Transcript Example

3. Polarization (极化强度矢量）
• Definition
In the volume of V ,
p

P
V
P
m
p0
Unit: C/m2
Uniform,
Nonuniform
• Relationship between P and s’
For uniform dielectrics
The induced charge distributes only on the surface of
dielectrics.
For displacement polarization
Consider the positive induced charge
through area dA due to polarization.
p

P
m
V
 nql
n, the number of molecular per unit
volume
dN  ndV  nldA cos 
dq '  qdN  nqldA cos 
 PdA cos 
 P  dA
 P  dA   q '
dq '  P  dA  P cos   dA
dq '
s'
 P cos   P  n  Pn
dA



2

2
, s 'e  0, 
, s 'e  0, 
4. Depolarization Field (退极化场）
E  E0  E '
Depolarization field E’
E ', E0 in the same direction.
At some place,
At another place, E ', E0 in the opposite direction
Example 1
A spherical dielectrics, uniform polarization P
se’=Pcos
Calculating the depolarization field E’ at the center.
s e '  Pn  P cos 
dq '
s e ' dA P cos  dA
dE ' 


2
2
4 0 R
4 0 R
4 0 R 2
dA  Rd  R sin  d
 R 2 sin  d d
dE ' 
P
4 0
cos  sin  d d
dE ' z  dE ' cos(   )  dE ' cos 

P
4 0
E 'z 
cos2  sin  d d
 dE ' z  
P
4 0


0
2
cos  sin  d  d  
2
0
P
3 0
Example 2
• Parallel plate:
s e '  P cos   P
se '
E' 
0
5. Polarization law of dielectrics (电介质的极化规律)
P  se '  E '  E
P( E ) function
For different materials, P ( E ) different and complicated,
which is determined by an experiment result.
• For general isotropic materials (各向同性)
P   e 0 E
e: Polarization coefficient
e=1+ e
(极化率)
6. Electric Displacement Vector D (电位移矢量)
and Gauss’ Law with Dielectrics.
In dielectrics: E  0
E0  P  s 'e  E '  E  E0  E '
very complicated
Introduce a new physical quantity:
D
Electric Displacement Vector (电位移矢量)
Or Electric Induction Vector (电感应矢量)
Electric Displacement Vector D
 0  E  dA   ( q0  q ')
Ins
 P  dA   q '
Ins
  E  dA   q   P  dA
0
0
Ins
 ( E  P)  dA   q
0
0
Ins
D  0E  P
D

dA

q

0

ins
 D  dA   q
0
Ins
 D  0
 E   / 0
Electric Displacement Vector D
D  0E  P
e
Dielectric constant(介电常数）
  0 E   e 0 E
e
Polarization Coefficient (极化率)
 (1   e ) 0 E
  e 0 E   E
   e 0
 e  1  e
 D  dA   q
0
ins
 D  0
Example
• Example 1
 D  dA   q
0
D1A  D2 A  s e 0A
E 1  0, D1   e1 0 E1  0, D1  0
 D  D2  s e 0   0 E0
E 
D   0 E  P   e 0 E
D
 e 0

 0 E0 E0

 e 0  e
In conductor
Example 2
D   0 E  P   e 0 E
 D  dA   q
0
4 r 2 D  q0
q0
D
4 r 2
D
E

 e 0
q0
4 0 e r
2

E0
e
The parallel-plate capacitor
charged with q, then:
 0 A V  Ed
qd  eqd
 0 A 0 A
E  qq 0eA
q
C
 0 A d
V
q
q
q
C
V
q
C'
  e 0 A d   e C
V '
q 2 q 2d
U

2C 2 0 A
2
1
1  q 
 Ad   0 E 2 Ω
  0 
2
2  0 A 
U 1
2
u   0E
Ω 2
energy density
Dielectrics and Gauss’ Law
 0 E0 A   0  E  dA  q
E0  q ( 0 A)
A dielectric
slab is be
Gauss’
law should
inserted, as:
amended
 
  

 0  E  dA  q  q  E

E
q

E
d
(
A


d

A
A
q
)


/

q
q
( 0 A)
00
e
0
e

q ’ is kinduced
surface
 instead
of  charge. .
e 0
0
1
E0 q qcontainedqwithin qthe Gauss
q
The charge
surface
q  q(  1)
E 

, 


is taken
the free
e
 e to
 ebe
 e 0 Acharge
 0 A only.
0 A
0A
Dielectrics and Gauss’ Law
 
 
 0  E  dA  q  q' p  dA

 
 0  ( E0  E ' )  dA  q  q'
 
