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Chapter 23: Gauss’s Law
Gauss’s Law
 
 E  dA  Qenc /  0
We got the same flux through our boxes and sphere no matter
what size or shape the closed surfaces are as long as they
fully enclose the charge.
S
closed surface = constant = qenclosed
o
• S is any closed surface.
• Qenc is the net charge enclosed within S.
• dA is an element of area on the surface of S.
• dA is in the direction of the outward normal.
Check Quiz, August 30, 2013
Find the flux through the Gaussian surface.
A.
q1
q3

q2
q5
o
q2=q3=-2 nC

C.
0
D.

q4=q5=4 nC

 6nC
o
q3=-2 nC
q4=q5=4 nC
qenc = -2nC + 4nC + 4 nC
 7 nC
B.
o
q1
q3
q6
 6nC
o
E. none of the above
q1 = +1 nC
D.
Gauss’s law says count only the charges inside the
surface!
 1nC
q6
q4
Answer
1 nC = 10-9C
q4
q2
q5
q6 = -6 nC
Gauss  Coulomb
dA r
Given a point charge,
draw a concentric sphere
and apply Gauss’s Law:
Spherical Charge densities d = r2sindrdd
 
qenc
   E  dA  4r 2 E 
r
o
S
Uniform Charge density o
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qenc    (r )d   od  r 3 o
3
V
V

E  E (r )rˆ and dA = da r
 
   E  dA  E (r ) 4 r 2
Non-uniform Charge density
For example, let o=A/r
r
Gauss    q /  0
 E (r )  q /(4 r  0 )  kq / r
2
qenc    (r )d  
V
2
!
V
A
r 2 sin drdd
d  A
r
r
r
 A(2 )(2)  rdr  2Ar 2
0
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Line of Charge
E
dA2
We know E is radial.
Gaussian Surface
Line of Charge Solution
dA1
dA3
++++++++++++++++++++++++++
L
Area 1: left endcap
Area 2: cylinder wall
Area 3: right endcap
Flux
=
enclosed charge
Line of Charge -- Direct Method
Mathematical Box
Box Integration
Mathematical Cylinder
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Mathematical Sphere
Cylinder Integration
Spherical Integration
Conductors
Definition -- Conductor: A material that allows free motion
of charges. Metals are good examples of conductors.
A free charge will move until it feels no net force.
Remember that
F = ma = qE.
Therefore charges will move in a conductor until
the electric field inside the conductor is ZERO.
All free charges on a conductor reside on their
SURFACES.
Conductors - - Example 1
We put +Q on an arbitrary conductor that is hollow
in the middle. How will it distribute itself?
Conductors - - Example 1 Solution
hollow conductor
qouter = +Q
qinner
+Q
qinner
Gaussian surface inside
the conductor
hollow conductor
qouter
+Q
+Q
Gaussian surface inside
the conductor
0
E field is 0 inside a conductor
We know:
qinner + qouter = +Q
0
qenclosed = qinner = 0
qouter = +Q
0
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Conductors - - Example 2
hollow, neutral conductor
qouter
qinner
Conductors - - Example 2 Solution
hollow conductor
qouter = -3Q
+3Q= qinner
0
Gaussian surface inside
the conductor
-3Q
-3Q
Gaussian surface inside
the conductor
Now we put a -3Q charge in the center. How will the
charges redistribute themselves?
E field is 0 inside a conductor
0
qenclosed = qinner + -3Q = 0
0
qinner = +3Q
Where did that +3Q come from? qinner + qouter = 0
qouter = -3Q
+3Q
Conductors - - Example 3
Now combine the two examples. +Q on the conductor and -3Q
inside the cavity. How will the charges distribute themselves?
+Q
Conductors - - Example 3 Solution
+Q
qinner = +3Q
hollow conductor
qouter = -2Q
qinner = +3Q
hollow conductor
qouter = -3Q + Q = -2Q
-3Q Charge left behind
-3Q
Gaussian surface inside
the conductor
From Example 1: +Q all on qouter
From Example 2: +3Q was attracted
from qouter leaving -3Q + Q on qouter
qouter = -3Q +Q = -2Q
Van de Graaff Demonstration
Gaussian surface inside
the conductor
when +3Q was pulled
to qinner
Charge initially
put on conductor
For this Example:
Inside cavity: qenc = -3Q, Ecavity = 0
Inside conductor: qenc = 0, qinner = +3Q, Econductor = 0
Outside conductor: qenc = -3Q + qinner + qouter = -2Q, Eoutside = 0
+3Q
-2Q
Fields in Good Conductors
Fact: In a steady state the electric field inside
a good conductor must be zero.
Why? If there were a field, charges would
move. Charges will move around until they
find the arrangement that makes the electric
field zero in the interior.
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Field at the Surface of a Conductor
Charges in Good Conductors
Fact: In a steady state, any net charge on a good
conductor must be entirely on the surface.
Why? If there were a charge in the interior, then
by Gauss’s Law there would be a field in the
interior, which we know cannot be true.
• Construct closed gaussian
surface, sides perpendicular to
metal surface, face area = A.
• Flux through left face is zero
because E=0.
• Field is perpendicular to
surface or charges would move,
therefore flux through sides is 0.
• So net outward flux is EA.
Field at the Surface of a Conductor
• Let σ stand for the surface
charge density (C/m2)
• Then
Qenc = σ A.
• Now Gauss’s Law gives us
  Qenc /  0
EA   A /  0
E   / 0
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