be written as the line integral of the electric field such that
From the equation above, the
incremental change in potential along
the integral path is
where
is the angle between the
direction of the integral path and the
electric field. The derivative of the
potential with respect to position along
the path may be written as
Note that the potential derivative is a maximum when = (when the
direction of the electric field is opposite to the direction of the path). Thus,
This equation shows that the
magnitude of the electric field is
equal to the maximum space rate of
change in the potential.
The
direction of the electric field is the
direction of the maximum decrease
in the potential (the electric field
always points from a region of
higher potential to a region of lower
potential).
The electric field can be written in terms of the potential as
where the operator “ ” (del) is the gradient operator. The gradient
operator is a differential operator which operates on a scalar function to
yield (1) the maximum increase per unit distance and (2) the direction of
the maximum increase. Since the electric field always points in the
direction of decreasing potential, the electric field is the negative of the
gradient of V.
The derivative with respect to l in the gradient operator above can be
generalized to a particular coordinate system by including the variation in
the potential with respect to the three coordinate variables. In rectangular
coordinates,
(Gradient operator in
rectangular coordinates)
The gradient operator is defined differently in rectangular, cylindrical and
spherical coordinates. The electric field expression as the gradient of the
potential in these coordinate systems are
Example (E as the gradient of V )
Given
(2, /2,0).
(a.) find E(r, , ) and (b.) E at
(a.)
(b.)
Summary of Electric Field / Potential Relationships
Electric Dipole
An electric dipole is formed by two point charges of equal magnitude
and opposite sign (+Q, Q) separated by a short distance d. The potential
at the point P due to the electric dipole is found using superposition.
If the field point P is moved a large distance from the electric dipole (in
what is called the far field, r d ) the lines connecting the two charges and
the coordinate origin with the field point become nearly parallel.
(Dipole far field potential, r
d)
The electric field produced by the electric dipole is found by taking
the gradient of the potential.
(Dipole electric field,
far field, r d)
If the vector dipole moment is defined as
the dipole potential and electric field may be written as
Note that the potential and electric field of the electric dipole decay faster
than those of a point charge.
V
E
point charge
~r
1
~r
2
electric dipole
~r
2
~r
3
For an arbitrarily located, arbitrarily oriented dipole, the potential can
be written as
Energy Density in the Electric Field
The amount of work necessary to assemble a group of point charges
equals the total energy (WE) stored in the resulting electric field.
Example (3 point charges)
Given a system of 3 point charges, we can determine the total energy
stored in the electric field of these point charges by determining the work
performed to assemble the charge distribution. We first define Vmn as the
absolute potential at Pm due to point charge Qn.
1. Bring Q1 to P1 (no energy required).
2. Bring Q2 to P2 (work = Q2V21).
3. Bring Q3 to P3 (work = Q3V31 +Q3V32).
WE = 0 + (Q2V21) + (Q3V31 +Q3V32)
(1)
If we reverse the order in which the charges are assembled, the total energy
required is the same as before.
1. Bring Q3 to P3 (no energy required).
2. Bring Q2 to P2 (work = Q2V23).
3. Bring Q1 to P1 (work = Q1V12 +Q1V13).
WE = 0 + (Q2V23) + (Q1V12 +Q1V13)
(2)
Adding equations (1) and (2) gives
2WE = Q1(V12 +V13) + Q2(V21 +V23) + Q3(V31 +V32) = Q1V1+ Q2V2 +Q3V3
where Vm = total absolute potential at Pm affecting Qm.
WE = ½(Q1V1+ Q2V2 +Q3V3)
In general, for a system of N point charges, the total energy in the electric
field is given by
For line, surface or volume charge distributions, the discrete sum total
energy formula above becomes a continuous sum (integral) over the
respective charge distribution. The point charge term is replaced by the
appropriate differential element of charge for a line, surface or volume
distribution: L dl, S ds or V dv. The overall potential acting on the point
charge Qk due to the other point charges (Vk) is replaced by the overall
potential (V) acting on the differential element of charge due to the rest of
the charge distribution. The total energy expressions become
Total Energy in Terms of the Electric Field
If a volume charge distribution V of finite dimension is enclosed by
a spherical surface So of radius ro, the total energy associated with the
charge is given by
Using the following vector identity,
the expression for the total energy may be written as
If we apply the divergence theorem to the first integral, we find
For each equivalent point charge ( V dv) that makes up the volume charge
distribution, the potential contribution on So varies as r 1 and electric flux
density (and electric field) contribution varies as r 2. Thus, the product of
the potential and electric flux density on the surface So varies as r 3. Since
the integration over the surface provides a multiplication factor of only r2,
the surface integral in the energy equation goes to zero on the surface So of
infinite radius. This yields
where the integration is applied over all space. The divergence term in the
integrand can be written in terms of the electric field as
such that the total energy (J) in the electric field is
The total energy in the previous integral can be written as the integral of
the electric field energy density (wE) throughout the volume.
