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Transcript of 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in...
![Page 1: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/1.jpg)
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W02D2Gauss’s Law
![Page 2: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/2.jpg)
Announcements
Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus
PS 2 due Week Three Tuesday Tues at 9 pm in boxes outside 32-082 or 26-152
W02D3 Reading Assignment Course Notes: Chapter Course Notes: Sections 3.6, 3.7, 3.10
Make sure your clicker is registered
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Outline
Electric Flux
Gauss’s Law
Calculating Electric Fields using Gauss’s Law
![Page 4: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/4.jpg)
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Gauss’s Law
The first Maxwell Equation!
A very useful computational technique to find the electric field when the source has ‘enough symmetry’.
E
![Page 5: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/5.jpg)
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Gauss’s Law – The Idea
The total “flux” of field lines penetrating any of these closed surfaces is the same and depends only on the amount of charge inside
![Page 6: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/6.jpg)
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Gauss’s Law – The Equation
Electric flux (the surface integral of E over
closed surface S) is proportional to charge
enclosed by the volume enclosed by S
ΦE =rE⋅d
rA
closedsurface S
“∫∫ =qenclosed
ε0
ΦE
![Page 7: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/7.jpg)
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Electric Flux Case I: E is a uniform vector field perpendicular to planar surface S of area A
ΦE =+EA
∫∫Φ AE
dE
Our Goal: Always reduce problem to finding a surface where we can take E out of integral and get simply E*Area
![Page 8: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/8.jpg)
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Case II: E is uniform vector field directed at angle to planar surface S of area A
ΦE =EAcosθ
Electric Flux
ΦE =
rE ⋅d
rA∫∫
drA =dAn̂
n̂
θ
![Page 9: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/9.jpg)
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Concept Question: FluxThe electric flux through the planar surface below (positive unit normal to left) is:
+q -qn̂
1. positive.
2. negative.
3. zero.
4. Not well defined.
![Page 10: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/10.jpg)
Concept Question Answer: Flux
The field lines go from left to right, opposite the assigned normal direction. Hence the flux is negative.
Answer: 2. The flux is negative.
+q -qn̂
![Page 11: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/11.jpg)
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Open and Closed Surfaces
A rectangle is an open surface — it does NOT contain a volume
A sphere is a closed surface — it DOES contain a volume
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Area Element: Closed Surface
Case III: not uniform, surface curved
For closed surface, is normal to surface and points outward ( from inside to outside)
if points out
if points in
dΦE > 0
dΦE < 0
drA
dΦE ≡rE⋅d
rA
ΦE = dΦE“∫∫ =
rE⋅d
rA“∫∫
rE
![Page 13: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/13.jpg)
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Group Problem Electric Flux: Sphere
Consider a point-like charged object with charge Q located at the origin. What is the electric flux on a spherical surface (Gaussian surface) of radius r ?
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Arbitrary Gaussian Surfaces
True for all surfaces such as S1, S2 or S3
Why? As area gets bigger E gets smaller
ΦE =rE⋅d
rA
closedsurfaceS
“∫∫ =Qε0
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Gauss’s Law
Note: Integral must be over closed surface
ΦE =rE⋅d
rA
closedsurface S
“∫∫ =qenclosed
ε0
![Page 16: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/16.jpg)
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Concept Question: Flux thru Sphere
The total flux through the below spherical surface is
+q+q
1. positive (net outward flux).
2. negative (net inward flux).
3. zero.
4. Not well defined.
![Page 17: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/17.jpg)
Concept Question Answer: Flux thru Sphere
We know this from Gauss’s Law:
No enclosed charge no net flux. Flux in on left cancelled by flux out on right
Answer: 3. The total flux is zero
+q+q
ΦE =rE⋅d
rA
closedsurface S
“∫∫ =qenclosed
ε0
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Concept Question: Gauss’s Law
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The grass seeds figure shows the electric field of three charges with charges +1, +1, and -1, The Gaussian surface in the figure is a sphere containing two of the charges. The electric flux through the spherical Gaussian surface is
1. Positive
2. Negative
3. Zero
4. Impossible to determine without more information.
![Page 19: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/19.jpg)
Concept Question Answer: Gauss’s Law
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Answer 3: Zero. The field lines around the two charged objects inside the Gaussian surface are the field lines associated with a dipole, so the charge enclosed in the Gaussian surface is zero. Therefore the electric flux on the surface is zero.
