Download EE3321 ELECTROMAGENTIC FIELD THEORY

Week 4
Gauss’ Law, Flux Density
Gaussian Surface
Electric Potential
Poisson’s Equation
Gauss’ Law
 As it turns out, Coulomb's law is
actually a special case of Gauss' Law, a
more fundamental description of the
relationship between the distribution
of electric charge in space and the
resulting electric field.
 Gauss's law is one of Maxwell's
equations, a set of four laws governing
electromagnetics.
 The law bears the name of Karl
Friedrich Gauss (1755-1855), one of the
greatest mathematicians of all time
who also made significant
contributions to theoretical physics.
Differential form of Gauss’ Law
 In differential form, Gauss' law states that the
divergence of the E field is proportional to the charge
density that produces it:
where
ρ is the total electric charge density (in units of C/m³)
єo is the electric constant (8.854 x 10-12 F/m)
Exercise
 Suppose that a charge is distributed in a sphere of
radius R =1 m. In the region 0 < R < 1 m the electric
field is given by E = 1 x 10 – 6 R aR. Find the charge
density in this region.
 Recall the expression for the divergence in spherical
coordinates.
 Find the charge density for R< 1 m
 Plot the charge density as a function of R.
Integral form of Gauss’ Law
 We can make use of the divergence
theorem to obtain the integral form of
Gauss’s law:
where
Q is the charge enclosed by the surface
of integration and is equal to the
volume integral of the charge density.
Flux Density
 We define the electric flux
density as
 Then, the electric flux is given
by
Electric Dipole
 Consider the case of an electric dipole. What is the
total flux coming out of the rectangular surface?
Gaussian Surface
 A Gaussian surface is a closed two-dimensional surface
through which a flux or electric field is to be calculated.
 The surface is used in conjunction with Gauss's law,
allowing one to calculate the total enclosed electric charge
by performing a surface integral.
 Gaussian surfaces are usually carefully chosen to exploit
symmetries of a situation to simplify the calculation of the
surface integral.
 If the Gaussian surface is chosen such that for every point
on the surface the component of the electric field along the
normal vector is constant, then the calculation will not
require difficult integration as the constant can be pulled
out of the integration sign.
Spherical Surface
 A spherical Gaussian surface is used
when finding the electric field or
the flux produced by any of the
following:
 a point charge
 a uniformly distributed spherical
shell of charge
 any other charge distribution with
spherical symmetry
 The spherical Gaussian surface is
chosen so that it is concentric with
the charge distribution.
Point Electric Charge
 What is the total flux out of a spherical surface around
a point charge?
 What is the divergence of D?
Exercise
 Let D = Do R aR for R≤ a. Determine the amount of
charge enclosed by a spherical surface of radius b for:
 b<a
 b=a
 b>a
Exercise
 Consider a charged
spherical shell of negligible
thickness, with a uniformly
charge density ρs and
radius a.
 Use Gauss's law to find the
magnitude of the resultant
electric field E inside the
charged shell.
Tesla Cage
 In a Tesla cage, the net flux is zero and the magnitude
of the electric field is also zero.
Cylindrical Gaussian Surface
 A cylindrical Gaussian surface is used when finding
the electric field or the flux produced by an infinitely
long line of uniform charge.
The Pillbox
 This Gaussian surface is used to find the electric field due
to an infinite plane of uniform charge.
Electric Potential
 The electrical potential difference is defined as the
amount of work needed to move a unit electric charge
from the second point to the first, or equivalently, the
amount of work that a unit charge flowing from the first
point to the second can perform.
 The potential difference between two points A and B is
the line integral of the electric field E
Line Integral
 The integration path is an
arbitrary path connecting
point A of known potential to
the observation point B as
shown below.
 The value of V is independent
of the integration path.
Line Integral
 The value of V is independent of the integration path.
Exercise
 Let E = Eo az (V/m) for z>0.
Find the potential between the
points A and B:
 A = (0, 0, 0) and B = (1, 1, 1)
 Steps: set up the integral
following a simple path
 A to (1, 0, 0)
 (1, 0, 0) to (1, 1, 0)
 (1, 1, 0) to B
Potential at Infinity
 The potential at infinity is said to be zero.
 If a test charge +q moves toward a charge +Q starting
at an infinite range it will gain potential at point P.
Gradient of V
 When the magnetic field is constant in time, it is
possible to express the electric field as the gradient of
the electrostatic potential.
Exercise
 A spherically charged shell of radius a, centered at the
origin, creates the potential V = – Voa/R (Volts) for
R>a.
 Determine the corresponding electric field E.
 Recall the expression of the gradient operator in
spherical coordinates.
 Notice that the potential V is independent of the
azimuth and elevation angles.
Exercise
 Consider the potential V = 4 ln ( a/r ) in cylindrical
coordinates. Find E.
 Recall the expression for the gradient operator in
cylindrical coordinates.
 Notice that the potential V is independent of the
azimuth angle and the elevation z.
Equipotential Lines and Surfaces
 An equipotential line is a line in
space where the potential is
constant.
 An equipotential surface is a
surface in space where the
potential is constant.
 No work is required for a charge
to move along an equipotential
surface. Work is required for a
charge to move to a different
equipotential surface.
Point Charge
 For a point charge, an equipotential line takes the
shape of a circle, and an equipotential surface is a
sphere centered on the charge.
Capacitor
 Consider two large parallel
plates separated by a short
distance L.
 Assuming that the plates have
a uniform charge distribution,
the electric field lines are
perpendicular to the plates and
the equipotential lines are
parallel to the plates.
Potential Well
 If a test charge approaches the
charged sphere it will
experience a decreasing
potential.
 The potential is symmetric and
decreases as the observation
gets closer to the sphere.
 The concentric orbits in the
figure represent lines of equal
potential.
Poisson’s Equation
 Poisson's equation relates the
potential V to the charge density ρ.
 To determine V one needs to make
use of boundary conditions.
 If the charge density is zero, then
Laplace's equation results.
Cathode Tube
Exercise
 Let d2V/dz2 = 0 with boundary conditions V(0) = 0 and
V(z=d)=100 V.
Find for V(z).
 Integrate Laplace’s Equation over z.
 Integrate again.
 Apply boundary conditions.
Homework
 Read sections 4-4 and 4-5
 Solve the following end-of-chapter problems:
4.22, 4.24, 4.28, 4.30, 4.34
Review “Hyperphysics”