Download Document

Lecture note download site:
http://ifts.zju.edu.cn/~zwma/ptwl_2
Last Chapter
•Electrostatic Phenomena
•Coulomb’s Law
r
F12 =
1
q1q2
ˆ
r
12
2
4pe o r12
q1
•Superposition
F1
F
Ftotal = F1 + F2 + ...
q
F2
Text Reference: Chapter 22.1 through 22.3
Examples: 22.1 – 22.5
q2
This Chapter
•Define Electric Field in terms of force
on a test charge
•How to think about fields
•Electric Field Lines
•Example Calculation: point charge,
Continuous Charge Distributions
Text Reference: Chapter 26.1, 2, 3, 5,
1
26-1 What is a Field?
A FIELD is something that can be defined anywhere in space
•It can be a scalar field (e.g., Temperature field)
•It can be a vector field (e.g., Velocity, Electric field)
•Fields represent physical quantities.
A Scalar Field
73
77
82
84
83
72
71
75
77
68
80
64 73
82
88
55
66
80 88
75
88
83 90 91
92
These isolated temperatures sample the scalar field
(you only learn the temperature at the point you choose,
but T is defined everywhere (x, y) )
A Vector Field
It may be more interesting to know which way the wind is blowing...
73
77
72
71
82
84
83
88
75
68 64
80
73
57 56 55
66
88
75 80
90
83
92
91
77
That would require a vector field
(you learn both wind speed and direction)
Velocity field
• Vector field
– Space field
– Space field varied with time
Velocity
field
r
V ( x, y , z )
r
V ( x, y , z ; t )
Gravitation field
r
r F
g=
m0
r
r r
M
g (r ) = G 2 rˆ
r
r
g
The procedure for measuring the
gravitational field
–use a test body of small mass m0
–release in the gravitational field
–measure F
–gravitational field:
m0 is small , does not disturb the mass
distribution.
The Moon can not be as a test mass.
Notes
• The force between gravitating bodies was
thought of as a direct and instantaneous
interaction, action at a distance(超距相互作
用). This view violates the special theory of
relativity
mass<=>mass
• A more modern interpretation
mass<=>field<=>mass
28-2 The Electric field
•Coulomb’s law
r
F=
1
Q1Q2
ˆ
r
4pe 0 r 2
r
E
1. Calculate the force that the
field exerts on the second
charge placed at particular
point in space.
2. Determine, by measurement
or calculation, the field
established by first charge at
every point in space.
r
r
F
E = lim
q0 0 q 0
r r
q0>0, E , F in the same direction
28-3 The Electric field of point charge
• A single charge
r
1 q0 q
F=
rˆ
2
4pe0 r
r
r F
1 q
E=
=
rˆ
2
q0 4pe0 r
r
E
The direction of
is the same as

the direction of F , along a radial
line from q, pointing outward if q is
positive, and inward if q is negative.
• A group of N point charges
Law of Superposition
q1
qN
P
q2
r
r r
r
E p = E1  E2  ...  EN
r
E1 =
1
q1
rˆ
2 1
4pe0 r1
r
E2 =
q2
rˆ
2 2
4pe0 r2
r
E3 =
q3
rˆ
2 3
4pe0 r3
........
r
EN =
1
1
qN
rˆ
2 N
4pe0 rN
1
The Electric Field
r
r
r r
F F
E =E 
q0 0q q 0
lim
With this concept, we can “map” the electric field
anywhere in space produced by any arbitrary:
Bunch of Charges
r
E=
q
1
 2i rˆi
4pe 0
+
+
+
-
-
+
Charge Distribution
r
E=
ri
+
-
F
+
1
dq
rˆ

