Download PHYS 1443 * Section 501 Lecture #1

PHYS 1442 – Section 004
Lecture #5
Wednesday January 29, 2014
Dr. Andrew Brandt
•
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•
•
•
Weds., Jan. 29, 2014
CH 17
Electric Potential due to Point Charges
Shape of the Electric Potential
Equi-potential Lines and Surfaces
Electron-volt
Capacitance
PHYS 1442-004, Dr. Andrew Brandt
1
Announcements
a) Avg was 81 on HW. Good job, you should dominate that material on the
test.
b) New HW assigned on ch 17
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
2
Electric Potential and Potential Energy
• The electric potential difference gives potential energy (or
the possibility to do work) based 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,
what is the total energy transformed?
C*J/C=J (Joules)
• Etot=5C*12V=60 What is the unit?
– If it is left on twice as long? Etot=10C*12V=120J.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
3
Example 17 – 2
Electrons in a 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
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
4
Example 17 – 2
• (b) Speed of the electron?
– The entire potential energy of the electron is transformed into
kinetic energy. Thus the equation is
1
K  me ve2  0  W  U 
2

(U  eVba )

  8.0  1016 J  8.0  1016 J
ve 
2  K

me
2  8.0 1016
7

4.2

10
m/ s
31
9.1 10
• (c) Speed of a proton that accelerates through V=-5000V?
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
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
5
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 often
can make problem solving easier.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
6
Example 17 – 3
Uniform electric field obtained from voltage:
Two parallel plates are charged to a voltage of
50 V. If the separation between the plates is
5.0 cm, calculate the magnitude of the electric
field between them, ignoring any fringe effects.
What is the relationship between electric field and the
potential for a uniform field? W   Fd  ( Eq)d
5cm
50V
V  U / q  W / q  Ed
V  Ed
Solving for E
Weds., Jan. 29, 2014
50V
V
50V
 1000V / m


E
2
d
5.0cm 5  10 m
PHYS 1442-004, Dr. Andrew Brandt
7
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 from a single point charge
depends only on the magnitude of the point charge
and the distance from it
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
8
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 from each of the charges to obtain the total potential
from multiple charges, since potential is a scalar quantity
r
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 a positive charge is large near the charge and
decreases to 0 at large distances.
• Potential for the negative charge is small (large magnitude but
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PHYS 1442-004, Dr. Andrew Brandt
9
negative) near the charge and increases with distance to 0
Shape of the Electric Potential
• So, what 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
A uniformly charged sphere would have the same potential as a single point charge.
Weds., Jan. 29, 2014
What does this mean?
PHYS 1442-004, Dr. Andrew Brandt
10
Uniformly charged sphere behaves like all the charge is on the single point in the center.
Example 17.5
Work to bring two positive charges close together: What
minimum work is required by an external force to bring a
charge q=3.00 μC from a great distance away ( r   ) to a
point 0.500 m 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  

q Q
q Q
W

0




4 0  rb
4 0 rb

8.99  10


9
we obtain


N  m 2 C 2  3.00  106 C 20.00  106 C
0.500m
  1.08J
In other words, the external force must input work of +1.08J to bring the charge
3.00C
from infinity to 0.500m
from the 20.0C charge.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
11
More on Example 17-5
Work to bring two positive charges close together: What
minimum work is required by an external force to bring a
charge q=3.00 μC from a great distance away ( r   ) to a
point 0.500 m 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  

q Q
q Q
W

0




4 0  rb
4 0 rb

8.99  10


9
we obtain


N  m 2 C 2  3.00  106 C 20.00  106 C
0.500m
  1.08J
In other words, the external force must input work of +1.08J to bring the charge
3.00C
from infinity to 0.500m
from the 20.0C charge.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
12
Electrostatic Potential Energy: Two charges
• What is the electrostatic potential energy of a configuration of
charges? (Choose V=0 at r=
– If there are no other charges around, a single point charge Q1 in
isolation has no potential energy and feels no electric force
• If a second point charge Q2 is to a distance r12 from Q1 ,the
Q1 1
potential at the position of Q2 is V  4 r
0 12
• The potential energy of the two charges relative to V=0 at r= 
1 Q1Q2
is
U  Q2V  4 r
0
12
-- This is the work that needs to be done by an external force to bring Q2
from infinity to a distance r12 from Q1.
– It is also a negative of the work needed to separate them to infinity.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
13
Electrostatic Potential Energy: Three Charges
• So what do we do for three charges?
• Work is needed to bring all three charges together
– There is no work needed to bring Q1 to a certain place without
the presence of any other charge
1 Q1Q2
– The work needed to bring Q2 to a distance to Q1 is U12  4 r
0
12
– The work need to bring Q3 to a distance to Q1 and Q2 is
U 3  U13  U 23
Q1Q3
1 Q2 Q3


