Gauss

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Gauss law (chap. 23)

• In principle one can simply apply the Coulombs
  law and find out the electrostatic field
  generated by any charge distributions.
• But it may be a very complex task for many
  situations.
• However, for certain charge distributions
  involving symmetry, we can save far more work by
  using a law called Gauss law, developed by C.F.
  Gauss (1777-1855).
• To begin with, we need to learn about the idea
  of
• electric flux??? .

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• consider the case of a uniform air stream flow
  through a square area.
• if the velocity vector of the air makes an angle
  with the loop, the volume flow rate (i.e. the
  volume of air flow through the loop per unit
  time) is given by

This is an example of flux.
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Using a vector notation for area A, its direction
is the normal the area A. Then the flux can be
rewritten as

• For the case of electric field, the flux is
  simply the electric field passes through a
  certain area A.
• if we are considering a small area ?A, the flux
  passes through it is

Note that ?F can be positive, negative or zero
depends on the direction of the vectors.
For an imaginary surface, which we call Gaussian
surface, the total flux passes through the whole
surface will the sum of all ?F
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or more accurately
It is surface integral which takes all over the
whole closed surface.

1. Note that the flux is a scalar and has an unit of
   Nm2/C.
2. Since the number of electric field lines is
   proportional to the electric field density, the
   electric flux through a Gaussian surface is then
   proportional to the net number of field lines
   passing through that surface.
3. The total flux on the left side of the Gaussian
   surface shown here is negative while those on the
   right side is positive.
4. The total flux through this surface is in fact
   equal to zero.

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• To compute this integral is in general difficult
  , but for surfaces with high symmetry, it becomes
  a simple task.

Example The flux passes through a cylindrical
surface in a uniform field can be calculated by
taking the surface integral.
Note that dA and E is always perpendicular to
each other on the surface b, therefore the
surface integral must be zero.
surface integral on surface a
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Since the field is uniform that it is a constant
on the surface a and b, the surface integral on
the surfaces is simply
This is not surprise because every field lines
enter the cylinder from one surface will
eventually leave it from another surface.
Therefore the net flux must be zero.
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Here we consider the case of non-uniform field
with a Gaussian cube
Ey4
Ex3x
The x component of the field changes with x while
Ey is constant. The flux through the right side
of the cube is
surface area4m
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Similar calculation applies to the left surface.
The only difference is now x1m, and dA-dA i. So
dAdAj
For the top surface, dAdAj.
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Gauss Law

• One can easily notice that every field lines
  enter a Gaussian surface must also leave the
  surface, therefore the number field lines enter
  the closed surface must equal to number of field
  lines that leave it, i.e. the net flux 0.
• UNLESS there are field lines terminated or
  created inside the surface, which means there are
  charges (positive if field lines are created and
  negative if field lines are terminated) inside
  the Gaussian surface.
• This observation leads to the Gauss law

for the case in vacuum (or air).
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1. The Gauss law is very powerful because it
   contains no details of the surface or the charge
   distribution.
2. Therefore it is very general and applies to all
   cases.

• For example, we can apply the Gauss law to the
  surfaces S1, S2, S3, and S4.
• The net flux through S1 is positive and is equal
  to q/e0.
• The net flux through S2 is negative.
• The net flux through S3 is zero and there is no
  charge inside.
• The net flux through S4 is also zero because the
  total charge is zero. There is as many lines
  entering the surface as leaving it.

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zero charge
q5e0
q-3e0
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Gauss law and Coulombs law
These two law must be related to each other
because both describe the electric field
distribution around a system of charges. In fact,
we can derive the Coulombs law starting from the
Gausss law.

• Consider a spherical Gaussian surface around a
  point charge q.
• apply the Gauss law and we have

note that E and dA has the same direction on
every point of the surface.
because of symmetry, EdA is the same on every
point and the integral becomes
It gives the Coulombs law at the end
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A charged isolated conductor

1. First we notice that there is no electric field
   inside a conductor, otherwise free electrons in
   the conductor have to flow and lead to current.
   But it shouldnt be the case for an isolated
   conductor.
2. Then we can draw a Gaussian surface very close to
   the surface of the conductor.
3. Since there is no net flux (no field inside)
   through the Gaussian surface, by Gauss law,
   total charge inside the conductor must be zero.
4. Therefore all charges must sit on the surface
   instead.
5. For (b), because there is no flux through the
   Gaussian surface, the net charge enclosed by it
   must be zero, i.e. there is also no charge on the
   inner surface.

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• While there is no field inside a conductor, the
  charges on the surface generate an electric field
  outside the surface
• consider a flat surface with uniform charge
  density s
• one can draw a cylindrical Gaussian surface as
  shown in (a).
• the net charge enclosed by the Gaussian surface
  is sA and the flux through the outside surface is
  EA
• note that the field on the left is zero and hence
  there is no flux on the left surface
• by Gauss law

Using symmetry, Gauss law provides an easy way
to find out the field around a system of charges.
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Since there is no field inside the conductor,
there is no flux through the Gaussian surface. By
Gauss law, the net charge enclosed by the
Gaussian surface must be zero, so the induced
charges on the inner surface 5µC.
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Cylindrical Symmetry????
Find the electric around a line charge There is
no flux through the top and bottom of the
Gaussian surface and the total flux through the
curve surface is
The result is same as we have found before using
Coulombs law, but the calculation here is much
simpler.
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Planar symmetry????

• non-conducting plane
• this time, the field on the left side is not
  zero.
• the total flux is now 2EA.

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Spherical symmetry???
For a charged spherical shell, the electric field
outside the shell is
by Gauss law on S2
make use of spherical symmetry
Inside the shell
by Gauss law, there is no charge enclosed by S1,
so there is no field inside the shell
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For a charged sphere with a uniform charge
density ?
The electric field outside of the sphere, of
course, is
When r lt R, then
where q is given by
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