Transcript Short Version : 20. Electric Charge, Force, & Fields

Short Version :
20. Electric Charge, Force, & Fields
20.1. Electric Charge
2 kinds of charges: + & .
Total charge = algebraic sum of all charges.
Like charges repel.
Opposite charges attract.
Q   qi   d 3 x   r 
i
All electrons have charge e.
e  1.60 1019 C = elementary charge
All protons have charge +e.
1st measured by Millikan on oil drops.
Theory (standard model) : basic unit of charge (carried by quark) = 1/3 e.
Quark confinement  no free quark can be observed.
 Smallest observable charge is e.
Conservation of charge: total charge in a closed region is always the same.
Coulomb’s law (force between 2 point charges) :
F12  k
q1 q2
rˆ
2
r
[q] = Coulomb = C
k   c 2107  N m 2 / C 2
c  2.99792458 108
k  9.0 109 N m2 / C 2
r12   4 , 3 m  4 ˆi  3 ˆj m
r12 
rˆ 
16  9 m  5 m
4ˆ 3ˆ
1

i j m
4
,
3
m
 
5
5
5
Conceptual Example 20.1. Gravity & Electric Force
The electric force is far stronger than the gravitational force,
yet gravity is much more obvious in everyday life.
Why?
Only 1 kind of gravitational “charge”
 forces from different parts of a source tend to reinforce.
2 kinds of electric charges
 forces from different parts of a neutral source tend to cancel out.
Making the Connection
Compare the magnitudes of the electric & gravitational
forces between an electron & a proton.
Fg  G
mem p
k e2
FE  2
r
r2
9 10 N  m / C   1.6 10 C 
FE
ke


6.67 1011 N  m2 / kg 2   9.11 1031 kg   1.67 1027 kg 
Fg G memp
9
2
 2.27  1039
2
2
19
2
Point Charges & the Superposition Principle
Extension of Coulomb’s law (point charges) to charge distributions.
Superposition principle:
Fnet
Fnet   Fi
i
F23
F13
Task: Find net force on q3 .
Independent of each other
Example 20.2. Raindrops
Charged raindrops are responsible for thunderstorms.
Two drops with equal charge q are on the x-axis at x = a.
Find the electric force on a 3rd drop with charge Q at any point on the y-axis.
y
F  F1  F2
F1
F2

Q
r
k qQ
 2 2  0 , sin  
r
r
y
q


x1 = a
q
x2 = a
r
a2  y2
k qQ
k qQ
cos

,
sin




  cos  , sin  
2
2
r
r
x
2
k qQ yˆ
j
3
r
2
k qQ y ˆ
j
2
2 3/2
a  y 
sin  
y
r
20.3. The Electric Field
Electric field E at r = Electric force on unit point charge at r.
E
1
F
q
F = electric force on point charge q.
E = F/q
g = F/m
[E]=N/C
=V/m
V = Volt
Implicit assumption:
q doesn’t disturb E.
Rigorous definition:
E  lim
Gravitational field
Electric field
q0
1
F
q
Force approach:
Charges interact at a distance (difficult to manage when many charges are present).
Fails when charge distributions are not known.
Field approach:
Charge interacts only with field at its position.
No need to know how field is generated.
Given E:
FqE
The Field of a Point Charge
E
1
F
qtest
Field at r from point charge q :
1
E
qtest
Field vectors for a negative point charge.
 k q qtest

2
 r
kq

ˆr   2 rˆ
r

20.4. Fields of Charge Distributions
Superposition principle 
E   Ei
(Discrete sources)
i

i
k qi
rˆi
2
ri
(Point charges)
The Electric Dipole
Electric dipole = Two point charges of equal magnitude but opposite charges
separated by a small distance..
Examples:
Polar molecules.
Heart muscle during contraction
 Electrocardiograph (EKG)
Radio & TV antennas.
H2O
Example 20.5. Modeling a Molecule
A molecule is modeled as a positive charge q at x = a,
and a negative charge q at x =  a.
Evaluate the electric field on the y-axis.
Find an approximate expression valid at large distances (y >> a).
y
Ey 
k  q   y  k q  y 
  2    0
2
r r r r
Ex 
k  q   a  k q  a 
2k qa




