Physics 2112 Unit 3: Electric Flux and Field Lines
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Transcript of Physics 2112 Unit 3: Electric Flux and Field Lines
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Physics 2112Unit 3: Electric Flux and Field
Lines
Today’s Concepts:A) Electric FluxB) Field Lines Gauss’ Law
Unit 3, Slide 1
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Electric Field Lines
Unit 3, Slide 2
+Q
First….a comment
What the point of all this flux stuff?
To create Guass’ Law so we can find the E field without doing an integral
Guass’ Law
Always true, sometimes not useful
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Electric Field Lines
Unit 3, Slide 3
Electric Field Lines
Electric Field is a “vector field”
“Density” represents magnitude
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Point Charge:Direction is radial3D Density area of sphere (1/R2)
Electric Field Lines
Unit 3, Slide 4
Electric Field Lines in 3D
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Just add E-Field vectors at every point
Electric Field Lines
Unit 3, Slide 5
-+
Electric Field Lines for multiple charges
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CheckPoint: Field Lines - Two Point Charges 1
Unit 3, Slide 6
Compare the magnitudes of the two chargeso |Q1| > |Q2|o |Q1| = |Q2| o |Q1| < |Q2| o There is not enough information to determine the relative magnitudes of
the charges.
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CheckPoint: Field Lines - Two Point Charges 1
Unit 3, Slide 7
Compare the magnitudes of the two chargesA. |Q1| > |Q2|B. |Q1| = |Q2| C. |Q1| < |Q2| D. There is not enough information to determine the relative magnitudes of
the charges.
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CheckPoint: Field Lines - Two Point Charges 3
Unit 3, Slide 8
Compare the magnitudes of the electric fields at points A and B.A. |EA| > |EB|B. |EA| = |EB|C. |EA| < |EB|D. There is not enough information to determine the relative magnitudes of
the fields at A and B.
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Flux
E
Flux Amount of electric field passing through an surface. Count field lines.
Unit 3, Slide 9
Same number of field lines but different areas.
Q
A
F = |E|*|A|*cosQ.
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Flux through surface S
Integral of on surface S
“I’m very confused by the general concepts of flux through surface areas. please help” Electric Flux “Counts Field Lines”
Unit 3, Slide 10
FS
S AdE
AdE
SAMPLE DATA: PLEASE REPLACE WITH OWN
FS
S AdE
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Unit 3, Slide 11
Example 3.1: Flux through a plate
|E|=12N/C
A 3cm X 3cm copper plate is placed in a constant horizontal electric field with a strength of 12N/C at an angle of 30o to the horizontal.
What is the electric flux through the plate?
Q
A30o
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Unit 3, Slide 12
TAKE s TO BE RADIUS !
CheckPoint: Flux from Uniformly Charged Rod
An infinitely long charged rod has uniform charge density of l, and passes through a cylinder (gray). The cylinder in case 2 has twice the radius and half the length compared to the cylinder in case 1.
Compare the magnitude of the flux through the surface of the cylinder in both cases.
A. Φ1 = 2 Φ2
B. Φ1 = Φ2
C. Φ1 = 1/2 Φ2
D. None of these
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Direction Matters:
FS
S AdE 0
For a closed surface, points outward
Unit 3, Slide 13
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E
E
E
E
E
E
E
E
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Direction Matters:
FS
S AdE 0
Unit 3, Slide 14
For a closed surface, points outwardAd
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E
E
E
E
E
E
E
E
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Gauss Law
Unit 3, Slide 15
0enclosed
surfaceclosed
SQAdE F
22 44
1 RRQAdE
osurfaceclosed
S
F
Gauss’ Law for simple point charge
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CheckPoint Results: Flux Point Charge (Sphere) 1
Unit 3, Slide 16
A positive charge (blue) is contained inside a spherical shell (black).
How does the electric flux through the two surface elements, dΦA and dΦB change when the charge is moved from position 1 to position 2? o dΦA increases and dΦB decreaseso dΦA decreases and dΦB increaseso Both dΦA and dΦB do not change
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CheckPoint: Flux from Point Charge (Sphere) 2
Unit 3, Slide 17
A positive charge (blue) is contained inside a spherical shell (black).
How does the flux ΦE through the entire surface change when the charge is moved from position 1 to position 2?
A. ΦE increasesB. ΦE decreasesC. ΦE does not change
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The flux through the right hemisphere is smaller for case 2.
1 2
Think of it this way:
The total flux is the same in both cases (just the total number of lines)
+Q+Q
Unit 3, Slide 18
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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.
Things to notice about Gauss Law
Unit 3, Slide 19
0enclosed
surfaceclosed
SQAdE F
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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 .” Things to notice about Gauss Law
Unit 3, Slide 20
0enclosedQAdE
0enclosedQEAAdE
0AQE enclosed
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A solid conducting sphere with net charge –Q and a radius r1. What is the electric field at within the sphere some distance r2 from the center? (r2<r1)
-Q
r1r2
A. -kQ/r12
B. -kQ/r22
C. ZeroD. None of the above
Unit 3, Slide 21
Question