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Gausss Law Applications

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Gauss's Law Applications. PHYS 2326-7 ... Applying Gauss's Law ... 4. Apply Gauss's Law = total flux through G surface equals charge enclosed divided by eo ... – PowerPoint PPT presentation

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Title: Gausss Law Applications


1
Gausss Law Applications
  • PHYS 2326-7

2
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

3
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)

4
General Case
5
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

6
General Equation
7
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

8
Example 1Plane with Uniform E field
  • What is the electric field just above a very
    large conducting plane with 10 uC / m2 charge
    density.

9
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

10
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

11
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

12
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

13
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
14
Solution
15
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)

16
Example 3
  • Given a charged conductor of arbitrary shape,
    find the Electric field just outside the surface

E
17
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
18
Solution
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