Gauss

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Gauss Law
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Class Objectives

• Introduce the idea of the Gauss law as another
  method to calculate the electric field.
• Understand that the previous method of
  calculating the electric field strength does not
  consider symmetry.
• Consider the different types of symmetry.

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Class Objectives

• Introduce the idea of the Gaussian surface.
• Define the properties of Gaussian surfaces.
• Show how to choose a Gaussian surface.
• Show how Gaussian surfaces can be used to take
  advantage of symmetry.

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Class Objectives

• Show that the results of Gausss law are the same
  as the standard results but quicker and easier!

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Student Objectives

• Be able to use Gauss law to calculate the
  electric field for various objects.
• Be able to determine the type of symmetry in the
  problem and hence the type of Gaussian surface to
  be used.

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Gauss Law

• Symmetry in problems arise naturally.

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Gauss Law

• Symmetry in problems arise naturally.
• Gauss law is an alternate method to coulombs
  law where there is symmetry.

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Gauss Law

• Outline of Gauss Law
• Central idea is a hypothetical closed surface
  called a Gaussian surface.
• The surface can be chosen any shape to complement
  the symmetry. Eg cylinder or sphere.
• Gauss law relates the electric field at a point
  on the closed surface to the net charge enclosed
  by the surface.

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Gauss Law

• The amount of charge enclosed is measured by the
  amount of flux passing through the surface.

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Gauss Law

• Example of a spherical Gaussian closed surface.

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Electric flux

• The electric flux is a measure of the number of
  field lines passing an area.
• The flux is represented by the symbol .

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Electric flux

• The electric flux is a measure of the number of
  field lines passing an area.
• The flux is represented by the symbol .
• Assuming an arbitrary Gaussian surface in a field.

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Electric flux

• If we divide the surface into small squares of
  area then the curvature of the surface can
  be ignored if the squares are taken small enough.

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Electric flux

• If we divide the surface into small squares of
  area then the curvature of the surface can
  be ignored if the squares are taken small enough.
• By convention we assign an area vector .
• The direction of which is perpendicular to the
  surface of the element directed away from its
  interior.

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Electric flux

• Because can be made arbitrarily small,
• may be taken as constant for a given square.
• The flux for a Gaussian surface is given as

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Electric flux

• Because can be made arbitrarily small,
• may be taken as constant for a given square.
• The flux for a Gaussian surface is given as
• The circle indicates that the integral is to be
  taken over the entire closed surface.

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Electric flux

• The follow diagrams gives three cases for the
  relative orientation of the field and dA. The
  angle measured is the smallest angle between the
  vectors.

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Electric flux

• Note that all of the elements on the surface do
  not contribution to the flux.

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Electric flux

• Note that all of the elements on the surface do
  not contribution to the flux.

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Gauss Law

• Guass Law is often written as
• Where is the sum of the enclosed charges.

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Gauss Law

• Choosing a Gaussian surface

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Gauss Law

• Choosing a Gaussian surface is not hard but
  subtle.

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Gauss Law

• Choosing a Gaussian surface is not hard but
  subtle.
• A simple surface must be chosen to take advantage
  of symmetry.
• Consider the following rules as a guide.

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Gauss Law

1. Choose the surface perpendicular to the field so
   the E and dA are parallel.

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Gauss Law

1. Choose the surface perpendicular to the field so
   the E and dA are parallel.
2. Choose so that points on the surface are equal
   distance away from the charge so the E doesnt
   vary.

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Gauss Law

1. Choose the surface perpendicular to the field so
   the E and dA are parallel.
2. Choose so that points on the surface are equal
   distance away from the charge so the E doesnt
   vary.
3. If this is not possible (1 and 2) choose a
   surface such that the dot product is zero. Ie.
   They are perpendicular.

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Gauss Law and Coulombs Law

• Consider a point charge q.

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Gauss Law and Coulombs Law

• Consider a point charge q.
• For symmetry we use a spherical Gaussian surface.

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Gauss Law and Coulombs Law

• the chosen Gaussian surface holds true for the
  first two guidelines.

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Gauss Law and Coulombs Law
dA

• the chosen Gaussian surface holds true for the
  first two guidelines.
• E and dA are parallel.
• Points on the surface are equal distance away
  from the charge so the E doesnt vary.

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Gauss Law

• From Gauss Law

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Gauss Law

• From Gauss Law
• Since the two vectors are parallel.

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Gauss Law

• From Gauss Law
• Since the two vectors are parallel.
• Also we note that the enclosed charge is simply
  q. So we can write that,

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Gauss Law

• The area of the closed surface is the area of a
  circle. So that

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Gauss Law

• The area of the closed surface is the area of a
  circle. So that
• Therefore

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Conductor in an Electric Field
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Conductor in an Electric Field

• For the cases to investigated, we will consider
  the cases where the conductor is in equilibrium
  (electrostatics).

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Conductor in an Electric Field

• For the cases to investigated, we will consider
  the cases where the conductor is in equilibrium
  (electrostatics).
• For a conductor in equilibrium we have following
  conditions

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Conductor in an Electric Field

1. The charge exists entirely on the surface of
   conductor(no charge is found within the body of
   the conductor).

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Conductor in an Electric Field

1. The charge exists entirely on the surface of
   conductor(no charge is found within the body of
   the conductor).
2. The electric field within the conductor is zero.

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Conductor in an Electric Field

1. The charge exists entirely on the surface of
   conductor(no charge is found within the body of
   the conductor).
2. The electric field within the conductor is zero.
   The charge distributes itself so as to get as far
   from each other as possible.

