Week 4 Gauss Law, Flux Density Gaussian Surface Electric
Potential Poissons Equation
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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.
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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 is the total electric charge
density (in units of C/m) o is the electric constant (8.854 x 10
-12 F/m)
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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 a R. 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.
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Integral form of Gauss Law We can make use of the divergence
theorem to obtain the integral form of Gausss law: where Q is the
charge enclosed by the surface of integration and is equal to the
volume integral of the charge density.
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Flux Density We define the electric flux density as Then, the
electric flux is given by
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Electric Dipole Consider the case of an electric dipole. What
is the total flux coming out of the rectangular surface?
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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.
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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.
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Point Electric Charge What is the total flux out of a spherical
surface around a point charge? What is the divergence of D?
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Exercise Let D = D o R a R for R a. Determine the amount of
charge enclosed by a spherical surface of radius b for: ba
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Exercise Consider a charged spherical shell of negligible
thickness, with a uniformly charge density s and radius a. Use
Gauss's law to find the magnitude of the resultant electric field E
inside the charged shell.
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Tesla Cage In a Tesla cage, the net flux is zero and the
magnitude of the electric field is also zero.
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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.
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The Pillbox This Gaussian surface is used to find the electric
field due to an infinite plane of uniform charge.
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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
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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.
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Line Integral The value of V is independent of the integration
path.
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Exercise Let E = E o a z (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
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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.
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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.
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Exercise A spherically charged shell of radius a, centered at
the origin, creates the potential V = V o a/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.
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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.
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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.
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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.
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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.
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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.
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Poissons Equation Poisson's equation relates the potential V to
the charge density . To determine V one needs to make use of
boundary conditions. If the charge density is zero, then Laplace's
equation results. Cathode Tube
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Exercise Let d 2 V/dz 2 = 0 with boundary conditions V(0) = 0
and V(z=d)=100 V. Find for V(z). Integrate Laplaces Equation over
z. Integrate again. Apply boundary conditions.
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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