24 Gauss

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24 Gauss law
24-1 A New Look at Coulombs Law
Gauss law relates the electric fields at points
on a (closed) Gaussian surface and the net
charge enclosed by that surface.
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24-2 Flux
(a)The rate Fis equal to vA
(b)
(c)
(d)A velocity field.Flux means the product
of an area and the field across that area.
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24-3 Flux of an Electric Field
A provisional definition for the flux of the
electric field for the Gaussian surface is
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Electric flux through a Gaussian surface
The electric flux F through a Gaussian surface
is proportional to the net number of electric
field lines passing through that surface.
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Sample Problem 24-1
What is the flux F of The electric field
through This closed surface?
Step one
Step two
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Step three
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Sample Problem 24-2
What is the electric flux through the right the
face, the left face,and the top face?
Right face
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Left face
Top face
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24-4 Gauss Law
Gauss law and Coulombs law, although expressed
in different forms, are equivalent ways of
describing relation between charge and electric
field in static situations. Gausss law is
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or
Surface S1
The electric field is outward for all point on
this surface.
Surface S2
The electric field is inward for all point on
this surface.
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Surface S3
This surface encloses no charge,and thus qenc0
Surface S4
This surface encloses no net charge, because the
enclosed positive and negative charges have
equal magnitudes.
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Sample Problem 24-3
What is the net electric flux through the
surface if Q1q43.1nC, q2q5-5.9nC, and
q3-3.1nC?
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24-5 Gauss Law and Coulombs Law
Gauss law as
Coulombs law
Gauss law is equivalent to Coulombs law.
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24-6 A Charged Isolated Conductor
If an excess charge is placed on an isolated
conductor,that amount of charge will move
entirely to the surface of the conductor .None
of the excess charge will be found within the
body of the conductor.
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An Isolated Conductor with a Cavity
There is no net charge on the cavity walls.
The Conductor Removed
The electric field is set up by the charges
and not by the conductor.The conductor simply
provides an initial pathway for the charges
to take up their position.
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The External Electric Field
Conducting surface
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Sample Problem 24-4
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Key idea
The electric flux through the Gaussian surface
must also be zero.The net charge enclosed by the
Gaussian surface must be zero.With a point charge
of -5.0µC within the shell,a charge of 5.0 µC
must lie on the inner wall of the shell.
Can you think of another key idea?
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24-7 Applying Gauss lawCylindrical Symmetry
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The electric field at any point due to an
infinite line of charge with uniform linear
charge density ?is perpendicular to the line of
charge and has magnitude
Where r is the perpendicular distance from the
line of charge to the point.
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Sample Problem 24-5
If air molecules break down (ionize) in an
electric field exceeding 3106N/C,what is
the column?
Key idea
The surface of the column of charge must be at
The radius r where the magnitude of is 3
106N/C,because air molecules within that Radius
ionize while those farther out do not.
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Can you think of another key idea?
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24-8 Applying Gauss lawPlanar Symmetry
nonconducting sheet
The electric field due to an infinite
nonconducting sheet with uniform surface charge
density sis perpendicular to the plane of the
sheet and has magnitude
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Two Conducting Plates
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Sample Problem 24-6
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Step one
Step two
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24-9 Applying Gauss lawSpherical Symmetry
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A shell of uniform charge attracts or repels a
charged particle that is outside the shell as if
all the shells charge were concentrated at the
center of the shell.
A shell of uniform charge exerts no
electrostatic force on a charged particle that
is located inside the shell.
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Spherical shell,field at r R
Spherical shell,field at r ltR
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Spherical distribution,field at r R
Uniform charge,field at r R
???