the parallel-plate capacitor charged with q, then: energy density
TRANSCRIPT
The parallel-plate capacitor
0
qC A d
V
charged with q, then: AqE 0
q
q
C
qU
2
2
A
dq
0
2
2
AdA
q2
002
1
ΩE 2
02
1
energy density
202
1E
Ω
Uu
0V Ed qd A
0'' e e
qC A d C
V
qC
V
0eq A 0 eqd A
Dielectrics and Gauss’ Law
0 dE A q
A dielectric slab is inserted,
)()( d 000 AqAqEqqAE
q ’ is induced surface charge.
A
q
A
q
A
q
A
qEE
eee 0000
0 ,
)1
1(
e
/d0 eqAE
d0 qAEe
kkee instead of instead of
. . The charge The charge qq contained within the Gauss surface contained within the Gauss surface is taken to be the is taken to be the free chargefree charge only. only.
0 0E A
0 0( )E q A
Gauss’ law should be amended as:
Dielectrics and Gauss’ Law
0 ep E
0 + D E p Electric displacement vectorElectric displacement vector
0 0+ eE E
0(1+ ) e E
0 E
'0 qqAdE
')'( 00 qqAdEE
qAdE
00
''0 qAdE
Adp
Adp
qAdpE
)( 00 qAdD
0D
E
0E
e
E
0
Electric polarization vectorElectric polarization vector
Chapter 31 DC Circuits
Circuits
Direct Circuits (DC)
Alternating Circuits (AC)
t
)(tV0)( VtV
ResistorBattery
Current Current
The direction of current is the direction of positive charge would move.
0( ) sin( )V t V t V(t)V(t)
ResistorBattery
Current Current
The direction of current is the direction of positive charge would move.
i
pump for chargepump for charge
A “A “pumppump” for charge, ” for charge, maintains the maintains the constant potential constant potential differencedifference between its between its two terminals. A two terminals. A source of energysource of energy to to raise the energy of raise the energy of electrons.electrons.
Conservation of Charge
Consider the flux of current density through a closed surface,
Therefore the Therefore the law of charge conservation is expressed as: is expressed as:
dVt
AjA d
dd
jjAjdi
d
SAji
d
For a steady current (e.g., the current in DC circuit), 0
d
ddV
t
0d SAj
Conservation of Charge
Consider the flux of current density through a closed surface,
dV
tAj
A d
dd
jj
0d
ddV
t
0d SAj
Consequently, at any junction Consequently, at any junction in an electric circuit,in an electric circuit,
AA11
AA33
AA22
the total the total current enteringcurrent entering the the junction must be equal to the junction must be equal to the total total current leavingcurrent leaving the the junction. junction.
i1
i2
i3
0132 iii 321 iii
ii11
ii22ii33
iiNN
In general,
N
n nii21
This is called junction rule (Kirchhoff’s first law).
A device that maintains a A device that maintains a constant potentialconstant potential differencedifference between two points in the circuit.between two points in the circuit.
Electromotive Force(EMF)
+ -
The EMF of a source is defined as the work on per unit positive charge,
q
W
q
Wε
d
d
Joule/Coulomb = Volt
Does this by moving charges from Does this by moving charges from low to high potential low to high potential by doing work.by doing work.
The EMF is a device transferring from variety of energy to electric energy.
Battery : Uses Uses chemical chemical energyenergy to do work to do work
Generator : Uses Uses mechanical energymechanical energy to do to do workwork
Solar Cell : Uses Uses lightlight to to do workdo work
Analysis of Circuits i
+
VHigh
VLow
-
Method of Method of potential potential differences differences :: differences differences in potential across each in potential across each circuit element. circuit element.
•Guess a direction for the current first.
• Passing through a Passing through a resistorresistor in direction of current in direction of current flow, from a high potential to a low potential givesflow, from a high potential to a low potential gives
V = Vfinal - Vinitial = Vlow - Vhigh = - iR (<0)
• Passing through Passing through batterybattery from “- ” to “+” , the potential from “- ” to “+” , the potential
increases soincreases so V = Vhigh - Vlow= (>0)
Making a complete loop gives total V =0 !!!
0nn iR
This is called This is called loop ruleloop rule (K (Kiechhoff’s second law)iechhoff’s second law)
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
+-
R
Real batteries have Real batteries have internalinternal resistance resistance. . So So what does our circuit really look like?what does our circuit really look like?
r
0 iRir
0)( Rri
irVTerm
)( Rri
Multi-loop DC Circuits
Examples:
It can be reduced to a simple one-loop circuit.
Example
• Find i1, i2, i3. i1
i3i2The junction rule at The junction rule at junction B gives:junction B gives:
321 iii
The loop rule leads to: 0223111 RiRi 0332322 RiRi
–There is another loop (around outside) but it gives no
new information, just the sum of the equations above:
0332111 RiRi
Example
• Find i1, i2, i3. i1
i3i2The junction rule at The junction rule at junction B gives:junction B gives:
321 iii
The loop rule leads to: 0223111 RiRi 0332322 RiRi
321 iii 0223111 RiRi
0332322 RiRi 321 ,, iii
Example
• Find i1, i2, i3. i1
i3i2The junction rule at The junction rule at junction B gives:junction B gives:
321 iii
The loop rule leads to: 0223111 RiRi 0332322 RiRi
321 iii 0223111 RiRi
0332322 RiRi
A763.076581 i
A132.076102 i
A895.076683 i
Note that i1 and i3 turned out negative!This means those two currents are flowing opposite to the directions assumed
Example
• Find i1, i2, i3. i1
i3i2The junction rule at The junction rule at junction B gives:junction B gives:
321 iii
321 iii 0223111 RiRi
0332322 RiRi
0 321 iii
312211 RiRi
233322 RiRi
Matrix method
212211 RiRi
32
31
3
2
1
32
21
0
0
0
111
i
i
i
RR
RRRI=VRI=VI=VRI=VR-1-1
Electric Fields in Circuits
Where does the electric field in wires come from?
