capacitor lecture

7/26/2019 capacitor lecture

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Fall 2008Physics 231 Lecture 4-1

Capacitance and Dielectrics


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CapacitorsDevice for storing electrical energy which can then be

released in a controlled manner

Consists of two conductors, carrying charges of qandq,

that are separated, usually by a nonconducting material - an

insulator

Symbol in circuits is

It takes work, which is then stored as potential energy in the

electric field that is set up between the two plates, to placecharges on the conducting plates of the capacitor

Since there is an electric field between the plates there is also a

potential difference between the plates


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Capacitors

We usually talk about

capacitors in terms of

parallel conducting

plates

They in fact can beany two conducting

obects


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Capacitance

abVQC =

!nits areVolt

Coulombfarad1

11 =

The capacitance is defined to be the ratio of the

amount of charge that is on the capacitor to the

potential difference between the plates at this point


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Calculating the Capacitance

We start with the simplest form " two parallel conducting

plates separated by vacuum

#et the conducting plates have area $ and be

separated by a distance d

The magnitude of the electric field

between the two plates is given by A

QE

00

==

We treat the field as being uniform

allowing us to write A

dQEdVab

0

==


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Calculating the Capacitance

%utting this all together, we have for the

capacitance

d

A

V

QC

ab0==

The capacitance is only dependent

upon the geometry of the capacitor


7/32Fall 2008Physics 231 Lecture 4-7

& farad Capacitor

'iven a & farad parallel plate capacitor having a

plate separation of &mm( What is the area of the

plates)

We start with

d

AC 0=

$nd rearrange to

solve forA, giving

( ) ( )

28

12

3

0

101.1

/1085.8

100.10.1

m

mF

mFdCA

=

==

This corresponds to a s*uare about &+km on a side



Series or %arallel Capacitors

Sometimes in order to obtain needed valuesof capacitance, capacitors are combined

in either

or

Series

%arallel



Capacitors in Series

Capacitors are often combined in series and the *uestionthen becomes what is the e*uivalent capacitance)

'iven what is

We start by putting a voltage,ab, across the capacitors




Capacitors become charged because of ab

If upper plate of C&gets a charge of ./,

Then the lower plate of C&gets a

charge of -/

What happens with C0)

Since there is no source of charge at point c, and we

have effectively put a charge of "/ on the lower plateof C&, the upper plate of C0gets a charge of ./

This then means that lower plate of C0has a charge of -/

Charge Conservation



We also have to have that the potential across C&plus thepotential across C0should e*ual the potential drop across

the two capacitors

21 VVVVV cbacab +=+=

We have

2

2

1

1C

QV

C

QV == and

Then21 C

Q

C

Q

Vab +=

Dividing through by Q, we have

21

11

CCQ

Vab +=




21

11

CCQ

Vab +=The left hand side is the inverse of

the definition of capacitance Q

V

C=

1

So we then have for the e*uivalent capacitance

21

111

CCCeq+=

If there are more than two capacitors in series,the resultant capacitance is given by

=i ieq CC

11

The e*uivalent capacitor will also have thesame voltage across it




Capacitors in %arallel

Capacitors can also be connected in parallel

'iven what is

$gain we start by putting

a voltage across a and b



The upper plates of both capacitors

are at the same potential

#ikewise for the bottom plates

We have that abVVV == 21

1ow

VCQVCQ

or

C

QV

C

QV

2211

2

2

21

1

1

==

==

and

and




The e*uivalent capacitor will have the same voltage acrossit, as do the capacitors in parallel

2ut what about the charge on the e*uivalent capacitor)

The e*uivalent capacitor will have the same total charge

21 QQQ +=!sing this we then have

21

21

21

CCC

or

VCVCVC

QQQ

eq

eq

+=

+=

+=




The e*uivalent capacitance is ust the sum of the

two capacitors

If we have more than two, the resultant capacitance is ustthe sum of the individual capacitances

=i

ieq CC




34ample &

C

C1 C2

C3a

b

a b

213

111

CCCC ++=

321

213 )(

CCC

CCCC

++

+=

Where do we start)

5ecogni6e that C&and C0are parallel with each other and

combine these to get C12

This C12is then in series with with C3

The resultant capacitance is then given by



Which of these configurations has the lowest overall

capacitance)

Three configurations are constructed using identical capacitors

a7Configuration $

b7Configuration 2

c7Configuration C

CC

C

C C

Configuration $Configuration 2 Configuration C

34ample 0

The net capacitance for $ is ust C

In 2, the caps

are in series andthe resultant is

given by

2

2111 CC

CCCC netnet==+=

In C, the caps are in parallel and

the resultant is given byCCCCnet 2=+=



$ circuit consists of three une*ual capacitors C&, C0, and C8

which are connected to a battery of emf The capacitorsobtain charges Q&Q0, Q8, and have voltages across their

plates V&, V0, and V8( Ceqis the e*uivalent capacitance of the

circuit(

Check all of the following that

apply9

a7Q&: Q0 b7Q0: Q8 c7V0: V8

d7E : V& e7V&; V0 f7Ceq< C&

34ample 8

$ detailed worksheet is available detailing the answers
http://lecture04examples.pdf/http://lecture04examples.pdf/