Thus, the energy density in an electric field is given by
Example (Energy density / total energy in an electric field)
Given V = (x y +xy +2z) volts, determine the electrostatic energy
stored in a cube of side 2m centered at the origin.
The electric field is found by taking the gradient of the potential function.
The energy density in the electric field is given by
The total energy within the defined cube is found by integrating the energy
density throughout the cube.
0
0
(Odd integrands / symmetric limits)
Electric Fields in Material Space
The charges considered up to this point have been assumed to be
stationary and located in free space (vacuum) or air. If we place charge
within a gas, solid or liquid material, the charge associated with the
material atoms will be affected. Also, under the influence of the applied
electric field, charges not bound by other forces (free charges) may be set
in motion (electric current).
Current (I ) - net flow of positive charge in a given direction (vector)
measured in units of Amperes (Ampere = Coulomb/second).
Note that, mathematically, the negative charge moving in the opposite
direction constitutes a positive component of the overall current flowing
in the ax direction.
Conductor - current carrying medium.
Insulator - non-conducting medium.
Material Classification Based on Conductivity
The conductivity (F) of a given material is a measure of the ability of
material to conduct current. Conductivity is measured in units of S/m or
®/m. The inverse of conductivity is resistivity (Dc = 1/F). For elements,
the structure of the element atom dictates the conductivity of the element.
Specifically, the element conductivity is related to the strength of the bonds
between the outer (valence) electrons and the atom nucleus.
Positive nucleus charge = Total negative electron charge
Centroid of the nucleus charge - atom center
Centroid of the overall electron charge - atom center
@
The atom is
electrically neutral.
(DV = 0, V = 0, E = 0)
If under the influence of an electric field, the bond between the
valence electron and the atom nucleus is broken, the electron becomes a
free electron or conduction electron. Materials are classified as
conductors, insulators, or semiconductors based on the strength of these
bonds between the valence electrons and the atom nucleus. The stronger
the bond between the valence electrons and the nucleus in a particular
material, the fewer free electrons are available for conduction.
The values of conductivity designated at the boundaries between
material types are defined differently by many authors. Since conductivity
is, in general, a function of temperature, comparisons of conductivity are
made at a constant temperature (reference temperature, usually To = 20oC).
The dependence of resistivity on temperature may be expressed as
where Dco is the material resistivity at the reference temperature To and "
is the temperature coefficient for the material. Certain conductors and
oxides exhibit superconductivity at temperatures near absolute zero (0K =
!273oC) where the resistivity of the material drops abruptly to zero.
Examples (Conductivity in S/m at T = 20oC)
Insulators
Porcelain (10!12)
Glass (10!12)
Mica (10!15)
Wax (10!17 )
Semiconductors
Silicon (4.4×10!4 )
Germanium (2.2)
Conductors
Silver (6.1×107 )
Copper (5.8×107 )
Gold (4.1×107 )
Aluminum (3.5×107 )
Carbon (3×104 )
Ideal Models
Perfect Insulator (F = 0)
Perfect conductor (F = 4)
Current Types
Currents that flow in conductors are only one of three different types
of currents. The three types of currents are:
(1)
(2)
(3)
Conduction current (current in a conductor)
Example - current in a copper wire.
Convection current (current through an insulator)
Example - electron beam in a CRT.
Displacement current (time-varying effect to be studied later)
Example - AC current in a capacitor.
Separate equations are necessary to define each of these three types of
currents given the different mechanisms involved.
For a current density J (A/m2) associated with any type of current, the
total current I passing through a given surface S is defined as
where ds = dsan, an is the unit normal to the surface and Jn is the
component of the current normal to the surface . The scalar result of this
integral is the magnitude of the total current flowing in the direction of the
unit normal. For the special case when the current density is uniform over
the surface S,
where A is the total area of the surface S. The total current in Amperes
(Coulomb/second) represents the amount of charge passing through the
surface per second. A total current of 1 mA means that a net charge of 1
mC is passing through the surface each second.
Convection Current
Convection current is a flow of charged particles through an
insulating medium (example: an electron beam in a cathode-ray tube).
Thus, the equation defining convection current density is independent of
the conductivity of the medium since the medium characteristics
(insulator) do not affect the current. The medium through which the
convection current flows is typically a very good insulator (very low
conductivity). Convection current is defined in terms of the free charge
density in the current (DV) and the vector drift velocity (u) of the charge in
the current. The drift velocity is the average velocity at which the charge
is moving.
