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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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