Gausss Law Applications

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Gausss Law Applications

• PHYS 2326-7

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Review Flux

• Electric field lines are tangent to the electric
  field vector at each point along the line
• Lines/unit area are proportional to the electric
  field
• For a uniform electric field perpendicular to a
  rectangular surface, FE the electric flux is FE
  EA (Nm2/C)
• Flux is proportional to the number of electric
  field lines penetrating a surface

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Non Perpendicular Surface

• Taking a square that is not perpendicular to the
  surface, the relationship becomes
• FE EAcos? (Nm2/C)
• where ? is the angle between the surfaces
  (normals to the surfaces)

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General Case
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Review

• Closed surface one that totally surrounds a
  region
• The net number of lines means the total leaving
  the surface minus the total entering the surface
• If more leave than enter net flux is positive
• If less net flux is negative

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General Equation
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Applying Gausss Law

• 1. Use symmetry of the charge to find direction
  of E field and where it is constant
• 2. Choose Gaussian surface to be perpendicular to
  E field where E is constant magnitude to give
  Flux EA
• 3. If possible put some G surface in conductor so
  E0 or where E is parallel to surface so EdA 0
• 4. Apply Gausss Law total flux through G
  surface equals charge enclosed divided by eo
• 5. Solve the resulting equation for magnitude of
  E field combine this with direction determined in
  step 1 for the vector direction

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Example 1Plane with Uniform E field

• What is the electric field just above a very
  large conducting plane with 10 uC / m2 charge
  density.

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Example 1

• Create a gaussian surface covering a portion of
  the charged surface
• Choose a cube of area a, centered half above and
  half below the surface
• Or, if you prefer, make it a cylinder

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Example 1

• The very large plane has a uniform E field near
  the surface that is normal to the surface.
• Constructing the Gaussian surface gives us the
  top and bottom surfaces perpendicular to the
  electric field and the 4 sides are parallel to
  the electric field
• Using Gausss Law to determine the flux

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Example 1

• The choice of surface gives
• E field parallel to the planes normal vector
  (perpendicular to the surface) EdA E dA
• E field is parallel to the side surfaces so EdA
  dot product is 0
• Since E is uniform over the cube side of area a
• F top flux bottom flux 0 0 0 0

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Example 1

• F top F bottom Ea Nm2/C
• F sides 0 Nm2/C for 4 sides of cube
• F F top F bottom 0
• F EaEa 2Ea q/eo Nm2/C
• E q/a 1/(2eo ) N/C
• q/a is charge per area s 10uC/m2
• E s/(2 eo) N/C 10E-6/(2 8.85E-12)
• 5.65E5 N/C
• Note in red is the eqn for near an infinite plane
  of charge

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Example 2

• An infinitely long distribution of charge is
  along the x-axis with ? charge/length.
• Construct a cylindrical Gaussian surface L x r to
  determine the Electric field

A3
X
A2
A1
L
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Solution
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Conductors in Equilibrium

• Electric field is zero inside the conductor
• An isolated conductor is charged its on the
  outside surface
• E-field is perpendicular to surface at the
  surface with magnitude s/e
• An irregular shaped conductor has greater charge
  density at smallest radius of curvature areas
  (chap. 25)

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Example 3

• Given a charged conductor of arbitrary shape,
  find the Electric field just outside the surface

E
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Solution

• Construct a small Gaussian surface at the point
  where E is to be determined.
• If G surface is small enough, E is uniform across
  the top area A and is parallel to the side of the
  cylinder
• The E field is 0 inside the conductor

E
A
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Solution