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LECTURE 4 POTENTIAL September 20, 2003

LECTURE 4 POTENTIAL September 20, 2003 Alternate Lecture Titles Back to Physics 2048 You can run but you can’t hide!

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Page 1: LECTURE 4 POTENTIAL September 20, 2003 Alternate Lecture Titles  Back to Physics 2048  You can run but you can’t hide!

LECTURE 4

POTENTIALSeptember 20, 2003

Page 2: LECTURE 4 POTENTIAL September 20, 2003 Alternate Lecture Titles  Back to Physics 2048  You can run but you can’t hide!

Alternate Lecture Titles

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

hide!

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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”

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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.

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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.

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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!

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I need some help.

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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.

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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!

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And also remember:

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

around a closed path is

ZERO!

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A nice landscape

mg

h

Work done by external force = mgh

How much work here by gravitational field?

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The gravitational case:

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Someone else’s path

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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!

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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.

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

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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.

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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

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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!

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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!

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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

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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!

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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

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UNITS OF POTENTIAL

VOLTCoulomb

Joules

q

UV

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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.

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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

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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

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Example

Electric Field = 2 N/C

1 C d= 100 meters

Joules

mCN4102

100)/(2C1qEdPE

Energy. potentialin Change

agent EXTERNALby doneWork

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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

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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.

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Example: A Set of Equipotenital Surfaces

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Back To YesteryearBack To Yesteryear

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Field Lines and Equipotentials

EquipotentialSurface

ElectricField

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Components

EquipotentialSurface

ElectricField

Enormal

Eparallel

x

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

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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

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ThereforeE

E

E

V=constant

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Field Lines are Perpendicular to the Equipotential Lines

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Equipotential

)(0 ifexternal VVqWork

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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

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Where

zyx

kji

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Typical Situation

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A Brief Review of Concept The creation of a charged particle

distribution creates ELECTRICAL POTENTIAL ENERGY = U.

If a system changes from an initial state i to a final state f, the electrostatic forces do work W on the system

This is the NEGATIVE of the work done by the field.

WUUU if

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Calculation

• An external force F is necessary to move the charge q from i to f. The work done by this external force is also equal to the change in potential energy of the charged particle. Note the (-) sign is because F and E are in opposite directions.

• Continuous case:

qi f

E

xqExFU

dxFU external

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For Convenience

It is often convenient to set up a particular reference potential.

For charged particles interacting with each other, we take U=0 when the particles are infinitely apart.

Consequently U=(-) of the work done by the field in moving a particle from infinity to the point in question.

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dF W

Keep in Mind

Force and Displacement are VECTORS!

Potential is a SCALAR.

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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.

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In Atomic Physics

It is 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.

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Coulomb Stuff

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

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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(

Remember This !!!

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For point charges

i i

i

r

qV

04

1

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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!!)

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Brief Discussion on

Math

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Integration

There are many applications of integration in physics.

We use it to add things up over an area, a line, or perhaps a volume.

We often use density functions. The Examples we will use will be

charge densities.

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Line of Charge

(x).position of

function aor constant becan

dx)(q

dxdq

coul/meter

.or lengthunit per Charge

x

dq

dx

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Area Charge Density

dAdq

area

chargedx

dydA=dxdy

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The Job

dx

dydA=dxdy

........crazy go and substitute

)2(2

1 2/122

22

xdxxRdy

xRy

dxdyA

dxdydAR

R

R

R

R

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Another Way to do dA

(note rhyme)

r

d

rd

dA= dr rdr drd

dr

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Take a look

222

0

2

2

00

22

2r

rd

rA

drdrA

rdrddAA

rdrddA

r

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Still another view

r

dr 2

0

2

2

rrdrA

rdrdAR

r

dr

r

dr

r

dr

r

dr

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When you have to integrate over dA

Pick a friendly coordinate system.. Use the appropriate dA Don’t forget the function or may

be functions of position. The coordinate system that you choose should match the symmetry of these two functions.

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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

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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

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The Potential From a Dipole

d

P

r(+)

r(-)

+

-

)()(

)()(

4

)()(4

1

)()()(

0

0

rr

rrqV

r

q

r

qV

VVVPV i

r

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Dipole - 2

d

P

r(+)

r(-)

+

-2

02

0

2

)cos(

4

1)cos(

4

)()(

)cos()()(

r

p

r

dqV

rrr

drr

Geometry

r

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Where is this going?Charges and

Forces

Electric Fields

Concept ofPotential

Batteriesand

Circuit Elements(R,C,L)

ElectricCircuits

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A Few Problems

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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

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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

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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

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Electrical Circuits Do Something! Include a power source Includes electrical

components resistors capacitors inductors transistors diodes tubes switches wires

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Battery Maintains a potential

difference between its two terminals. The potential Difference – the

“voltage” is maintained by an electrochemical reaction which takes place inside of the battery.

Has two terminals+ is held at the higher potential - is held at the lower potential

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SymbolDC Power Source

Low Potential High Potential

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SWITCH

schematic symbol