Download Wednesday, June 10, 2009

PHYS 1442 – Section 001
Lecture #3
Wednesday, June 10, 2009
Dr. Jaehoon Yu
•
Chapter 16
–
•
Electric Field Lines
Chapter 17
–
–
–
–
–
Electric Potential Energy
Electric Potential
Electric Potential and Electric Field
Equi-potential Lines
The Electron Volt, a Unit of Energy
Today’s homework is homework #2, due 9pm, next Thursday!!
Wednesday, June 10, 2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
1
Announcements
• Your five extra credit points for e-mail subscription
extended till midnight tonight! Please take a full
advantage of the opportunity.
– Seven of you have subscribed so far. Thank you!!!
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– Fantastic job!!
– Remember, the due is 9pm tomorrow.
• Reading assignments: CH17 – 6 and 17 – 10
• Summer clinic hours extended to cover
– Mon – Fri: 11am – 6pm
– Saturdays: 11am – 4pm
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
2
Reminder: Special Project – Angels & Demons
• Compute the total possible energy released from an annihilation
of x-grams of anti-matter and the same quantity of matter, where
x is the last two digits of your SS#. (20 points)
– Use the famous Einstein’s formula for mass-energy equivalence
• Compute the power output of this annihilation when the energy is
released in x ns, where x is again the last two digits of your SS#.
(10 points)
• Compute how many cups of gasoline (8MJ) this energy
corresponds to. (5 points)
• Compute how many months of electricity usage it corresponds to
(3.6GJ). (5 points)
• Due by the beginning of the class Monday, June 15.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
3
Reminder: Special Project – Magnitude of Forces
• What is the magnitude of the Coulomb force one
proton exerts to another 1m away? (10 points)
• What is the magnitude of the gravitational force one
proton exerts to another 1m away? (10 points)
• Which one of the two forces is larger and by how
many times? (10 points)
• Due at the beginning of the class Monday, June 22.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
4
The Electric Field
• Both gravitational and electric forces act over
a distance without touching objects  What
kind of forces are these?
– Field forces
• Michael Faraday developed an idea of field.
– Faraday argued that the electric field extends
outward from every charge and permeates
through all of space.
• Field by a charge or a group of charges can
be inspected by placing a small test charge in
the vicinity and measuring the force on it.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
5
The Electric Field
• The electric field at any point in space is defined as the
force exerted on a tiny positive test charge divide by the
magnitude of the test charge
F
F  qE
– Electric force per unit charge
E
q
• What kind of quantity is the electric field?
– Vector quantity. Why?
• What is the unit of the electric field?
– N/C
• What is the magnitude of the electric field at a distance r
from a single point charge Q?
F
1 Q
kQq r 2
kQ

