physics 212 - university of illinoisphysics 212 lecture 3, slide 12 “the flux is directly...
TRANSCRIPT
Physics 212 Lecture 3, Slide 1
Physics 212Lecture 3
Today's Concepts:
Electric Flux and Field Lines
Gauss’s Law
Music
Who is the Artist?
A) Professor LonghairB) Henry ButlerC) David EganD) Dr. JohnE) Allen Toussaint
WHY?
Mardi Gras is 4 weeks from today
Physics 212 Lecture 13
Physics 212 Lecture 3, Slide 3
Our Comments
• SmartPhysics scores will not be visible in the main gradebook for a few weeks
• You will need to understand integrals in this course!!!
• Forces and Fields are Vectors
• Always Draw a Picture First… What do the Forces/Fields Look Like?
Prelecture
Infinite line of charge
E
Lecture
Finite line of charge
(constant λ)
E
Worked Example
Finite line of charge
(non-constant λ)
E
Homework
Arc of charge
(constant λ)
E2 2
0
1ˆ ˆ
4
dq dqE k r r
r rπε= =∫ ∫�
WORKS FOR ALL !!
The “1/4 shell problem” is not
special !!
It’s the same as all the others!
Physics 212 Lecture 3, Slide 4
Electric Field Lines
Direction & Density of Lines represent
Direction & Magnitude of E
Point Charge:Direction is radialDensity α 1/R2
07
Physics 212 Lecture 3, Slide 5
Electric Field Lines
Dipole Charge Distribution:Direction & Density
much more interesting…
simulation08
Physics 212 Lecture 3, Slide 6
Preflight 3
Checkpoint 3.1
simulation09
The following three questions pertain to the electric field
lines due to the two charges shown. Compare the
magnitude of the two charges
A. |Q1| > |Q2| B. |Q1| = |Q2| C. |Q1| < |Q2|
D. There isn’t enough information to determine the relative
magnitude of the charges
“more field lines emanating from the charge 1, so must be greater in magnitude.”
Physics 212 Lecture 3, Slide 7
Checkpoint 3.3
10simulation
What do we know about the signs of the charges from
looking at the picture?
A. Q1 and Q2 have the same sign
B. Q1 and Q2 have the opposite sign
C. There isn’t enough information to determine the relative
signs of the charges
“The field lines leave one and go to the other..”
Physics 212 Lecture 3, Slide 8
Preflight 3
Checkpoint 3.5
12
Compare the magnitude of the electric field at point A and
B.
A. |EA| > |EB| B. |EA| = |EB| C. |EA| < |EB|
D. There isn’t enough information to determine the relative
magnitude of the electric field at points A and B
“Since the electric field lines are denser near
point B the magnitude of the electric field is greater at point B than at point A.”
Physics 212 Lecture 3, Slide 9
Point Charges
+2q
-q
What charges are inside the red circle?
+Q +Q
-Q
+2Q
-Q -2Q
+Q
-Q
A B C D E
“Telling the difference between positive and negative charges while looking at field lines.
Does field line density from a certain charge give information about the sign of the charge?”
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Physics 212 Lecture 3, Slide 10
Which of the following field line pictures best represents the electric field from two charges that have the same sign but different magnitudes?
A B
C D
simulation15
Physics 212 Lecture 3, Slide 11
Electric Flux “Counts Field Lines”
S
S
E dAΦ = ⋅∫��
Flux through surface S
Integral of on surface S
E dA⋅��
“I still feel a little unclear about flux integrals. Can you go over some more example problems?”
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Physics 212 Lecture 3, Slide 12
“The flux is directly proportional to
the charge enclosed by the
Gaussian surface, therefore Case
1 will have 2 times the flux of of
Case 2 because it has two charges enclosed, versus 1 in Case 2. “
“The flux is the measure of field
lines passing through the surface.
Once calculated, it can be shown
that the surface area of both cylinders are the same.”
Checkpoint 1
(A)Φ1=2Φ2
Φ1=Φ2
(B)Φ1=1/2Φ2
(C)none(D)
TAKE s TO BE RADIUS !
L/2
An infinitely long charged rod has uniform
charge density λ and passes through a
cylinder (gray). The cylinder in Case 2 has
twice the radius and half the length compared
with the cylinder in Case 1.
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Physics 212 Lecture 3, Slide 13
Checkpoint 1
(A)Φ1=2Φ2
Φ1=Φ2
(B)Φ1=1/2Φ2
(C)none(D)
TAKE s TO BE RADIUS !