4 0 r13
4 0 r23
1
• So the total electrostatic potential of the three charge
system is
U  U12  U13  U 23
Weds., Jan. 29, 2014
1  Q1Q2 Q1Q3 Q2 Q3 





4 0  r12
r13
r23 
PHYS 1442-004, Dr. Andrew Brandt
V  0 at r   
14
Equi-potential Surfaces
• Electric potential can be visualized using equipotential lines in
2-D or equipotential surfaces in 3-D
• Any two points on equipotential surfaces (lines) have the same
potential
• What does this mean in terms of the potential difference?
– The potential difference between the two points on an equipotential
surface is 0.
• How about the potential energy difference?
– Also 0.
• What does this mean in terms of the work to move a charge
along the surface between these two points?
– No work is necessary to move a charge between these two points.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
15
Equi-potential Surfaces
•
An equipotential surface (line) must be perpendicular to the electric field. Why?
– If there are any parallel components to the electric field, it would
require work to move a charge along the surface.
•
•
•
Since the equipotential surface (line) is perpendicular to the electric field, we can draw
these surfaces or lines easily.
There can be no electric field inside a conductor in static case, thus the entire volume of
a conductor must be at the same potential.
So the electric field must be perpendicular to the conductor surface.
Parallel
Plate
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Point
charges
PHYS 1442-004, Dr. Andrew Brandt
Just like a topographic map
16
Electrostatic Potential Energy: electron Volt
• What is the unit of electrostatic potential energy?
– Joules
• Joules is a very large unit in dealing with electrons, atoms or
molecules
• For convenience a new unit, electron volt (eV), is defined
– 1 eV is defined as the energy acquired by a particle carrying the
charge equal to that of an1eV
electron
across
a
19
 1.6 (q=e)
 1019when
C  1V it moves
1.6  10 J
potential difference of 1V.
– How many Joules is 1 eV then?
• eV however is not a standard SI unit. You must convert the
energy to Joules for computations.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
17
Capacitors (or Condensers)
• What is a capacitor?
– A device that can store electric charge without letting the charge flow
• What does it consist of?
– Usually consists of two oppositely charged conducting objects (plates or
sheets) placed near each other without touching
– Why can’t they touch each other?
• The charges will neutralize each other
• Can you give some examples?
– Camera flash, surge protectors, computer keyboard, binary circuits…
• How is a capacitor different than a battery?
– Battery provides potential difference by storing energy (usually chemical
energy) while the capacitor stores charge but very little energy.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
18
Capacitors
• A simple capacitor consists of a pair of parallel plates
of area A separated by a distance d.
– A cylindrical capacitors are essentially parallel plates
wrapped around as a cylinder.
• Symbols for a capacitor and a battery:
– Capacitor -||– Battery (+) -|l- (-)
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Circuit
Diagram
PHYS 1442-004, Dr. Andrew Brandt
19
Capacitors
• What do you think will happen if a battery is connected
(voltage is applied) to a capacitor?
– The capacitor gets charged quickly, one plate positive and the other
negative with an equal amount. of charge
• Each battery terminal, the wires and the plates are
conductors. What does this mean?
– All conductors are at the same potential.
– the full battery voltage is applied across the capacitor plates.
• So for a given capacitor, the amount of charge stored in the
capacitor is proportional to the potential difference Vba
between the plates. How would you write this formula?
Q  CVba
C is a property of a capacitor so does not depend on Q or V.
– C is a proportionality constant, called capacitance of the device.
PHYS 1442-004, Dr. Andrew Brandt
20
Normally use F or pF.
– What is the unit? C/V or Farad (F)
Weds., Jan. 29, 2014
Determination of Capacitance
• C can be determined analytically for capacitors w/ simple
geometry and air in between.
• Let’s consider a parallel plate capacitor.
– Plates have area A each and separated by d.
• d is smaller than the length, so E is uniform.
– For parallel plates E=s/0, where s is the surface charge density.
• E and V are related Vba  Ed
• So from the formula:
– What do you notice?
Weds., Jan. 29, 2014
0 A
Q
Q
C