(a)Ceq= (3/2)C (b)Ceq= (2/3)C (c)Ceq= 3C

o

o

C CC

Ceq

CC

C

C C1

CCC

111

1

+=2

1

CC = C

CCCeq

2

3

2=+=

34ample =

What is the e*uivalentcapacitance, Ce*, of the

combination shown)



3nergy Stored in a Capacitor

3lectrical %otential energy is stored in a

capacitorThe energy comes from the work that is done in charging

the capacitor

#et qand vbe the intermediate charge and potential on

the capacitor

The incremental work done in bringing an incremental

charge, dq, to the capacitor is then given by

C

dqqdqvdW ==


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3nergy Stored in a Capacitor

The total work done is ust the integral of this

e*uation from 0to Q

C

Qdqq

CW

Q

2

1 2

0

==

!sing the relationship between capacitance, voltage and

charge we also obtain

VQVCC

QU2

1

2

1

2

22

===

where Uis the stored potential energy


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34ample >

d

A

- - - - -

+ + + +

1ow suppose you pull the plates furtherapart so that the final separation is d&

Which of the *uantities Q, C, V, U,Echange)

?ow do these *uantities change)

$nswers9 Cdd

C

1

1 = Vd

dV

11 =

Suppose the capacitor shown here is charged

to Qand then the battery is disconnected

Ud

dU 11=

Charge on the capacitor does not change

Capacitance Decreases

oltage Increases

%otential 3nergy Increases

/9

C9

9

!9

39 3lectric @ield does not change


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Suppose the battery AV7 is keptattached to the capacitor

$gain pull the plates apart from dto d&

1ow which *uantities, if any, change)

?ow much do these *uantities change)

$nswers9

34ample B

C

d

dC

1

1= E

d

dE

1

1=U

d

dU

1

1=

/9

C9

9

!9

39

oltage on capacitor does not change

Capacitance Decreases

Charge Decreases

%otential 3nergy Decreases

3lectric @ield Decreases

Q

d

dQ

1

1=


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3lectric @ield 3nergy Density

The potential energy that is stored in the capacitor can be

thought of as being stored in the electric field that is in the

region between the two plates of the capacitor

The *uantity that is of interest is in fact the ener!densit!

dAu"ensit!Ener!

VC 2

2

1

==

whereAand dare the area of the capacitor plates and their

separation, respectively


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3lectric @ield 3nergy Density

!singdAC 0= and dEV= we then have

2

02

1Eu =

3ven though we used the relationship for a parallel capacitor,

this result holds for allcapacitors regardless of configuration

#$is re%resents t$e ener! densit! of t$e electric field ineneral


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Dielectrics

ost capacitors have a nonconducting material betweentheir plates

This nonconducting material, a dielectric, accomplishes

three things

&7Solves mechanical problem of keeping the plates

separated

07Increases the ma4imum potential difference allowed

between the plates

87Increases the capacitance of a given capacitor overwhat it would be without the dielectric


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DielectricsSuppose we have a capacitor of value C+that is charged to

a potential difference of +and then removed from thecharging source

We would then find that it has a charge of00VCQ =

We now insert the dielectric material into the capacitor

We find that the potential difference decreasesby a factor&

&

VV 0=

r e*uivalently the capacitance has increasedby a factor of&

0C&C =

This constant&is known as the dielectric constant and is

dependent upon the material used and is a number

greater than &


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%olari6ation

Without the dielectric in

the capacitor, we have

The electric field points undiminished from the positive

to the negative plate

Q+++++++++++++++

- - - - - - - - - - - - - - -

E0V0

With the dielectric

in place we have

The electric field between the plates of the capacitor is

reduced because some of the material within the dielectric

rearranges so that their negative charges are oriented

towards the positive plate

- - - - - - - - - - - - - - -

+++++++++++++++Q

V E+

- +

-

+

-+

-

+

-

+

-

+

-


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%olari6ation

These rearranged charges set up an

internal electric field that opposes

the electric field due to the charges

on the platesThe net electric field is given by

&

EE 0=


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5edefinitionsWe now redefine several *uantities using the dielectric

constant

Capacitance9dA

dA&&CC

=== 00

'it$ t$e last t'o relations$i%s $oldin for a %arallel

%late ca%acitor

3nergy Density 2202

1

2

1EE&u

==

We define the permittivity of the dielectric as being

0&=


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Two identical parallel plate capacitors are given the same charge

Q, after which they are disconnected from the battery( $fter C2

has been charged and disconnected it is filled with a dielectric(

Compare the voltages of the two capacitors(

a7V&< V0 b7V&: V0 c7V&; V0

34ample E

We have that /&: /0and that C0: FC&

We also have that C : /G or : /GC

Then&

&

&

C

QV and

&

&

&

0

0

0

&V

&&C

Q

C

QV =

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