PHY 184

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PHY 184

• Spring 2007
• Lecture 7

Title Using Gauss law
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Announcements

• Homework set 3 opened this morning.
• Starting Monday, we will have clicker
  quizzes/questions giving extra credit (up to 5).
• Helproom Hours for the Honors Students start next
  week.

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Review - Gauss Law

• Gauss Law
• Gauss Law says that the electric flux through a
  closed surface is proportional to the net charge
  enclosed by this surface.
• If we add the definition of the electric flux we
  get another expression for Gauss Law

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Gauss Law for Various Charge Distributions

• We have applied Gauss Law to a point charge and
  showed that we get Coulombs Law.
• Now lets look at more complicated distributions
  of charge and calculate the resulting electric
  field.
• We will use a charge density to describe the
  distribution of charge.
• This charge density will be different depending
  on the geometry.

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Cylindrical Symmetry

• Lets calculate the electric field from a
  conducting wire with charge per unit length ?
  using Gauss Law
• We start by assuming a Gaussian surface in the
  form of a right cylinder with radius r and length
  L placed around the wire such that the wire is
  along the axis of the cylinder

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Cylindrical Symmetry (2)

• From symmetry we can see that the electric field
  will extend radially from the wire.
• How?
• If we rotate the wire along its axis, the
  electric field must look the same
• Cylindrical symmetry
• If we imagine a very long wire, the electric
  field cannot be different anywhere along the
  length of the wire
• Translational symmetry
• Thus our assumption of a right cylinder as a
  Gaussian surface is perfectly suited for the
  calculation of the electric field using Gauss
  Law.

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Cylindrical Symmetry (3)

• The electric flux through the ends of the
  cylinder is zero because the electric field is
  always parallel to the ends.
• The electric field is always perpendicular to the
  wall of the cylinder so
• and now solve for the electric field

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Planar Symmetry

• Assume that we have a thin, infinite
  non-conducting sheet of positive charge
• The charge density in this case is the charge
  per unit area, ?
• From symmetry, we can see that the electric field
  will be perpendicular to the surface of the sheet

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Planar Symmetry (2)

• To calculate the electric field using Gauss Law,
  we assume a Gaussian surface in the form of a
  right cylinder with cross sectional area A and
  height 2r, chosen to cut through the plane
  perpendicularly.
• Because the electric field is perpendicular to
  the planeeverywhere, the electric field will be
  parallel to the walls of the cylinder and
  perpendicular to the ends of the cylinder.
• Using Gauss Law we get
• so the electric field from an
  infinitenon-conducting sheet with charge density
  ?

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Planar Symmetry (3)

• Assume that we have a thin, infinite conductor
  (metal plate) with positive charge
• The charge density in this case is also the
  charge per unit area, ?, on either surface there
  is equal surface charge on both sides.
• From symmetry, we can see that the electric field
  will be perpendicular to the surface of the sheet

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Planar Symmetry (4)

• To calculate the electric field using Gauss Law,
  we assume a Gaussian surface in the form of a
  right cylinder with cross sectional area A and
  height r, chosen to cut through one side of the
  plane perpendicularly.
• The field inside the conductor is zero so the end
  inside the conductor does not contribute to the
  integral.
• Because the electric field is perpendicular to
  the plane everywhere,the electric field will be
  parallel to the walls of the cylinder and
  perpendicular to the end of the cylinder outside
  the conductor.
• Using Gauss Law we get
• so the electric field from an
  infiniteconducting sheet with surface charge
  density ? is

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Spherical Symmetry

• Now lets calculate the electric field from
  charge distributed as a spherical shell.
• Assume that we have a spherical shell of charge Q
  with radius R (gray).
• We will assume two different spherical Gaussian
  surfaces
• r gt R (blue) i.e. outside
• r lt R (red) i.e. inside

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Spherical Symmetry (2)

• Lets start with the Gaussian surface outside the
  sphere of charge, r gt R (blue)
• We know from symmetry arguments that the electric
  field will be radial outside the charged sphere
• If we rotate the sphere, the electric field
  cannot change
• Spherical symmetry
• Thus we can apply Gauss Law and get
• so the electric field is

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Spherical Symmetry (3)

• Lets lets take the Gaussian surface inside the
  sphere of charge, r lt R (red)
• We know that the enclosed charge is zero so
• We find that the electric field is zero
  everywhere inside spherical shell of charge
• Thus we obtain two results
• The electric field outside a spherical shell of
  charge is the same as that of a point charge.
• The electric field inside a spherical shell of
  charge is zero.

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Spherical Symmetry (4)

• Next, lets calculate the electric field from
  charge distributed uniformly throughout a sphere.
• Assume that we have a solid sphere of charge Q
  with radius R with constant charge density per
  unit volume ?.
• We will assume two different spherical Gaussian
  surfaces
• r2 gt R (outside)
• r1 lt R (inside)

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Spherical Symmetry (5)

• Lets start with a Gaussian surface with r1 lt R.
• From spherical symmetry we know that the electric
  field will be radial and perpendicular to the
  Gaussian surface.
• Gauss Law gives us
• Solving for E we find

volume
area
inside
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Spherical Symmetry (6)
In terms of the total charge Q
inside
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Spherical Symmetry (7)

• Now consider a Gaussian surface with radius r2 gt
  R.
• Again by spherical symmetry we know that the
  electric field will be radial and perpendicular
  to the Gaussian surface.
• Gauss Law gives us
• Solving for E we find

area
full charge
outside
same as a point charge!
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Electric Field from a Ring of Charge
We cant solve it by Gauss Law! Use the method
of integration
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Pointed Surfaces
E field is always perpendicular to the surface of
a charged conductor.
Now consider a sharp point. The field lines are
much closer together, i.e., the field is much
stronger and looks much more like the field of a
point charge.
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Demo - Electrostatic Wind
The strong electric field at the sharp tip
polarizes atmospheric molecules. If a molecule
is ionized, the ion will be repelled from the
point. The recoil force causes the whirl to spin.