L/2
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
cylinder in Case 1.
Definition of Flux:
∫ ⋅≡Φsurface
AdE��
E constant on barrel of cylinderE perpendicular to barrel surface(E parallel to dA)
Case 1
sE
0
12πε
λ=
LsA )2(1 π= 0
1ε
λL=Φ
Case 2
)2(2 0
2s
Eπε
λ=
sLLsA ππ 22/))2(2(2 ==0
2
)2/(
ε
λ L=Φ
RESULT: GAUSS’ LAWΦ proportional to charge enclosed !
“The flux is the measure of field
lines passing through the surface.
Once calculated, it can be shown
that the surface area of both cylinders are the same.”
WHAT IS WRONG WITH THIS REASONING??
THE E FIELD AT THE BARREL SURFACE
IS NOT THE SAME IN THE TWO CASES !!
barrelbarrel
EAAdE =∫=Φ�
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Physics 212 Lecture 3, Slide 14
Direction Matters:
0S
S
E dAΦ = ⋅ >∫��
dA
dA
dA dA
dA
E
E
E
E
E
E
EE
E
For a closed surface, A points outward
29
Physics 212 Lecture 3, Slide 15
Direction Matters:
0S
S
E dAΦ = ⋅ <∫��
dA
dA
dA dA
dA
E
E
E
E
E
E
EE
E
For a closed surface, A points outward
30
Physics 212 Lecture 3, Slide 16
Trapezoid in Constant Field
0ˆE E x=
�
Label faces:1: x = 0
2: z = +a3: x = +a
4: slanted
0E�
y
z
Define Φn = Flux through Face n
A
B
C
Φ1 < 0
Φ1 = 0
Φ1 > 0
A
B
C
Φ2 < 0
Φ2 = 0
Φ2 > 0
A
B
C
Φ3 < 0
Φ3 = 0
Φ3 > 0
A
B
C
Φ4 < 0
Φ4 = 0
Φ4 > 0
x
1 2
3
dA
E
0<⋅ AdE��
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Physics 212 Lecture 3, Slide 17
Trapezoid in Constant Field + Q0ˆE E x=
�
Label faces:1: x = 0
2: z = +a3: x = +a
Define Φn = Flux through Face nΦ = Flux through Trapezoid
A
B
C
Φ1 increases
Φ1 decreases
Φ1 remains same
Add a charge +Q at (-a,a/2,a/2)
+Q
How does Flux change?
A
B
C
Φ increases
Φ decreases
Φ remains same
A
B
C
Φ3 increases
Φ3 decreases
Φ3 remains same
x
y
z
0E�
x
1 2
3
36
Physics 212 Lecture 3, Slide 18
Q
Gauss Law
dA
dA
dA dA
dA
E
E
E
E
E
E
EE
E
o
enclosed
surfaceclosed
SQ
AdEε
=∫ ⋅≡Φ��
41
Physics 212 Lecture 3, Slide 19
Checkpoint 2.3
(A)
Φ increases(B)
Φ decreases
(C)
Φ stays same
43
“The flux throughout the entire
surface does not change as the
charge moves, for it is still in the
sphere. However, if it was moved
outside of the sphere, the fluxes
would be different because one
would be zero!“
Physics 212 Lecture 3, Slide 20
“charge is closer, field is larger, more
lines through a so increase in flux.“
“as long as the charge is within the
shell, the flux will remain the same”
Checkpoint 2.1
(A)
dΦA increases
dΦB decreases
(B)
dΦA decreases
dΦB increases
(C)
dΦA stays same
dΦB stays same
44
Physics 212 Lecture 3, Slide 21
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.
1 2
Think of it this way:
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Physics 212 Lecture 3, Slide 22
Things to notice about Gauss’s Law
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.
o
enclosed
surfaceclosed
SQ
AdEε
=∫ ⋅≡Φ��
47
Physics 212 Lecture 3, Slide 23
Things to notice about Gauss’s Law
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
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…
“Gausses law and what concepts are most important that we know .”
o
enclosed
surfaceclosed
QAdE
ε=∫ ⋅��
o
enclosed
surfaceclosed
QEAAdE
ε==∫ ⋅
��
0
enclosedQE
Aε=
48
Physics 212 Lecture 3, Slide 24
Final Thoughts….
• Keep plugging away– Prelecture 4 and Checkpoint 4 due Thursday
– Homework 2 due next Tuesday … 7 questions mostly Gauss’ law – try them today!
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