Chapter 23 Gausss Law

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Chapter 23 Gausss Law

• The electric flux through a closed surface is
  proportional to the charge enclosed

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FLUX
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FLUX
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Definition of Flux

• The amount of field, material or other physical
  entity passing through a surface.
• Surface area can be represented as vector defined
  normal to the surface it is describing
• Defined by the equation

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Electric Flux

• The amount of electric field passing through a
  surface area
• The units of electric flux are N-m2/C

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Carl Friedrich Gauss 1777-1855
Yes its the same guy that gave you the Gaussian
distribution and
To give you some perspective he was born 50 years
after Newton died (1642-1727). Predicted the
time and place of the first asteroid CERES (Dec.
31, 1801). Had the unit of magnetic field named
after him and of course had much to do with the
development of mathematics
Ceres t 4.6 year , d4.6 Au
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Gauss Law in E M

• Uses symmetry to determine E-field due to a
  charge distribution
• Method Considers a hypothetical surface
  enclosing some charge and calculates the E-field
• The shape of that surface is
  EVERYTHING

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KEY TO USING Gausss Law

• The shape of the surrounding surface is one that
  MIMICS the symmetry of the charge distribution ..

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Spherical Coordinates

• Surface Integrals

right (green)
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The electric flux passing through a spherical
surface surrounding a point charge
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The electric flux passing through a spherical
surface surrounding a point charge
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Gausss Law

• The previous example is true in general
• The total flux passing through a closed surface
  is proportional to the charge enclosed within
  that surface.

Note The area vector points outward
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The Gaussian Surface

• An imaginary closed surface created to enable
  the application of Gausss Law

What is the total flux through each surface?
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Solving problems with Gausss Law

• Charge densities
• It is convenient to define charge
  densities for linear, surface and volume charge
  distributions
• 2. Symmetry and coordinate systems
• Choose that coordinate system that most
  nearly matches the symmetry of the charge
  distribution. For example, we chose spherical
  coordinates to determine the flux due to a point
  charge because of spherical symmetry.

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Solving problems with Gausss Law

• 3. Determining the Electric Field from Gausss
  Law - We will use Gausss Law to determine the
  electric field for problems for which the
  Electric Field can be shown to be constant in
  magnitude in direction for a particular Gaussian
  surface.
• Consider three examples (1) the long straight
  line of charge, (2) the infinite plane sheet of
  charge, and (3) a charged sphere.

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Example - Long straight line of charge
Looking at the diagram (b), we can determine that
the problem has a cylindrical symmetry. Therefore
cylindrical coordinates are appropriate.
There are three surfaces to consider. The upper
and lower circular surfaces have normals parallel
to the z axis which are perpendicular to the
electric field, thus contribute zero to the flux.
The integral to be evaluated is that of the
cylinder of height l. The charge enclosed is ll.
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Example - Long straight line of charge
Since the field has radial symmetry, it is also
constant at a fixed distance of r.
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Gauss Law Determining the E-field near the
surface of a nonconducting (insulating) sheet
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Results of other geometries
Uniformly charged dielectric infinite plane sheet
Uniformly charged dielectric (insulating) sphere
Can you derive these results?
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Conclusions Gauss LAW

• Only the charge enclosed within a volume defined
  by a closed surface contributes to the net
  electric flux through the surface.
• That net flux through the surface is proportional
  to the charge enclosed within the volume.

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Conclusions

• 3. Gaussian surface is an imaginary closed
  surface necessary to solve a problem using
  Gausss Law
• 4. Gausss Law can be used to determine the
  electric field of a charge distribution if there
  is a high degree of symmetry

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Conclusion Gauss Law and Conductors

• 5. Applying Gausss Law to the interior of an
  electrostatically charged conductor we conclude
  that the electric field within the conductor is
  zero
• 6. Any Net charge on a conductor must reside on
  its surface

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Shell Theorems Conductors

• A shell of uniform charge attracts or repels a
  charge particle that is outside the shell as
  though all charge is concentrated at the center.
• If a charged particle is located inside such a
  shell, there is no electrostatic force on the
  particle from the shell

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Electric Field at surface of conductor is
perpendicular to the surface and proportional to
the charge density at the surface
Shown that the excess charge resides on the outer
surface of conductor surface. Unless the surface
is spherical the charge density s (chg. per unit
area) varies However the E-field just outside of
a conducting surface is easy to determine using
Gauss Law
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Applied Gauss Law to Determine E-field in Cases
where have

• Spherical symmetry
• Cylindrical symmetry
• Planar symmetry