Potencial elétrico e capacitores

Baseado no 8.02T MIT-opencourse

1

!g = !GM

r2r

!Fg = m!g

!E = keq

r2r

!Fe = q !E

Gravidade x eletricidade

Massa M Carga(+/-q)

Campos

Forças

2

Energia potencial x potencial

Gravidade

3

Gravidade: força e trabalho

!Fg = !GMm

r2r

Força exercida em m devido a M

4

Gravidade: força e trabalho

!Fg = !GMm

r2r

Força exercida em m devido a M

Wg =! B

A

!Fg · d!s

Trabalho exercido pela gravidade ao mover m de A a B

integral de trajetória

4

12P04 !

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Wg =! B

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!Fg · d!s

= =! B

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#

Trabalho realizado pela gravidade terrestre

Trabalho realizado pela gravidade ao mover m de A a B

Trabalho depende apenas dos pontos A e B!

5

Forças conservativas

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Forças conservativas:

6

Forças conservativas

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Forças conservativas:

6

Forças conservativas

Mecânica: !EcinWA!B =

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Forças conservativas:

6

!Ug = UB ! UA = !! B

A

!Fg · d!s = !Wg = Wext

Energia potencial x potencial

7

!Ug = UB ! UA = !! B

A

!Fg · d!s = !Wg = Wext

!Fg = !GMm

r2r " Ug = G

Mm

r+ U0

Energia potencial x potencial

7

!Ug = UB ! UA = !! B

A

!Fg · d!s = !Wg = Wext

!Fg = !GMm

r2r " Ug = G

Mm

r+ U0

Energia potencial x potencial

U0: constante que depende do pto de referência

Apenas tem significado físico

!Ug ! !Vg

7

!Ug = UB ! UA = !! B

A

!Fg · d!s = !Wg = Wext

!Fg = !GMm

r2r " Ug = G

Mm

r+ U0

!Vg =!Ug

m= !

! B

A(!Fg/m) · d!s = !

! B

A!g · d!s

Energia potencial x potencial

U0: constante que depende do pto de referência

Apenas tem significado físico

!Ug ! !Vg

Definição da diferença de potencial gravitacional

7

!Ug = UB ! UA = !! B

A

!Fg · d!s = !Wg = Wext

!Fg = !GMm

r2r " Ug = G

Mm

r+ U0

!Vg =!Ug

m= !

! B

A(!Fg/m) · d!s = !

! B

A!g · d!s

Energia potencial x potencial

U0: constante que depende do pto de referência

Apenas tem significado físico

!Ug ! !Vg

Definição da diferença de potencial gravitacional

!Fg ! !gCampoForça

!Ug ! !VgPotencialEnergia

7

Potencial gravitacional

Potencial de planeta +sol

8

Gravidade x eletricidade

!E = keq

r2r

!Fe = q !E

Carga(+/-q)Massa M

!g = !GM

r2r

!Fg = m!g

!Ug = !! B

A

!Fg · d!s

!Vg = !! B

A!g · d!s

Ambas as forças são conservativas, então:

!U = !! B

A

!Fe · d!s

!V = !! B

A

!E · d!s

9

!V = !! B

A

!E · d!s

Potencial e energia

Unidades: Joules/Coulomb

=Volts

10

!V = !! B

A

!E · d!s

Potencial e energia

Unidades: Joules/Coulomb

=Volts

Wext = !U = UB ! UA

= q!VJoules

Trabalho realizado pela gravidade ao mover m de A a B:

10

Potencial

V (!r) = V0 + !V = V0 !! B

A

!E · d!s

Cargas geram potenciais

11

Potencial

V (!r) = V0 + !V = V0 !! B

A

!E · d!s

Cargas geram potenciais

28P04 !

Potential Landscape

Positive Charge

Negative Charge

q positiva

q negativa

11

Potencial

V (!r) = V0 + !V = V0 !! B

A

!E · d!s

Cargas geram potenciais

U(!r) = qV (!r)Cargas sentem potenciais

28P04 !