2
4pe 0 r
+ +++ + +
+ + +++
+
“Net” E at origin
These charges or this charge distribution
“source” for the electric field throughout space
Example: Electric Field
What is the electric field at the origin due to this set of charges?
y
1) Notice that the fields from the top-right
and bottom left cancel at the origin?
a
+q
a
2) The electric field, then, is just the field
from the top -left charge. It points away
a
from the top-left charge as shown.
+q
3) Magnitude of E-field at the origin is:
E = kq2
2a
The x and y components of the field at (0,0) are:
kq cosq
kq sinq
=
Ex
=
E
y
2a2
2a2
kq 1
kq 1
= 2
= - 2a2
2a 2
2
a
a 2
+q
a
Q
x
Example: Electric Field
Now, a charge, Q, is placed at the origin. What is the net force
y
on that charge?
a
a
+q
+q
q 1
q
1
Ex = k 2
E y = k 2
a 2
a
a
2a 2
2a 2
x
Q
a
+q
Qq
Fx = QE x = k
2 2a 2
Qq
Fy = QE y = k
2 2a 2
Note: If the charge Q is positive, the force will be in
the direction of the electric field
If the charge Q is negative, the force will be against
the direction of the electric field
F is
F is
Reality of Electric Fields
•The electric field has been introduced as a mathematical
convenience, just as the gravitational field of Physics I
•There is MUCH MORE to electric fields than this!
IMPORTANT FEATURE: E field propagates at speed of light
• NO instantaneous action at a distance (we will explain this when
we discuss electromagnetic waves)
• i.e., as charge moves, resultant E-field at time t depends upon
where charge was at time t - dt
• For now, we avoid these complications by restricting ourselves to
situations in which the source of the E-field is at rest.
(electrostatics)
Mathematics of Fields
z
•Scalar Fields
–Number associated with each point in space
–May be time dependent (future)
–Expressed as a function g(x, y, z, t)
•Vector Fields
–Vector associated with each point in space
–May be time dependent (future)
–Expressed as a function G(x, y, z, t)
•Physical Fields
y
x
Pick a
coordinate
system…
This specifies
the formula:
g(x, y, z, t)
–Obey a simple rule
–Created by sources
G(x, y, z, t)
–Continuous and well behaved
–What field looks like depends on rule and sources
Two Opposite Charges: Dipole(偶极矩)
Dipoles are central to our
existence!
Molecular Force Model
Basis of Attraction
Ion-dipole
Ion and polar molecule
Dipole-dipole
Partial charges of polar
molecules
Induced dipoles of
polarizable molecules
London
dispersion
y
The Electric Dipole (p=2Qa)
+Q
a
q
a
-Q
r
E
x
E
What is the E-field generated by
this arrangement of charges?
Calculate for a point along the x-axis: (x, 0)
Ex = ??
Symmetry
Ex(x,0) = 0
Ey = ??
E
Electric Dipole
y
+Q
a
x
a
-Q
What is the Electric Field
generated by this charge
arrangement?
Now calculate for a point along the y-axis: (0,y)
Ex = ??
Coulomb Force
Radial
Ey = ??
y
Electric Dipole
+Q
a
a
-Q
For points along x-axis:
For r >>a,
r
x
Case of special interest:
(antennas, molecules)
r>>a
For points along y-axis:
For r >>a
Continued (p=2Qa)
x  a
1
p
1
2Qa
E=
=
2
2 3/ 2
4pe 0 ( x  a )
4pe 0 x 3 (12  (a / x) 2 ) 3 / 2
Taylor expansion
n(n  1) 2
(1  y ) = 1  ny 
y  ...
2!
f ' ( 0)
f ' ' ( 0) 2
f ( x ) = f ( 0) 
x
x  ....
1!
2!

f ( n ) (0) n
=
x
n!
n =0
p
3 a 2
E =
[1  ( )( )  ....]
3
2 x
4pe 0 x
n
y
Continued (p=2Qa)
+Q
a
q
a
-Q
r
E
p
x
E
3 a 2
E =
[1  ( )( )  ....]
3
2 x
4pe 0 x
• The first degree approximation
E=
p
4pe 0 x 3
• A more general result for the field at any
point in the xz plane can also be
calculated (Problem 2).
• The field of an electric quadrupole
(Problem 3).
1
E 3
r
1
E 4
r
26-4 Electric Fields from
Continuous Charge Distributions
Examples:
• line of charge
• charged plates
• electron cloud in
atoms, …
• Principles (Coulomb’s Law + Law of Superposition)
remain the same.
Only change:



Charge Densities
• How do we represent the charge “Q” on an extended
object? total charge
small pieces
Q
of charge
dq
• Line of charge:
l = charge per
unit length
dq = l dx
• Surface of charge:
s = charge per
unit area
dq = s dA
• Volume of Charge:
r = charge per
unit volume
dq = r dV
How We Calculate (Uniform) Charge
Densities:
Take total charge, divide by “size”
Examples:
10 coulombs distributed over a 2-meter rod.
10C
λ=
= 5 C/m
2m
14 pC (pico = 10-12) distributed over a shell of radius 1 μm.
14 1012 C 14
2
σ=
=
C/m
4π(10-6 m) 2 4π
14 pC distributed over a sphere of radius 1 mm.
14 1012 C (3) 14 3
3
ρ= 4
=

10
C/m
-3
3
π(10
m)
4π
3
Electric field from an infinite line charge
Approach:
“Add up the electric field contribution from each bit of
charge, using superposition of the results to get the final field.”
In practice:
• Use Coulomb’s Law to find the E-field per segment of charge
• Plan to integrate along the line…
– x: from  to  OR
q: from p/2 to p/2
q
+++++++++++++++++++++++++++++
x
Any symmetries ? This may help for easy cancellations
Infinite Line of Charge
Charge density = l
We need to add up the E-field
contributions from all
segments dx along the line.
Infinite Line of Charge
We use Coulomb’s Law to find dE:
What is dq in terms of dx?
Therefore,
What is r’ in terms of r ?
Infinite Line of Charge
We still have x and q variables.
We are dealing with too
many variables. We
must write the integral in
terms of only one
variable (q or x). We will
use q.
x and q are not independent!
x = r tan q
dx = r sec2 q dq
Infinite Line of Charge
• Components:
• Integrate:
Infinite Line of Charge
• To find the total field E, we
must integrate over all
charges along the line. If we
integrate over q, we must write
r’ and dq in terms of q and dq .
•
The electric field due to dq is:
•
Solution: After the appropriate
change of variables, we integrate and
find:
Ex = 0
2l
Ey =
4pe 0 r
1
Infinite Line of Charge
Conclusion:
• The Electric Field produced by an infinite line of
charge is:
- everywhere perpendicular to the line
- is proportional to the charge density
- decreases as 1 / r
- next lecture: Gauss’ Law makes this trivial!!
Summary
Electric Field Lines
Electric Field Patterns
Dipole
Point Charge
Infinite
Line of Charge
~ 1/R3
~ 1/R2
~ 1/R
Coming up:
Electric field Flux
and
Gauss’ Law
A Uniform Ring of Charge
lds
lds
dE =
=
2
4pe 0 r
4pe 0 ( z 2  R 2 )
Ex = E y = 0
Ez  0
dE z = dE cosq
lds
z
=
 2
2
2
2 1/ 2
4pe 0 ( z  R ) ( z  R )
zlds
=
4pe 0 ( z 2  R 2 ) 3 / 2
A Uniform Ring of Charge
zlds
E z =  dE z = 
4pe0 ( z 2  R 2 )3 / 2
zq
=
2
2 3/ 2
4pe0 ( z  R )
zq
z  R
Ez =
z0
Ez = 0
4pe0 z
3
=
q
4pe0 z 2
A Uniform Disk of Charge
dq = 2p  d  s
zdq
z 2psd
dE =
=
2
2 3/ 2
2
2 3/ 2
4pe 0 ( z   )
4pe 0 ( z   )
sz R
d
E =  dE =

2e 0 0 ( z 2   2 ) 3 / 2
sz d ( z   )
=

4e 0 0 ( z 2   2 ) 3 / 2
s
1
=
(1 
)
2
2e 0
R
R
2
1
2
z2
A Uniform Disk of Charge
s
E=
(1 
2e 0
R  z
z  R
1
R2
1 2
z
1
2
)
R
1 2
z
0
E=
s
Infinite sheet
2e 0
1 R2 3 R4
= 1