 0  E0  dA  q
 
 
 0  E 'dA  q'    p  dA
p   e 0 E
Electric polarization vector
  
 ( 0 E0  p)  dA  q
D   0 E +p
 
 D  dA  q
Electric displacement vector
  0 E + e 0 E  (1+ e ) 0 E   0 E

E

D
0


E0



E0
e
Chapter 31
DC Circuits
Current
Circuit
s
V(t)
Resistor
Battery
The direction of current is
defined by the moving direction
of positive charge.
Current
Direct Circuits (DC)
V (t )
V (t )  V0
t
V (t )  V0 sin(t   )
Alternating Circuits (AC)
Current
Resistor
Battery
The direction of current is
defined by the moving direction
of positive charge.
Current
i
pump for charge
A “pump” for charge,
maintains the
constant potential
difference between its
two terminals. A
source of energy to
raise the energy of
electrons.
Conservation of Charge
Consider the flux of current
density through a closed surface,
 
i di
  j ddAA

j
j
S
Therefore the law of charge
conservation is expressed as:
For a steady current (e.g.,
the current in DC circuit),
 
d
A j  dA   dt  dV
d
dV  0

dt 

S
 
j  dA  0
Conservation of Charge
Consider the flux of current
density through a closed surface,
Consequently, at any junction j
in an electric circuit,
A2 i2
A1
i1
i3
A3
the total current entering the
junction must be equal to the
total current leaving the
junction.
ii21 i3 i2 i1i03
j
 
d
A j  dA   dt  dV
d
dV  0

dt 

S
 
j  dA  0
In general,
i1  n2 in
N
i1
i2
i3
iN
This is called junction rule
(Kirchhoff’s first law).
Electromotive Force(EMF)
A device that maintains a constant potential difference
between two points in the circuit.
Does this by moving charges from
low to high potential by doing work.
The
The EMF
EMF is
 aofdevice
a source is
transferring
from
variety
of unit
defined as the
work
on per
energy
electric energy.
positivetocharge,
+
-
W
dW
Battery
: Uses
chemical

ε

energy q
to do work
dq
Generator
Solar Cell
: Uses
: Uses
mechanical
light to
energy
do work
to do work = Volt
Joule/Coulomb
Analysis of Circuits
i
VHigh

+
-
VLow
Method of potential
This
is called: loop
rule
differences
differences
(Kiechhoff’s
law)
in potential second
across each
circuit element.
   iR
n
n
0
•Guess a direction for
Making a complete loop
the current first.
gives total V =0 !!!
• Passing through a resistor in direction of current
flow, from a high potential to a low potential gives
V = Vfinal - Vinitial = Vlow - Vhigh = - iR (<0)
• Passing through battery from “- ” to “+” , the potential
increases so
V = Vhigh - Vlow=  (>0)
It can be used the following two rules:
1. Junction Rule
At any junction in an electric circuit, the total
current entering the junction must be the same
as the total current leaving the junction.
2. Loop Rule
The algebraic sum of all differences in potential
around a complete circuit loop must be zero.
What is “±” for the differences potential of resistor?
What is “±” for the differences potential of EMF?
Analysis of Circuits
i

+
VTerm    ir
r
R
-
Real batteries have internal resistance.
So what does our circuit really look like?
  ir  iR  0
  i (r  R)  0
i

(r  R)
Multi-loop DC Circuits
Examples:
It can be reduced to a simple one-loop circuit.
Example
• Find i1, i2, i3.
The junction rule at
junction B gives:
i1
i2
i3
i1  i2  i3
The loop rule leads to:
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
–There is another loop (around outside) but it gives no
new information, just the sum of the equations above:
 i1R1  1   2  i3 R3  0
Example
• Find i1, i2, i3.
i1
The junction rule at
junction B gives:
i2
i3
i1  i2  i3
The loop rule leads to:
i1  i2  i3
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
i1 , i2 , i3
Example
• Find i1, i2, i3.
i1
The junction rule at
junction B gives:
i2
i3
i1  i2  i3
The loop rule leads to:
i1  i2  i3
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
i1   58 76  0.763 A
i2  10 76  0.132 A
i3   68 76  0.895 A
This
Notemeans
that i1 and
those
i3 two
turned
currents
out negative!
are flowing
opposite to the directions assumed
Example
• Find i1, i2, i3.
i1
The junction rule at
junction B gives:
i2
i3
i1 method
i2  i3
Matrix
i1  i2  i3
 i1R1  1   3  i2 R2  0
i2 R2   3   2  i3 R3  0
I=VR-1 RI=V

i1  i2  i3  0
i1iR
R22
11 
23
1R
1 1i2i2R
i2 R2  i3 R3  3   2
 1  1  1   i1   0 
R R
 i      
0
2
 1
 2   1 3 
 0 R2  R3  i3   2   3 
Electric Fields in Circuits
 


j  sE, E  j
In a conductor,
Where does the electric field in wires come from?
A tiny amounts of charge on the surface of wires
provide the electric field
Energy Transfers in an Electric Circuit
The
As potential
the battery
difference
moves a
2
i
R
quantity of
between
two
charge
terminals
dq from
of the
its
i
negativeis,terminal toleftits positive
resistor
right