 2 
E 
2
4

r
q
q
r
0
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
6
Direction of the Electric Field
• If there are more than one charge, the individual field due to
each charge is added vectorially to obtain the total field at
any point.
ETot  E1  E2  E3  E4  ....
• This superposition principle of electric field has been verified
by experiments.
• For a given electric field E at a given point in space, we can
calculate the force F on any charge q, F=qE.
– What happens to the direction of the force and the field depending
on the sign of the charge q?
– The two are in the same directions if q>0
– The two are in opposite directions if q<0
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
7
Field Lines
• The electric field is a vector quantity. Thus, its magnitude can be
expressed in the length of the vector and the arrowhead pointing to
the direction.
• Since the field permeates through the entire space, drawing vector
arrows is not an ideal way of expressing the field.
• Electric field lines are drawn to indicate the direction of the force
due to the given field on a positive test charge.
– Number of lines crossing unit area perpendicular to E is proportional to the
magnitude of the electric field.
– The closer the lines are together, the stronger the electric field in that region.
– Start on positive charges and end on negative charges.
Earth’s G-field lines
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
8
Electric Fields and Conductors
• The electric field inside a conductor is ZERO in the static
situation. (If the charge is at rest.) Why?
– If there were an electric field within the conductor, there would be a
force on free electrons inside of it.
– The electrons will move until they reached positions where the
electric field becomes zero.
– Electric field, however, can exist inside a non-conductor.
• Consequences of the above
– Any net charge on a conductor distributes
itself on the surface.
– Although no field exists inside a conductor,
the fields can exist outside the conductor
due to induced charges on either surface
– The electric field is always perpendicular to
the surface outside of the conductor.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
9
Example 16-10
• Shielding, and safety in a storm. A hollow metal
box is placed between two parallel charged plates.
What is the field like in the box?
• If the metal box were solid
– The free electrons in the box would redistribute themselves
along the surface so that the field lines would not penetrate
into the metal.
• The free electrons do the same in hollow metal boxes
just as well as it did in a solid metal box.
• Thus a conducting box is an effective device for
shielding.  Faraday cage
• So what do you think will happen if you were inside a
car when the car was struck by a lightening?
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
10
Motion of a Charged Particle in an Electric Field
• If an object with an electric charge +q is at a point in
space where electric field is E, the force exerting on
the object is F  qE.
• What do you think will happen
to the charge?
– Let’s think about the cases like
these on the right.
– The object will move along the
field line…Which way?
– The charge gets accelerated.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
11
Example
• Electron accelerated by electric field. An electron (mass m =
9.1x10-31kg) is accelerated in the uniform field E (E=2.0x104N/C)
between two parallel charged plates. The separation of the
plates is 1.5cm. The electron is accelerated from rest near the
negative plate and passes through a tiny hole in the positive
plate. (a) With what speed does it leave the hole? (b) Show that
the gravitational force can be ignored. Assume the hole is so
small that it does not affect the uniform field between the plates.
The magnitude of the force on the electron is F=qE and is
directed to the right. The equation to solve this problem is
F  qE  ma
F qE

The magnitude of the electron’s acceleration is a 
m
m
Between the plates the field E is uniform, thus the electron undergoes a
uniform acceleration
eE 1.6  10 C  2.0  10
a

m
 9.110 kg 
19
Wednesday, June
e 10,
2009
4
N /C
  3.5 10
31
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
15
m s2
12
Example cont’d
Since the travel distance is 1.5x10-2m, using one of the kinetic eq. of motions,
v2  v02  2ax v  2ax  2  3.5 1015 1.5 102  1.0 107 m s
Since there is no electric field outside the conductor, the electron continues
moving with this speed after passing through the hole.
• (b) Show that the gravitational force can be ignored. Assume the hole is
so small that it does not affect the uniform field between the plates.
The magnitude of the electric force on the electron is



Fe  qE  eE  1.6  1019 C 2.0  104 N / C  3.2  1015 N
The magnitude of the gravitational force on the electron is


FG  mg  9.8 m s 2  9.1 1031 kg  8.9  1030 N
Thus the gravitational force on the electron is negligible compared to the
electromagnetic force.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
13
Gauss’ Law
• Gauss’ law states the relationship between electric
charge and electric field.
– More general and elegant form of Coulomb’s law.
• The electric field by the distribution of charges can be
obtained using Coulomb’s law by summing (or
integrating) over the charge distributions.
• Gauss’ law, however, gives an additional insight into
the nature of electrostatic field and a more general
relationship between the charge and the field
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
14
Electric Flux
• Let’s imagine a surface of area A through which a uniform
electric field E passes
• The electric flux is defined as
–  E=EA, if the field is perpendicular to the surface
–  E=EAcosθ, if the field makes an angle θ to the surface
• So the electric flux is defined as  E  E  A.
• How would you define the electric flux in words?
– Total number of field lines passing through the unit area
perpendicular to the field. N E  EA   E
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
15
Example of Flux
• Electric flux. (a) Calculate the electric
flux through the rectangle in the figure
(a). The rectangle is 10cm by 20cm
and the electric field is uniform with
magnitude 200N/C. (b) What is the flux
in figure if the angle is 30 degrees?
The electric flux is
 E  E  A  EA cos
So when (a) θ=0, we obtain
 E  EA cos  EA   200 N / C   0.1 0.2m 2  4.0 N  m 2 C