 ....
2
4
2
2z
8z
R
1 2
z
1
s 1 R2 3 R4
s 1 R2
q
E =
( 2 
 ....) 
=
4
2
2e 0 2 z
8z
2e 0 2 z
4pe 0 z 2
Point Charge
26-5 Ways to Visualize the E Field
(Electric Field Lines（电力线）)
Consider the E-field of a positive point charge at the origin
vector map
field lines
+ chg
+ chg
+
+
Rules for Vector Maps
+ chg
+
•Direction of arrow indicates direction of field
•Length of arrows  local magnitude of E
Rules for Field Lines
+
-
•Lines leave (+) charges and return to (-) charges
•Number of lines leaving/entering charge  amount
of charge
•Tangent of line = direction of E
•Local density of field lines  local magnitude of E
• Field lines at two white dots differs by a factor of 2
since r differs by a factor of 2 (in 2D). (l=2pr)
•Local density of field lines should differ by a factor of 4
(in 3D). (S=4pr2)
Other ways to Visualize the E Field
Consider a point charge at the origin
Field Lines
+ chg
Graphs
Ex, Ey, Ez as a function of (x, y, z)
Er, Eq, Ef as a function of (r, q, f)
Ex(x,0,0)
+
x
r
1 q
E=
4pe 0 r 2
Appendix A
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
–Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3A Er
r
f
x
Q
Er
Er
Fixed
r>0
0
f
2p
0
(a)
3B
f
2p
f
2p
(c)
Ex
Ex
0
0
2p
(b)
Ex
Fixed
r>0
f
0
f
2p
0
f
2p
Appendix “ACT”
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
– Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3A Er
r
f
x
Q
Er
Er
Fixed
r>0
0
f
(a)
2p
0
f
(b)
2p
0
f
2p
(c)
• At fixed r, the radial component of the field is a constant,
independent of f!!
• For r>0, this constant is > 0. (note: the azimuthal component
Ef is, however, zero)
“ACT”
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
–Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3B
Ex
Fixed
r>0
f
(a)
2p
f
x
Q
Ex
Ex
0
r
0
f
(b)
2p
0
f
2p
(c)
• At fixed r, the horizontal component of the field Ex is given by:
Electric Dipole Field Lines
• Lines leave positive charge
and return to negative charge
What can we observe about E?
• Ex(x,0) = 0
• Ex(0,y) = 0
• Field largest in space between two charges
• We derived:
... for r >> a,
Field Lines From Two Like Charges
• There is a zero halfway
between the two charges
• r >> a: looks like the field
of point charge (+2q) at origin
3
Electric Field inside a Conductor
• A two electron atom, e.g., Cu
– heavy ion core
– two valence electrons
2+
• An array of these atoms
– microscopically crystalline
– ions are immobile
– electrons can move easily
• Viewed macroscopically:
– neutral
There is never a net electric field inside
a conductor – the free charges always
move to exactly cancel it out.
2+
2+ 2+
2+
2+
2+ 2+
2+
2+
2+ 2+
2+
2+
2+ 2+
2+
26-6 A point Charge in An Electric
Field
• In the preceding sections, we have
considered the first part of the
charge <=>field <=>charge interaction
• Given a collection of charges, what is the
resulting electric fields
r
r
r
r
1. F = qE +Newton’s Law  F = ma
Motion of object
r
E = const.
Deflecting electrode system of an ink-jet printer
Page 598 Problem 26-6
++++++++++++
E
----------------
Drop
generator
Charging
unit
paper
An ink drop : m=1.3x10-10kg
q= -1.5x10-13C,
v=18m/s
L=1.6cm
E=1.4x106 N/C
Deflecting electrode system of an ink-jet printer
Page 598 Problem 26-6
paper
++++++++
E
y
1 2
y = at
2
L = vt
qE  mg = ma
---------------
Deflecting
plates
qE
qE
a=
g 
m
m
1 2 1 qE L2
y = at = 
 2  0.64mm
2
2 m v
Deflecting electrode system of an ink-jet printer
Page 598 Problem 26-6
• One letter, about 100
drops
• In the printer, there are
100000 drops/s
2. Measuring the elementary charge
(e=1.60217646210-19C)
• Milikan oil drop Experiment (1910-1913),
1923 Nobel Prize
Let us assume that this carries a
charge q, which we take to be
negative. (assume drag force is η)
E=0
E≠0
mg= ηv
qE=mg+ηv’
q= η(v+v’)/E
The charge q on the drop can be found from
measurements of v and v’
• Milikan found that the value of q were all
consistent with the relation
q=ne, n=0,+1,-1,+2,-2,…
e=1.6010-19C
• =>charge is quantized
1923 Noble prize
3. Motion in no uniform electric
field of point charge
E


