lecture topic 4 potential september 19, 2005 alternate lecture titles back to physics 2048 you can...
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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