Transcript Chapter 24
From last time…
y
dE
Continuous
charge distributions
+ ++ +++ ++ ++ + ++ +++ ++ ++
Motion of charged
particles
This Friday’s honors lecture:
Biological Electric Fields
Dr. J. Meisel, Dept. of Zoology, UW
Sep. 27, 2007
Physics 208 Lecture 8
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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.
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Physics 208 Lecture 8
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Electric Flux
Electric flux E through a surface:
(component of E-field surface) X (surface area)
Proportional to
# E- field lines
penetrating surface
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Why perpendicular component?
Suppose surface make angle surface normal
E E|| sˆ E nˆ
A Anˆ
nˆ
Component || surface
Component surface
Only component
‘goes through’ surface
sˆ
E = EA cos
E =0 if E parallel A
E = EA (max) if E A
E E A
Flux SI units are N·m2/C
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Total flux
E not constant
add up small areas
where it is constant
Surface not flat
add up small areas
where it is ~ flat
Ei E iA i cos E i Ai
Add them all up: E
E dA
surface
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Quick quiz
Two spheres of radius R and 2R are centered on
the positive charges of the same value q.
The flux through these spheres compare as
R
A) Flux(R)=Flux(2R)
q
B) Flux(R)=2Flux(2R)
2R
C) Flux(R)=(1/2)Flux(2R)
D) Flux(R)=4Flux(2R)
q
E) Flux(R)=(1/4)Flux(2R)
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Flux through spherical surface
Closed spherical surface,
positive point charge at center
E-field surface,
directed outward
E-field magnitude on sphere:
1 q
E
4o R 2
R
Net flux through surface:
E constant, everywhere surface
E
E dA
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1 q
2
EdA E dA EA
4
R
q / o
2
4o R
Physics 208 Lecture 8
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Quick quiz
Two spheres of the same size enclose different
positive charges, q and 2q. The flux through
these spheres compare as
A) Flux(A)=Flux(B)
A
B) Flux(A)=2Flux(B)
q
B
2q
C) Flux(A)=(1/2)Flux(B)
D) Flux(A)=4Flux(B)
E) Flux(A)=(1/4)Flux(B)
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Gauss’ law
net electric flux through closed surface
= charge enclosed /
E
E dA
Qenclosed
o
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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
Sep. 27, 2007
E dA Q
Physics 208 Lecture 8
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/o
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Field outside
uniformly-charged sphere
Field direction: radially out from charge
Gaussian surface:
Surface area where E dA 0 :
Sphere of radius r
4 r 2
Value of E dA on this area:
E
Flux thru Gaussian surface:
E 4 r 2
Charge enclosed:
Q
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Q
Gauss’ law: E 4 r Q /o E
4 o r 2
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Quick Quiz
Which Gaussian surface could be used to calculate
E-field from infinitely long line of charge?
None of
these
A.
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B.
C.
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E-field from line of charge
Field direction: radially out from line charge
Gaussian surface:
Area where E dA 0 :
Cylinder of radius r
2rL
Value of E dA on this area:
E
Flux thru Gaussian surface:
E2rL
Charge enclosed:
L
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Linear charge density C/m
1
Gauss’ law: E2 rL L /o E
2o r
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Quick Quiz
Which Gaussian surface could be used to calculate
E-field from infinite sheet of charge?
Any of
these
A.
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B.
C.
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Field from infinite plane of charge
Field direction: perpendicular to plane
Gaussian surface:
Area where E dA 0 :
Cylinder of radius r
2 r 2
Value of E dA on this area:
E
Flux thru Gaussian surface:
E 2 r 2
Charge enclosed:
r 2
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Surface charge C/m2
Gauss’ law: E2 r r /o E
2o
2
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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
Ein
Etot =0
Etot = E+Ein= 0
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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.
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Induced dipole
What charge is
induced on
sphere to make
zero electric
field?
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Conductors: charge on surface only
Choose a gaussian surface inside (as
close to the surface as desired)
E=0
There is no net flux through the
gaussian surface (since E=0)
Any net charge must reside on the
surface (cannot be inside!)
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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 law
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Field lines
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