Transcript electric potential

ELECTRIC POTENTIAL
February 3. 2006
Wassup?

Today
– Quick Review of Exam
– Start New Topic – Electric Potential
– New WebAssign on board.
– Read the chapter for Monday

Monday, Wednesday
– More of the same
Friday – QUIZ and some problems
 Monday – Finish Topic (??) + Problems

Potential

We will be dealing with
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Work
Energy
We do work if we try to move a charge in an
electric field.
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
Does the FIELD do work?
Does the FIELD have a brain? Muscles?
How Much Work?
W=qeE ??
W=mgd ???
Please Recall
Work in moving an object from A to B
B
W   F  dr
A
The PHY 2048 Brain Partition
To move the mass m from the ground to
a point a distance h above the ground
requires that work be done on the particle.
h
h
B
m
Reference “0”
W   mgdy  mgh
0
W is the work done by an external force.
mgh represents this amount of work and
is the POTENTIAL ENERGY of the mass
at position h above the ground.
A
The reference level, in this case, was chosen
as the ground but since we only deal with
differences between Potential Energy Values,
we could have chosen another reference.
Let’s Recall Some more PHY2048
A mass is dropped from a height h above the
ground. What is it’s velocity when it strikes
the ground?
We use conservation of energy to compute the answer.
h
B
m
A
1 2
mgh  (0)  (0)  mv
2
and
v  2 gh
Result is independent of the mass m.
Using a different reference.
y
y=h
E  PE  KE
B
m
y=b (reference level)
y=0
A
Still falls to here.
1 2
mg (h  b)  0  mg (b)  mv
2
1 2
mgh  mgb  mgb  mv
2
v  2 gh
Energy Methods
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
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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!!!

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).
FIELD
You can also think of this as the
doing the negative of this amount of work on the
particle.
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
Work done by external force = mgh
How much work here by
gravitational field?
h

mg
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…
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A conservative force is NOT a Republican.
An External Agent is NOT 007.
Electric 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=PE
HIGH U
LOWER U
E

q
F
A
B
d
Work done by FIELD is Fd
Negative of the work done by the FIELD is -Fd
Change in Potential Energy is also –Fd.
The charge sort-of “fell” to lower potential energy.
Gravity
Negative of the work done by the FIELD is –mg D h = D U

mg
Bottom Line:
Things tend to fall down and lower
their potential energy. The change,
Uf – Ui is NEGATIVE!
Electrons have those *&#^
negative signs.
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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 point
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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:
F
E
q
VECTOR
We define ELECTRICAL POTENTIAL as the
POTENTIAL ENERGY PER UNIT CHARGE:
U
V
q
SCALAR
UNITS OF POTENTIAL
U
Joules
V 
 VOLT
q Coulomb
Watch those #&@% (-) signs!!
The electric potential difference DV between two points I and f in
the electric field is equal to the energy PER UNIT CHARGE
between the points:
U i DU
W
DV  V f  Vi 



q
q
q
q
Uf
Where W is the work done BY THE FIELD in moving the charge from
One 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.
DK  K f  K i  Wapplied  W field
if
DK  0
Wapplied  W field
also
DU  U f  U i  Wapplied
Furthermore…
DU Wapplied
DV 

q
q
so
Wapplied  qDV
If we move a particle through a potential
difference of DV, the work from an external
“person” necessary to do this is qDV
Example
Electric Field = 2 N/C