Note that the electric field E is clearly NOT zero on the surface of the sphere. It is only the INTEGRAL over the spherical surface of E dotted into dA that is zero.
![Page 20: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/20.jpg)
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Choosing Gaussian Surface
Desired E: Perpendicular to surface and uniform on surface. Flux is EA or -EA.
Other E: Parallel to surface. Flux is zero
ΦE =rE⋅d
rA
closedsurfaceS
“∫∫ =qenclosed
ε0
True for all closed surfaces
Useful (to calculate electric field ) for some closed surfaces for some problems with lots of symmetry.
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Symmetry & Gaussian Surfaces
Source Symmetry Gaussian Surface
Spherical Concentric Sphere
Cylindrical Coaxial Cylinder
Planar Gaussian “Pillbox”
Desired E: perpendicular to surface and constant on surface. So Gauss’s Law useful to calculate electric field from highly symmetric sources
![Page 22: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/22.jpg)
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Virtual ExperimentGauss’s Law Applet
Bring up the Gauss’s Law Applet and answer the experiment survey questions
http://web.mit.edu/viz/EM/visualizations/electrostatics/flux/closedSurfaces/closed.htm
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Applying Gauss’s Law1. Based on the source, identify regions in which to calculate electric field.
2. Choose Gaussian surface S: Symmetry
3. Calculate
4. Calculate qenc, charge enclosed by surface S
5. Apply Gauss’s Law to calculate electric field:
rE⋅d
rA
closedsurface S
“∫∫ =qenclosed
ε0
ΦE =
rE ⋅d
rA
S“∫∫
![Page 24: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/24.jpg)
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Examples:Spherical Symmetry
Cylindrical SymmetryPlanar Symmetry
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Group Problem Gauss: Spherical Symmetry
+Q uniformly distributed throughout non-conducting solid sphere of radius a. Find everywhere.
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Concept Question: Spherical Shell
We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at right (r<a) what does the electric field do?
a
Q
1. Zero
2. Uniform but Non-Zero
3. Still grows linearly
4. Some other functional form (use Gauss’ Law)
5. Can’t determine with Gauss Law
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P04 -
Concept Question Answer: Flux thru Sphere
Spherical symmetry Use Gauss’ Law with spherical surface.
Any surface inside shell contains no charge No flux
E = 0!
Answer: 1. Zero
a
Q
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Demonstration Field Inside Spherical Shell
(Grass Seeds):
![Page 29: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/29.jpg)
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Worked Example: Planar Symmetry
Consider an infinite thin slab with uniform positive charge density . Find a vector expression for the direction and magnitude of the electric field outside the slab. Make sure you show your Gaussian closed surface.
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Gauss: Planar Symmetry
Symmetry is Planar
Use Gaussian Pillbox
rE =±Ex̂
x̂Note: A is arbitrary (its size and shape) and should divide out Gaussian
Pillbox
![Page 31: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/31.jpg)
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Gauss: Planar Symmetry
NOTE: No flux through side of cylinder, only endcaps
Total charge enclosed: qenclosed= A
=E 2A( ) =
qenclosed
ε0
= Aε0
ΦE =
rE⋅d
rA
S“∫∫ =E dA
S“∫∫ =EAEndcaps
E =
2ε0
++++++++++++
E
E
x
A
![Page 32: 1 W02D2 Gauss’s Law. Announcements Math Review Tuesday Week Three Tues from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday Tues at 9 pm.](https://reader036.fdocuments.net/reader036/viewer/2022062409/5697bfe11a28abf838cb3b0d/html5/thumbnails/32.jpg)
Concept Question: Superposition
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Three infinite sheets of charge are shown above. The sheet in the middle is negatively charged with charge per unit area , and the other two sheets are positively charged with charge per unit area . Which set of arrows (and zeros) best describes the electric field?
2
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Concept Question Answer: Superposition
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Answer 2 . The fields of each of the plates are shown in the different regions along with their sum.
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Group Problem: Cylindrical Symmetry
An infinitely long rod has a uniform positive linear charge density .Find the direction and magnitude of the electric field outside the rod. Clearly show your choice of Gaussian closed surface.
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Electric Field for Charged Infinite Plane
1) Dipole: E falls off like 1/r3
1) Spherical charge: E falls off like 1/r2
1) Line of charge: E falls off like 1/r
(infinite)
4) Plane of charge: E uniform on
(infinite) either side of plane