V

V

V
 iR
R work
R
R
terminal, it does
i
As a dq
moves
through
the
dW  dq
resistor, it experiences a potential
energy change: dU  dq  VR  iRdq
TheThis
power
delivered
by transferred
the source of
(battery)
energy
must be
to EMF
the resistor,
is
then: as Joule Heating.
known
P
 i
The power
reads:
PEMF transferred
dW dt   dto
q dthe
t resistor
EMF
PR  dU dt  iR dq dt  i R  (V ) / R
2
2
In a real battery with internal
resistance r, the potential
difference between the terminals is,
VBattary    ir
The charge dq passing through the battery gains
potential energy :
dU  dq  VBattary  dq(  ir )
The power delivered by this battery is:
PBattery  dU dt  i  i r
2
RC Circuits
Combine Resistor and Capacitor in Series
At tt == 0,
0, i(0)
q(0)
V
(0)
00
At
===
/R
a
c
Switch at position (a)
At
At t=
t=,, VV
i(q(

()))==
=0.0.63C
0.63
37/R
c
(t)=?
R
C
b

At
At tt ==t ,
,
q()
==q0C
q d
dt Vi()
c()
C
0 RC  0 C  q
q(t )
t
C qq
ε dqq dqq
Cε
VC 

ln(
))Cln(
ln(
C)qln(
 Cq )
ln
 
ln(1
C
RC
RC
CC
R RCdt RC
dt
t

VR  i(t ) R
dt
dq
e RC  1 q / C

q

t /tRC
RC
C


q
q((tt)) CC((11ee / ))
 - iR-  0
C
  RC
qq((tt))
ddqq  t/tRC
t/ tRC
/
/
V
(
(
t
t
)
)






(
1
(
1


e
e
)
i (t ))   ee
cc
CC
ddtt RR
Then switch turn to b, C discharges:
dq
a
q
i
 iR  0
dt
C
R
b

t dt
dt dqq dq
q dq
 

R  0  0
C C dt
RC qq0 q
RC
  RC
/
 t/tRC
qq((tt))qq0e0 e
 t/tRC
/
i((tt))  ee
RR
t

 ln q  ln q0
RC
Time to charge capacitor
Example
a

How long time dose voltage
b 12
1F

R
of C to /2 ?
C
q(t )  C (1  e t / )   RC
Voltage on capacitor: VC (t )  q(t ) C   (1  e
V
0.5   (1  e

/2
0 8.32 s
t
t /
) e
t /
t 
)
1 2
 t   ln( 0.5) 0.69
t  0.69  0.69 RC  8.32 s
Example
a

How long time dose voltage
b 12
1F

R
of C reduce to /2 ?
C
q(t )  C e
  RC
VC (t )  q C   e
Voltage on capacitor:
V
0.5  e

t /
e
t /
t 
1 2
 t   ln( 0.5)  0.69
t  (12)(1F)0.69  8.32 s
/2
0 8.32 s
t /
t
Example
a

b
How long time dose voltage
12
1F
R

of C reduce to /2 ?
C q  Ri  q '  0 q  q'  q0
C
C
q0  q
q
dq
 Ri 
 0 2q  CR  q0  0
C
C
dt
2t
V

q0 q CR t CR
V
qC(t ) (1 e (1) e )

2 C 2
 
/2
t 
 (1  e t CR )
2
2
0
t
Example
a

b
U t 0
Energy?
12
1F

2t
CR
q0
q  (1  e )
2
2t

q0 CR
i
e
CR
R
U t 
C
1 q02

2C
1 (q0 / 2) 2 1 (q0 / 2) 2 1 q02



2 C
2 C
4C
U t   U t 0
dU R  dq  VR  iRdq
dq
dU R  iR dt
dt

U R   i 2 Rdt
0

1 q0 2
UR  ( )  e
R C 0

4t
CR
2
0
1q
dt 
4C
Example
100V
t  10s, VC  1.06V
R?
100 F
q
 iR  0
C
q0 t t/ /RC
RC V
V

VCC  C
ee
 10 / RC  ln 0.0106
q
dq

R0
C
dt
q  q0 e  t / RC
t 10
 100e 10 / RC  1.06
R  2.2 10 4 
Exercises
P719~722 11, 13, 25, 47
Problems
P724
15
Example
C
A
C' 
d
0 A
d
0 A
d /2
q
q
C

d
d
V
E E
2
2
0 A
q


( q /  0 A) d
d
1
1
1


C
C' C'
C
0 A
d
Example
C
A
d
0 A
d
q
q
C

V
Ed
 0 e A
q


( q /  e 0 A) d
d
Example
C
0 A
d
q
q
C

V
Ed / 2  E ' d / 2
q

(q /  0 A) d / 2  (q /  e 0 A)d / 2
A
d
2 0 A

d  d /e
Example
C
0 A
d
q  q'
q  q'
C

V
( 2q /  0 A) d
A
q'   e q
d
q   eq
C
( 2q /  0 A)d
(1   e ) 0 A
C
2d
Example
C' 
0 A
C'' 
2d
C  C 'C ' '
A
(1   e ) 0 A
C
2d
d
 0 e A
2d