1 mC
d= 100 meters
Work done by EXTERNAL agent
 Change in potential Energy.
DPE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
One Step More
Work done by EXTERNAL agent
 Change in potential Energy.
DPE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
DPE
Change in POTENTIAL 
q
2 10 4 Joules
J
DV 
 200  200 Volts
6
110 C
C
The Equipotential Surface
DEFINED BY
DV  0
It takes NO work to move a charged particle
between 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 Yesteryear
Field Lines and Equipotentials
Electric
Field
Equipotential
Surface
Components
Enormal
Electric
Field
Dx
Eparallel
Work to move a charge a distance
Dx along the equipotential surface
Is Q x Eparallel X Dx
Equipotential
Surface
BUT
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
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
This an EQUIPOTENTIAL Surface
No work is needed since DV=0 for such a
surface.
Consequently Eparallel=0
E must be perpendicular to the equipotential
surface
Therefore
E
E
E
V=constant
Field Lines are Perpendicular to the
Equipotential Lines
Equipotential
Work external  q0 (V f  Vi )
Consider Two Equipotential
Surfaces – Close together
Work to move a charge q from a to b:
dWexternal  Fapplied  ds   qEds
b
a
E
ds
also
dWexternal  q (V  dV )  V   qdV
 Eds  dV
V+dV and
dV
V
E
ds
Vector...E  V
Where



  i  j k
x
y
z
I probably won’t ask about this.
Typical Situation
Keep in Mind
W  F d
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Force and Displacement are
VECTORS!
Potential is a SCALAR.
UNITS
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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
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
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
Consider a unit charge (+) being brought from
infinity to a distance r from a
Charge q:
q
x
r
1
q
E
2
40 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.
The math….
()1 dr
DW   Fexternaldx  (1)q 
2
4

r
0

r
and
DW
q
dr
q r 1 r
V


2

Q( 1)
40  r
40 (1) 
r
1
q
V
40 r
For point charges
qi
V

40 i ri
1
Example: Find potential at P
q1
q2
1 1
V
(q1  q2  q3  q4 )
40 r
d  1.3m
d
r
 0.919m
2
d
P
r
q3
q4
q1=12nC q2=-24nC q3=31nC q4=17nC Sq=36 x 10-9C
V=350 Volts (check the arithmetic!!)
An Example
finite line of charge
x
dx
r
1
dx
dV 
40 (d 2  x 2 )1/ 2
d
1
dx
V
40 0 (d 2  x 2 )1/ 2
L
P
and
What about a rod
that goes from –L to +L??

1
L  ( L2  x 2 )1/ 2
V
ln
40
d
At P
Using table of integrals

Example (from text)
z
R
disk
=charge
per
unit
area

V
2 0
z
2

d

z
dz
R z
2
V

Ez  

z
2 0
 
z
1 
Ez 
2
2
2 0 
z R
2
 R2  z




Which was the result we obtained earlier

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?
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
(b) If this entire charge undergoes a potential
difference of 12 V, how much energy is involved?
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
qEd  54volts

0.1e  6
E

 5.65e3
2 0 2 x8.85e  12
Ed  DV
54
3
d
 9 x10 meters
5.65e3
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)
Energy = qDV= 2.3 e+09 x 43C=98.9 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
E=(1/2)Mv2 v= sqr(2E/M)= 14,100 m/s
POTENTIAL PART 5
Capacitance
Encore By Special Request
Where for art
thou, 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
1
2
3
5
6
s
4
qi
  2q  2q 3q  3q  5q  5q 
 k


r
d
/
2
1
.
12
d


i
i
q
q
  8q 10q 


V  k


k

8

8
.
93

0
.
93
k

d
1
.
12
d
d
d


q
V  8.35 x109
d
V  k
Text gets 8.49 … one of us is right!
2
d 2 5d 2
d 
2
2
2
s  d    d 

2
4
4
 
d
s
5  1.12d
2
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.
3
1
4
2
diagonal  a 2  1.71a
3
1
4
2
W=qDV
V1  0
q
V2  k 

 a 
W1  0
q2
W2  qV2  k
a
q 
q
1 
q
q
V3  k 

k
1 
  0.293k
a 
a
2
 a a 2
q2
W3  qV3  0.293k
a
q 
q
q
q q
V4  k   
  k 1  1  .707   1.29k
a
a
a a a 2
q2
W4  (q )V4  1.29k
a
Add them up ..
W  W1  W2  W3  W4
2
q2
q2
q
W  k 0  1  0.293  1.29   2.58k
 2.32 x1010
a
a
a