The convection current density is defined as
The total convection current is found by integrating the current density
over the cross section of the convection current.
Conduction Current
Conduction current is different from convection current in that the
current medium is a conductor rather than an insulator. A simple example
of conduction current is the current flowing in a conducting wire. If a
voltage V is applied to a cylindrical conductor (conductivity = F, length =
l, cross-sectional area = A), a conduction current results.
The potential difference between the ends of the conductor means that an
electric field exists within the conductor (pointing from the region of
higher potential to the region of lower potential). The conduction current
can be defined in the same way as convection current using the free charge
density (DV) and the vector drift velocity (u).
In a conductor, there is an abundance of free electrons. The drift velocity
in a conductor may be written as the product of the electric field (E) and
the conductor mobility (:).
The mobility of a material is a measure of how efficiently free
carriers can move through the material. Since typical conductors (metals)
are dense materials, the free electrons accelerated under the influence of
the electric field frequently collide with atom nuclei and other electrons.
The resulting particle motion looks somewhat random but has a net
component of motion in the direction opposite to the electric field (average
velocity of all like carriers = drift velocity of that carrier). Inserting the
drift velocity formula into the current density equation yields the
conduction current density in terms of the electric field:
such that the conductivity is
If the current density in the conductor is uniform, the corresponding
electric field is also uniform (J = FE). The voltage between the ends of the
wire can be expressed as the line integral of the electric field.
Thus, the voltage and the uniform electric field may be written as
The uniform current density is then
where
Resistance of a cylinder (length = l, crosssectional area = A, conductivity = F) carrying
a uniform current density
If the current density is not uniform, the resistance formula becomes
The power density inside the conductor is found by forming the dot
product of the vector electric field and the vector current density.
The total power dissipated in the conductor is found by integrating the
power density throughout the conductor.
Example (Conduction current)
A copper wire (F = 5.8 × 10!7 ®/m, DV = !1.4 × 1010 C/m3, radius =
1 mm, length = 20 cm) carries a current of 1 mA. Assuming a uniform
current density, determine
(a.) the wire resistance.
(b.) the current density.
(c.) the electric field within the wire.
(d.) the drift velocity of the electrons in the wire.
Perfect Conductor (F = 4)
R=0
E=0
@
Equipotential volume
Perfect Insulator (F = 0)
R=4
J=0
Polarization in Dielectrics
Nonconducting materials are commonly designated as insulators or
dielectrics. When an electric field is applied to a dielectric atom, an effect
known as polarization results. With no electric field applied, the centroid
of the (negative) electron charge is coincident with the centroid of the
(positive) nucleus charge such that the atom is electrically neutral. When
an electric field is applied to the atom, the positively charged nucleus is
displaced in the direction of the electric field while the centroid of the
negative electron charge is displaced in the direction opposite to the
electric field. The dielectric atom is thus polarized and may be modeled
as an equivalent electric dipole.
If a voltage V is applied to a cylindrical insulator (conductivity = F,
length = l, cross-sectional area = A), the insulator is polarized. If the
electric field is assumed to be uniform, then the electric field within the
insulator is E = V/l.
The polarization within the dielectric produces an additional electric
flux density component which is included in the electric flux density
equation as the vector polarization P.
The polarization P is defined as the dipole moment per unit volume such
that
where n is the number of dipoles in the volume v. Assuming that the
polarization vector P is proportional to the electric field E, we may write
where Pe is defined as the electric susceptibility (unitless). Inserting this
definition of P into the electric flux equation gives
where
Note that the electric susceptibility Pe and the relative permittivity ,r are
both measures of the polarization within a given material. The larger the
value of Pe or ,r for the material, the more polarization within the material.
For free space (vacuum), there is no polarization such that
P=0
Y
Pe = 0 or ,r = 1
The amount of polarization found in air is extremely small, so that we
typically model our atmosphere with the free space permittivity.
The magnitude of the polarization in a dielectric increases with the
magnitude of the applied electric field (the equivalent dipole moments
grow with the electric field magnitude). For a good insulator, the bonds
between the atom nuclei and the valence electrons are very strong and can
withstand very large electric fields. The electric field level at which these
bonds are broken, and the insulator begins to conduct (breakdown), is
designated as the dielectric strength. Some typical values of dielectric
strengths for some common insulators are:
Mica
Glass
Air
70 MV/m
35 MV/m
3 MV/m
The total charge density (DT) in an insulating material consists of the
free conduction charge density (Dv) plus the bound polarization charge
density (Dvp).
From our previous definition of the differential form of Gauss’s law, we
see that the divergence of the electric flux density yields the free charge
density.