Vba Qd  0 A
d
C only depends on the area
(A) and the separation (d) of
the plates and the permittivity
of the medium between them.
PHYS 1442-004, Dr. Andrew Brandt
21
Example 17-7
Capacitor calculations: (a) Calculate the capacitance of a capacitor whose
plates are 20 cm x 3.0 cm and are separated by a 1.0 mm air gap. (b) What
is the charge on each plate if the capacitor is connected to a 12 V battery? (c)
What is the electric field between the plates? (d) Estimate the area of the
plates needed to achieve a capacitance of 1F, given the same air gap.
(a) Using the formula for a parallel plate capacitor, we obtain
0 A
C

d

 8.85  1012 C 2 N  m2

0.2  0.03m2
12 2

53

10
C N  m  53 pF
3
1  10 m
(b) From Q=CV, the charge on each plate is


Q  CV  53  1012 C 2 N  m 12V   6.4  1010 C  640 pC
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
22
More 17-7
(c) Using the formula for the electric field in two parallel plates
Q
s
6.4  1010 C
4
4


1.2

10
N
C

1.2

10
V m
E 
 0 A 0 6.0  103 m 2  8.85  1012 C 2 N  m 2
Or, since V  Ed we can obtain
12V
V
4

1.2

10
V m
E 
3
d 1.0  10 m
(d) Solving the capacitance formula for A, we obtain
0 A
C
d
Solve for A
1F  1  103 m
Cd
8 2
2

10
m

100
km
A

0
9  1012 C 2 N  m2


About 40% the area of Arlington (256km2).
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
23
PHYS 1444 – Section 003
Lecture #8
Thursday Sep. 20, 2012
Ian Howley
•
Weds., Jan. 29, 2014
Chapter 24
•
Capacitors and Capacitance
•
Capacitors in Series and Parallel
•
Energy Stored in Capacitors
•
Dielectrics
24
PHYS 1442-004, Dr.
Andrew Brandt
Capacitor Cont’d
• A single isolated conductor can be said to have a
capacitance, C.
• C can still be defined as the ratio of the charge to absolute
potential V on the conductor.
– So Q=CV.
• The potential of a single conducting sphere of radius rb can
be obtained as
1 1
Q
  
where ra  
4

r
0 b
 rb ra 
• So its capacitance is C  Q / V  4 0 rb
Q
V
4 0
• Although it has capacitance, a single conductor is not considered to be
a capacitor, as a second nearby charge is required to store charge
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
25
Electric Energy Storage
• A charged capacitor stores energy.
– The stored energy is the work done to charge it.
• The net effect of charging a capacitor is removing one type of
charge (+ or -) from a plate and moving it to the other plate.
– Battery does this when it is connected to a capacitor.
• Capacitors do not charge immediately.
– Initially when the capacitor is uncharged, no work is necessary to
move the first bit of charge. Why?
• Since there is no charge, there is no field that the external work needs to
overcome.
– When some charge is on each plate, it requires work to add more
charge due to electric repulsion.
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
26
Example 24 – 7
Energy store in a capacitor: A camera flash unit stores
energy in a 150F capacitor at 200V. How much electric
energy can be stored?
Use the formula for stored energy.
Umm.. Which one?
What do we know from the problem?
C and V
1
So we use the one with C and V: U  CV 2
2


1
1
2
2
6
U  CV  150  10 F  200V   3.0 J
2
2
How do we get J from
Weds., Jan. 29, 2014
FV2?
C 2
J


FV    V  CV  C    J
V 
C
2
PHYS 1442-004, Dr. Andrew Brandt
27
Electric Energy Density
• The energy stored in a capacitor can be considered as being
stored in the electric field between the two plates
• For a uniform field E between two plates, V=Ed and C=0A/d
• Thus the stored energy is
1
1
1  0 A 
2
2
2
U  CV  
Ed


