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Physics 212
Lecture 3
Today's Concepts:
Electric Flux and Field Lines
Gauss’s Law
Physics 212 Lecture 3, Slide 1
Introduce a new constant: 0
q
E  k 2 rˆ
r
k
1
4 0
k = 9 x 109 N m2 / C2
0 = 8.85 x 10-12 C2 / N m2
1
q
E
r
2 ˆ
4 0 r
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Physics 212 Lecture 3, Slide 2
Plan for Today
• A little more about electric field lines
• Electric field lines and flux
– An analogy
• Introduction to Gauss’s Law
– Gauss’ Law will make it easy to calculate electric fields for some geometries.
06
Physics 212 Lecture 3, Slide 3
Electric Field Lines
Direction & Density of Lines
represent
Direction & Magnitude of E
Point Charge:
Direction is radial
Density  1/R2
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Physics 212 Lecture 3, Slide 4
Electric Field Lines
Dipole Charge Distribution:
Direction & Density
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Physics 212 Lecture 3, Slide 5
Checkpoint
Preflight 3
Field lines are are denser near Q1 so  Q1  >  Q2 
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Physics 212 Lecture 3, Slide 6
Checkpoint
The electric field lines connect the charges. A test charge will move
towards one charge and away from the other. So charges 1 and 2
have opposite signs.
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Physics 212 Lecture 3, Slide 7
Checkpoint
Preflight 3
Density of lines is greater at B than at A.
Therefore, magnitude of field at B is greater
than at A.
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Physics 212 Lecture 3, Slide 8
Point Charges
-q
+2q
What charges are inside the red circle?
13
-Q
-Q
-2Q
+Q
+Q
+2Q
+Q
A
B
C
D
-Q
E
Physics 212 Lecture 3, Slide 9
Which of the following field line pictures best represents the
electric field from two charges that have the same sign but
different magnitudes?
15
A
B
C
D
Physics 212 Lecture 3, Slide 10
Electric Flux “Counts Field Lines”
 S   E  dA
S
Flux through
surface S
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Integral of E  dA
on surface S
Physics 212 Lecture 3, Slide 11
Electric Field/Flux Analogy:
Velocity Field/Flux
 S   v  dA  flowrate
S
Flux through
surface S
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Integral of v  dA
on surface S
Physics 212 Lecture 3, Slide 12
Checkpoint
An infinitely long charged rod has uniform
charge density l and passes through a
cylinder (gray). The cylinder in Case 2 has
twice the radius and half the length compared
TAKE s TO BE RADIUS !
with the cylinder in Case 1.
L/2
1=22
(A)
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1=2
(B)
1=1/22
(C)
none
(D)
Physics 212 Lecture 3, Slide 13
Checkpoint
An infinitely long charged rod has uniform
charge density l and passes through a cylinder
(gray). The cylinder in Case 2 has twice the
radius and half the length compared with the
TAKE s TO BE RADIUS !
cylinder in Case 1.
Definition of Flux:

 
 E  dA
surface
L/2
E constant on barrel of cylinder
E perpendicular to barrel surface
(E parallel to dA)

  E  dA  EAbarrel
barrel
1=22
(A)
Case 1
l
E1 
2 0 s
A1  ( 2s ) L
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1RESULT:
=2

GAUSS’
1=1/22LAW none
(D) !
(B)
(C) enclosed
 proportional
to charge
Case 2
1 
lL
0
E2 
l
2 0 ( 2 s )
A2  ( 2 ( 2 s )) L / 2  2sL
2 
l ( L / 2)
0
Physics 212 Lecture 3, Slide 14
Direction Matters:
E
E
E
dA
For a closed surface,
A points outward
dA
E
E
dA
E
dA
dA
E
E
E
 S   E  dA  0
S
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Physics 212 Lecture 3, Slide 15
Direction Matters:
E
E
E
dA
For a closed surface,
A points outward
dA
E
E
dA
E
dA
dA
E
E
E
 S   E  dA  0
S
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Physics 212 Lecture 3, Slide 16
Trapezoid in Constant Field
E  E0 xˆ
y
Label faces:
1: x = 0
2: z = +a
3: x = +a
4: slanted
3
1
2
x
Define n = Flux through Face n
dA
 
E  dA  0
z
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E
A
1 < 0
A
2 < 0
A
3 < 0
A
4 < 0
B
1 = 0
B
2 = 0
B
3 = 0
B
4 = 0
C
1 > 0
C
2 > 0
C
3 > 0
C
4 > 0
Physics 212 Lecture 3, Slide 17
Trapezoid in Constant Field + Q
E  E0 xˆ
y
E0
Label faces:
1: x = 0
2: z = +a
3: x = +a
3
+Q
36
2
x
Define n = Flux through Face n
 = Flux through Trapezoid
z
1
Add a charge +Q at (-a,a/2,a/2)
How does Flux change?
A
1 increases
A
3 increases
A
 increases
B
1 decreases
B
3 decreases
B
 decreases
C
1 remains same
C
3 remains same C
 remains same
Physics 212 Lecture 3, Slide 18
Gauss Law
E
E
Q
E
E
dA
dA
E
dA
E
dA
dA
E
E
E
S 
  Qenclosed
 E  dA 
closed
surface
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o
Physics 212 Lecture 3, Slide 19
Checkpoint
What happens to total flux through
the sphere as we move Q ?
(A)
 increases
(B)
 decreases
(C)
 stays same
The same amount of charge is still enclosed by the
sphere, so flux will not change.
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Physics 212 Lecture 3, Slide 20
Checkpoint
(A)
dA increases
dB decreases
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(B)
dA decreases
dB increases
(C)
dA stays same
dB stays same
Physics 212 Lecture 3, Slide 21
Think of it this way:
1
2
The total flux is the same in both cases (just the total number of lines)
The flux through the right (left) hemisphere is smaller (bigger) for case 2.
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Physics 212 Lecture 3, Slide 22
Things to notice about Gauss’s Law
S 
  Qenclosed
 E  dA 
o
closed
surface
If Qenclosed is the same, the flux has to
be the same, which means that the
integral must yield the same result
for any surface.
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Physics 212 Lecture 3, Slide 23
Things to notice about Gauss’s Law
  Qenclosed
 E  dA 
closed
surface
o
In cases of high symmetry it may be possible to bring E outside
the integral. In these cases we can solve Gauss Law for E
 
Qenclosed
 E  dA  EA 
closed
surface
o
Qenclosed
E
A 0
So - if we can figure out Qenclosed and the area of the
surface A, then we know E !
This is the topic of the next lecture…
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Physics 212 Lecture 3, Slide 24
– Prelecture 4 and Checkpoint 4 due Thursday
– Homework 2 due next Monday
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Physics 212 Lecture 3, Slide 25