If we insert the expression for the electric flux density in terms of the
polarization and the free charge density in terms of the total charge density,
we find
Equating terms yields
The divergence of the polarization vector gives the negative of the bound
polarization charge density.
Media Classifications
The electrical properties of a given medium are defined by three
constants: conductivity (F), permittivity (,), and permeability (:). The
permeability will be defined later when we study magnetic fields. The
following media classifications are made based on the characteristics of the
medium constants.
Linear medium - electrical properties do not vary with field magnitude.
Homogeneous medium - electrical properties do not vary with position.
Isotropic medium - electrical properties do not vary with field direction.
Otherwise, the medium is nonlinear, inhomogeneous, or anisotropic.
Continuity Equation
The continuity equation defines the basic conservation of charge
relationship between current and charge. That is, a net current in or out of
a given volume must equal the net increase of decrease in the total charge
in the volume. If we define a surface S enclosing a volume V, the net
current out of the volume (Iout) is defined by
where ds = dsan and an is the
outward pointing normal. If the
current I is a DC current, then the
net current out of the volume is
zero (as much current flows out as flows in). For a time-varying current,
the net current out of the volume may be non-zero and can be expressed in
terms of the change in the total charge within the volume (Q).
The previous equation is the integral form of the continuity equation. The
differential form of the continuity equation can be found by applying the
divergence theorem to the surface integral and expressing the total charge
in terms of the charge density.
The second and last terms in the equation above yield integrals that are
valid for any volume V that we may choose.
Since the previous equation is valid for any volume V, we may equate the
integrands of the integrals (the only way for the integrals to yield the same
value for any volume V is for the integrands to be equal). This yields the
continuity equation.
The continuity equation is given in differential form and relates the current
density at a given point to the charge density at that point. For steady
currents (DC currents), the charge density does not change with time so
that
the divergence of the current density is always zero.
The continuity equation is the basis for Kirchhoff’s current law.
Given a circuit node connecting a system of N wires (assuming DC
currents) enclosed by a spherical surface S, the integral form of the
continuity equation gives
The integral form of the continuity equation (and thus Kirchhoff’s current
law) also holds true for time-varying (AC) currents if we let the surface S
shrink to zero around the node.
Relaxation Time
If some amount of charge is placed inside a volume of conducting
material, the Coulomb forces on the individual charges cause them to
migrate away from each other (assuming the charge is all positive or all
negative). The end result is a surface charge on the outer surface of the
conductor while the inside of the conductor remains charge-neutral. The
time required for the conductor to reach this charge-neutral state is related
to a time constant designated as the relaxation time. The relaxation time
can be determined by inserting the relationship for the current density in
terms of electric field
into the continuity equation
which yields
The divergence of the electric field is related to the charge density by
Inserting this result into previous equation yields
or
The solution to this homogeneous, first order PDE is
where Tr is the relaxation time given by.
The relaxation time is a time constant that describes the rate of decay of the
charge inside the conductor. After a time period of Tr, the charge has
decayed to 36.8 percent (1/e) of its original value.
Example (Relaxation time)
Determine the relaxation time for copper (,r = 1, F = 5.8×107 ®/m)
and fused quartz (,r = 5, F = 10!17 ®/m).
Copper
Fused Quartz
Electric Field Boundary Conditions
A knowledge of the behavior of electric fields at a media interface
between distinct materials is necessary to solve many common problems
in electromagnetics. The fundamental boundary conditions involving
electric fields relate the tangential components of electric field and the
normal components of electric flux density on either side of the media
interface.
Tangential Electric Field
In order to determine the boundary condition on the tangential
electric field at a media interface, we evaluate the line integral of the
electric field along a closed incremental path that extends into both regions
as shown below.
The closed line integral of the electric field yields a result of zero such that
If we take the limit of this integral as )y = 0, the integral contributions on
the vertical paths vanish.
The integrals along the upper and lower paths on either side of the interface
reduce to
where the electric field components are assumed to be constant over the
paths of length )x. Dividing the result by )x gives
or
The tangential components of electric field
are continuous across a media interface.
If region 1 is a dielectric and region 2 is a perfect conductor (F2 = 4), then
Et 2 = 0 and
The tangential component of electric field on
the surface of a perfect conductor is zero.
Normal Electric Flux Density
In order to determine the boundary condition on the normal electric
flux density at a media interface, we apply Gauss’s law to an incremental
volume that extends into both regions as shown below.
The application of Gauss’s law to the closed surface above gives
If we take the limit as the height of the volume )z = 0, the integral
contributions on the four sides of the volume vanish.