Potential Landscape

Positive Charge

Negative Charge

q positiva

q negativa

11

26P04 !

Potential Created by Pt Charge

!!"# ˆˆ !drdr "#!

B

B AA

V V V d$ # % # % &' $ "!

!

2

ˆ

rkQ

!$ #!

2 2

ˆB B

A A

drkQ d kQr r

# % & # %' '!

"!

1 1

B A

kQr r

( )# %* +

, -

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r

kQrV #)(ChargePoint

!V = VB ! VA = !! B

A

!E · d!s

= !! B

A

"k

Q

r2r

#· d!s = !

! B

Ak

Q

r2dr

=$k

Q

r

%rB

rA

= kQ

"1rB! 1

rA

#

Potencial criado por uma carga pontual

12

26P04 !

Potential Created by Pt Charge

!!"# ˆˆ !drdr "#!

B

B AA

V V V d$ # % # % &' $ "!

!

2

ˆ

rkQ

!$ #!

2 2

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A A

drkQ d kQr r

# % & # %' '!

"!

1 1

B A

kQr r

( )# %* +

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r

kQrV #)(ChargePoint

!V = VB ! VA = !! B

A

!E · d!s

= !! B

A

"k

Q

r2r

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! B

Ak

Q

r2dr

=$k

Q

r

%rB

rA

= kQ

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rA

#

Vcarga pontual(r) = kQ

r

V (r =!) = 0

Potencial criado por uma carga pontual

12

Potencial: princípio da superposição

Soma direta.Potencial é um

escalar!

Potencial devido a um conjunto de cargas:

13

Potencial: princípio da superposição

Soma direta.Potencial é um

escalar!

Potencial devido a um conjunto de cargas:

Potencial devido a uma distribuição contínua de cargas:

densidadelinear de carga

densidadesuperficial de carga

densidadevolumétrica de carga

13

Calculando E a partir de V

30P04 !

Deriving E from V

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! (

Ex = Rate of change in V

with y and z held constant

!V = !! B

A

!E · d!s

A = (x, y, z), B = (x + !x, y, z)!!s = !xı

14

!V = !! (x+!x,y,z)

(x,y,z)

!E · d!s " !E · !!s = ! !E · (!xı) = !Ex!x

Calculando E a partir de V

30P04 !

Deriving E from V

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B

A

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x

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x x

! (' # ) #

! (

Ex = Rate of change in V

with y and z held constant

!V = !! B

A

!E · d!s

A = (x, y, z), B = (x + !x, y, z)!!s = !xı

14

!V = !! (x+!x,y,z)

(x,y,z)

!E · d!s " !E · !!s = ! !E · (!xı) = !Ex!x

Ex ! "!V

!x# "!V

!x

Calculando E a partir de V

30P04 !

Deriving E from V

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x

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! (

Ex = Rate of change in V

with y and z held constant

!V = !! B

A

!E · d!s

A = (x, y, z), B = (x + !x, y, z)!!s = !xı

14

Calculando E a partir de V

!E = !!

"V

"xi +

"V

"yj +

"V

"zk

"

= !!

"

"xi +

"

"yj +

"

"zk

"V

15

Calculando E a partir de V

!E = !!

"V

"xi +

"V

"yj +

"V

"zk

"

= !!

"

"xi +

"

"yj +

"

"zk

"V

!! =!

"

"xi +

"

"yj +

"

"zk

"

Operadorgradiente

15

Calculando E a partir de V

!E = !!

"V

"xi +

"V

"yj +

"V

"zk

"

= !!

"

"xi +

"

"yj +

"

"zk

"V

!! =!

"

"xi +

"

"yj +

"

"zk

"

!E = !!"VOperadorgradiente

15

V !"#$%&'%'( &'( &( )*'!+!,%( #-&$.%(/*,%'( &0*".( '*/%( 1!$%#+!*"2( '&3( x2( 4!+-( 2(

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7 8V x! ! "!"