And when (b) θ=30 degrees, we obtain



2
2
200
N
/
C

0.1

0.2
m
cos30

3.5
N

m
C
 E  EA cos30  

Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
16
Electric Potential Energy
• Concept of energy is very useful solving mechanical problems
• Conservation of energy makes solving complex problems easier.
• When can the potential energy be defined?
– Only for a conservative force.
– The work done by a conservative force is independent of the path. What
does it only depend on??
• The difference between the initial and final positions
– Can you give me an example of a conservative force?
• Gravitational force, Spring force
• Is the electrostatic force between two charges a conservative
force?
– Yes. Why?
– The dependence of the force to the distance is identical to that of the
gravitational force.
• The only thing matters is the direct linear distance between the object not the
path.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
17
Electric Potential Energy
• What does this mean in terms of energies?
– The electric force is a conservative force.
– Thus, the mechanical energy (K+U) is conserved
under this force.
– A charged object has only the electric potential
energy at the positive plate.
– The electric potential energy decreases and turns
into kinetic energy of the charge object as the
electric force works on the charged objects and the
charged object gains speed.
• Point of greatest potential energy for
– Positively charged object
– Negatively charged object
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
PE= U
KE= 0
ME= U
U+K
0
K
K
18
Electric Potential
• How is the electric field defined?
– Electric force per unit charge: F/q
• We can define electric potential (potential) as
– The electric potential energy per unit charge
– This is like the voltage of a battery…
• Electric potential is written with the symbol V
– If a positive test charge q has potential energy Ua at
a point a, the electric potential of the charge at that
point is
Ua
Va 
q
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
19
Electric Potential
• Since only the difference in potential energy is meaningful,
only the potential difference between two points is
measurable
• What happens when the electric force does “positive work”?
– The charge gains kinetic energy
– Electric potential energy of the charge decreases
• Thus the difference in potential energy is the same as the
negative of the work, Wba, done on the charge by the electric
field to move the charge from point a to b.
• The potential difference Vba is
U b  U a Wba
Vba  Vb  Va 

q
q
PHYS 1442-001, Summer
2009 Dr.
– Electric potential is independent
of the
test charge!!
Wednesday, June 10,
2009
Jaehoon Yu
20
A Few Things about Electric Potential
• What does the electric potential depend on?
– Other charges that creates the field
– What about the test charge?
• No, the electric potential is independent of the test charge
• Test charge gains potential energy by existing in the potential created by other
charges
• Which plate is at a higher potential?
– Positive plate. Why?
• Since positive charge has the greatest potential energy on it.
– What happens to the positive charge if it is let go?
• It moves from higher potential to lower potential
– How about a negative charge?
• Its potential energy is higher on the negative plate. Thus, it moves from negative
plate to positive. Potential difference is the same.
Zero point of electric potential
unit of the electric potential is Volt (V).
can be chosen arbitrarily.
• The
• From the definition, 1V = 1J/C.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
Often the ground, a conductor
21
connected to Earth, is zero.
Example 17 – 1
A negative charge: Suppose a negative charge, such
as an electron, is placed at point b in the figure. If the
electron is free to move, will its electric potential energy
increase or decrease? How will the electric potential
change?
• An electron placed at point b will move toward the positive plate
since it was released at its highest potential energy point.
• It will gain kinetic energy as it moves toward left, decreasing its
potential energy.
• The electron, however, moves from the point b at a lower
potential to point a at a higher potential. ΔV=Va-Vb>0.
• This is because the potential is generated by the charges on the
plates not by the electron.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
22
Electric Potential and Potential Energy
• What is the definition of the electric potential?
– The potential energy difference per unit charge
• OK, then, how would you express the potential energy that a
charge q would obtain when it is moved between point a and
b with the potential difference Vba?
U b  U a  q Vb  Va   qVba
– In other words, if an object with charge q moves through a
potential difference Vba, its potential energy changes by qVba.
• So based on this, how differently would you describe the
electric potential in words?
– A measure of how much energy an electric charge can acquire in a
given situation
– A measure of how much work a given charge can do.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
23
Comparisons of Potential Energies
• Let’s compare gravitational and electric potential energies
m
•
2m
What are the potential energies of the rocks?•
– mgh and 2mgh
•
– QVba and 2QVba
Which rock has a bigger potential energy? •
– The rock with a larger mass
•
Why?
– It’s got a bigger mass.
Wednesday, June 10,
What are the potential energies of the charges?
Which object has a bigger potential energy?
– The object with a larger charge.
•
Why?
– It’s got a bigger charge.
PHYS 1442-001, Summer 2009 Dr.
24
The potential
is the same but the heavier
rockYuor larger charge can do a greater work.
2009
Jaehoon
Electric Potential and Potential Energy
• The electric potential difference gives potential energy or
possibility to do work depending on the charge of the object.
• So what is happening in batteries or generators?
– They maintain a potential difference.
– The actual amount of energy used or transformed depends on how
much charge flows.
– How much is the potential difference maintained by a car’s
battery?
• 12Volts
– If for a given period, 5C charge flows through the headlight lamp,
Joules
what is the total energy transformed?
• Etot=5C*12V=60 Umm… What is the unit?
– If it is left on twice as long? Etot=10C*12V=120J.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
25
Some Typical Voltages
Sources
Thundercloud to ground
Approximate Voltage
108 V
High-Voltage Power Lines
Power supply for TV tube
Automobile ignition
106 V
104 V
104 V
Household outlet
Automobile battery
Flashlight battery
Resting potential across nerve membrane
102 V
12 V
1.5 V
10-1 V
Potential changes on skin (EKG and EEG)
10-4 V
Wednesday, June 10, 2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
26
Example 17 – 2
Electrons in TV tube: Suppose an electron in the picture tube of a
television set is accelerated from rest through a potential difference
Vba=+5000V. (a) What is the change in potential energy of the
electron? (b) What is the speed of the electron (m=9.1x10-31kg) as a
result of this acceleration? (c) Repeat for a proton (m=1.67x10-27kg)
that accelerates through a potential difference of Vba=-5000V.
• (a) What is the charge of an electron?
–
e  1.6  1019 C
• So what is the change of its potential energy?


