LECTURE Topic 4

POTENTIALSeptember 19, 2005

Alternate Lecture Titles

Back to Physics 2048 You can run but you can’t

hide!

The PHY 2048 Brain Partition

hm

A

B

To move the mass m from the ground toa point a distance h above the groundrequires that work be done on the particle.

h

mghmgdyW0

W is the work done by an external force.mgh represents this amount of work andis the POTENTIAL ENERGY of the massat position h above the ground.

The reference level, in this case, was chosenas the ground but since we only deal withdifferences between Potential Energy Values,we could have chosen another reference.

Reference “0”

Let’s Recall Some more PHY2048

hm

A

B

A mass is dropped from a height h above theground. What is it’s velocity when it strikesthe ground?

We use conservation of energy to compute the answer.

ghv

and

mvmgh

2

2

1)0()0( 2

Result is independent of the mass m.

Using a different reference.

y=hm

A

B

y

y=b (reference level)

y=0 ghv

mvmgbmgbmgh

mvbmgbhmg

KEPEE

2

2

12

1)(0)(

2

2

Still falls to here.

Energy Methods

Often easier to apply than to solve directly Newton’s law equations.

Only works for conservative forces. One has to be careful with SIGNS.

VERY CAREFUL!

I need some help.

THINK ABOUT THIS!!!THINK ABOUT THIS!!!

When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external

force (you). You can also think of this as the

FIELDFIELD doing the negative of this amount of work on the particle.

Let’s look at it:move a mass from yi to yf

yf

yi

Ex

tern

al

Fie

ld

Change in potential energy due to external force:

)()()(

)()(

.

PEyymgW

PEyymgW

distforceW

if

if

Negative of the work done BY THE FIELD.

Keep it!Keep it!

Move It!

Move the charge at constant velocity so it is in mechanical equilibrium all the time.

Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.

And also remember:

The net work done by a conservative (field)force on a particle moving

around a closed path is

ZERO!

A nice landscape

mg

h

Work done by external force = mgh

How much work here by gravitational field?

The gravitational case:

Someone else’s path

IMPORTANT

The work necessary for an external agent to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!

The Electric Field Is a conservative field.

No frictional losses, etc. Is created by charges. When one (external agent) moves a test

charge from one point in a field to another, the external agent must do work.

This work is equal to the increase in potential energy of the charge.

It is also the NEGATIVE of the work done BY THE FIELD in moving the charge from the same points.

A few things to remember… A conservative force is NOT a Republican. An External Agent is NOT 007.

Electric Potential EnergyElectric Potential Energy When an electrostatic force acts

between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system.

Example: NOTATION U=PEU=PE

A B

dd

E

q

F

Work done by FIELD is Fd

Negative of the work done by the FIELD is -Fd

Change in Potential Energy is also –Fd.Change in Potential Energy is also –Fd.The charge sort-of “fell” to lower potential energy.

HIGH U LOWER U

Gravity

mg

Negative of the work done by the FIELD is –mg h = U

Bottom Line: Things tend to fall down and lower their potential energy. The change, Uf – Ui is NEGATIVE!

Electrons have those *&#^ negative signs.

Electrons sometimes seem to be more difficult to deal with because of their negative charge.

They “seem” to go from low potential energy to high.

They DO! They always fall AGAINST the field! Strange little things. But if YOU were

negative, you would be a little strange too!

An Important ExampleDesigned to Create Confusionor Understanding … Your Choice!

E

e

A sad and confusedElectron.

Initial position

Final position

d

The change in potential energyof the electron is the negative of the work done by the field in moving the electronfrom the initial position to the finalposition.

!

)()()(

)(

negative

yyEeWU

yyFW

if

if

FORCE

negativecharge

Force againstThe directionof E

An important point In calculating the change in potential

energy, we do not allow the charge to gain any kinetic energy.

We do this by holding it back. That is why we do EXTERNAL work. When we just release a charge in an

electric field, it WILL gain kinetic energy … as you will find out in the problems!

Remember the demo!

