Transcript q - UCF Physics

LECTURE Topic 4
POTENTIAL
September 19, 2005
Alternate Lecture Titles
 Back to Physics 2048
 You can run but you can’t
hide!
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)
1 2
mg (h  b)  0  mg (b)  mv
2
1 2
mgh  mgb   mgb  mv
2
v  2 gh
y=0
A
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!!!
 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
FIELD doing the negative of this
amount of work on the particle.
Let’s look at it:
move a mass from yi to yf
Change in potential energy due to external force:
yf
Negative of the work done BY THE FIELD.
External
Field
yi
W  force  dist .
W  mg  ( y f  yi )  ( PE )


W   (mg )  ( y f  yi )  ( PE )
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
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…
 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  h =  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.
 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 Example
Designed to Create Confusion
or Understanding … Your Choice!
The change in potential energy
Final position of the electron is the negative of the work
done by the field in moving the electron
from the initial position to the final
position.
FORCE
d
W  F  ( y f  yi )
E
U  W  (e)  ( E )  ( y f  yi )
Initial position
e
A sad and confused
Electron.
negative!
negative
charge
Force against
The direction
of 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:
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 V between two points I and f in
the electric field is equal to the energy PER UNIT CHARGE
between the points:
U i U
W
V  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.
K  K f  K i  Wapplied  W field
if
K  0
Wapplied  W field
also
U  U f  U i  Wapplied
Furthermore…
U Wapplied
V 

q
q
so
Wapplied  qV
If we move a particle through a potential
difference of V, the work from an external
“person” necessary to do this is qV
Example
Electric Field = 2 N/C

1 mC
d= 100 meters
Work done by EXTERNAL agent
 Change in potential Energy.
PE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
One Step More
Work done by EXTERNAL agent
 Change in potential Energy.
PE  qEd  1mC  2( N / C ) 100m
 2 10  4 Joules
PE
Change in POTENTIAL 
q
2 10 4 Joules
J
V 
 200  200 Volts
6
110 C
C
The Equipotential Surface
DEFINED BY
V  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
x
Eparallel
Work to move a charge a distance
x along the equipotential surface
Is Q x Eparallel X x
Equipotential
Surface
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
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
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
U  U f  U i  W
 This is the NEGATIVE of the work done by the
field.
Calculation
E
q
i
f
• 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.
U  Fx  qEx
• Continuous case:
U   Fexternaldx
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.
Keep in Mind
W  F d
 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
Consider a unit charge (+) being brought from
infinity to a distance r from a
Charge q:
q
x
r
1
q
E
40 r 2
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
W   Fexternaldx  (1)q 
2
4

r
0

r
and
W
q
V

Q( 1)
40
1
q
V
40 r
dr
q r 1 r
 r 2   40 (1) 
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!!)
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
dx
dq
Charge per unit length  m or .
m  coul/meter
dq  dx
q    ( x)dx
 can be constant or a function
of position (x).
Area Charge Density s
dA=dxdy
dy
dx
charge
s
area
dq  sdA
The Job
dA  dxdy
dA=dxdy
dy
dx
R R
A

 R R
y R x
1 2
2 1 / 2
dy  R  x
(2 xdx)
2
substitute and go crazy ........
2
R
dxdy

2

Another Way
to do dA
(note rhyme)
dr
rdq
dq
r
q
dA= dr rdq
r drdq
Take a look
dA  rdrd q
A   dA   rdrd q
r

A     rdr dq
0
0

2
2  r 
r2 
2
A    dq  2    r
0
2
2
2
Still another view
dA  2rdr
r
R
dr
A  2  rdr  r
0
2
When you have to integrate
over dA



Pick a friendly coordinate system..
Use the appropriate dA
Don’t forget the function s or  may
be functions of position. The
coordinate system that you choose
should match the symmetry of these
two functions.
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

1
L  ( L2  x 2 )1/ 2
V
ln
40
d
At P
Using table of integrals

Example (from text)
z
R
disk
s=charge
per
unit
area
s
V
2 0
z
2

d

z
dz
R z
2
V
s
Ez  

z
2 0
s 
z
1 
Ez 
2
2
2 0 
z R
2
 R2  z




Which was the result we obtained earlier

The Potential From a Dipole
P
r(+)
+
V ( P)  Vi  V ()  V ()
r
r(-)
d
q
-
1  q
q 


V

40  r () r () 
q  r ( )  r (  ) 