The integrals over the upper and lower surfaces on either side of the
interface reduce to
where the electric flux density is assumed to be constant over the upper and
lower incremental surfaces. Evaluation of the surface integrals yields
Dividing by )x )y gives
where the charge density DS is assumed to be uniform.
The difference in the normal component of electric flux
density across the media interface is equal to the charge
density on the interface.
On a charge-free interface (DS = 0), such that
The normal components of electric flux density
are continuous across a charge-free media interface.
If region 1 is a dielectric and region 2 is a perfect conductor (F2 = 4), then
Dn2 = 0 and
The normal component of electric flux density on
the surface of a perfect conductor equals the
surface charge density.
The following statements describe the characteristics of a perfect
conductor under static conditions:
(1)
(2)
(3)
(4)
(5)
(6)
E = 0 inside the conductor.
Dv = 0 inside the conductor [free charge, if present, lies on the
outer surface of the conductor (Ds)].
The conductor is an equipotential volume.
Tangential E on the surface of the conductor is zero.
Normal D on the surface of the conductor equals Ds.
The electric field lines are normal to the surface of the
conductor.
Example (Polarization/Boundary conditions)
A dielectric cylinder (region 1) of radius D =3 and permittivity
,r1=2.5 is surrounded by another dielectric
(region 2) of permittivity ,r2 =10. Given an
electric field inside the cylinder of
determine (a.) P1 and Dvp1 (b.) E2 and D2.
(a.)
(b.)
Example (Boundary conditions)
Determine E and D everywhere for the charge-free boundary shown
below given E1 (or D1).
In general, we may determine the relationship between the electric
field and electric flux vectors in the two regions in terms of the two angles
21 and 22 measured with respect to the normal to the interface.
According to the geometry of the field and flux components, we see
that
Dividing the first equation by the second gives
The electric field and electric flux density boundary conditions on the
charge-free boundary are
such that
Given both media characteristics and the direction of the field in one of the
regions, the direction of the field in the other region can be determined
using this formula.
Electrostatic Boundary Value Problems
Many problems in electrostatics take the form of boundary value
problems where the charge density or potential is known in certain regions
or at certain boundaries. The governing partial differential equation
defining potential in terms of its source (charge density) is Poisson’s
equation. The derivation of Poisson’s equation begins with the
differential form of Gauss’s law.
Inserting the relationship between electric field and electric flux density
gives
Assuming a homogeneous medium, the permittivity may be treated like a
constant and brought outside the divergence operator.
Substituting the electric field definition in terms of the gradient
yields
Dividing both sides of the equation by the permittivity yields Poisson’s
equation:
The divergence of the gradient of a scalar is defined as the Laplacian
operator and designated by L 2 in operator notation.
Using the definition of the Laplacian operator, Poisson’s equation can be
written as
A special case of Poisson’s equation is Laplace’s equation in which the
source term (charge density) is zero.
Thus, Poisson’s equation governs the potential behavior in regions where
free charge exists, while Laplace’s equation governs the potential behavior
in regions where no free charge exists.
Uniqueness
A unique solution to a given boundary value problem in a specific
region is ensured if
(1.) We use the proper governing D.E. (there are an infinite
number of solutions to the D.E.).
(2.) We use the proper boundary conditions (there are an infinite
number of solutions to the boundary conditions).
There is only one (unique) solution to the governing differential equation
that also satisfies the given boundary conditions. By defining the proper
physics in the problem (proper boundary conditions), we ensure a unique
solution to the governing differential equation.
Laplace / Poisson Equation Solutions
The techniques used to solve Laplace’s and Poisson’s equation are
dependent on the dimensional complexity of the problem. For onedimensional (1-D) problems [potential is a function of one variable only],
where the governing partial differential equation (PDE) reduces to an
ordinary differential equation (ODE), we can simply integrate once to
determine the electric field and twice to determine the potential. For 2-D
and 3-D problems, procedures such as the separation of variables technique
are necessary.
Example (1-D solution / Poisson’s equation / semiconductor junction)
p-n junction
NA - acceptor dopant concentration
ND - donor dopant concentration
Depletion region (wp # z # wn)
n-type silicon
p-type silicon
A
Homogeneous medium (silicon permittivity ,=11.8,o)
We assume no variation in the p-n junction potential and electric field in
the x or y directions (1D problem) Y V(z), Ez(z)
Poisson’s equation is the governing differential equation that relates
the charge density in the p-n junction depletion region to the potential
distribution in the p-n junction.
Assuming a 1D solution (no variation in the x or y directions), Poisson’s
equation reduces to
or
The regions away from the depletion region are charge neutral (E = 0 and
V = constant). Thus, the p-n junction boundary conditions are
The electric field within the p-n junction is found by integrating Poisson’s
equation with respect to z.
where C1 and C2 are constants of integration. The first derivative of V(z)
with respect to z is equal to !Ez (z) according to
The integral of Poisson’s equation may then be written as
Application of the electric field boundary conditions at the edges of the
depletion region determines the unknown constants.
