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Transcript
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
04
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
07
Physics 212 Lecture 3, Slide 4
Electric Field Lines
Dipole Charge Distribution:
Direction & Density
08
Physics 212 Lecture 3, Slide 5
Checkpoint
Preflight 3
Field lines are are denser near Q1 so  Q1  >  Q2 
09
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.
10
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.
12
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
18
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
20
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)
23
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
26
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
29
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
30
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
31
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
41
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.
43
Physics 212 Lecture 3, Slide 20
Checkpoint
(A)
dA increases
dB decreases
44
(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.
45
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.
47
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…
48
Physics 212 Lecture 3, Slide 24
– Prelecture 4 and Checkpoint 4 due Thursday
– Homework 2 due next Monday
50
Physics 212 Lecture 3, Slide 25