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

qq

/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

qq

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