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(

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"

:( ;"( +-!'( #&'%2( :rdV E dr% # (;6( !'(

@"*4"2(+-%"(! (/&3(=%(*=+&!"%1(&'(

9 >V r"

(

( ?r

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1!/%"'!*"':(Q%"%$&0!K&+!*"(+*(+-$%%(1!/%"'!*"'(!'('+$&!.-+6*$4&$1:(

2V% #.!"

(

8"339:(

(

R%6%$$!".( +*( D!.<$%( A:B:H2( 0%+( +-%( )*+%"+!&0( &+( &( )*!"+( 9 2 >P x y =%( 9 2 >V x y :(S*4(/<#-( !'(

#-&".%1(&+(&("%!.-=*$!".()*!"+(V 9 2P x dx >y dy/ / T(U%+(+-%(1!66%$%"#%(=%(4$!++%"(&'(

(

( N8

Superfícies equipotenciais

Superfícies de mesma energiaV=constante

•E perpendicular às equipotenciais:

• Nenhum trabalho é necessário para mover uma carga ao longo de uma superfície equipotencial

• Componente tangencial de E é zero ao longo das equipotenciais

!E = !!"V

16

V !"#$%&'%'( &'( &( )*'!+!,%( #-&$.%(/*,%'( &0*".( '*/%( 1!$%#+!*"2( '&3( x2( 4!+-( 2(

+-%"(+-%$%(!'(&("*"5,&"!'-!".(#*/)*"%"+(*6(!

7 8V x! ! "!"

(!"(+-%(*))*'!+%(1!$%#+!*"( 9 :(;"(+-%(

#&'%(*6(.$&,!+32(!6(+-%(.$&,!+&+!*"&0()*+%"+!&0(!"#$%&'%'(4-%"(&(/&''(!'(0!6+%1(&(1!'+&"#%(h2(

+-%(.$&,!+&+!*"&0(6*$#%(/<'+(=%(1*4"4&$1:(

8>xE# $

(

;6(+-%(#-&$.%(1!'+$!=<+!*"()*''%''%'(')-%$!#&0('3//%+$32(+-%"(+-%($%'<0+!".(%0%#+$!#(6!%01(!'(

&( 6<"#+!*"( *6( +-%( $&1!&0( 1!'+&"#%( r2( !:%:2( ?rE%! "

"

:( ;"( +-!'( #&'%2( :rdV E dr% # (;6( !'(

@"*4"2(+-%"(! (/&3(=%(*=+&!"%1(&'(

9 >V r"

(

( ?r

dVE

dr

& '% % #( )

* +! " ?"!"

( 9A:B:C>(

(

D*$(%E&/)0%2(+-%(%0%#+$!#()*+%"+!&0(1<%(+*(&()*!"+(#-&$.%(q(!'( 89 > 7 FV r q r,-% :(G'!".(+-%(

&=*,%(6*$/<0&2(+-%(%0%#+$!#(6!%01(!'('!/)03(H

8?9 F >q r,-%! # "

!"

:((

(

(

$%&%'()"*+,-./(*.+(!01,23/-./,*45(

((

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#-&$&#+%$!K%1( =3( #*"'+&"+ 9 2 >V x y &$%( #&00%1( %L<!)*+%"+!&0( #<$,%':( ME&/)0%'( *6(

%L<!)*+%"+!&0(#<$,%'(&$%(1%)!#+%1(!"(D!.<$%(A:B:N(=%0*4:(

(

((

6,71"-($%&%'(ML<!)*+%"+!&0(#<$,%'(

(

;"( +-$%%( 1!/%"'!*"'( 4%( -&,%( %L<!)*+%"+!&0( '<$6&#%'( &"1( +-%3( &$%( 1%'#$!=%1( =3(

9 2 2 >V x y z O#*"'+&"+:( I!"#%( (4%( #&"( '-*4( +-&+( +-%( 1!$%#+!*"( *6( !!"