AN IMPORTANT DEFINITION

Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE:

We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE:

q

FE

q

UV

VECTOR

SCALAR

UNITS OF POTENTIAL

VOLTCoulomb

Joules

q

UV

Watch those #&@% (-) signs!!

The electric potential difference V between two points I and f in the electric field is equal to the energy PER UNIT CHARGE between the points:

q

W

q

U

q

U

q

UVVV if

if

Where W is the work done BY THE FIELD in moving the charge fromOne point to the other.

BREAK

Start September 21 (Winter??)

Let’s move a charge from one point

to another via an external force.

The external force does work on the particle.

The ELECTRIC FIELD also does work on the particle.

We move the particle from point i to point f.

The change in kinetic energy is equal to the work done by the applied forces.

appliedif

fieldapplied

fieldappliedif

WUUU

also

WW

K

if

WWKKK

0

Furthermore…

VqW

so

q

W

q

UV

applied

applied

If we move a particle through a potential difference of V, the work from an external

“person” necessary to do this is qV

Example

Electric Field = 2 N/C

1 C d= 100 meters

Joules

mCN4102

100)/(2C1qEdPE

Energy. potentialin Change

agent EXTERNALby doneWork

One Step More

Joules

mCN4102

100)/(2C1qEdPE

Energy. potentialin Change

agent EXTERNALby doneWork

Volts 200200101

102

q

PE POTENTIALin Change

6

4

C

J

C

JoulesV

The Equipotential SurfaceDEFINED BY

0VIt takes NO work to move a charged particlebetween two points at the same potential.

The locus of all possible points that require NO WORK to move the charge to is actually a surface.

Example: A Set of Equipotenital Surfaces

Back To YesteryearBack To Yesteryear

Field Lines and Equipotentials

EquipotentialSurface

ElectricField

Components

EquipotentialSurface

ElectricField

Enormal

Eparallel

x

Work to move a charge a distancex along the equipotential surfaceIs Q x Eparallel X x

BUT

This an EQUIPOTENTIAL Surface No work is needed since V=0 for

such a surface. Consequently Eparallel=0 E must be perpendicular to the

equipotential surface

ThereforeE

E

E

V=constant

Field Lines are Perpendicular to the Equipotential Lines

Equipotential

)(0 ifexternal VVqWork

Consider Two EquipotentialSurfaces – Close together

V

V+dV

dsab

Work to move a charge q from a to b:

VVectords

dVE

and

dVEds

qdVVdVVqdW

also

qEdsdsFdW

external

appliedexternal

E...

)(E

Where

zyx

kji

I probably won’t ask about this.

Typical Situation

dF W

Keep in Mind

Force and Displacement are VECTORS!

Potential is a SCALAR.

UNITS 1 VOLT = 1 Joule/Coulomb For the electric field, the units of N/C can be

converted to: 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM) Or

1 N/C = 1 V/m So an acceptable unit for the electric field is

now Volts/meter. N/C is still correct as well.

In Atomic Physics It is sometimes useful to define an

energy in eV or electron volts. One eV is the additional energy that

an proton charge would get if it were accelerated through a potential difference of one volt.

1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6 x 10-19 Joules.

Nothing mysterious.

Coulomb Stuff: A NEW REFERENCE

204

1

r

qE

Consider a unit charge (+) being brought from infinity to a distance r from a Charge q:

q r

To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.

x

The math….

r

qV

rq

r

drq

Q

WV

and

r

drqdxFW

rr

r

external

0

1

02

0

20

4

1

)1(44)1(

4

1)()1(

This thing must

be positive anyway.

r final<r initia

lneg sig

n

For point charges

i i

i

r

qV

04

1

Example: Find potential at Pq1 q2

q3 q4

d

rP

md

r

md

qqqqr

V

919.02

3.1

)(1

4

14321

0

q1=12nC q2=-24nC q3=31nC q4=17nC q=36 x 10-9C

V=350 Volts (check the arithmetic!!)

An Examplefinite line of charge

d

r

x

dx

d

xLLV

and

xd

dxV

xd

dxdV

L

2/122

0

02/122

0

2/1220

)(ln

4

1

)(4

1

)(4

1

P

At P Using table of integrals

What about a rodthat goes from –L to +L??

