hour 1: gauss’ law hour 2: gauss’ law · gauss’s law the first maxwell equation a very useful...
TRANSCRIPT
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1P05 -
Class 05: Outline
Hour 1:Gauss’ Law
Hour 2:Gauss’ Law
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2P05 -
Six PRS Questions On Pace and Preparation
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3P05 -
Last Time:Potential and E Field
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4P05 -
E Field and Potential: Creating
A point charge q creates a field and potential around it:
2ˆ;e e
q qk V kr r
= =E r Use superposition for systems of charges
;B
B A AV V V V d= −∇ ∆ ≡ − = − ⋅∫E E s
They are related:
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5P05 -
E Field and Potential: Effects
q=F EIf you put a charged particle, q, in a field:
To move a charged particle, q, in a field:
W U q V=∆ = ∆
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6P05 -
Two PRS Questions:Potential & E Field
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7P05 -
Gauss’s Law
The first Maxwell Equation A very useful computational technique
This is important!
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8P05 -
Gauss’s Law – The Idea
The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside
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9P05 -
Gauss’s Law – The Equation
0Ssurface
closed εin
Eqd =⋅=Φ ∫∫ AE
Electric flux ΦE (the surface integral of E over closed surface S) is proportional to charge inside the volume enclosed by S
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10P05 -
Now the Details
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11P05 -
Electric Flux ΦE
Case I: E is constant vector field perpendicular to planar surface S of area A
∫∫ ⋅=Φ AE dE
E EAΦ = +
Our Goal: Always reduce problem to this
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12P05 -
Electric Flux ΦE
Case II: E is constant vector field directed at angle θ to planar surface S of area A
cosE EA θΦ =
∫∫ ⋅=Φ AE dE
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13P05 -
PRS Question:Flux Thru Sheet
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14P05 -
Gauss’s Law
0Ssurface
closed εin
Eqd =⋅=Φ ∫∫ AE
Note: Integral must be over closed surface
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15P05 -
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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16P05 -
Area Element dA: Closed Surface
For closed surface, dA is normal to surface and points outward
( from inside to outside)
ΦE > 0 if E points out
ΦE < 0 if E points in
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17P05 -
Electric Flux ΦE
Case III: E not constant, surface curved
Ed dΦ = ⋅E A
S
Ad E
∫∫ Φ=Φ EE d
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18P05 -
Example: Point ChargeOpen Surface
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19P05 -
Example: Point ChargeClosed Surface
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20P05 -
PRS Question:Flux Thru Sphere
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21P05 -
Electric Flux: Sphere Point charge Q at center of sphere, radius r
E field at surface:
20
ˆ4
Qrπε
=E r
Electric flux through sphere:
20S
ˆ ˆ4
Q dArπε
= ⋅∫∫ r rS
E dΦ = ⋅∫∫ E A
rA ˆdAd =0
Qε
=22
0
44
Q rr
ππε
=20 S4
Q dArπε
= ∫∫
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22P05 -
Arbitrary Gaussian Surfaces
0Ssurface
closed εQdE =⋅=Φ ∫∫ AE
For all surfaces such as S1, S2 or S3
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23P05 -
Applying Gauss’s Law1. Identify regions in which to calculate E field.2. Choose Gaussian surfaces S: Symmetry3. Calculate
4. Calculate qin, charge enclosed by surface S5. Apply Gauss’s Law to calculate E:
0Ssurface
closed εin
Eqd =⋅=Φ ∫∫ AE
∫∫ ⋅=ΦS
AE dE
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24P05 -
Choosing Gaussian SurfaceChoose surfaces where E is perpendicular & constant.
Then flux is EA or -EA.
Choose surfaces where E is parallel.Then flux is zero
OR
EA
EA−
EExample: Uniform Field
Flux is EA on topFlux is –EA on bottomFlux is zero on sides
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25P05 -
Symmetry & Gaussian Surfaces
Symmetry Gaussian Surface
Spherical Concentric Sphere
Cylindrical Coaxial Cylinder
Planar Gaussian “Pillbox”
Use Gauss’s Law to calculate E field from highly symmetric sources
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26P05 -
PRS Question:Should we use Gauss’ Law?
