Transcript Chapter 24

Chapter 24
Gauss’s Law (cont.)
Dr. Jie Zou
PHY 1361
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Outline
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Gauss’s law (24.2)
Application of Gauss’s law to various
charge distributions (24.3)
Dr. Jie Zou
PHY 1361
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Gauss’s law
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Gauss’s law describes a general
relationship between the net electric
flux through a closed surface and the
charge enclosed by the surface.
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Closed surface: often called a gaussian
surface.
Dr. Jie Zou
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Let’s begin with one example.
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A spherical gaussian surface of
radius r surrounding a point
charge q.
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 E   E  dA   EdA  E  dA 
Dr. Jie Zou
The magnitude of the electric
field everywhere on the surface
of the sphere is E = keq/r2.
The electric field is  to the
surface at every point on the
surface.
Net electric flux through such
gaussian surface is
ke q
1
q
2


4

r

4

k
q

4

q

e
r 2 PHY 1361
40
0
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A non-spherical closed surface
surrounding a point charge
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Dr. Jie Zou
As we discussed in the previous section,
the electric flux is proportional to the
number of electric field lines passing
through a surface.
The number of lines through S1 is equal
to the number of lines through the
nonspherical surfaces S2 and S3.
The net flux through any closed surface
surrounding a point charge q is given by
q/0 and is independent of the shape
of that surface.
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A point charge located outside
a closed surface
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Any electric field line that enters the
surface leaves the surface at another
point.
The net electric flux through a closed
surface that surrounds no charge is
zero.
Revisit Example 24.2
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Example: Problem #14, P. 762

Calculate the total electric flux through
the paraboloidal surface due to a
constant electric field of magnitude E0 in
the direction shown in the figure.
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Now let’s consider a more
general case.
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The net electric flux through S is
E = q1/0.
The net electric flux through S’ is
E = (q2 + q3)/ 0.
The net electric flux through S” is
E = 0.
Charge q4 does not contribute to
the flux through any surface
because it is outside all surfaces.
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Gauss’s law
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Gauss’s law: the net flux through any closed
q
surface is
 E   E  dA  in
0
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qin = the net charge inside the gaussian surface.
E = the (total) electric field at any point on the
surface, which includes contributions from charges
both inside and outside the surface.
Pitfall prevention: Zero flux is not zero field.
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Application of Gauss’s law to
various charge distributions
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Example 24.5 A spherically
symmetric charge distribution:
An insulating solid sphere of
radius a has a uniform volume
charge density  and carries a
total positive charge Q.
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(A) Calculate the magnitude of the
electric field at a point outside the
sphere.
(B) Find the magnitude of the
electric field at a point inside the
sphere.
Answer: (A) E = keQ/r2, r > a;
(B) E  k e
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Q
r , r < a.
a3
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Application of Gauss’s law to
various charge distributions
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Dr. Jie Zou
Example 24.7 A cylindrical
symmetric charge distribution:
Find the electric field a distance r
from a line of positive charge of
infinite length and constant
charge per unit length .
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Answer: E 
 2k e
20 r
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r
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Homework
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Ch. 24, P. 762, Problems: #12, 14, 29.
Dr. Jie Zou
PHY 1361
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