Part 4. Electromagnetism
Applications:
Bio-EM:
• Computer microchips
• Heartbeat pacing
4 fundamental laws:
• Cell phones
• Nerve impulses
• Motors & generators
• Osmosis thru cell membranes
Maxwell’s eqs.
Electromagnetism
20. Electric Charge, Force, & Fields
21. Gauss’s Law
22. Electric Potential
23. Electrostatic Energy & Capacitors
24. Electric Current
25. Electric Circuits
26. Magnetism: Force & Field
27. Electromagnetic Induction
28. Alternating-Current Circuits
29. Maxwell’s Equations & EM Waves
20. Electric Charge, Force, & Field
1.
2.
3.
4.
5.
Electric Charge
Coulomb’s Law
The Electric Field
Fields of Charge Distributions
Matter in Electric Fields
What holds your body together?
What keeps a skyscraper standing?
What keeps a car on the road as it turns?
What governs the electronics in computers?
What provides the tension in a climbing rope?
What enables the photosynthesis of plants?
Ans: electric forces.
All macroscopic phenomena are governed by gravity & EM forces.
Notable effective forces of electrical origin:
• Tension
• Normal Forces
• Compression
• Friction
• Most forces in chemistry & biology
What’s the fundamental criterion for initiating a lightning strike?
Ans. E > 3 MV/m
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.
20.2. Coulomb’s Law
“Insulators” can be charged by rubbing.
Examples:
Rubbed balloon sticks to clothing.
2 rubbed balloons repel each other.
Socks from dryer cling to clothings.
Bits of styrofoam cling to hand.
Walk across carpet & feel shock touching doorknob.
Ground or low energy state of matter tends to be charge neutral.
Triboelectric Series
Most positively charged
+
Air
Human skin
Leather
Rabbit's fur
Glass
Quartz
Mica
Human hair
Nylon
Wool
Lead
Cat's fur
Silk
Aluminium
Paper (Small positive charge)
Cotton (No charge)
0
Steel (No charge)
Wood (Small negative charge)
Lucite
Amber
Sealing wax
Acrylic
Polystyrene
Rubber balloon
Resins
Hard rubber
Nickel, Copper
Sulfur
Brass, Silver
Gold, Platinum
Acetate, Rayon
Synthetic rubber
Polyester
Styrene (Styrofoam)
Orlon
Plastic wrap
Polyurethane
Polyethylene (like Scotch tape)
Polypropylene
Vinyl (PVC)
Silicon
Teflon
Silicone rubber
Ebonite

Most negatively charged
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
GOT IT? 20.1.
Charge q1 is at x = 1 m, y = 0.
What is the unit vector r in F12 if q2 is located at
(a) the origin,
(b) x = 0, y = 1 m ?
Explain why you can answer without knowing the sign of either charge.
r12  r2  r1
(a)
r12   0  1, 0  0     1, 0
(b)
r12   0  1, 1  0     1, 1 

rˆ    1, 0   i
1
  1, 1 
2
1
1

i
j
2
2
rˆ 
Example 20.1. Force Between Two Charges
A 1.0 C charge is at x = 1.0 cm, & a 1.5 C charge is at x = 3.0 cm.
What force does the positive charge exert on the negative one?
How would the force change if the distance between the charges tripled?
kq q
F12  12 2 rˆ
r
9.0 10


9
Nm2 / C 2  1.0 106 C   1.5 106 C 
Distance tripled  force drops by
 0.020 m 
1/32.
2
F12  
ˆi
34 ˆ
i N  3.8 ˆi N
9
 34 ˆi N
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


Fg G me m p
6.67 1011 N  m2 / kg 2   9.11 1031 kg   1.67 1027 kg 
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
r
k qQ
 2 2  0 , sin  
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
a
2
y

2 3/2
ˆj
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
Example 20.3. Thunderstorm
A charged raindrop carrying 10 C experiences a force of 0.30 N in the +x direction.
What’s the electric field at its location?
What would the force be on a 5.0 C drop at the same point?
F
E
q
0.30 ˆi N

10  106 C
 30 ˆi kN / C


F  q E   5.0 C  30 ˆi kN / C  0.15 ˆi N / C
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)
Example 20.4. Two Protons
Two protons are 3.6 nm apart.
Find the electric field at a point between them,
1.2 nm from one of them.
r̂1
1.2 nm
3.6 nm
P
r̂2
x
2.4 nm
Find the force on an electron at this point.
E  E1  E2 
 
keˆ ke ˆ
i  2 i
2
r1
r2
1 1
 ke  2  2  ˆi
 r1 r2 


1
1
 ˆi
  9.0 10 N m / C 1.6 10 C  

2
2
 1.2 109   2.4 109  


9
2
19
2


F  e E   1.6 1019 C  750 ˆi MN / C  0.12 ˆi  nN 
 750 ˆi  MN / C 
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
k p
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  
dEy  
Line
P
r
r
y
dq
Line
k dq  y 
k  dx  y 

 
 
r 2  r  Line r 2  r 
k y
x
dq
x



y
dx
2
x

2 3/2

k y 
 y 2


x

2
2
y  x  
 1  1 
 k y   2    2   2 k 
y
 y  y 
2k 
E
ρˆ 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
GOT IT? 20.3.
An electron, a proton, a deuteron (1p, 1n), a 3He nucleus (2p, 1n),
a 4He nucleus (2p, 2n), a 13C nucleus (6p, 7n), & an 16O nucleus
(8 p, 8 n) all find themselves in the same electric field.
Rank order their accelerations from lowest to highest assuming
1.p & n have the same mass.
2.The mass of a composite particle is the sum of the masses of
its constituents.
e
q/m
1800
Ans:
2H
p
1/1
13C,
1/2
3He
2/3
4He
2/4
C13
6/13
(16O, 4He, deuteron), 3He, p, e.
16O
8/16
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
d
d   p E
2
f

sin  d
i
U  W   p E cos  f   p  E
t // tangent
 p E  cos  f  cos i 
( 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.