Sep. 27, 2007Physics 208 Lecture 81 From last time… This Friday’s honors lecture: Biological...
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Transcript of Sep. 27, 2007Physics 208 Lecture 81 From last time… This Friday’s honors lecture: Biological...
Sep. 27, 2007 Physics 208 Lecture 8 1
From last time…
This Friday’s honors lecture:Biological Electric FieldsDr. J. Meisel, Dept. of Zoology, UW
Continuous charge distributions
y
++++++++++++++++++++
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dr E
Motion of charged particles
Sep. 27, 2007 Physics 208 Lecture 8 2
Exam 1 Wed. Oct. 3, 5:30-7 pm, 2103 Ch (here)
Covers Chap. 21.6, 22-27+ lecture, lab, discussion, HW
Sample exams on website.Solutions on website.HW4 (assigned today) covers exam material.
Sep. 27, 2007 Physics 208 Lecture 8 3
Electric Flux Electric flux E through a surface:
(component of E-field surface) X (surface area) Proportional to
# E- field lines penetrating surface
Sep. 27, 2007 Physics 208 Lecture 8 4
Why perpendicular component? Suppose surface make angle surface normal
E = EA cos E =0 if E parallel A E = EA (max) if E A
Flux SI units are N·m2/C
€
r E = E||ˆ s + E⊥ˆ n
€
ˆ n
€
ˆ s
€
r A = A ˆ n
Component || surfaceComponent surface
Only component‘goes through’ surface
€
E =r E •
r A
Sep. 27, 2007 Physics 208 Lecture 8 5
Total flux E not constant
add up small areas where it is constant
Surface not flat add up small areas
where it is ~ flat
€
δEi = E iδA icosθ =
r E i • δ
r A i
Add them all up:
€
E =r E • d
r A
surface∫
Sep. 27, 2007 Physics 208 Lecture 8 6
q
Quick quizTwo spheres of radius R and 2R are centered on
the positive charges of the same value q. The flux through these spheres compare as
A) Flux(R)=Flux(2R)
B) Flux(R)=2Flux(2R)
C) Flux(R)=(1/2)Flux(2R)
D) Flux(R)=4Flux(2R)
E) Flux(R)=(1/4)Flux(2R)
q
R
2R
Sep. 27, 2007 Physics 208 Lecture 8 7
Flux through spherical surface Closed spherical surface,
positive point charge at center E-field surface,
directed outward E-field magnitude on sphere:
Net flux through surface: E constant, everywhere surface
€
E = 14πεo
qR2
R
€
E =r E • d
r A ∫ = EdA∫ = E dA∫ = EA = 1
4πεo
qR2
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 4πR2( ) = q /εo
Sep. 27, 2007 Physics 208 Lecture 8 8
q
Quick quizTwo spheres of the same size enclose different
positive charges, q and 2q. The flux through these spheres compare as
A) Flux(A)=Flux(B)
B) Flux(A)=2Flux(B)
C) Flux(A)=(1/2)Flux(B)
D) Flux(A)=4Flux(B)
E) Flux(A)=(1/4)Flux(B)
2q
A B
Sep. 27, 2007 Physics 208 Lecture 8 9
Gauss’ law
net electric flux through closed surface = charge enclosed /
€
E =r E • d
r A ∫ = Qenclosed
εo
Sep. 27, 2007 Physics 208 Lecture 8 10
Using Gauss’ law Use to find E-field from known charge distribution.
Step 1: Determine direction of E-field
Step 2: Make up an imaginary closed ‘Gaussian’ surface Make sure E-field at all points on surface either
perpendicular to surface or parallel to surface
Make sure E-field has same value whenever perpendicular to surface
Step 3: find E-field on Gaussian surface from charge enclosed. Use Gauss’ law:
electric flux thru surface = charge enclosed / o
€
=r E • d
r A ∫ = Qin /εo
Sep. 27, 2007 Physics 208 Lecture 8 11
Field outside uniformly-charged sphere
Field direction: radially out from charge Gaussian surface:
Sphere of radius r Surface area where
Value of on this area:
Flux thru Gaussian surface:
Charge enclosed:
€
r E • d
r A ≠ 0 :
€
4π r2
€
r E • d
r A
€
E
€
Q
€
E 4π r2
Gauss’ law:
€
E 4π r2 = Q /εo ⇒ E = 14πεo
Qr2
Sep. 27, 2007 Physics 208 Lecture 8 12
Quick QuizWhich Gaussian surface could be used to calculate
E-field from infinitely long line of charge?
None ofthese
A. B. C. D.
Sep. 27, 2007 Physics 208 Lecture 8 13
E-field from line of charge Field direction: radially out from line charge Gaussian surface:
Cylinder of radius r Area where
Value of on this area:
Flux thru Gaussian surface:
Charge enclosed:
€
r E • d
r A ≠ 0 :
€
2πrL
€
r E • d
r A
€
E
€
λL
€
E2πrL
Gauss’ law:
€
E2π rL = λL /εo ⇒ E = 12πεo
λr
Linear charge density λ C/m
Sep. 27, 2007 Physics 208 Lecture 8 14
Quick QuizWhich Gaussian surface could be used to calculate
E-field from infinite sheet of charge?
Any ofthese
A. B. C. D.
Sep. 27, 2007 Physics 208 Lecture 8 15
Field from infinite plane of charge Field direction: perpendicular to plane Gaussian surface:
Cylinder of radius r Area where
Value of on this area:
Flux thru Gaussian surface:
Charge enclosed:
€
r E • d
r A ≠ 0 :
€
2π r2
€
r E • d
r A
€
E
€
ηπ r2
€
E2πr2
Gauss’ law:
€
E2π r2 = ηπ r2 /εo ⇒ E = η2εo
Surface charge η C/m2
Sep. 27, 2007 Physics 208 Lecture 8 16
Properties of Conductor in Electrostatic Equilibrium
In a conductor in electrostatic equilibrium there is no net motion of charge Property 1: E=0 everywhere inside the conductor
Conductor slab in an external field E: if E 0 free electrons would be accelerated
These electrons would not be in equilibrium
When the external field is applied, the electrons redistribute until they generate a field in the conductor that exactly cancels the applied field.Etot = E+Ein= 0
Ein
Etot =0
Sep. 27, 2007 Physics 208 Lecture 8 17
Induced dipole What charge is
induced on sphere to make zero electric field?
+
Sep. 27, 2007 Physics 208 Lecture 8 18
Conductors: charge on surface only Choose a gaussian surface inside (as
close to the surface as desired)
There is no net flux through the gaussian surface (since E=0)
Any net charge must reside on the surface (cannot be inside!)
E=0
Sep. 27, 2007 Physics 208 Lecture 8 19
Field’s Magnitude and Direction
E-field always surface: Parallel component of E would put
force on charges Charges would accelerate This is not equilibrium
Apply Gauss’s lawField lines