Example (from text)zR

220

22

0

22

0

12

2

2

Rz

zE

zRzdz

d

z

VE

zRzV

z

z

Which was the result we obtained earlier

disk=charge per unit area

The Potential From a Dipole

d

P

r(+)

r(-)

+

-

)()(

)()(

4

)()(4

1

)()()(

0

0

rr

rrqV

r

q

r

qV

VVVPV i

r

Dipole - 2

d

P

r(+)

r(-)

+

-2

02

0

2

)cos(

4

1)cos(

4

)()(

)cos()()(

r

p

r

dqV

rrr

drr

Geometry

r

Where is this going?Charges and

Forces

Electric Fields

Concept ofPotential

Batteriesand

Circuit Elements(R,C,L)

ElectricCircuits

A Few Problems

A particular 12 V car battery can send a total charge of 81 A · h (ampere-hours) through a circuit, from one terminal to the other. (a) How many coulombs of charge does this represent?

(b) If this entire charge undergoes a potential difference of 12 V, how much energy is involved?

Sometimes you need to look things up …1 ampere is 1 coulomb per second.

81 (coulombs/sec) hour = 81 x (C/s) x 3600 sec = 2.9 e +5

qV=2.9 e 05 x 12=3.5 e+6

An infinite nonconducting sheet has a surface charge density = 0.10 µC/m2 on one side. How far apart are equipotential surfaces whose potentials differ by 54 V?

d

54V

metersxe

d

VEd

eex

eE

voltsqEd

3

0

109365.5

54

365.51285.82

61.0

2

54

In a given lightning flash, the potential difference between a cloud and the ground is 2.3x 109 V and the quantity of charge transferred is 43 C.

(a) What is the change in energy of that transferred charge? (GJ)

(b) If all the energy released by the transfer could be used to accelerate a 1000 kg automobile from rest, what would be the automobile's final speed?m/s

Energy = qV= 2.3 e+09 x 43C=98.9 GJ

E=(1/2)Mv2 v= sqr(2E/M)= 14,100 m/s

POTENTIAL PART 5Capacitance

Encore By Special Request

Where for artthou, oh Potential?

In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

Continuing

dd

s

ddd

dds

12.152

4

5

42

222

222

s

1 2 3

45 6

d

qxV

d

qk

d

qk

d

q

d

qkV

d

qqqq

d

qqk

r

qkV

i i

i

91035.8

93.093.8812.1

108

12.1

5533

2/

22

Text gets 8.49 … one of us is right!

Derive an expression in terms of q2/a for the work required to set up the four-charge configuration in the figure, assuming the charges are initially infinitely far apart.

1 2

3 4

aadiagonal 71.12

1 2

3 4

0

0

1

1

W

V

W=qV

a

qkqVW

a

qkV

2

22

2

a

qkqVW

a

qk

a

qk

a

q

a

qkV

2

33

3

293.0

293.02

11

2

a

qkVqW

a

qk

a

qk

a

q

a

q

a

qkV

2

44

4

29.1)(

29.1707.112

Add them up ..

a

qx

a

qk

a

qkW

WWWWW2

1022

4321

1032.258.229.1293.010

Capacitors

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!

Two Charged Plates(Neglect Fringing Fields)

d

Air or Vacuum

Area A

- Q +QE

V=Potential Difference

Symbol

More on Capacitorsd

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!

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

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

Continuing…

d

AC

sod

AVq

V

qC

0

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!

Units of 0

mpFmF

andm

Farad

Voltm

CoulombVoltCoulombm

Coulomb

Joulem

Coulomb

Nm

Coulomb

/85.8/1085.8 120

2

2

2

2

0

pico

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

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 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 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

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

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.

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 Materials (Water)

Apply an Electric Field

Some LOCAL ordering Large 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..

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

MaterialDielectric Strength

BreakdownKV/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 C0

qC0V

THE EXTRA CHARGE COMES FROM THE BATTERY!

Remember – We hold V constant with the battery.

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!

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

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

C0

00

00

00

Vd

AVCq

d

AC

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

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

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

0

00

0arg

'

'

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 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