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
1
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
2
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
3
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
Sep. 27, 2007
Physics 208 Lecture 8
4
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 iA i cos  E i  Ai
Add them all up: E 
 E  dA
surface
Sep. 27, 2007
Physics 208 Lecture 8
5
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)
Sep. 27, 2007
Physics 208 Lecture 8
6

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
4o R 2
R
Net flux through surface:

E constant, everywhere  surface
E 
 E  dA  
Sep. 27, 2007
 1 q 
2
EdA  E  dA  EA  
4

R
 q / o


2 
4o R 
Physics 208 Lecture 8
7
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)
Sep. 27, 2007
Physics 208 Lecture 8
8
Gauss’ law

net electric flux through closed surface
= charge enclosed / 
E 
 E  dA 
Qenclosed
o

Sep. 27, 2007
Physics 208 Lecture 8
9
Using Gauss’ law

Use to find E-field from known charge distribution.
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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
in
/o
10
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

Sep. 27, 2007
Q
Gauss’ law: E 4  r  Q /o  E 
4 o r 2
2
Physics 208 Lecture 8
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11
Quick Quiz
Which Gaussian surface could be used to calculate
E-field from infinitely long line of charge?
None of
these
A.
Sep. 27, 2007
B.
C.
Physics 208 Lecture 8
D.
12
E-field from line of charge

Field direction: radially out from line charge

Gaussian surface:


Area where E  dA  0 :


Cylinder of radius r
2rL
Value of E  dA on this area:
 
E
Flux thru Gaussian surface:
 E2rL
 Charge enclosed:


L

Sep. 27, 2007
Linear charge density  C/m
1 
Gauss’ law: E2 rL  L /o  E 
2o r
Physics 208 Lecture 8
13
Quick Quiz
Which Gaussian surface could be used to calculate
E-field from infinite sheet of charge?
Any of
these
A.
Sep. 27, 2007
B.
C.
Physics 208 Lecture 8
D.
14
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

Sep. 27, 2007
Surface charge  C/m2

Gauss’ law: E2 r   r /o  E 
2o
2
Physics 208 Lecture 8
2
15
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
Sep. 27, 2007
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.
Physics 208 Lecture 8
16
Induced dipole

What charge is
induced on
sphere to make
zero electric
field?
Sep. 27, 2007
+
Physics 208 Lecture 8
17
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!)
Sep. 27, 2007
Physics 208 Lecture 8
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Field’s Magnitude and Direction
E-field always  surface:



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Parallel component of E would put
force on charges
Charges would accelerate
This is not equilibrium
Apply Gauss’s law
Sep. 27, 2007
Field lines
Physics 208 Lecture 8
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