To determine the potential within the p-n junction, we integrate a second
time with respect to z.
Application of the potential boundary condition [V(0)=0] gives
Separation of Variables
The direct integration technique used in 1-D problems (ODE’s) is
not applicable to 2-D and 3-D problems (PDE’s). The separation of
variables technique is applicable to certain separable coordinate systems
including rectangular, cylindrical and spherical coordinates.
Example (Separation of variables technique)
The 3-D potential in a charge-free region in rectangular coordinates
is characterized by Laplace’s equation:
Note that Laplace’s equation is a 3-D, 2nd order PDE. The separation of
variables technique is based on the assumption that the solution to the PDE
may be written as the product of functions of only one variable. Thus, the
3-D solution for the potential is assumed to be
Inserting the assumed solution into Laplace’s equation yields
Dividing this equation by the assumed solution gives
The three terms on the left hand side of the equation above are each
dependent on only one variable. Thus, we may write
The individual functions must add to zero for all values of x, y, and z in the
3-D region of interest. Thus, each of these three functions must be
constants, such that
and
Note that the original 3-D 2nd order PDE has been transformed into three
1-D 2nd order ODE’s subject to the separation equation. The general
solutions to the three separate ODE’s are of the following form:
or linear
combinations
of these
Once each of the three individual functions is determined according to the
boundary conditions of the problem, the overall solution is simply the
product of these three functions.
Capacitors and Capacitance
A capacitor is an energy storage device that stores energy in an
electric field. A capacitor consists of two conductors separated by an
insulating medium. If the capacitor conductors are assumed to be initially
uncharged (neutral), the application of a voltage (potential difference)
between the conductors causes a charge separation (+Q on one conductor
and !Q on the opposite conductor). This charge separation produces an
electric field within the insulating medium between the conductors
(permittivity = ,) such that energy is stored in the capacitor.
The magnitude of the charge stored on either conductor is proportional to
the voltage applied between the conductors. The ratio of the total charge
magnitude on either conductor to the potential difference between the
conductors defines the capacitance.
The capacitance of a capacitor depends only on the geometry of the
conductors (conductor shape, separation distance, etc.) and the permittivity
of the insulating medium between the capacitor conductors. According to
the definition of capacitance, the charge on a capacitor conductor increases
at the same rate as the capacitor voltage (e.g., if the capacitor voltage is
doubled, the charge on each conductor is doubled).
Example (Capacitor)
Determine E and V between two perfectly conducting plates of
infinite extent in a homogeneous dielectric (F =0, , = ,r ,o ). The plates are
separated by a distance d with a potential difference of Vo between the
plates.
Due to symmetry, the charge distribution on the plates must be uniform.
The electric field can be determined by superposition using the electric
field expression for a uniformly charged plate of infinite extent.
Between the plates (0 # z # d),
Above the plates (z > d ),
Below the plates (z < d ),
Thus, the electric field is uniform between the plates and zero elsewhere.
The potential between the plates varies linearly given a uniform electric
field according to the definition of the electric field as the gradient of the
potential. The potential between the plates is a function of z only (1-D
problem) such that
Integrating this equation gives
Assuming that the bottom plate is used as a voltage reference (ground),
then the absolute potentials on the capacitor plates are
such that the potential between the plates is given by
and the electric field between the plates is
Note that the magnitude of the electric field between the plates is simply
the ratio of the voltage between the plates to the plate separation distance.
This capacitor problem can also be solved as a boundary value
problem. The governing differential equation is Laplace’s equation given
the ideal dielectric between the plates of the capacitor (Dv =0).
With no variation in the potential in the x or y directions (by symmetry),
Laplace’s equation in rectangular coordinates reduces to the following 1-D
form:
The boundary conditions on the potential are
Integrating both sides of Laplace’s equation with respect to z gives
Integrating again gives
Application of the capacitor boundary conditions yields
The potential between the plates is then
which is the same result found before. Note that the result of the first
integration was the electric field between the plates.
Ideal Parallel Plate Capacitor
A commonly encountered capacitor geometry is the parallel plate
capacitor. The parallel plate capacitor is formed by two large flat
conducting plates of area A separated by a small distance d. The volume
between the plates is filled with a homogeneous insulating medium of
permittivity 0.
The charge and electric field characteristics in an actual parallel plate
capacitor can be approximated using the ideal parallel plate capacitor
model.
