Transcript Document

Chapter 19
Electric Charge and
Electric Field
▲ When a comb has been passed through hair, it
can attract paper scraps. Why does this happen?
▲ Why atoms and molecules can be hold together
to form liquids and solids?
▲ What really happens in an electric circuit?
▲ How do electric motors and
generators works?
▲ And what is light?
2
Electromagnetism
J. C. Maxwell
1831-1879
☆Electromagnetic Field
☆Gauss’s law
☆Ampère’s law
☆Faraday’s law
☆Maxwell’s equations
Then God said:











  E  0 /  0
B
 E  
t
B  0
  B  0 J   0 0
E
t
and there was light.
3
Electric charge
1) Two types: positive & negative
Unlike charges attract; like charges repel.
2) Quantized: elementary charge e  1.6 1019 C
3) Law of conservation of electric charge:
The net amount of electric charge produced in any
process is zero.
▲ Insulators, conductors, semiconductors
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Coulomb’s law (1)
Electric forces between two point charges:
Q1Q2
1 Q1Q2
F k 2 r
r
2
r
4 0 r
where r is unit vector
Q1
r
F
Q2
r
in SI units: k  8.988 109  9.0 109 N  m2 / C 2
0 
1
4 k
 8.85 1012 C 2 / N  m 2
 0 : permittivity of free space
5
Coulomb’s law (2)
Q1Q2
1 Q1Q2
F k 2 r
r
2
r
4 0 r
1) It describes force at rest, or electrostatics
2) It is valid for two point charges
3) Principle of superposition:
If several charges are present, the net force on
any one of them will be the vector sum of forces
due to each of the others.
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Electric field
Charges interact each other by electric field
Charges
create
electric field
interact on
Q1
Q2
Field is a special form of matter
How to describe?
by using a test charge
Fb
Electric field (strength):
.
b
E  F / q or E  lim F / q
q 0
Fc
.c
.a
Fa
Q
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Properties of electric field
E  F /q
F  qE
1) The field doesn't depend on the test charge q
2) It is a vector field, magnitude & direction
1
Q
r
3) For a point charge E 
2
4 0 r
4) For more charges, total field
E   Ei
i
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Electric equilibrium
Example1:Two charges, -Q and -3Q, are a distance
l apart. How can we place a third charge nearby to
reach an equilibrium?
Q
x
Solution: Position?
Q
Q
1
3Q

0  x
2
2
4 0 x
4 0 (l  x)
1
1
l
3Q
x
3 1
l  0.366Q
2
What is the Charge?
Q1
1 3Q
63 3



0
 Q1 
Q  0.402Q
4 0 x 2 4 0 l 2
2
1
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Continuous charge distribution
The electric field can be calculated by integral
① Divide it into infinitesimal charges dQ
1
dQ
r
② Contribution from dQ: dE 
2
4 0 r
③ Consider all the components dEx , dE y , dEz
④ Finish the integration:
Ex   dEx , E y   dE y , Ez   dEz
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A line of charge
Example2: Charge is distributed uniformly along a
line. Find the electric field at any given point P
around it. (Charge per unit length is λ)
Solution: ① x-y axes
dE
y
dEy
② dQ   dx
 dx
③ dE 
4 0 r 2
④ dE  dE  cos 
x
dE y  dE  sin 
P
dEx
a
r

 o
x
dx
dq
x
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⑤ Ex   dx cos 
 4 0 r 2



cos θdθ


4 0 a
2
dE
dEx  dE  cos 
y
dEy
dE y  dE  sin 
 dx
dE 
4 0 r 2
P
dEx
1


(sin θ2  sin θ1 )
4 0 a
 dx
Ey  
sin 
2
4 0 r



sin θdθ


4 0 a
2
a
r

1
 o
x
dx
dq
2
x
r=a/sin , x=- a·cot ,
dx=ad /sin2
1


(cos θ1  cos θ2 )
4 0 a
E  Ex i  E y j
12
y

Ex 
(sin θ2  sin θ1 )
4 0 a

Ey 
(cos θ1  cos θ2 )
4 0 a
Discussion:
P
a
2
1

o
x
1) If it is very long or infinite, 1 =0, 2 =
Ex  0,

Ey 
2 0 a
→ useful result
2) For surface distribution:
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A plane of charge
Question: Charge is distributed uniformly on an
infinite plane. Find the electric field at any given
point P around it. (Charge per unit area is σ)
E  Ey  