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Conductor in an Electric Field

1. The external electric field is perpendicular to
   the surface of the conductor.

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Conductor in an Electric Field

1. The external electric field is perpendicular to
   the surface of the conductor. If not it would
   cause the charges to move along the surface of
   the conductor.

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Conductor in an Electric Field

• The external electric field is perpendicular to
  the surface of the conductor. If not it would
  cause the charges to move along the surface of
  the conductor.
• Note unless the conductor is spherical, the
  charge does not distribute itself uniformly.

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Conductor in an Electric Field

• When a conductor is place in an external field,
  the mobile electrons experience a force pushing
  them in the opposite direction to the field

E
e
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Conductor in an Electric Field

• Thus making the top of the conductor positive and
  bottom negative.

E

e
_ _ _ _ _ _ _ _ _ _ _
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Conductor in an Electric Field

• Thus making the top of the conductor positive and
  bottom negative. This sets up an internal
  electric which grows in strength as more
  electrons move.

E

Electric field caused by pd between the top and
bottom of the conductor
Eint
e
_ _ _ _ _ _ _ _ _ _ _
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Conductor in an Electric Field

• Electrons move until the two fields have the same
  magnitude. Hence the net electric field E0.

E

Eint
e
_ _ _ _ _ _ _ _ _ _ _
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Conductor in an Electric Field

• Electrons move until the two fields have the same
  magnitude. Hence the net electric field E0. The
  magnitude of the charge on the top bottom of
  the conductor are the same.

E
E

Eint
e
Eint
_ _ _ _ _ _ _ _ _ _ _
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Conductor in an Electric Field

• Let us now consider the electric field just
  outside the surface of a conductor.

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Conductor in an Electric Field

• Let us now consider the electric field just
  outside the surface of a conductor.

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Conductor in an Electric Field

• Let us now consider the electric field just
  outside the surface of a conductor. We use a
  cylindrical Gaussian surface.

A
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Conductor in an Electric Field

• Assume the conductor has charge per unit area
  and total charge Q with uniform charge
  distribution.

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Conductor in an Electric Field

• Recall Gauss Law

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Conductor in an Electric Field

• Recall Gauss Law
• We note that

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Conductor in an Electric Field

• Recall Gauss Law
• We note that
• We can therefore write that

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Conductor in an Electric Field

• Integrating as before we get that

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Conductor in an Electric Field

• Integrating as before we get that
• Hence

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Conductor in an Electric Field

• Integrating as before we get that
• Hence
• The electric field.

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Conductor in an Electric Field

• Cylindrical rod

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Conductor in an Electric Field

• Consider an infinitely long cylindrical rod with
  uniform linear charge density ?. What is the
  electric field a distance r from its axis.

r

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Conductor in an Electric Field

• We choose a cylindrical Gaussian surface.

E
E
r

h
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• Writing Gauss law

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• Writing Gauss law
• Hence,

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• Writing Gauss law
• Hence,

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• Writing Gauss law
• Hence,
• The area enclosed is that of a cylinder.

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• Writing Gauss law
• Hence,
• The area enclosed is that of a cylinder.

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• The electric field due to a line of charge is

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• Charged sphere

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• We previously looked at the properties of a
  conductor.
• For a spherical conductor of charge Q with
  uniform charge density, what is the electric
  field outside and inside the conductor?

Q
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• We know the charge resides on the surface.
• Consider a Gaussian surface inside the sphere.

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• We know the charge resides on the surface.
• Consider a Gaussian surface inside the sphere.
• No charge is enclosed by the surface.

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• We know the charge resides on the surface.
• Consider a Gaussian surface inside the sphere.
• No charge is enclosed by the surface.
• Therefore E 0 as expected.

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• Now consider a Gaussian surface outside the
  sphere.

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• Now consider a Gaussian surface outside the
  sphere.
• The total charge enclosed is Q.

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• Now consider a Gaussian surface outside the
  sphere.
• The total charge enclosed is Q.
• Therefore from Gauss Law,

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• Now consider a Gaussian surface outside the
  sphere.
• The total charge enclosed is Q.
• Therefore from Gauss Law,

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• Hence,
• As expected!!

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• Uniform Charge non-conducting Sphere

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• Consider a solid non-conducting sphere of radius
  R with uniform charge distribution ?. What is the
  electric field outside and inside the sphere?

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• The total charge is given by

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• The total charge is given by
• Case r gt R (outside the sphere)

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• The total charge is given by
• Case r gt R (outside the sphere)
• Gauss Law

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• The total charge is given by
• Case r gt R (outside the sphere)
• Gauss Law
• The charge enclosed is the total charge.

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• The total charge is given by
• Case r gt R (outside the sphere)
• Gauss Law
• The charge enclosed is the total charge.
• E is constant on the surface, therefore

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• Integrating over the surface we have,

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• Integrating over the surface we have,

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• Integrating over the surface we have,
• That is the sphere acts like a point charge (as
  expected).

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• Case r lt R (inside the sphere)

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• Case r lt R (inside the sphere)
• Again we use Gauss Law.
• However the total charge is not enclosed.

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• Case r lt R (inside the sphere)
• Again we use Gauss Law.
• However the total charge is not enclosed.
• The total charge enclosed is

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• Case r lt R (inside the sphere)
• Again we use Gauss Law.
• However the total charge is not enclosed.
• The total charge enclosed is
• Therefore,

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• Case r lt R (inside the sphere)
• Again we use Gauss Law.
• However the total charge is not enclosed.
• The total charge enclosed is
• Therefore,

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• Case r lt R (inside the sphere)
• Again we use Gauss Law.
• However the total charge is not enclosed.
• The total charge enclosed is
• Therefore,