V
40  r ()r () 
Dipole - 2
P
r(+)
r
+
r(-)
d
q
-
Geometry
r ()  r ()  d cos(q )
r (  ) r ( )  r 2
q d cos(q )
1 p cos(q )
V

2
40
r
40
r2
Where is this going?
Charges and
Forces
Electric Fields
Concept of
Potential
Batteries
and
Circuit Elements
(R,C,L)
Electric
Circuits
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?
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
s
0.1e  6
E

 5.65e3
2 0 2 x8.85e  12
Ed  V
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 = qV= 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
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
Symbol
DC Power Source
Low Potential
Potential
High
SWITCH
schematic symbol
POTENTIAL PART 5
Capacitance
What’s Portending?
 No new WebAssign, as promised.
 Old ones are still in the queue.
 Exam #1 On MONDAY..




Charge
Electric Field
Gauss
Potential
 Today .. Capacitors
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
2
1
s
5
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

6
d
1
.
12
d
d
d


q
V  8.35 x109
d
3
V  k
2
d 2 5d 2
d 
2
s  d    d 

4
4
2
d
s
5  1.12d
2
2
2
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.
3
1
4
2
diagonal  a 2  1.71a
3
1
4
2
W=qV
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
0  1  0.293  1.29  2.58k  2.32 x1010
W k
a
a
a
Chapter 26
 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
E
-Q
Area A
V=Potential Difference
+Q
Symbol
More on Capacitors
Gauss
d
q
 E  dA  
Air or Vacuum
-Q
E
+Q
Area A
V=Potential Difference
0
 EA  0 A   EA  
Gaussian
Surface
Q
0
Q   0 EA
Q
(Q / A) s
E


0 A
0
0
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.
q
C
V
or
q  CV
UNITS
 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
q
C
V
or
q  CV
Continuing…
q
C
V
 0 AV
q
d
so
0 A
C
d
 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
Coulomb 2 Coulomb 2
 0  

2
m  Joule
Nm
Coulomb 2

m  Coulomb  Volt
Coulomb Farad


m
m  Volt
and
 0  8.85 10 12 F / m  8.85 pF / m
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!
q CV (55 10 15 F )(5.3V )
6
n 


1
.
8

10
electrons
19
e
e
1.6  10 C
TWO Types of Connections
SERIES
PARALLEL
Parallel Connection
q1  C1V1  C1V
C1
C2
C3
V
q2  C2V
q3  C3V
QE  q1  q2  q3
V
CEquivalent=CE
QE  V (C1  C2  C3 )
therefore
C E  C1  C2  C3
Series Connection
q
V
-q
C1
q
-q
C2
The charge on each
capacitor is the same !
C2
V2
-q
-q
q
Series Connection Continued
V C1
q
V1
V  V1  V2
q
q
q


C C1 C 2
or
1
1
1


C C1 C 2
More General
Series
1
1

C
i Ci
Parallel
C   Ci
i
Example
C1
C2
(12+5.3)pf
V
C3
C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
More on the Big C
E=0A/d
+dq
+q
-q
 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
So….
dW  dq  Ed
Gauss
s
q 1
E

0 A 0
q 1
dW 
d  dq
A 0
Q
W U  
0
d
q2 d
q2 1
qdq 

A 0
2 A 0
2 ( A 0 )
d
or
Q 2 C 2V 2 1
U

 CV 2
2C
2C
2
Not All Capacitors are Created
Equal
 Parallel
Plate
 Cylindric
al
 Spherica
l
Spherical Capacitor
Gauss
q
 E  dA  
4r E 
2
0
q
0
q
E (r ) 
2
4r  0
surprise ???
Calculate Potential Difference V
positive. plate
Eds

V
negative. plate
q 1
V  
 2 dr
40  r 
b
a
(-) sign because E and ds are in OPPOSITE directions.
Continuing…
q
b
dr
q
1
V

