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Chapter 23. Gauss Law23.1. What is Physics?23.2. Flux23.3. Flux
of an Electric Field23.4. Gauss' Law23.5. Gauss' Law and Coulomb's
Law23.6. A Charged Isolated Conductor23.7. Applying Gauss' Law:
Cylindrical Symmetry23.8. Applying Gauss' Law: Planar Symmetry23.9.
Applying Gauss' Law: Spherical Symmetry
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What is Physics? Gaussian surface is a hypothetical (any
imaginary shape) closed surface enclosing the charge
distribution.
Gauss' law relates the electric fields at points on a (closed)
Gaussian surface to the net charge enclosed by that surface.
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Gaussian surface Let us divide the surface into small squares of
area A, each square being small enough to permit us to neglect any
curvature and to consider the individual square to be flat. We
represent each such element of area with an area vector
Magnitude is the area A. Direstion is perpendicular to the
Gaussian surface and directed away from the interior of the
surface.
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Flux The rate of volume flow through the loop is :
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Flux of an Electric FieldThe electric field for a surface is The
electric flux through a Gaussian surface is proportional to the net
number of electric field lines passing through that surface.The
electric field for a gaussian surface is SI Unit of Electric Flux:
Nm2/C
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Problem 1 The drawing shows an edge-on view of two planar
surfaces that intersect and are mutually perpendicular. Surface 1
has an area of 1.7 m2, while surface 2 has an area of 3.2 m2. The
electric field E in the drawing is uniform and has a magnitude of
250 N/C. Find the electric flux through (a) surface 1 and (b)
surface 2.
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Sample Problem 2 Figure 23-4 shows a Gaussian surface in the
form of a cylinder of radius R immersed in a uniform electric field
E , with the cylinder axis parallel to the field. What is the flux
of the electric field through this closed surface?
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Sample Problem 3 A nonuniform electric field given by pierces
the Gaussian cube shown in Fig. (E is in newtons per coulomb and x
is in meters.) What is the electric flux through the right face,
the left face, the top face, and the Gaussian surface?
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Gauss LawFor a point charge:
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Gauss LawFor charge distribution Q: The electric flux through a
Gaussian surface times by 0 ( the permittivity of free space) is
equal to the net charge Q enclosed :The net charge qenc is the
algebraic sum of all the enclosed charges.Charge outside the
surface, no matter how large or how close it may be, is not
included in the term qenc.
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Check Your Understanding The drawing shows an arrangement of
three charges. In parts (a) and (b) different Gaussian surfaces are
shown. Through which surface, if either, does the greater electric
flux pass?
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Sample Problem Figure 23-7 shows five charged lumps of plastic
and an electrically neutral coin. The cross section of a Gaussian
surface S is indicated. What is the net electric flux through the
surface if q1=q4=3.1 nC, q2=q5=-5.9 nC, and q3=-3.1 nC?
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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. For an Isolated Conductor with a
Cavity, There is no net charge on the cavity walls; all the excess
charge remains on the outer surface of the conductor
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The External Electric Field of a Conductoraccording to Gauss'
law,If is the charge per unit area,
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Sample Problem Figure 23-11a shows a cross section of a
spherical metal shell of inner radius R. A point charge of q is
located at a distance R/2 from the center of the shell. If the
shell is electrically neutral, what are the (induced) charges on
its inner and outer surfaces? Are those charges uniformly
distributed? What is the field pattern inside and outside the
shell?
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Applying Gauss' Law: Cylindrical Symmetry Figure 23-12 shows a
section of an infinitely long cylindrical plastic rod with a
uniform positive linear charge density .
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Applying Gauss' Law: Planar Symmetry Figure 23-15 shows a
portion of a thin, infinite, nonconducting sheet with a uniform
(positive) surface charge density
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Applying Gauss' Law: Spherical Symmetry A shell of uniform
charge attracts or repels a charged particle that is outside the
shell as if all the shell's charge were concentrated at the center
of the shell. If a charged particle is located inside a shell of
uniform charge, there is no electrostatic force on the particle
from the shell.
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Any spherically symmetric charge distribution with the volume
charge density For r>R, the charge produces an electric field on
the Gaussian surface as if the charge were a point charge located
at the center,
For r
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Checkpoint The figure shows two large, parallel, nonconducting
sheets with identical (positive) uniform surface charge densities,
and a sphere with a uniform (positive) volume charge density. Rank
the four numbered points according to the magnitude of the net
electric field there, greatest first.
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Conceptual QuestionsTwo charges, +q and q, are inside a Gaussian
surface. Since the net charge inside the Gaussian surface is zero,
Gauss law states that the electric flux through the surface is also
zero; that is =0. Does the fact that =0 imply that the electric
field E at any point on the Gaussian surface is also zero? Justify
your answer.
The drawing shows three charges, labeled q1, q2, and q3. A
Gaussian surface is drawn around q1 and q2. (a) Which charges
determine the electric flux through the Gaussian surface? (b) Which
charges produce the electric field at the point P? Justify your
answers.
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(3) A charge +q is placed inside a spherical Gaussian surface.
The charge is not located at the center of the sphere. (a) Can
Gauss law tell us exactly where the charge is located inside the
sphere? Justify your answer. (b) Can Gauss law tell us about the
magnitude of the electric flux through the Gaussian surface?
Why?