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27P05 -
Gauss: Spherical Symmetry+Q uniformly distributed throughout non-conducting solid sphere of radius a. Find E everywhere
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28P05 -
Gauss: Spherical Symmetry
Symmetry is Spherical
Use Gaussian Spheres
rE ˆE=
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29P05 -
Gauss: Spherical SymmetryRegion 1: r > aDraw Gaussian Sphere in Region 1 (r > a)
Note: r is arbitrary but is the radius for which you will calculate the E field!
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30P05 -
Gauss: Spherical SymmetryRegion 1: r > aTotal charge enclosed qin = +Q
S
E dA EA= =∫∫S
E dΦ = ⋅∫∫ E A
( )24E rπ=
2
0 0
4 inE
q Qr Eπε ε
Φ = = =
204
QErπε
= 20
ˆ4
Qrπε
⇒ =E r
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31P05 -
Gauss: Spherical Symmetry
Gauss’s law:
QarQ
a
rqin ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
= 3
3
3
3
3434
π
π
Region 2: r < aTotal charge enclosed:
( )24E E rπΦ =
304
Q rEaπε
= 30
ˆ4Q r
aπε⇒ =E r
OR inq Vρ=
3
30 0
inq r Qaε ε
⎛ ⎞= = ⎜ ⎟
⎝ ⎠
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32P05 -
PRS Question:Field Inside Spherical Shell
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33P05 -
Gauss: Cylindrical SymmetryInfinitely long rod with uniform charge density λFind E outside the rod.
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34P05 -
Gauss: Cylindrical SymmetrySymmetry is Cylindrical
Use Gaussian Cylinder
rE ˆE=
Note: r is arbitrary but is the radius for which you will calculate the E field!
is arbitrary and should divide out
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35P05 -
Gauss: Cylindrical Symmetryλ=inqTotal charge enclosed:
( )0 0
2 inqE r λπε ε
= = =
S SE d E dA EAΦ = ⋅ = =∫∫ ∫∫E A
02E
rλπε
=0
ˆ2 rλπε
⇒ =E r
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36P05 -
Gauss: Planar SymmetryInfinite slab with uniform charge density σFind E outside the plane
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37P05 -
Gauss: Planar Symmetry
Symmetry is Planar
Use Gaussian Pillbox
xE ˆE±=x̂
GaussianPillbox
Note: A is arbitrary (its size and shape) and should divide out
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38P05 -
Gauss: Planar SymmetryAqin σ=Total charge enclosed:
NOTE: No flux through side of cylinder, only endcaps
( )0 0
2 inq AE A σε ε
= = =
S SE Endcapsd E dA EAΦ = ⋅ = =∫∫ ∫∫E A +
+++++++++++ σ
E E
x
A
{ }0
ˆ to rightˆ- to left2
σε
⇒ = xE x02
E σε
=
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39P05 -
PRS Question:Slab of Charge
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40P05 -
Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). Find E everywhere.
x̂
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41P05 -
PRS Question:Slab of Charge
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42P05 -
Potential from E
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43P05 -
Potential for Uniformly Charged Non-Conducting Solid SphereFrom Gauss’s Law
20
30
ˆ,4
ˆ,4
Q r Rr
Qr r RR
πε
πε
⎧ >⎪⎪= ⎨⎪ <⎪⎩
rE
r
B
B AA
V V d− = − ⋅∫E sUse
( )0
BV V=
− ∞Region 1: r > a
204
r Q drrπε∞
= −∫0
14
Qrπε
=
Point Charge!
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44P05 -
Potential for Uniformly Charged Non-Conducting Solid Sphere
( ) ( )R r
RdrE r R drE r R
∞= − > − <∫ ∫
Region 2: r < a( )
0
DV V=
− ∞
2
20
1 38
Q rR Rπε⎛ ⎞
= −⎜ ⎟⎝ ⎠
2 30 04 4
R r
R
Q Qrdr drr Rπε πε∞
= − −∫ ∫
( )2 23
0 0
1 1 14 4 2
Q Q r RR Rπε πε
= − −
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45P05 -
Potential for Uniformly Charged Non-Conducting Solid Sphere
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46P05 -
Group Problem: Charge SlabInfinite slab with uniform charge density ρThickness is 2d (from x=-d to x=d). If V=0 at x=0 (definition) then what is V(x) for x>0?
x̂
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47P05 -
Group Problem: Spherical ShellsThese two spherical shells have equal but opposite charge.
Find E everywhere
Find V everywhere (assume V(∞) = 0)