( )
2

40 a r
40 r
q 1 1
q ba
V
  


40  a b  40  ab 
q
ab
C   40
V
ba
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 is
REDUCED
We can measure the C of a
capacitor (later)
C0 = Vacuum or air Value
C = With dielectric in place
C=kC0
(we show this later)
How to Check This
C0
V
V0
Charge to V0 and then disconnect from
The battery.
Connect the two together
C0 will lose some charge to the capacitor with the dielectric.
We can measure V with a voltmeter (later).
Checking the idea..
q0  C0V0
V
q1  C0V
q2  CV
q0  q1  q2
C0V0  C0V  CV
 V0 
C  C0   1
V

Note: When two Capacitors are the same (No dielectric), then V=V0/2.
Some k values
Material
Air
Dielectric
Strength
1
Breakdown
KV/mm
3
Polystyrene
2.6
24
Paper
3.5
16
Pyrex
4.7
14
Strontium
Titanate
310
8
Messing with Capacitor
The battery means that the
potential difference across
the capacitor remains constant.
+
+
V
-
-
For this case, we insert the
dielectric but hold the voltage
constant,
+
+
q=CV
V
Remember – We hold V
constant with the battery.
-
since C  kC0
qk kC0V
THE EXTRA CHARGE
COMES FROM 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
V0
-
+
q=C0Vo
When the dielectric is inserted, no charge
is added so the charge must be the same.
qk
V
-
qk  kC0V
q0  C0V0  qk  kC0V
or
V
V0
k
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..
q
Consider this virgin capacitor.
No dielectric experience.
Applied Voltage via a battery.
++++++++++++
V0
C0
-q
------------------
C0 
0 A
d
q  C0V0 
0 A
d
V0
Remove the Battery
q
++++++++++++
V0
The Voltage across the
capacitor remains V0
q remains the same as
well.
-q
-----------------The capacitor is fat (charged),
dumb and happy.
Slip in a Dielectric
Almost, but not quite, filling the space
Gaussian Surface
q
V0
-q
++++++++++++
- - - - - - - -
-q’
+ + + + + +
+q’
------------------
E0
E
E’ from induced
charges
in..small..gap
q
 E  dA 
0
q  s 
  
E0 
0 A  0 
A little-q’sheet
from the past..
+q’
- -q
-
+
q+
+
Esheet
s
q'


2 0 2 0 A
q'
0
2xEsheet
0
q'
Esheet / dialectric  2 

2 0 A  0 A
Some more sheet…
Edielectricch arg e
q
E 0
0 A
so
q  q'
E
0 A
 q'

0 A
A Few slides back
Case II – No Battery
+
q0
V0
-
+
q=C0Vo
When the dielectric is inserted, no charge
is added so the charge must be the same.
qk
V
-
qk  kC0V
q0  C0V0  qk  kC0V
or
V
V0
k
From this last equation
V
V0
k
and
V  Ed
V0  E0 d
thus
V 1 E
 
V0 k E0
E
E0
k
A Bit more…..
 q 


0 A 
V0
E0

k 

V
E  q  q' 


 0 A 
therefore
q
q  q' 
k
Important Result
 Electric Field is
Reduced by the
presence of the
material .
 The material
reduces the field
by a factor k.
E
E0
k
k is the DIELECTRIC CONSTANT
of the material
Another look
Vo
+
-
Parallel  Plate
0 A
C0 
d
 0 AV0
Q0  C0V0 
d
Electric  Field
V0
E0 
d
Q0  0V0
s0 

A
d
Add Dielectric to Capacitor
Vo
 Original Structure
+
-
+
V0
 Disconnect Battery
+
 Slip in Dielectric
Note: Charge on plate does not change!
What happens?
so +
so -
si 
si 
E0
V0 1
E

k
d k
and
V0
V  Ed 
k
Potential Difference is REDUCED
by insertion of dielectric.
Q
Q
C 
 kC0
V V0 / k 
Charge on plate is Unchanged!
Capacitance increases by a factor of k
as we showed previously
SUMMARY OF RESULTS
V
V0
E
E0
k
C  kC0
k
APPLICATION OF GAUSS’ LAW
E0 
q
0 A
q  q ' E0
E

0 A
k
E
q
k 0 A
and
q  q' 
q
k
New Gauss for Dielectrics
k
E

d
A


sometimes
  k 0
q free
0
The Insertion Process With A
Battery
Vo
+
F
-
--------