, jEEj
In a conductor,
A tiny amounts of charge on the surface of wires provide the electric field
Energy Transfers in an Electric Circuit
i
As the battery moves a quantity of charge dq from its negative terminal to its positive terminal, it does work
qW dd
The power delivered by the source of EMF (battery) is then:
tqtWPEMF dddd iPEMF
The potential difference between two terminals of the resistor is,
iRVVV rightR
leftRR
As a dq moves through the resistor, it experiences a potential energy change: qiRVqU R ddd
This energy must be transferred to the resistor, known as Joule Heating.
The power transferred to the resistor reads:
RitqiRtUPR2dddd RV /)( 2
Ri2
i
irVBattary
The charge dq passing through the battery gains potential energy :
)(ddd irqVqU Battary
The power delivered by this battery is:
riitUPBattery2dd
In a real battery with internal resistance r, the potential difference between the terminals is,
RC Circuits
Combine Resistor and Capacitor in Series
C
a
Rb
Switch at position (a) VC(t)=?
C
tqV
)(C
RtiVR )(
0- C
qiR-
RC
q
t
q
R
ε
d
d
qC
q
RC
t
dd
t
q
RC
qCε
d
d
qt
qC
q
RC
t00
dd
)ln()ln( qCCRC
t
Cqe RC
t
/1
)1()( / RCteCtq RCte
Rt
qti /
d
d)(
)1()(
)( / RCtc e
C
tqtV RC
)1()( / teCtq /
d
d)( te
Rt
qti )1(
)()( / t
c eC
tqtV
ln( ) ln( )t
C q CRC
ln C q
C
ln(1 ) q
C
At t = 0, q(0) = 0
At t=, q() = 0.63C
At t = , q() = C
At t = 0, i(0) = /R
At t=, i() = 0. 37RAt t = , i() = 0
At t = 0, Vc(0) = 0
At t=, Vc() = 0.63
At t = , Vc() =
Then switch turn to b, C discharges:
0 iRC
q
RCt
RCt
eR
ti
eqtq
/
/0
)(
)(
C
a
Rbt
qi
d
d
0 Rdt
dq
C
q dt dq
RC q
q
q
t
q
dq
RC
dt00
0lnln qqRC
t
/
/0
)(
)(
t
t
eR
ti
eqtq
RC
Time to charge capacitorTime to charge capacitor
a
Rb
CC
12
1F
How long time dose voltage
of C to /2 ?
RC)1()( / teCtq
Voltage on capacitor: )1()()( tC eCtqtV
)1(5.0 / te 21/ te
Example
)5.0ln( t 69.0sRCt 32.869.069.0 t
V
0
2/
s32.8
R
CC
a
b 12
1F
How long time dose voltage
of C reduce to /2 ?
RC/( ) tq t C e
Voltage on capacitor: ( ) tCV t q C e
/5.0 te 21/ te
Example
)5.0ln( t 69.0st 32.869.0)F1)(12(
t
V
0
2/
s32.8
How long time dose voltage
of C reduce to /2 ?'
0q q
RiC C
0' qqq R
CC
a
b 12
1F
00
C
qqRi
C
q02 0 q
dt
dqCRq
)1(2
20 CR
t
eq
q
Example
)1(2
)( CRtC e
C
qtV
t
V
0
2/)1(22
CRte t
C
qU
t
20
0 2
1
Energy?
R
CC
a
b 12
1F
Example
C
q
C
q
C
qU
t
20
20
20
4
1)2/(
2
1)2/(
2
1
0
ttUU
0
2RdtiU R
qiRVqU RR ddd
dtdt
qiRU R
dd )1(
2
20 CR
t
eq
q
CR
t
eCR
qi
20
C
qdte
C
q
RU CR
t
R
20
0
420
4
1)(
1
Example
V100
?R
0 iRC
q
VVst C 06.1 ,10
F1000 R
dt
dq
C
q
RCteqq /0
RCtC e
C
qV /0 06.1100 /10
10
RC
teV
0106.0ln/10 RC 4102.2R
RCtC eV /
Example
RCteR
ti /)(
ExercisesP719~722 11, 13, 25, 47 ProblemsP724 15
Example
A
d
d
AC 0
d
A
dAq
q
dE
dE
q
V
qC
0
0 )/(
22
2/' 0
d
AC
'
1
'
11
CCC
d
AC 0
Example
A
d
d
AC 0
d
A
dAq
qEd
q
V
qC
e
e
0
0 )/(
Example
A
d
d
AC 0
2/)/(2/)/(
2/'2/
00 dAqdAq
qdEEd
q
V
qC
e
edd
A
/
2 0
Example
A
d
d
AC 0
dAq
V
qqC
)/2(
''
0
qq e'
dAq
qqC e
)/2( 0
d
AC e
2
)1( 0
Example
A
d
d
AC
2' 0
''' CCC
d
AC e
2
)1( 0
d
AC e
2'' 0