Title: Chapter 23. Gauss
1Chapter 23. Gauss Law
- 23.1. What is Physics?     Â
- 23.2. Flux     Â
- 23.3. Flux of an Electric Field     Â
- 23.4. Gauss' Law     Â
- 23.5. Gauss' Law and Coulomb's Law     Â
- 23.6. A Charged Isolated Conductor    Â
- 23.7. Applying Gauss' Law Cylindrical
Symmetry      - 23.8. Applying Gauss' Law Planar Symmetry    Â
- 23.9. Applying Gauss' Law Spherical Symmetry
2What 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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3Gaussian 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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4Flux
- The rate of volume flow through the loop is
5Flux of an Electric Field
- The electric field for a surface is
- The electric field for a gaussian surface is
The electric flux F through a Gaussian surface is
proportional to the net number of electric field
lines passing through that surface.
SI Unit of Electric Flux Nm2/C
6Problem 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.
7Sample 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 F of the
electric field through this closed surface?
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8Sample 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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9Gauss Law
10Gauss Law
- For charge distribution Q
The electric flux through a Gaussian surface
times by e0 ( 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.
11Check 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?
12Sample 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 q1q43.1 nC, q2q5-5.9 nC, and
q3-3.1 nC?
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13A 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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14The External Electric Field of a Conductor
If s is the charge per unit area,
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15Sample 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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16Applying 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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17Applying Gauss' Law Planar Symmetry
- Figure 23-15 shows a portion of a thin,
infinite, nonconducting sheet with a uniform
(positive) surface charge density s
18Applying 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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19- Any spherically symmetric charge distribution
with the volume charge density ?
- For rgtR, the charge produces an electric field on
the Gaussian surface as if the charge were a
point charge located at the center, - For rltR, the electric field is
20Checkpoint
- 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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21Conceptual Questions
- Two 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 F0. Does the fact that F 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.
22- (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?