Assumptions for the ideal parallel plate capacitor model:
(1) The plate surface charge densities are uniform
(Ds =±Q/A).
(2) The electric field between the plates is uniform (E=V/d).
(3) The electric field outside the volume between the plates
is zero.
In an actual parallel plate capacitor, the surface charge densities are not
uniform since the charge density grows large at the edges of the plates.
This crowding of charge at the conductor edges causes an effect known as
electric field fringing (nonuniform electric field).
The amount of fringing in the electric field near the edges of the
capacitor plates is small for closely-spaced large plates so that the ideal
parallel plate capacitor model is accurate for most capacitors. The ideal
parallel plate capacitor model becomes more accurate as the capacitor plate
area grows larger and the capacitor plate separation grows smaller.
The equation for the capacitance of the ideal parallel plate capacitor
is determined by starting with the capacitance definition in terms of charge
and potential.
The uniform electric field in the ideal parallel plate capacitor means that
the electric flux density within the capacitor is also uniform.
According to the electric flux boundary condition on the surface of either
plate, the electric flux density component normal to the plate surface is
equal to the uniform charge density on the plate.
The uniform electric field within the capacitor may be written in
terms of the potential V using
Inserting the equations for Q and V into the capacitance equation yields
Note that the capacitance of the ideal parallel plate capacitor is directly
proportional to the plate area and the insulator permittivity and inversely
proportional to the plate separation distance.
The total energy stored in the capacitor may be found by integrating
the energy density associated with the capacitor electric field.
Given the uniform electric field in the volume between the plates of the
ideal parallel plate capacitor, the energy density is also uniform, such that
The total energy results above can be rearranged into the normal circuits
equation for the total energy of a capacitor.
The last three equations on the right hand side of the total energy equation
above are valid for any capacitor.
Coaxial Capacitor
A coaxial capacitor is formed by two concentric conducting cylinders
(inner radius = a, outer radius = b, length = L) separated by an insulating
medium (,). Uniform charge distributions on both capacitor conductors
are assumed for the idealized model of a coaxial capacitor (just like the
ideal parallel plate capacitor). In an actual coaxial capacitor, the charge
densities grow large close to the ends of the conductors (sharp edges)
where electric field fringing is seen. The ideal coaxial capacitor model is
accurate for an actual coaxial capacitor which is long and has a small
cross-sectional area (fringing is negligible).
Assuming a voltage V is applied to the initially uncharged conductors of
the coaxial capacitor (from the inner conductor to the outer conductor,
using the outer conductor as the voltage reference), a total charge of +Q is
produced on the surface of the inner cylinder and a total charge of !Q is
produced on the inside surface of the outer cylinder. The uniform surface
charge densities on the inner and outer capacitor conductors (Dsa and Dsb,
respectively) are
Note that the surface charge density on the inner conductor is larger than
that on the outer conductor. This produces a nonuniform electric field (ED)
within the coaxial capacitor. By symmetry, ED is a function of D only.
To determine the capacitance of the coaxial capacitor, we
(1) determine the electric field E in terms of the charge Q using
Gauss’s law,
(2) find the potential V in terms of Q by evaluating the line integral
of the electric field, and
(3) determine the capacitance using the capacitance definition.
(C=Q/V ).
Applying Gauss’s law on a closed cylindrical surface S of radius D such
that (a #D# b) yields
The closed surface includes the endcaps on either end of the cylindrical
surface, but there is no electric flux component normal to the endcaps so
that Gauss’s law gives
The voltage V across the capacitor conductors is found by evaluating the
line integral of the electric field along the contour C from the outer
conductor to the inner conductor.
The ratio of charge to potential for the capacitor gives the capacitance:
A convenient parameter is the per-unit-length capacitance for the coaxial
capacitor. The per-unit-length capacitance is found by dividing the
capacitance C by the capacitor length. This yields
The units on per-unit-length capacitance are F/m. The overall capacitance
of a particular length coaxial capacitor is found by multiplying the perunit-length capacitance by the length L.
Spherical Capacitor
A capacitor can also be formed using two concentric spherical
conductors (inner radius = a, outer radius = b) separated by an insulating
medium (,). The ideal spherical capacitor is characterized by uniform
charge densities on the surfaces of both the inner and outer conductors.
The uniform surface charge densities on the inner and outer conductors of
the ideal spherical capacitor (Dsa and Dsb, respectively) are
The surface charge density on the inner conductor is larger than that on the
outer conductor given the larger surface area of the outer conductor. This
produces a nonuniform electric field (Er) within the spherical capacitor
which is a function of r only, by symmetry. The process used to determine
the capacitance of the ideal spherical capacitor is the same as that used for
the ideal cylindrical capacitor.