U  qVba  eVba  1.6 1019 C  5000V   8.0  1016 J
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
27
Example 17 – 2
• (b) Speed of the electron?
– The entire potential energy of the electron turns to its kinetic
energy. Thus the equation is
1
K  me ve2  0  W  U  eVba 
2
19

  1.6  10
ve 
2  eVba

me

C 5000V  8.0  1016 J
2  8.0 1016
7

4.2

10
m/ s
31
9.1 10
• (C) Speed of a proton?
1
K  m p v 2p  0  W  U   e   Vba   eVba  8.0 1016 J
2
2  8.0 1016
2  eVba
5
vp 

9.8

10
m/ s

27
mp
1.67 10
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
28
Electric Potential and Electric Field
• The effect of a charge distribution can be
described in terms of electric field or electric
potential.
– What kind of quantities are the electric field and the
electric potential?
• Electric Field: Vector
• Electric Potential: Scalar
– Since electric potential is a scalar quantity, it is often
easier to handle.
• Well other than the above, what are the
connections between these two quantities?
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
29
Electric Potential and Electric Field
• The potential energy is expressed in terms of a
conservative force
Ub  U a  

b
a
F  dl
• For the electrical case, we are more interested in
the potential difference:
Ub  U a

Vba  Vb  Va 
q

b
a
F
 dl  
q

b
a
E  dl
– This formula can be used to determine Vba when the
electric field is given.
• When the field is uniform
Vb  Va  

b
a
E  dl   E

b
a
dl   Ed
or
Vba   Ed
Wednesday,
June 10, field in termsPHYS
1442-001, SummerV/m
2009 Dr. Can you derive this from N/C?
30
Unit
of the electric
of potential?
2009
Jaehoon Yu
Example 17 – 3
Uniform electric field obtained from voltage:
Two parallel plates are charged to a voltage of
50V. If the separation between the plates is
5.0cm, calculate the magnitude of the electric
field between them, ignoring any fringe effect.
5cm
50V
What is the relationship between electric field and the
potential for a uniform field?
V   Ed
Solving for E
Wednesday, June 10,
2009
50V
V
50V
 1000V / m