!'( &04&3'(

)%$)%"1!#<0&$( +*( +-%( %L<!)*+%"+!&0( +-$*<.-( +-%( )*!"+:( P%0*4( 4%( .!,%( &( )$**6( !"( +4*(

1!/%"'!*"':(Q%"%$&0!K&+!*"(+*(+-$%%(1!/%"'!*"'(!'('+$&!.-+6*$4&$1:(

2V% #.!"

(

8"339:(

(

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(

( N8

Superfícies equipotenciais

Superfícies de mesma energiaV=constante

•E perpendicular às equipotenciais:

• Nenhum trabalho é necessário para mover uma carga ao longo de uma superfície equipotencial

• Componente tangencial de E é zero ao longo das equipotenciais

!E = !!"V

Gravidade: mapa topográfico mostra superfícies equipotenciais :Vg=gz

!"#$%&'%#&()#*$'+$#,-)%'(#.()/0$*-&+/1#*$1/.$2#$*-33/&)4#5$/*$+'00'6*7$

$

8)9 !"#$ #0#1(&)1$ +)#05$ 0).#*$ /&#$ %#&%#.5)1-0/&$ ('$ ("#$ #,-)%'(#.()/0*$ /.5$ %').($ +&'3$

"):"#&$('$0'6#&$%'(#.()/0*;$

$

8))9 <=$*=33#(&=>$("#$#,-)%'(#.()/0$*-&+/1#*$%&'5-1#5$2=$/$%').($1"/&:#$+'&3$/$+/3)0=$

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%#&%#.5)1-0/&$('$("#$+)#05$0).#*;$

$

8)))9 !"#$ (/.:#.()/0$ 1'3%'.#.($'+$ ("#$#0#1(&)1$ +)#05$/0'.:$ ("#$#,-)%'(#.()/0$ *-&+/1#$ )*$

4#&'>$ '("#&6)*#$ .'.?@/.)*").:$6'&A$6'-05$2#$5'.#$ ('$3'@#$ /$ 1"/&:#$ +&'3$'.#$

%').($'.$("#$*-&+/1#$('$("#$'("#&;$

$

8)@9 B'$6'&A$)*$&#,-)&#5$('$3'@#$/$%/&()10#$/0'.:$/.$#,-)%'(#.()/0$*-&+/1#;$

$

C$ -*#+-0$ /./0':=$ +'&$ #,-)%'(#.()/0$ 1-&@#*$ )*$ /$ ('%':&/%")1$ 3/%$ 8D):-&#$ E;F;G9;$ H/1"$

1'.('-&$0).#$'.$("#$3/%$&#%&#*#.(*$/$+)I#5$#0#@/()'.$/2'@#$*#/$0#@#0;$J/("#3/()1/00=$)($)*$

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H/&("$)*$ >$("#*#$1-&@#*$1'&&#*%'.5$('$:&/@)(/()'./0$#,-)%'(#.()/0*;$

8 > 9 1'.*(/.(z f x y! !

gV g! z

$

$ $

!"#$%&'()*)+$C$('%':&/%")1$3/%$

$

$

,-./01&'()23'45"67%/18'9:.%#&;'<7;'

'

L'.*)5#&$/$.'.?1'.5-1().:$&'5$'+$0#.:("$ ! $"/@).:$/$-.)+'&3$1"/&:#$5#.*)(=" ;$D).5$("#$

#0#1(&)1$%'(#.()/0$/( >$/$%#&%#.5)1-0/&$5)*(/.1#$P y $/2'@#$("#$3)5%').($'+$("#$&'5;$

$

$$

!"#$%&'()*)*$C$.'.?1'.5-1().:$&'5$'+$0#.:("$ $/.5$-.)+'&3$1"/&:#$5#.*)(=! " ;$$

$ MN

16

Equipotenciais

Carga pontual Dipolo elétrico Placas paralelas

17

Equipotenciais e linhas de campo

18

!"#$% -

Conductors in Equilibrium

Conductors are equipotential objects:

1) E = 0 inside

2) Net charge inside is 0

3) E perpendicular to surface

4) Excess charge on surface

$!