F = qE,  Newton' s law
F = qE  const
+
+ + +
+ +
+
+
+
+
+ + + +
F ( z)
d 2z q
= E( z)
2
dt
m
z
E=
4pe0 ( z 2  R 2 ) 3 / 2
d 2z q
z
=
2
dt
m 4pe0 ( z 2  R 2 ) 3 / 2
26-7 A Dipole in an Electric field
• Torque
To introduce the dipole
moment vector
(偶极矩矢量)
r
r
p = qd


p = qd
The net force F = F  F = 0
d
d
The net torque  = F  sin q  F  sin q = Fd sin q = qEd sin q
2
2
r r
 = pE
r
Energy
The work done by the external field

E
make
q
The motion of a dipole in
a uniform field can
therefore be interpreted
either from the
perspective of force or
energy.
r
p


q0  q
q
W =  dw =    dq =   pE sin qdq
q0
q0
= pE (cosq  cosq 0 )
W =  U
U = U (q )  U (q 0 ) =  pE (cosq  cosq 0 )
q 0 = 90
 
U (q ) =  pE cosq =  p  E
Problem 7 page 601
• H2O vapor
p=6.2x10-30C·m
H2O
CO2
p≠0
p=0
Dipole-molecule
non-dipole molecule
有极分子
无极分子
• The maximum torque on a molecule of H2O
E=1.5x104N/C for a typical lab electric field
τ=pEsinθ=6.2  10-30  1.5  104xsin(90o)
=9.3  10-26N·m
• θ0=π→0
W=pE(cosθ-cosθ0)=2pE=1.910-25J
• RT: T=300, internal energy
εint=3kT/2=6.3  10-21J
εint>>εelect align effect is negligible
• + charge center → - charge center : distance
q=10e
O: 8个 H2(2 electron)
d=p/10e=6.2x10-30/10x1.6x10-19=3.9pm
• 4% of H-O bond distance in the molecule
26-8 The Nuclear Model of The atom
• Today, we know the structure of the atom.
atom: positive + negative charge
nucleus: +Ze, 99.995%M
Electron: -Ze,
• But, in the early years of the 20th century
these facts were not known
• There was much speculation about the structure
of the atom.
• Thomson model: (plum pudding)
The positive charge is distributed
more or less uniformly throughout
the entire spherical volume of the
atom. The electrons are imbedded
throughout the diffuse spherical of
positive charge like raisins in a plum
pudding.
•Alpha particle are scattered by the electric
field of the atom.
Q
E=
4pe0 r 2
Qr
E=
4pe0 R 3
Emax =
Q
4pe 0 R 2
rR
rR
= 1.2  1013 N / C
R = 1.0  10 10 m
Q = 79e
U = 6 Mev = 9.6  10 13 J
2U
v=
= 1.7  10 7 m / s
m
F = qEmax = ma
q
a = Emax
m
q
2R
3
v = at = Emax
= 6.6 10 m / s
m
v
3
1 v
1 6.6  10
q = tg
= tg (
)  0.02
7
v
1.7 10
The experiment was done by Rutherford
1
10 4
particles was reversed
Rutherford to comment:
Au foil
“It was quite the most
incredible event that ever
happened to me in my life. It
was almost as incredible as if
you had fired a 15-inch shell
at a piece of tissue paper and
it came back and hit you.”
Rutherford atomic model
rnuclear  7 10
15
ratom  1.0 10
10
ratom
4
 10
rnuclear
m
m
Homework
• Page 606 (Exercises)
11,16,18,
• Page 609 (Problems)
4, 8, 12,