September 29, 2008 CAPACITORS How did you do? A. Great B. OK C. Poor D. Really bad E. I absolutely...

September 29, 2008

CAPACITORS

How did you do?

A. GreatB. OKC. PoorD. Really badE. I absolutely

flunked!

Calendar of the DayCalendar of the DayExams will be returned within a week.

If you did badly in the exam you need to have a plan to succeed. Let me know if you want any help with this.

Quiz on Friday – Potential or Capacitance.WebAssign will appear shortly if it hasn’t

done so already.There is a WA on board for potential.Quizzes are in the bin on the third floor

through the double doors.

Two +q charges are separated by a distance d. What is the potential at a point midway between the charges on the line connecting them

A. ZeroB. Kq/dC. Kq/dD. 2Kq/dE. 4kq/d

Capacitors

A simple Capacitor

Remove the battery Charge Remains on the

plates. The battery did WORK to

charge the plates That work can be

recovered in the form of electrical energy – Potential Difference

WIRES

TWO PLATES

Battery

WIRES

INSIDE THE DEVICE

Two Charged Plates(Neglect Fringing Fields)

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

Symbol

ADDED CHARGE

Where is the charge?

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

------

++++++

AREA=A

=Q/A

One Way to Charge:

Start with two isolated uncharged plates.Take electrons and move them from the + to

the – plate through the region between.As the charge builds up, an electric field

forms between the plates.You therefore have to do work against the

field as you continue to move charge from one plate to another.

Capacitor

More on Capacitors

000

0

0

0

)/(

0

AQ

A

QE

EAQ

QEAAEA

qd

Gauss

AE

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

GaussianSurface

Same result from other plate!

DEFINITION - CapacityThe 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

UNITSUNITS

A capacitor which acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD

One Farad is one Coulomb/Volt

CVq

orV

qC

The two metal objects in the figure have net charges of +79 pC and -79 pC, which result in a 10 V potential difference between them.

(a) What is the capacitance of the system? [7.9] pF(b) If the charges are changed to +222 pC and -222 pC, what does the capacitance become? [7.9] pF(c) What does the potential difference become?[28.1] V

NOTEWork to move a charge from one side of a capacitor to the other is qEd.

Work to move a charge from one side of a capacitor to the other is qV

Thus qV=qEdE=V/d As before

Continuing…

d

AC

sod

AVEAAq

V

qC

0

00

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!

Units of Units of 00

mpFmF

andm

Farad

Voltm

CoulombVoltCoulombm

Coulomb

Joulem

Coulomb

Nm

Coulomb

/85.8/1085.8 120

2

2

2

2

0

Simple Capacitor CircuitsBatteries

Apply potential differencesCapacitorsWires

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)Probably less these days!

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(

October 1, 2008

Cap-II

Note:I do not have the grades yet. Probably by

Friday.Quiz on Friday … Potential or Capacitors.Watch WebAssign for new stuff.

Last Time

We defined capacitance:C=q/VQ=CV

We showed thatC=0A/d

AndE=V/d

TWO Types of Connections

SERIES

PARALLEL

Parallel Connection

VCEquivalent=CE

321

321

321

33

22

1111

)(

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

V C1 C2

q -q q -q

More GeneralMore General

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

(12+5.3)pf

More on the Big CWe 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

0

2

0

2

0 0

0

00

2

1

22

)(

1

22

1

1

CVC

VC

C

QU

ord

Aq

A

dqqdq

A

dUW

dqdA

qdW

A

qE

Gauss

EddqdW

Q

Not All Capacitors are Created Equal

Parallel Plate

CylindricalSpherical

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.

A Thunker

If a drop of liquid has capacitance 1.00 pF, what is its radius?

STEPS

Assume a charge on the drop.Calculate the potentialSee what happens

In the drawing below, find the equivalent capacitance of the combination. Assume that C1 = 8 µF, C2 = 4 µF, and C3 = 3  µF.

5.67µF

In the diagram, the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 20 µF.

What is the charge on( a) capacitor C1 and (b) capacitor C2?

In the figure, capacitors C1 = 0.8 µF and C2 = 2.8 µF are each charged to a potential difference of V = 104 V, but with opposite polarity as shown. Switches S1 and S2 are then closed.

(a) What is the new potential difference between points a and b? 57.8 VWhat are the new charges on each capacitor?(b)46.2µC (on C1)(c)162µC (on C2)

Anudder ThunkerAnudder ThunkerFind the equivalent capacitance between points a and b in the combination of capacitors shown in the figure.

V(ab) same across each

DIELECTRIC

Polar Materials (Water)

Apply an Electric Field

Some LOCAL ordering Larger Scale Ordering

Adding things up..

- +Net effect REDUCES the field

Non-Polar Material

Non-Polar Material

Effective Charge isREDUCED

We can measure the C of a capacitor (later)

C0 = Vacuum or air Value

C = With dielectric in place

C=C0

(we show this later)

How to Check This

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).

Checking the idea..

00

0

000

210

2

01

000

1 CV

VCC

CVVCVC

qqq

CVq

VCq

VCq

V

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

Messing with Capacitors

+

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 C0

qC0V

THE EXTRA CHARGE COMES FROM THE BATTERY!

Remember – We hold V constant with the battery.

Another CaseWe 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.

No Battery

0

0000

0

VV

or

VCqVCq

VCq

+

-

+

-

q0

q

q0 =C0Vo

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

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!

The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.

A Closer Look at this stuff..

00

00

00

Vd

AVCq

d

AC

Consider this virgin capacitor.No dielectric experience.Applied Voltage via a battery.

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

000

0

....

A

qE

qd

gapsmallin

AE

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

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

V0

q

-q

- - - - - - - -

+ + + + + +

-q’

+q’

E0

E

E’ from inducedcharges

Gaussian Surface

A little sheet from the past..

A

q

A

qE

A

qE

dialectricsheet

sheet

00/

00

'

2

'2

2

'

2

+++

---q-q

-q’ +q’

0 2xEsheet 0

Some more sheet…

A

qqE

so

A

qE

A

qE echdielectric

0

00

0arg

'

'

A Few slides backNo 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

Another look

d

V

A

Qd

VE

FieldElectricd

AVVCQ

d

AC

PlateParallel

0000

00

00000

00

+

-

Vo

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

00 /

CV

Q

V

QC

+

-

ii

oo

Potential Difference is REDUCEDby insertion of dielectric.

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