"#E

E = !/"0

Condutores

•E perpendicular à superfície do condutor

• E=0 dentro do condutor

• Condutores são objetos equipotenciais

19

Potencial em um condutor

No condutor E=0: variação do potencial = 0

Campo elétrico = variação do potencial V constante no condutor

20

Potencial em um condutor

No condutor E=0: variação do potencial = 0

Campo elétrico = variação do potencial V constante no condutor

Mas qual o valor de V ?

Valor que ele tem na superfície

V é uma função contínua

20

Capacitores

21

Capacitance and Dielectrics

5.1 Introduction

A capacitor is a device which stores electric charge. Capacitors vary in shape and size,

but the basic configuration is two conductors carrying equal but opposite charges (Figure

5.1.1). Capacitors have many important applications in electronics. Some examples

include storing electric potential energy, delaying voltage changes when coupled with

resistors, filtering out unwanted frequency signals, forming resonant circuits and making

frequency-dependent and independent voltage dividers when combined with resistors.

Some of these applications will be discussed in latter chapters.

Figure 5.1.1 Basic configuration of a capacitor.

In the uncharged state, the charge on either one of the conductors in the capacitor is zero.

During the charging process, a charge Q is moved from one conductor to the other one,

giving one conductor a charge Q! , and the other one a charge . A potential

difference is created, with the positively charged conductor at a higher potential than

the negatively charged conductor. Note that whether charged or uncharged, the net charge

on the capacitor as a whole is zero.

Q"

V#

The simplest example of a capacitor consists of two conducting plates of area A , which

are parallel to each other, and separated by a distance d, as shown in Figure 5.1.2.

Figure 5.1.2 A parallel-plate capacitor

Experiments show that the amount of charge Q stored in a capacitor is linearly

proportional to , the electric potential difference between the plates. Thus, we may

write

V#

|Q C V |$ # (5.1.1)

2

C =Q

|!V |

Capacitores

Dois condutores com cargas iguais e opostas separados por uma distância d e com uma diferença de potencial ∆V entre eles.

Armazenamento de Energia!

Unidade: Coulomb/Volt

Farad

=

22

Capacitor de placas paralelas

!"#$% -

Parallel Plate Capacitor

top

bottom

V d! " # $% E S!!

$

Qd

A&"Ed"

d

A

V

QC $

&"

!"

C depends only on geometric factors A and d

Integral de trajetória para encontrar V

23

!V = !! d

0

!E · d!s = Ed ="

#0d =

Q

A#0d

Capacitor de placas paralelas

!"#$% -

Parallel Plate Capacitor

top

bottom

V d! " # $% E S!!

$

Qd

A&"Ed"

d

A

V

QC $

&"

!"

C depends only on geometric factors A and d

Integral de trajetória para encontrar V

23

!V = !! d

0

!E · d!s = Ed ="

#0d =

Q

A#0d

C =Q

|!V | =A!0

d

Capacitor de placas paralelas

!"#$% -

Parallel Plate Capacitor

top

bottom

V d! " # $% E S!!

$

Qd

A&"Ed"

d

A

V

QC $

&"

!"

C depends only on geometric factors A and d

Integral de trajetória para encontrar V

23

Energia necessária para carregar capacitor

!!"#$ -

Energy To Charge Capacitor

1. Capacitor starts uncharged.

2. Carry +dq from bottom to top.

Now top has charge q = +dq, bottom -dq

3. Repeat

4. Finish when top has charge q = +Q, bottom -Q

+q

-q

• Capacitor inicialmente descarregado

• +dq sai da placa inferior e vai para a superior

• Uma placa fica com +dq e a outra com -dq

• Processo ocorre até uma placa ter +Q e a outra -Q

24

dW = dq!V = dqq

V=

1C

qdq

W =!

