Applications of Gauss Law

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Applications of Gauss Law
In cases of strong symmetry, Gauss's law may be
readily used to calculate E. Otherwise it is not
generally useful and integration over the charge
distribution is required. But when the symmetry
permits it, Gauss's law is the easiest way to go!
The KEY TO ITS APPLICATION is the choice of
Gaussian surface. Keep in mind that this is not a
surface of the charge distribution itself, but
rather an imaginary surface constructed for
application of Gauss's law.

• To make Gauss's law useful
• All sections of the gaussian surface should be
  chosen so that they are either parallel or
  perpendicular to E. For this we need to have
  already determined the direction of E everywhere
  on the surface by symmetry arguments.
• E should be constant on each surface. Again,
  this requires sufficient symmetry to "see" the
  constancy of field strength.
• And once again, make sure that the surface is
  closed!

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Gauss Law and Coulomb law
The angle ? between E and dA is zero at any
point on the surface, we can re-write Gauss Law
as
E has the has same value at all points on the
surface
E is can be moved out
Integral is the sum of surface area
A spherical Gaussian surface centered on a point
charge q
Coulombs Law
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Cylindrical Symmetry
Evaluate the electric field that arises from an
infinite line of charge 1) Understand the
geometry Pick a point P for evaluation of E.
2) Understand the symmetry
The only variable on which E may depend is R, the
distance from the line of charge. There is no
angular dependence because the line is
cylindrically symmetric, i.e. it does not matter
at which angle about the line you view it. There
is no axial dependence because the line of charge
is infinitely long, i.e. there is no preferred
position along the line. Similar arguments
determine that the direction of E is directly
away from the line, perpendicular to it at all
locations. There may be no azimuthal or axial
components because the line is infinite and has
no extent in the transverse directions.
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Cylindrical Symmetry
3) Construct the Gaussian surface
A cylindrical tube is chosen to match the
symmetry of the line of charge, centered about
it, with a radius equal to the distance between
the point of evaluation and the line of charge.
The electric field anywhere on this surface has
the same direction as the infinitesimal area
vector and has a constant value everywhere on the
surface. This surface is denoted by 2. Since a
gaussian surface must be closed, the tube is then
capped with flat endcaps, 1 and 3. The electric
field does not have the same value at all points
on the endcaps, but the field vector is
perpendicular to the area vector so there is no
flux through the endcaps.
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Cylindrical Symmetry
4) Examine the Gaussian surface
The flux through the endcap 1 is zero since A1 is
perpendicular to E everywhere on this surface.
Similarly, the flux through surface 3 is zero.
Since the electric field is everywhere constant
on surface 2 and points in the same direction as
the surface vector at any point, the flux through
surface 2 will be the field magnitude E times the
area of the tube wall.
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Cylindrical Symmetry
5) Evaluate the electric flux through the
Gaussian surface
6) Evaluate the charge enclosed by the Gaussian
surface
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Cylindrical Symmetry
7) Apply Gauss's law for the result!
Source http//www.physics.udel.edu/watson/phys20
8/line-gauss/line-gauss1.html
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For comparisonLine of Charge by Direct
Integration 1
Evaluate the electric field that arises from an
infinite line of charge.
Pick a point P for evaluation of E. The only
variable which matters is R, the distance from
the line of charge. There is no angular
dependence because the line is cylindrically
symmetric, i.e. it does not matter at which angle
about the line you view it. There is no axial
dependence because the line of charge is
infinitely long, i.e. there is no preferred
position along the line. The infinite line of
charge is certainly a physically-impossible
situation to set up (infinite amount of charge)
but one whose consideration is useful in
interpreting finite distributions, e.g. checking
the limiting behavior of solutions.
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For comparisonLine of Charge by Direct
Integration 2
Span the charge distribution
Assuming a uniform charge distribution along a
line, the variable which appears to be the
easiest to use to span the distribution is the
variable x, which follows along the line from
negative infinity to positive infinity. Thus the
charge infinitesminal dq will be the product of
charge density and dx.
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For comparisonLine of Charge by Direct
Integration 3
Evaluate the contribution from the infinitesimal
charge element
Considering a positive test charge at point P,
the direction of the electric field is shown. The
strength of the electric field contribution from
the infinitesimal charge shown is proportional to
dq and inversely proportional to the square of
the distance separating point P and the charge
element.
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For comparisonLine of Charge by Direct
Integration 4
Exploit symmetry as appropriate
By examining the conjugate infinitesimal on the
other side of the origin, we see that the
horizontal components of the two contributions
will balance. We need only consider the vertical
contribution to the electric field by each
infinitesimal charge.
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For comparisonLine of Charge by Direct
Integration 5
Set up the integral
In setting up the integral we wish to express all
factors entering the electric field contribution
as functions of the integration variable x. After
finding the magnitude of the infinitesimal
contribution to the field, we consider its
projection along the vertical axis. After the
integrand is determined, we evaluate the integral
between the limits of integration.