dx

2 0 r


2 0



2 0
  d
2

2
= ·dx
cos 
1
dE
E
dx
y

dE
P. r
a 
o
x
dx
x
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Uniform charged ring
Example3: A thin ring of radius R holds a total
charge Q distributed uniformly. Find the electric
field at point P on the axis, x from its center.
dQ
Solution: dE 
2
4 0 r
dQ
E  Ex  
cos 
2
Ring 4 0 r
y
dQ
R
o
r

P
x
x
dE
Q cos 
Q x


2
4 0 r
4 0 ( x 2  R 2 )3/ 2
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Q cos 
Q x
E

2
4 0 r
4 0 ( x 2  R 2 )3/ 2
R
o
P
x
E
Discussion:
.
1) x=0 or x >> R, E=?
o
2) At what position along the axis, E= Emax ?
3) If there is a small gap in the circle, E o = ?
4) If there is only a semi-circle, E o = ?

2 0 R
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Uniform charged disk
Example5: Charge is distributed uniformly over a
thin disk of radius R. Determine E at point P on the
axis, x from its center. (Charge per unit area is σ)
Solution: dQ   2 rdr
xdQ
dE 
4 0 ( x 2  r 2 )3 2
E
R
o
R
o
P
r
x
x
dr

  x  2 rdr

[1 
2
2 3 2
2 0
4 0 ( x  r )
x
x2  R2
]
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
E
[1 
2 0
R
x
x R
2
2
]
o
P
x
x
Discussion:
When R →∞:
Parallel-plate
capacitor

E
2 0
+
infinite plane
E=0

E
0
-
E=0
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Field lines
Visualize the electric field → electric field lines
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Properties of field lines
1) The field point in the direction tangent to the
field line at any point.
2) The magnitude of E is proportional to the
number of lines crossing unit area ⊥ the lines.
3) Field lines start on + charges,
end on – charges.
4) Field lines never cross each other;
and there are no closed field lines.
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Electric dipoles (1)
Combination of two equal charges of opposite
sign is called an electric dipole.
(Q  Q)
Dipole moment: p  Ql
Ea  2E cos 


Q
l2
2
4 0 ( y  )
4
2
l
4 0 ( y  )3/2
4
2
l

p
Ea  E  E
l
y 
4
2
E
p
4 0 y 3
(y
.a
E
2

E
l)
Q
y
l

Q
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Electric dipoles (2)
Eb 
Q
l
4 0 ( x  ) 2
2
2p

4 0 x 3
o
Q
Eb  E  E

(x
Q
l
4 0 ( x  ) 2
2
l)
E b
.
x
Q
E
l.

Ea 
E
2 px
l2 2
4 0 ( x  )
4
2
p
4 0 y
3
(y
l)
Discussion:
In which cases: E  r 3 / r 2 / r 1 / r 0 ?
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Dipole in external field
E
Q
1) The total force:
F  F  F  0
2) Torque on the dipole:
l

F
1
1
  F  l sin θ  F  l sin θ
2
2
o.
θ

F

p
Q
Uniform field
 QE  l sin θ  pE sin θ or   p  E
Discussion: what is the torque on the dipole when
  0,  / 2,  ?
23
Vibrating charge
Thinking: Negative charge -Q is distributed on a
ring uniformly. A positive charge q is placed from
the center of ring a small distance x. Show that it
will undergo SHM when released, and what is T ?
Q  x
E
4 0 ( x 2  R 2 )3/ 2
R
x
o q, m
x
-Q
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