dW =! Q

0

1C

qdq

W =1C

Q2

2

Trabalho realizado para carregar capacitor

!!"#$ -

Energy To Charge Capacitor

1. Capacitor starts uncharged.

2. Carry +dq from bottom to top.

Now top has charge q = +dq, bottom -dq

3. Repeat

4. Finish when top has charge q = +Q, bottom -Q

+q

-q

25

U =1C

Q2

2=

12C|!V |2

Energia armazenada no capacitor

C =Q

|!V |

26

U =1C

Q2

2=

12C|!V |2

5.4.1 Energy Density of the Electric Field

One can think of the energy stored in the capacitor as being stored in the electric field

itself. In the case of a parallel-plate capacitor, with 0 /C A d!" and | |V Ed# " , we have

$ % $22 00

1 1 1| |

2 2 2E

AU C V Ed E Ad

d%2!

!" # " " (5.4.3)

Since the quantity Ad represents the volume between the plates, we can define the electric

energy density as

2

0

1

Volume 2

EE

Uu !" " E (5.4.4)

Note that is proportional to the square of the electric field. Alternatively, one may

obtain the energy stored in the capacitor from the point of view of external work. Since

the plates are oppositely charged, force must be applied to maintain a constant separation

between them. From Eq. (4.4.7), we see that a small patch of charge

Eu

( )q A&# " # experiences an attractive force 2

0( ) / 2F A& !# " # . If the total area of the

plate is A, then an external agent must exert a force 2

ext 0/ 2F A& !" to pull the two plates

apart. Since the electric field strength in the region between the plates is given by

0/E & !" , the external force can be rewritten as

20ext

2F E A

!" (5.4.5)

Note that is independent of d . The total amount of work done externally to separate

the plates by a distance d is then

extF

2

0ext ext ext

2

E AW d F d

!' (" ) " " *

+ ,- F s d.!

!

(5.4.6)

consistent with Eq. (5.4.3). Since the potential energy of the system is equal to the work

done by the external agent, we have . In addition, we note that the

expression for is identical to Eq. (4.4.8) in Chapter 4. Therefore, the electric energy

density can also be interpreted as electrostatic pressure P.

2

ext 0/Eu W Ad E!" " / 2

Eu

Eu

Interactive Simulation 5.2: Charge Placed between Capacitor Plates

This applet shown in Figure 5.4.2 is a simulation of an experiment in which an aluminum

sphere sitting on the bottom plate of a capacitor is lifted to the top plate by the

electrostatic force generated as the capacitor is charged. We have placed a non-

13

Energia armazenada no capacitor

C =Q

|!V |

26

U =1C

Q2

2=

12C|!V |2

5.4.1 Energy Density of the Electric Field

One can think of the energy stored in the capacitor as being stored in the electric field

itself. In the case of a parallel-plate capacitor, with 0 /C A d!" and | |V Ed# " , we have

$ % $22 00

1 1 1| |

2 2 2E

AU C V Ed E Ad

d%2!

!" # " " (5.4.3)

Since the quantity Ad represents the volume between the plates, we can define the electric

energy density as

2

0

1

Volume 2

EE

Uu !" " E (5.4.4)

Note that is proportional to the square of the electric field. Alternatively, one may

obtain the energy stored in the capacitor from the point of view of external work. Since

the plates are oppositely charged, force must be applied to maintain a constant separation

between them. From Eq. (4.4.7), we see that a small patch of charge

Eu

( )q A&# " # experiences an attractive force 2

0( ) / 2F A& !# " # . If the total area of the

plate is A, then an external agent must exert a force 2

ext 0/ 2F A& !" to pull the two plates

apart. Since the electric field strength in the region between the plates is given by

0/E & !" , the external force can be rewritten as

20ext

2F E A

!" (5.4.5)

Note that is independent of d . The total amount of work done externally to separate

the plates by a distance d is then

extF

2

0ext ext ext

2

E AW d F d

!' (" ) " " *

+ ,- F s d.!

!