E
2
d
5.0cm 5  10 m
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
31
Electric Potential due to Point Charges
• What is the electric field by a single point charge Q
at a distance r?
Q
1 Q
E
4 0 r
k
2
r2
• Electric potential due to the field E for moving from
point ra to rb in radial direction away from the
charge Q is
Vb  Va  


rb
ra
Q
4 0
Wednesday, June 10,
2009
E  dl  

rb
ra
Q
4 0

rb
ra
rˆ
ˆ 
 rdr
2
r
1
Q 1 1
dr 
  
2
4 0  rb ra 
r
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
32
Electric Potential due to Point Charges
• Since only the differences in potential have physical
meaning, we can choose Vb  0 at rb   .
• The electrical potential V at a distance r from a single
point charge is
1 Q
V 
4 0 r
• So the absolute potential by a single point charge
can be thought of as the potential difference by a
single point charge between r and infinity
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
33
Properties of the Electric Potential
• What are the differences between the electric potential and
the electric field?
1 Q
– Electric potential
V
4 0 r
• Electric potential energy per unit charge
• Inversely proportional to the distance
• Simply add the potential by each of the charges to obtain the total potential
from multiple charges, since potential is a scalar quantity
1 Q
– Electric field
E 
2
4

r
0
• Electric force per unit charge
• Inversely proportional to the square of the distance
• Need vector sums to obtain the total field from multiple charges
• Potential for the positive charge is large near the charge and
decreases towards 0 at a large distance.
• Potential for the negative charge is large negative near the
Wednesday, June 10,
PHYS 1442-001, Summer 2009 Dr.
34
2009
Jaehoon 0
Yu at a large distance.
charge and increases towards
Shape of the Electric Potential
• So, how does the electric potential look like as a function of
distance?
– What is the formula for the potential by a single charge?
1 Q
V 
4 0 r
Positive Charge
Negative Charge
Uniformly charged sphere would have the potential the same as a single point charge.
Wednesday, June 10,
PHYS 1442-001, Summer 2009 Dr.
35
What does
this mean? Uniformly charged sphere behaves
2009
Jaehoon Yulike all the charge is on the single point in the center.
Example 23 – 6
Work to bring two positive charges close together: What
minimum work is required by an external force to bring a
charge q=3.00μC from a great distance away (r=infinity) to a
point 0.500m from a charge Q=20.0 μC?
What is the work done by the electric field in terms of potential
energy and potential?
q Q Q
W   qVba  
  
4 0  rb ra 
Since rb  0.500m, ra  
we obtain

q Q
8.99  109 N  m2 C 2    3.00  106 C  20.00  106 C 
q Q

W 

 1.08J
  0  
0.500
m
4 0  rb
4

r

0 b
Electric force does negative work. In other words, the external force must work
+1.08J to bring the charge 3.00mC from infinity to 0.500m to the charge 20.0mC.
Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
36
Electric Potential by Charge Distributions
• Let’s consider that there are n individual point
charges in a given space and V=0 at r=infinity.
• Then the potential due to the charge Qi at a point a,
Qi 1
distance ria from Qi is
Via 
4 0 ria
• Thus the total potential Va by all n point charges is
n
n
Qi 1
Via 
Va 
i 1 4 0 ria
i 1


• For a continuous charge
distribution, we obtain
Wednesday, June 10,
2009
V 
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
1
4 0

dq
r
37
Example 23 – 8
• Potential due to a ring of charge: A thin
circular ring of radius R carries a uniformly
distributed charge Q. Determine the electric
potential at a point P on the axis of the ring a
distance x from its center.
• Each point on the ring is at the same distance from the point P.
What is the distance?
r  R2  x2
• So the potential at P is
1
dq
1
What’s this?
V 

dq 
4 0 r
4 0 r
Q
1
dq 
2
2
2
2
4 0 x  R
4 0 x  R



Wednesday, June 10,
2009
PHYS 1442-001, Summer 2009 Dr.
Jaehoon Yu
38