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LECTURE 4
POTENTIALSeptember 20, 2003
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!
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!
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.
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
Typical Situation
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
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
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.
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 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.
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
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 !!!
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!!)
Brief Discussion on
Math
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.
Line of Charge
(x).position of
function aor constant becan
dx)(q
dxdq
coul/meter
.or lengthunit per Charge
x
dq
dx
Area Charge Density
dAdq
area
chargedx
dydA=dxdy
The Job
dx
dydA=dxdy
........crazy go and substitute
)2(2
1 2/122
22
xdxxRdy
xRy
dxdyA
dxdydAR
R
R
R
R
Another Way to do dA
(note rhyme)
r
d
rd
dA= dr rdr drd
dr
Take a look
222
0
2
2
00
22
2r
rd
rA
drdrA
rdrddAA
rdrddA
r
Still another view
r
dr 2
0
2
2
rrdrA
rdrdAR
r
dr
r
dr
r
dr
r
dr
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.
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
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
Electrical Circuits Do Something! Include a power source Includes electrical
components resistors capacitors inductors transistors diodes tubes switches wires
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
SymbolDC Power Source
Low Potential High Potential
SWITCH
schematic symbol