(5.4.6)

consistent with Eq. (5.4.3). Since the potential energy of the system is equal to the work

done by the external agent, we have . In addition, we note that the

expression for is identical to Eq. (4.4.8) in Chapter 4. Therefore, the electric energy

density can also be interpreted as electrostatic pressure P.

2

ext 0/Eu W Ad E!" " / 2

Eu

Eu

Interactive Simulation 5.2: Charge Placed between Capacitor Plates

This applet shown in Figure 5.4.2 is a simulation of an experiment in which an aluminum

sphere sitting on the bottom plate of a capacitor is lifted to the top plate by the

electrostatic force generated as the capacitor is charged. We have placed a non-

13

Energia armazenada no capacitor

C =Q

|!V |

5.4.1 Energy Density of the Electric Field

One can think of the energy stored in the capacitor as being stored in the electric field

itself. In the case of a parallel-plate capacitor, with 0 /C A d!" and | |V Ed# " , we have

$ % $22 00

1 1 1| |

2 2 2E

AU C V Ed E Ad

d%2!

!" # " " (5.4.3)

Since the quantity Ad represents the volume between the plates, we can define the electric

energy density as

2

0

1

Volume 2

EE

Uu !" " E (5.4.4)

Note that is proportional to the square of the electric field. Alternatively, one may

obtain the energy stored in the capacitor from the point of view of external work. Since

the plates are oppositely charged, force must be applied to maintain a constant separation

between them. From Eq. (4.4.7), we see that a small patch of charge

Eu

( )q A&# " # experiences an attractive force 2

0( ) / 2F A& !# " # . If the total area of the

plate is A, then an external agent must exert a force 2

ext 0/ 2F A& !" to pull the two plates

apart. Since the electric field strength in the region between the plates is given by

0/E & !" , the external force can be rewritten as

20ext

2F E A

!" (5.4.5)

Note that is independent of d . The total amount of work done externally to separate

the plates by a distance d is then

extF

2

0ext ext ext

2

E AW d F d

!' (" ) " " *

+ ,- F s d.!

!

(5.4.6)

consistent with Eq. (5.4.3). Since the potential energy of the system is equal to the work

done by the external agent, we have . In addition, we note that the

expression for is identical to Eq. (4.4.8) in Chapter 4. Therefore, the electric energy

density can also be interpreted as electrostatic pressure P.

2

ext 0/Eu W Ad E!" " / 2

Eu

Eu

Interactive Simulation 5.2: Charge Placed between Capacitor Plates

This applet shown in Figure 5.4.2 is a simulation of an experiment in which an aluminum

sphere sitting on the bottom plate of a capacitor is lifted to the top plate by the

electrostatic force generated as the capacitor is charged. We have placed a non-

13

Densidade de energia

Energia armazenada no campo!

26

Aumentando a capacitância

27

Dielétricos (visão microscópica)

Dielétricos polares

Dielétricos com momento de dipolo permanente

Ex: água

28

Dielétricos (visão microscópica)

Dielétricos polares

Dielétricos com momento de dipolo permanente

Ex: água

28

Dielétricos não polares (visão microscópica)

Dielétricos com momento de dipolo induzido pelo campo elétrico

Ex: CH4

29

Dielétricos não polares (visão microscópica)

Dielétricos com momento de dipolo induzido pelo campo elétrico

Ex: CH4

29

Dielétricos (visão macroscópica)

1

1

Volume

N

i

i!

! "P p!

!

(5.5.2)

In the case of our cylinder, where all the dipoles are perfectly aligned, the magnitude of

is equal to P!

Np

PAh

! (5.5.3)

and the direction of is parallel to the aligned dipoles. P!

Now, what is the average electric field these dipoles produce? The key to figuring this

out is realizing that the situation shown in Figure 5.5.4(a) is equivalent that shown in

Figure 5.5.4(b), where all the little ± charges associated with the electric dipoles in the

interior of the cylinder are replaced with two equivalent charges, PQ# , on the top and

bottom of the cylinder, respectively.

