Transcript HW Ch27.1-27.3

Chapter 28
Gauss’s Law
28.1 to 28.4
Symmetry
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An object is
symmetric if a group
of geometrical
transformations don’t
cause any physical
change.
Translate, rotate, and
reflect
Cylindrically and Spherically Symmetric
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The symmetry of
the electric field
must match the
symmetry of the
charge
distribution.
Cylindrically
symmetric
charge
distribution
cannot have a
component
parallel to the
cylinder axis.
It cannot have a
component
tangent to the
circular cross
section.
Three types of
Symmetries
Gaussian Surface
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Definition: A closed surface that divides space into
distinct inside and outside regions.
The electric field must be perpendicular to the surface
Same Magnitude at each point
Appropriate
Gaussian
Surface
Inappropriate
Gaussian
Surface
Types of Flux
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If the net amount of flux lines go
out of Gaussian surface, then
positive charge inside.
If the net amount of flux lines go
into a Gaussian surface, then
negative charge inside
If the net amount of flux lines is
equal to zero, then no charge
inside.
Page 855, Stop-to Think 27.2
Calculating Electric Flux
 
e  E  A
  EperA
  EA cos 
 is flux
E is Electric Field Strength
A is Area of surface that field lines
cross
 is the angle in which field lines
make with perpendicu lar line
to surface.
Electric Flux through a
Closed Surface
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Closed surface completely
surrounds a region.
If the electric field is
everywhere tangent to a
surface, the electric flux
through the surface is e=0
If the electric field is
everywhere perpendicular to a
surface and has the same
magnitude E at every point, the
electric flux through the surface
is e=EA
Try Stop to Think 27.3
 
 e   E  dA
dA surface area element
Examples
Flux for rod of charge
 e  Erod 2rl
Flux for a point charge
 e  E point 4r 2
Gauss’s Law
   E  dA 
q
o
Nm
 is flux measured in
C
E is Electric Field measured in N/C
dA is Surface area of Gaussian Surface
q is enclosed surface
2
 o is the permittivi ty constant 8.85 x 10 C / N  m
-12
2
2
Determine Electric Field
Equation for a point
charge from Gauss’s Law
q
 E  dA  
E  dA 
E  4r
E 
2
0
q
0

1
4 0
q
0
q
r2
Practice Problems
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Stop to Think 28.1 to 28.4
Page 875, #6, 8,17,20,23