 


r2  r  r2  r 
r3
E2
E
Q=1
E1
r

r
y
q

x1 = a

q
x2 = a
x
a

2k qa
2
y
2k qa
y
3

2 3/2
(y >> a)
Dipole ( q with separation d ):
E
qd
r3
for r >> d = 2a
Typical of neutral, non-spherical, charge distributions ( d ~ size ).
Dipole moment :
p = q d.
On perpendicular bisector:
On dipole axis:
d = vector from q to +q
y
kp
k
E   3 ˆi   3 p
y
y
E2
E
2k pˆ
E
i
3
x
Q=1
E1
r
r
y
(Prob 習題
51)
q


x1 = d/2
p
q
x2 = d/2
x
Continuous Charge Distributions
All charge distributions are ultimately discrete ( mostly protons & electrons ).
Continuum approximation: Good for macroscopic bodies.
Volume charge density  [ C/m3 ]
Surface charge density  [ C/m2 ]
Line charge density  [ C/m ]
E  d E

k dq
rˆ
2
r
Example 20.6. Charged Ring
A ring of radius a carries a uniformly distributed charge Q.
Find E at any point on the axis of the ring.
By symmetry, E has only axial (x-) component.
Ex  
dEx  
Ring

E
a
kQx
2
x

2 3/2
ˆi
Ring
kx
 a2  x

2 3/2

Ring
k dq  x 
 
r2  r 
dq 
a
kQx
2
x

2 3/2
On axis of uniformly charged ring
Example 20.7. Power Line
A long electric power line running along the x-axis
carries a uniform charge density  [C/m].
Find E on the y-axis, assuming the wire to be infinitely long.
dEy
y
By symmetry, E has only y- component.
dE
dE
Ey  
dE y 

Line
P
r
r
y
dq
Line
k  dx  y 
k dq  y 

 
 
r 2  r  Line r 2  r 
k y
x
dq



x
y
dx
2
x

2 3/2

k y 
 y 2
 1  1 
 k y   2    2   2 k 
y
 y  y 
E
2k 

ρˆ


x

2
2
y  x  
Perpendicular to an infinite wire
20.5. Matter in Electric Fields
Point Charges in Electric Fields
Newton’s 2nd law 
a
q
E
m
(point charge in field E)
 Trajectory determined by charge-to-mass ratio q/m.
Constant E
 constant a.
E.g., CRT, inkjet printer, ….
Uniform field between charged plates (capacitors).
Example 20.8. Electrostatic Analyzer
Two curved metal plates establish a field of strength E = E0 ( b/r ),
where E0 & b are constants.
E points toward the center of curvature, & r is the distance to the center.
Find speed v with which a proton entering vertically from below will leave
the device moving horizontally.
Too fast, hits
outer wall
For a uniform circular motion:
Too slow, hits
inner wall
v2
b
m  e E0
r
r

v
e
E0 b
m
Dipoles in Electric Fields
Uniform E:
F  q E   q  E  0
Total force:
Torque about center of dipole:
τ
d
 d
pˆ   q E      pˆ   q E   d q pˆ  E
2
 2
τ  pE
Work done by E to rotate dipole :
W 
f
i
 qE sin   qE sin  
Potential energy of
dipole in E (i = /2)
p  q d pˆ = dipole moment

f
W   F  tˆ r d
i
t // tangent
f
d
d   p E  sin  d  p E  cos  f  cos i 
i
2
U  W   p E cos f   p  E
( U = 0 for p  E )
Non-uniform field:
Total force:
F  q E   q  E  0
Example: dipole-dipole interaction
| F  | > | F+ |
Force on  end of B is stronger;
hence net force is toward A
c.f. Van der Waals interaction,
long range part.
Application: Microwave Cooking & Liquid Crystals
Microwave oven:
GHz EM field vibrates (dipolar) H2O molecules in food
 heats up.
Liquid Crystal Display (LCD)
dipolar molecules aligned
but positions irregular
Exploded view of a TN (Twisted Nematic) liquid crystal cell showing the
states in an OFF state (left), and an ON state with voltage applied (right)
Conductors, Insulators, & Dielectrics
Bulk matter consists of point charges: e & p.
Conductors: charges free to move (  electric currents ),
e.g., e (metal), ion ( electrolytes ), e+ion (plasma).
Insulators: charges are bounded.
Dielectrics: insulators with intrinsic / induced dipoles.
internal field from dipoles
Induced dipole
Alignment of intrinsic dipoles.