Applying Gauss’s law on a closed spherical surface S of radius r such
that (a #r# b) yields
Inserting the spherical capacitor electric flux density (D= ,Er ar) gives
The voltage V across the capacitor conductors is found by evaluating the
line integral of the electric field along the contour C from the outer
conductor to the inner conductor.
The ratio of charge to potential for the capacitor gives the capacitance:
Resistance, Capacitance and Relaxation Time
If the medium between the conductors of a capacitor is not a perfect
insulator, there is a finite resistance between the conductors (the resistance
between the conductors is infinite when the medium is a perfect insulator).
Assuming the medium between the capacitor conductors is a homogenous
medium characterized by (,, F), the capacitance between the conductors is
given by
while the resistance between the conductors is given by
If we take the product of the resistance and the capacitance, we find
where Tr is the relaxation time. Thus, given the capacitance for a particular
capacitance geometry, the corresponding resistance can be determined
easily according to
For the previously considered capacitors, we find the following resistance
values:
Ideal parallel plate capacitor
Ideal coaxial capacitor
Ideal spherical capacitor
The equivalent circuit for these capacitors is shown below.
Capacitors with Inhomogeneous Dielectrics
Each of the previously considered capacitor geometries has contained
a homogeneous insulating medium between the capacitor conductors. For
some geometries with inhomogeneous dielectrics, the capacitance can be
shown to be a simple series or parallel combination of homogeneous
dielectric capacitances.
We may use the analogies of total current in a resistor and total
electric flux in a capacitor to identify series and parallel combinations of
capacitors. The equations for the total current I in a resistor (in terms of
current density J) and the total electric flux R in a capacitor (in terms of
electric flux density D) follow the same form.
Thus, the total current I in a resistor and the total electric flux R in a
capacitor are analogous quantities and can be used to visualize the series
and parallel combinations in capacitors with inhomogeneous dielectrics.
Series (common current/flux, distinct voltages)
Parallel (common voltage, distinct currents/fluxes)
Example (Equivalent series capacitances)
Boundary condition
Y
D1 = D2
(normal D is continuous)
The common electric flux through the two dielectric regions denotes a
series capacitance combination. Even though no conductor exists on the
interface between the dielectrics, this configuration can be viewed as two
capacitors in series by placing two total charges of +Q and !Q (net charge
= 0) on the interface.
The overall capacitance of the inhomogeneous dielectric capacitor (C) is
found using the homogeneous dielectric capacitance equation applied to
the two dielectric regions.
The series combination of these two capacitances is
Example (Equivalent parallel capacitances)
Boundary condition
Y
E1 = E2
(tangential E is continuous)
Since the electric field is equal in both regions, the electric flux density is
distinct in the two regions (parallel capacitors). The capacitances of the
individual regions are
The overall capacitance (C) of the parallel combination is
Conductors in Electric Fields
(Induced Charges)
When a conductor is placed in an applied electric field, charges are
induced on the surface of the conductor that produce a secondary electric
field (induced electric field). The total electric field is the superposition of
the applied electric field and the induced electric field. For a perfect
conductor (F = 4, PEC - perfect electric conductor), the induced electric
field exactly cancels the applied field to yield a total field of zero inside the
conductor. The applied electric field separates the charge on the conductor
(positive charge forced in the direction of the applied field, negative charge
forced in the direction opposite to that of the applied field).
Thus, the charge distribution induced on the surface of the PEC produces
an induced electric field that exactly cancels the applied electric field
inside the PEC. The total field outside the PEC is the sum of the applied
electric field and the induced electric field due to the induced surface
charge.
Image Theory
(Method of Images)
Given a charge distribution and/or a current distribution over a PEC
ground plane, image theory may be used to determine the total fields above
the ground plane without ever having to determine the surface charges
and/or currents induced on the ground plane. Image theory is based on the
electric field boundary condition on the surface of the perfect conductor
(the tangential electric field is zero on the surface of a PEC). Using image
theory, the ground plane boundary condition is satisfied by replacing the
ground plane by equivalent image currents or charges located an equal
distance below the ground plane.
Example (Image theory / point charge over ground)
The total electric field at some arbitrary point P located on the ground
plane is
The vectors pointing from the point charges +Q and !Q to the field point
P (r+ and r!, respectively) are
The corresponding unit vectors r+ and r! are
where
The total electric field on the ground plane due to the original point charge
and its image charge is
The corresponding electric flux density on the ground plane is
According to the boundary condition on the ground plane surface, the
surface charge density on the ground plane is
The procedure for handling a point charge over ground may be
expanded to line, surface or volume charge distributions over ground using
superposition. Each differential element of charge in the charge
distribution may be “imaged” point by point.