February 14, 2005Capacitors1 Welcome Back Exam returned Wed or Friday. Problem 1 – We did it in class Problem 2 - A Web Assign Problem Problem 3

February 14, 2005 Capacitors 1

Welcome Back

Exam returned Wed or Friday. Problem 1 – We did it in class Problem 2 - A Web Assign Problem Problem 3 - Superposition

Quiz on Friday (Capacitors) New WebAssign Posted … Wait until

Wednesday to try.


Chapter 25

Capacitors

Battery


Remember distributed Charges


Infinite Metal Plates

-------

++++++++


Addemup

-------

++++++++

E=0E=0

0

E


Not quite infinite …


Capacitor

Composed of two metal plates. Each plate is charged

one positive one negative

Stores Charge Can store a LOT of charge and can be

dangerous!


A Simple Electric Circuit


Don’t ask questions because I don’t know the answers!

Zn Metal Cu Metal

AqueousSolution

of


What’s Next?

Zn(solid) Zn2+ +2e-

Electrons Hang Around Zn ion goes into the solution.

Cu2+(solution) +2e- Cu depositing on Cu electrode





Gauss on Capacitors

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

GaussianSurface

000

0

0

0

)/(

0

AQ

A

QE

EAQ

QEAAEA

qd

Gauss

AE

Same result from other plate!


Two Charged Plates(Neglect Fringing Fields)

d

Air or Vacuum

Area A

- Q +QE


Symbol


Noted

Air or Vacuum

Area A

- Q +QE


+

Consider a +q chargeat the (-) plate.

Move it to the (+)plate

Work to do this isW=Fd=qEdalsoW=q(Vf-Vi)=qV

ThereforeEd=VE=V/d


Device The Potential Difference

is APPLIED by a battery or a circuit.

The charge q on the capacitor is found to be proportional to the applied voltage.

The proportionality constant is C and is referred to as the CAPACITANCE of the device.

CVq

orV

qC

DEFINITION


UNITS

Faradvolt

coulomb

V

QC


Look again

d

AC

so

CVVd

A

d

VAAEq

E

Aq

0

000

0


Continuing…

d

AC 0

The capacitance of a parallel plate capacitor depends only on the Area and separation between the plates.

C is dependent only on the geometry of the device!

Diversion on Capacitors


Two Metal Plates a Capacitor Make.


More is better!


Implementation - Variable


How do you do that?


Roll it up, Scottie


Stacked Disks, etc.


Units of 0

mpFmF

andm

Farad

Voltm

CoulombVoltCoulombm

Coulomb

Joulem

Coulomb

Nm

Coulomb

/85.8/1085.8 120

2

2

2

2

0

pico

Coulomb

JouleVolt


Simple Capacitor Circuits

Batteries Apply potential differences

Capacitors Wires

Wires are METALS. Continuous strands of wire are all at the

same potential. Separate strands of wire connected to

circuit elements may be at DIFFERENT potentials.


Size Matters! A Random Access Memory stores

information on small capacitors which are either charged (bit=1) or uncharged (bit=0).

Voltage across one of these capacitors ie either zero or the power source voltage (5.3 volts in this example).

Typical capacitance is 55 fF (femto=10-15) Question: How many electrons are stored

on one of these capacitors in the +1 state?


Small is better in the IC world!

electronsC

VF

e

CV

e

qn 6

19

15

108.1106.1

)3.5)(1055(


TWO Types of Connections

SERIES

PARALLEL


Parallel Connection

V

VCEquivalent=CE

C1 C2 C3

321

321

321

33

22

111

)(

CCCC

therefore

CCCVQ

qqqQ

VCq

VCq

VCVCq

E

E

E


Series Connection

V C1 C2

q -q q -q

The charge on eachcapacitor is the same !


Series Connection Continued

21

21

21

111

CCC

or

C

q

C

q

C

q

VVV

VC1 C2

q -q q -q

V1 V2


For Bunches of Capacitors

ii

i i

CC

Parallel

CC

Series

11



Example

C1 C2

V

C3

C1=12.0 fC2= 5.3 fC3= 4.5 d

(12+5.3)pf

series


More on the Big C

• We move a charge dq from the (-) plate to the (+) one.

• The (-) plate becomes more (-)

• The (+) plate becomes more (+).

• dW=Fd=dq x E x d+q -q

E=0A/d

+dq


So….

2222

00

2

00

2

0 0

0

00

2

1

22

|)(

1

2|

2

1

1

CVC

VC

C

QU

ord

Aq

A

dqqdq

A

dUW

dqdA

qdW

A

qE

Gauss

EddqdW

QQQ


Not All Capacitors are Created Equal

• Parallel Plate

• Cylindrical• Spherical


Spherical Capacitor

???

4)(

4

02

0

2

0

surprise

r

qrE

qEr

qd

Gauss

AE


Calculate Potential Difference V

drr

qV

EdsV

a

b

platepositive

platenegative

20

.

.

1

4

(-) sign because E and ds are in OPPOSITE directions.


Continuing…

ab

ab

V

qC

ab

abq

ba

qV

r

q

r

drqV

b

a

0

00

02

0

4

4

11

4

)1

(44

Lost (-) sign due to switch of limits.


Real Materials

Consist of atoms or molecules bonded together.