Substituting these terms, dEy becomes
Setting up the integrating over the charge
distributing
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For comparisonLine of Charge by Direct
Integration 6
Solve the integral by trig substitution
Looking at the integrand, you may now realize
that trig substitution is required. All that
means is that we should have integrated over the
angle variable theta rather than the linear
variable x. The figure below shows the
relationship between the two variables for the
quantities involved
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For comparisonLine of Charge by Direct
Integration 7
Expressing the integrand in terms of theta and
switching the limits of integration, the integral
may be solved and the result finally obtained!
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A Charged Isolated Conductor
Gauss Law permits us to prove an important
theorem about conductors.
If excess charged is placed on an isolated
conductor, the charge will move to the surface of
the conductor. None of the excess charge will be
found within the body pf he conductor.
The electric field inside a conductor in static
equilibrium must be zero. Otherwise, the field
would exert forces on the conduction electrons
producing motion or currents which is not a
static equilibrium state. If we place a Gaussian
surface just inside the surface of the charged
conductor, the E field is zero for all points on
the Gaussian surface. Therefore the flux through
the Gaussian is zero and according to Gauss Law
the net charge enclosed the Gaussian surface must
also be zero. Therefore the excess charged must
be outside the Gaussian surface and must lie on
the actual surface of the conductor.
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External Electric Field of a Conductor
Consider a section of the surface that is small
enough to to neglect any curvature and assume the
section is flat.
Embed a tiny cylindrical Gaussian surface with
one end cap full inside the conductor and the
other fully outside and cylinder is perpendicular
to the surface.
The electric field just outside the surface must
also be perpendicular to that surface otherwise
surface charges would be subject to motion.
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External Electric Field of a Conductor
We now sum the flux through the Gaussian
surface. There is no flux through the internal
endcap. Why ?? There is no flux through the
curved surface. Why ?? The only flux through the
Gaussian surface is that through the external
endcap where E is perpendicular. Assuming a cap
area of A, the flux through the cap is EA. The
charge qenc enclosed by the Gaussian surface lies
on the conductors surface area A. If ? is the
charge per unit area , the qenc is equal to ? A.
Therefore Gauss Law becomes
?oEA ? A
E ? / ?o (conducting
surface)
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Planar Symmetry
Two different views of a very large thin plastic
sheet, uniformly charged on one side with surface
density of ?. A closed cylindrical Gaussian
surface passes throught the sheet and is
perpendicular. From symmetry , E must be
perpedicular to the surface and endcaps. The
surface charge is positive so E emanates from
the surface. Since the field lines do not piece
the curved surface there is no flux.
What happens is we magically make the sheet a
conductor ?? Will the E field change ?
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Planar Symmetry
Figure (a) show a cross-section of a thin,
infinite conducting plate with excess positive
charge. We know the this excess charge lies on
the surface of the plate. If there is no external
electric field to force the charge into some
particular distribution, it will spread out onto
the two surfaces with an uniform charge density
of ?1. From the previous slide we know this
charge sets up an electric field
which points
1
away from the plate. Figure (b) is a identical
plate with negative charge. In this case E points
inward.
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Planar Symmetry
Suppose we move the two plates in the figures (a)
and (b) close together and parallel. Since the
plates are conductors, the excess charge on one
plate attracts the excess charge on the other and
all the excess charge moves into the inner faces
of the plates. The new surface charge density has
now doubled,
therefore the electric field between the plates is
The field points away from positive charged plate
and toward the negative plate. Since no excess is
left on the outer faces, the electric field
outside is zero.
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Spherical Symmetry
Lets apply Gausss Law to a hollow charged sphere
What is electric field inside the hollow charged
sphere ?
x
Consider the point mark X in the figure located
on an interior equator By symmetry Ey0, what
about Ex ?? (Since X can be viewed at any angle,
the E-field must be in the radial direction.
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Spherical Symmetry
x
x
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Spherical Symmetry
Instead of deriving the solution through a long
integration Lets use Gauss Law. 1st choose an
Gaussian surface inside the shell that coincides
with the point x.
Since fields lines can not cross, the spherical
symmetry requires E field to be in the radially
direction and uniform on the Gaussian surface,
therefore Gauss Law
x
r
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Spherical Symmetry
Outside the hollow sphere
Now lets consider the electric field a a point
outside the charged sphere. We define a Gaussian
surface around the sphere for which rgtR. Again
since fields lines can not cross, the spherical
symmetry requires E field to be in the radially
direction and uniform on the Gaussian surface,
therefore Gauss Law becomes
x
Same as point charge
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Spherical Symmetry
Using Gauss Law we were able to prove two shell
theorems that were presented in the 1st lecture
without proof. A shell of uniform charge
attracts and repels a charged particle that is
outside the shell as if all shells charge were
concentrated at the center. If a charged particle
is located inside a shell of uniform charge,
there is no net electric force on the particle
from the shell.
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