Figure 5.5.4 (a) A cylinder with uniform dipole distribution. (b) Equivalent charge

distribution.

The equivalence can be seen by noting that in the interior of the cylinder, positive charge

at the top of any one of the electric dipoles is canceled on average by the negative charge

of the dipole just above it. The only place where cancellation does not take place is for

electric dipoles at the top of the cylinder, since there are no adjacent dipoles further up.

Thus the interior of the cylinder appears uncharged in an average sense (averaging over

many dipoles), whereas the top surface of the cylinder appears to carry a net positive

charge. Similarly, the bottom surface of the cylinder will appear to carry a net negative

charge.

How do we find an expression for the equivalent charge PQ in terms of quantities we

know? The simplest way is to require that the electric dipole moment PQ produces,

PQ h , is equal to the total electric dipole moment of all the little electric dipoles. This

gives , or PQ h Np!

P

NpQ

h! (5.5.4)

18

QP = Carga induzida

30

Dielétricos em capacitores

31

Dielétricos em capacitores

C =Q

|!V |Aumento da capacitância com diminuição de ∆V

31

Dielétricos em capacitores

C =Q

|!V |Aumento da capacitância com diminuição de ∆V

∆V diminui porque a polarização do dielétrico diminui o campo elétrico

31

Constante dielétrica κ

dielétricos diminuem o campo elétrico original por um fator κ

Constante dielétrica

32

Constante dielétrica κ

Constantes dielétricas Vácuo 1.0 Papel 3.7 Vidro Pyrex 5.6 Água 80

dielétricos diminuem o campo elétrico original por um fator κ

Constante dielétrica

32

Lei de Gauss num dielétrico

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Lei de Gauss sem dielétricos

33

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Lei de Gauss num dielétrico

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Lei de Gauss sem dielétricos

33

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Lei de Gauss num dielétrico

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Lei de Gauss sem dielétricos

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

QP = Carga induzida

33

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Lei de Gauss num dielétrico

The capacitance becomes

00

0 0| | | |

ee

QQC

V VC

!!" " "

# # (5.5.17)

which is the same as the first case where the charge Q0 is kept constant, but now the

charge has increased.

5.5.4 Gauss’s Law for Dielectrics

Consider again a parallel-plate capacitor shown in Figure 5.5.7:

Figure 5.5.7 Gaussian surface in the absence of a dielectric.

When no dielectric is present, the electric field 0E!

in the region between the plates can be

found by using Gauss’s law:

0 0

0 0

,S

Qd E A E

$

% %& " " ' "(( E A!" "

#

We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced

charge PQ of opposite sign on the surface, and the net charge enclosed by the Gaussian

surface is PQ Q) .

Figure 5.5.8 Gaussian surface in the presence of a dielectric.

22

Lei de Gauss sem dielétricos

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

permissividade elétrica do meio

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

Gauss’s law becomes

0

P

S

Q Qd EA

!

"# $ $%% E A

!" "

# (5.5.18)

or

0

PQ QE

A!

"$ (5.5.19)

However, we have just seen that the effect of the dielectric is to weaken the original field

by a factor . Therefore, 0E e&

0

0 0

P

e e

E Q QQE

A A& & ! !

"$ $ $ (5.5.20)

from which the induced charge PQ can be obtained as

1

1P

e

Q Q&

' ($ ")

* +, (5.5.21)

In terms of the surface charge density, we have

1

1P

e

- -&

' ($ ")

* +, (5.5.22)

Note that in the limit , 1e& $ 0PQ $ which corresponds to the case of no dielectric

material.

Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be

rewritten as

0eS

Q Qd

& ! !# $ $%% E A

""

# (5.5.23)

where 0e! & !$ is called the dielectric permittivity. Alternatively, we may also write

S

d Q# $%% D A

""

# (5.5.24)

where 0! &$D

!""

E is called the electric displacement vector.

23

QP = Carga induzida

33