Some atoms and molecules do not have dipole moments when isolated.

Some do.

Two types to consider: Polar Non-Polar


Polar Molecule

E


Polar Materials



Apply an Electric Field

Some LOCAL ordering Large Scale Ordering


Adding things up..

- E +Net effect REDUCES the field


Non-Polar Material


Non-Polar Material

Effective Charge isREDUCED

Electric Fieldin the dielectric

is reduced0

netE


Effect of Capacitor MaterialDielectric

Effective Charge isREDUCED

Electric Fieldin the dielectric

is reduced0

netE


We can measure the C of a capacitor (later)

C0 = Vacuum or air Value

C = With dielectric in place

C=C0

Definition


Dielectric Constant

0C

C


How to Check

Charge to V0 and then disconnect fromthe battery.C0 V0

Connect the two togetherV

C0 will lose some charge to the capacitor with the dielectric.We can measure V with a voltmeter (later).

Q


Checking the idea..

V

100

000

210

2

01

000

V

VCC

CVVCVC

qqq

CVq

VCq

VCq

Note: When two Capacitors are the same (No dielectric), then V=V0/2.


Some values

Material

Dielectric

Strength

Breakdown

KV/mm

Air 1 3

Polystyrene 2.6 24

Paper 3.5 16

Pyrex 4.7 14

Strontium Titanate

310 8


Messing with Capacitor

+

V-

+

V-

+

-

+

-

The battery means that thepotential difference acrossthe capacitor remains constant.

For this case, we insert the dielectric but hold the voltage constant,

q=CV

since C kC0

qk kC0V

THE EXTRA CHARGE COMES FROM THE BATTERY!

Remember – We hold V constant with the battery.

WHERE IS THIS NEW CHARGE?

Hang on … we will get there.

But there is more capacity so there is more charge for the same applied voltage


Another Case

We charge the capacitor to a voltage V0.

We disconnect the battery. We slip a dielectric in between the

two plates. We look at the voltage across the

capacitor to see what happens.


Case II – No Battery

+

-

+

-

q0

q

q=C0Vo

When the dielectric is inserted, no chargeis added so the charge must be the same.

0

0000

0

VV

or

VCqVCq

VCq

V0

V


Another Way to Think About This

There is an original charge q on the capacitor.

If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material.

If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp!


q0

http://www.ucar.edu/communications/quarterly/summer95/vandegraaf.gif


A Reminder of days past

d

AC

definitionV

QC

A

QdV

equatingd

VE

A

QE

0

0

00

)(

:


A Closer Look at this stuff..Consider this capacitor.No dielectric.Applied Voltage via a battery.

00

00

00

Vd

AVCq

d

AC

C0

++++++++++++

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

V0

q

-q


Remove the Battery

++++++++++++

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

V0

q

-q

The Voltage across thecapacitor remains V0

q remains the same aswell.

The capacitor is fat (charged),dumb and happy.


Slip in a DielectricAlmost, but not quite, filling the space

++++++++++++

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

V0

q

-q

- - - - - - - -

+ + + + + +

-q’

+q’

E0

E

E’ from inducedcharges

Gaussian Surface

000

0

....

A

qE

qd

gapsmallin

AE


A little sheet from the past..

+++

---q-q

-q’ +q’

A

q

A

qE

A

qE

dialectricsheet

sheet

00/

00

'

2

'2

2

'

2

0 2xEsheet 0


Some more sheet…

A

qqE

so

A

qE

A

qE echdielectric

0material dielectricin

00

0arg

'

'

(original field)


A Few slides backCase II – No Battery

+

-

+

-

q0

q

q=C0Vo

When the dielectric is inserted, no chargeis added so the charge must be the same.

0

0000

0

VV

or

VCqVCq

VCq

V0

V


From this last equation

0

00

00

0

1

EE

E

E

V

V

thus

dEV

EdV

and

VV


A Bit more…..

qqq

therefore

Aqq

Aq

E

E

V

V

'

'

0

000


Important Result (We already know)

• Electric Field is Reduced by the presence of the material .

• The material reduces the field by a factor .

0EE

is the DIELECTRIC CONSTANTof the material


Another look

+

-

Vo

d

V

A

Qd

VE

FieldElectricd

AVVCQ

d

AC

PlateParallel

0000

00

00000

00


Add Dielectric to Capacitor

• Original Structure

• Disconnect Battery

• Slip in Dielectric

+

-

Vo

+

-

+

-

V0

Note: Charge on plate does not change!


What happens?

0

00 1

VEdV

andd

VEE

+

-

ii

oo

Potential Difference is REDUCEDby insertion of dielectric.

00 /

CV

Q

V

QC

Charge on plate is Unchanged!

Capacitance increases by a factor of as we showed previously


SUMMARY OF RESULTS

0

0

0

EE

CC

VV


APPLICATION OF GAUSS’ LAW

qqq

and

A

qE

E

A

qqE

A

qE

'

'

0

0

0

00


New Gauss for Dielectrics

0

0

sometimes

qd freeAE


The Insertion Process With A Battery

+

-

Vo --------F

Documents

February 14, 2005Capacitors1 Welcome Back Exam returned Wed or Friday. Problem 1 – We did it in class Problem 2 - A Web Assign Problem Problem 3