Transcript W02D2_Presentation_answers

W02D2
Gauss’s Law
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pm in 26-152Vector Calculus
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W02D3 Reading Assignment Course Notes: Chapter
Course Notes: Sections 3.6, 3.7, 3.10
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Outline
Electric Flux
Gauss’s Law
Calculating Electric Fields using Gauss’s
Law
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Gauss’s Law
The first Maxwell Equation!
A very useful computational technique to find the
electric field E when the source has ‘enough
symmetry’.
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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
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Gauss’s Law – The Equation
r r qenclosed
 E  “ E  d A 
closed
surface S
0
Electric flux  E (the surface integral of E over
closed surface S) is proportional to charge
enclosed by the volume enclosed by S
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Electric Flux
Case I: E is a uniform vector field
perpendicular to planar surface S of area A
 
 E   E  dA
 E  EA
Our Goal: Always reduce
problem to finding a
surface where we can
take E out of integral and
get simply E*Area
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Electric Flux
Case II: E is uniform vector field directed
at angle  to planar surface S of area A
n̂
r r
 E   E d A
r
dA  dA nφ
 E  EAcos
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Concept Question: Flux
The electric flux through the planar surface
below (positive unit normal to left) is:
+q
1.
2.
3.
4.
n̂
positive.
negative.
zero.
Not well defined.
-q
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Concept Question Answer: Flux
Answer: 2. The flux is negative.
+q
n̂
E
-q
The field lines go from left to right,
opposite the assigned normal direction.
Hence the flux is negative.
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
r
Case III: E not uniform,
r surface curved
For closed surface, dA is normal to surface
and points outward ( from inside to outside)
r r
d E  E  dA
d E  0 if E points out
d E  0 if E points in
r r
E  “ dE  “ E d A
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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
E 
“
r r Q
EdA 
closed
surface S
0
True for all surfaces such as S1, S2 or S3
Why? As area gets bigger E gets smaller
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Gauss’s Law
E 
“
closed
surface S
r r qenclosed
EdA 
0
Note: Integral must be over closed surface
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Concept Question: Flux thru Sphere
The total flux through the below spherical
surface is
+q
1.
2.
3.
4.
positive (net outward flux).
negative (net inward flux).
zero.
Not well defined.
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Concept Question Answer: Flux thru Sphere
Answer: 3. The total flux is zero
+
q
We know this from Gauss’s Law:
E 
“
closed
surface S
r r qenclosed
EdA 
0
No enclosed charge  no net flux.
Flux in on left cancelled by flux out on right
Concept Question: Gauss’s Law
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.
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Concept Question Answer: Gauss’s Law
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.
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Choosing Gaussian Surface
E 
“
r r qenclosed
EdA 
closed
surface S
0
True for all closed surfaces
Useful (to calculate electric field ) for
some closed surfaces for some
problems with lots of symmetry.
Desired E: Perpendicular to surface and uniform on surface.
Flux is EA or -EA.
Other E:
Parallel to surface. Flux is zero
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Symmetry & Gaussian Surfaces
Desired E: perpendicular to surface and constant
on surface. So Gauss’s Law useful to calculate
electric field from highly symmetric sources
Source Symmetry Gaussian Surface
Spherical
Concentric Sphere
Cylindrical
Coaxial Cylinder
Planar
Gaussian “Pillbox”
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Virtual Experiment
Gauss’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 Law
1. Based on the source, identify regions in which to calculate
electric field.
2. Choose Gaussian surface S: Symmetry
3. Calculate
r r
 E  “ E  d A
S
4. Calculate qenc, charge enclosed by surface S
5. Apply Gauss’s Law to calculate electric field:
“
closed
surface S
r r qenclosed
EdA 
0
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Examples:
Spherical Symmetry
Cylindrical Symmetry
Planar Symmetry
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Group Problem Gauss: Spherical Symmetry
+Q uniformly distributed throughout non-conducting
solid sphere of radius a. Find E 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?
1.
2.
3.
4.
5.
a
Q
Zero
Uniform but Non-Zero
Still grows linearly
Some other functional form (use Gauss’ Law)
Can’t determine with Gauss Law
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Concept Question Answer: Flux thru Sphere
Answer: 1. Zero
Q
a
Spherical symmetry
 Use Gauss’ Law with spherical surface.
Any surface inside shell contains no charge
 No flux
E = 0!
P04 -
Demonstration
Field Inside Spherical Shell
(Grass Seeds):
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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
r
E   E xφ
Use Gaussian Pillbox
x̂
Note: A is arbitrary (its
size and shape) and
should divide out
Gaussian
Pillbox
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Gauss: Planar Symmetry
Total charge enclosed:
qenclosed   A
NOTE: No flux through side of cylinder, only endcaps
r r
 E  “ E  d A  E “ dA  EAEndcaps
S
 E 2 A
S
qenclosed
0

E
2 0
A

0
E
+
+
+
+
+
+
+
+
+
+
+
+
x
A
E

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Concept Question: Superposition
Three infinite sheets of
charge are shown above.
The sheet in the middle is
negatively charged with
charge per unit area 2 , and
the other two sheets are
positively charged with
area
charge per unit
.
Which set of arrows (and
zeros) best describes the
electric field?
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Concept Question Answer: Superposition
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:
(infinite)
E falls off like 1/r
4) Plane of charge:
(infinite)
E uniform on
either side of plane
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