E
Ad
0
 
2 d 
2
2
• Since Ad is the gap volume, we can obtain the energy
density, stored energy per unit volume, as
1
2
u  0 E
2
Valid for plates
with a vacuum gap
Electric energy stored per unit volume in any region of space
PHYS 1442-004, Dr. Andrew Brandt
is proportional to the square of the electric field in that region.
Weds., Jan. 29, 2014
28
Dielectrics
• Capacitors generally have an insulating sheet of
material, called a dielectric, between the plates to
– Increase the breakdown voltage above that in air
– Allows the plates get closer together without touching
• Increases capacitance ( recall C=0A/d)
– Also increases the capacitance by the dielectric constant
C  KC0
• Where C0 is the intrinsic capacitance when the gap is vacuum,
and K or  is the dielectric constant
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
29
Dielectrics
• The value of dielectric constant varies depending on
material (Table 24 – 1)
K for vacuum is 1.0000
K for air is 1.0006 (this is why permittivity of air and
vacuum are used interchangeably.)
• The maximum electric field before breakdown occurs
is the dielectric strength. What is its unit?
V/m
• The capacitance of a parallel plate capacitor with a
dielectric (K) filling the gap is
A
C  KC0  K  0
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
d30
Dielectrics
• A new quantity, the permittivity of dielectric, is defined
as =K0
• The capacitance of a parallel plate with a dielectric
medium filling the gap is
A
C 
d
• The energy density stored in an electric field E in a
dielectric is
1
1 2
2
u  K0 E   E
2
2
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
Valid for any space
with dielectric of
permittivity .
31
Effect of a Dielectric Material
• Let’s consider the two cases below:
Case #1 :
constant V
Case #2 :
constant Q
• Constant voltage: Experimentally observed that the total charge on
each plate of the capacitor increases by K as the dielectric material
is inserted between the gap  Q=KQ0
– The capacitance increased to C=Q/V0=KQ0/V0=KC0
• Constant charge: Voltage found to drop by a factor K  V=V0/K
– The capacitance increased to C=Q0/V=KQ0/V0=KC0
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
32
Capacitors in Series or Parallel
• Capacitors are used in many electric circuits
• What is an electric circuit?
– A closed path of conductors, usually wires connecting capacitors and
other electrical devices, in which
• charges can flow
• there is a voltage source such as a battery
• Capacitors can be connected in various ways.
– In parallel
Weds., Jan. 29, 2014
and
in Series
PHYS 1442-004, Dr. Andrew Brandt
or in combination
33
Capacitors in Parallel
• Parallel arrangement provides the same
voltage across all the capacitors.
– Left hand plates are at Va and right hand
plates are at Vb
– So each capacitor plate acquires charges
given by the formula
• Q1=C1V, Q2=C2V, and Q3=C3V
• The total charge Q that must leave battery is then
 Q=Q1+Q2+Q3=V(C1+C2+C3)
• Consider that the three capacitors behave like a single
“equivalent” one
For capacitors in parallel the capacitance is the
 Q=CeqV= V(C1+C2+C3)
sum of the individual capacitors
Weds., Jan.
2014
PHYS 1442-004, Dr. Andrew
Brandt
• Thus
the29,equivalent
capacitance
in parallel
is
Ceq  C1  C2  34
C3
Capacitors in Series
• Series arrangement is more “interesting”
 When battery is connected, +Q flows to the left plate of
C1 and –Q flows to the right plate of C3
 This induces opposite sign charges on the other plates.
 Since the capacitor in the middle is originally neutral,
charges get induced to neutralize the induced charges
 So the charge on each capacitor is the same value, Q. (Same charge)
• Consider that the three capacitors behave like an equivalent one
 Q=CeqV  V=Q/Ceq
• The total voltage V across the three capacitors in series must be equal to
the sum of the voltages across each capacitor.
 V=V1+V2+V3=(Q/C1+Q/C2+Q/C3)
• Putting all these together, we obtain:
• V=Q/Ceq=Q(1/C1+1/C2+1/C3)
1
1
1
1



• Thus the equivalent capacitance is Ceq C1 C2 C3
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
The total capacitance is smaller than the smallest C!!!
35
Example 24 – 5
Equivalent Capacitor: Determine the
capacitance of a single capacitor that will
have the same effect as the combination
shown in the figure. Take C1=C2=C3=C.
We should do these first!!
How?
These are in parallel so the equivalent capacitance is:
Ceq  C3  C2  2C
Now the equivalent capacitor is in series with C1.
1
1
1
1
1
3
2
C



Solve for Ceq
 
Ceq 
Ceq Ceq1 C2 2C C 2C
3
Weds., Jan. 29, 2014
PHYS 1442-004, Dr. Andrew Brandt
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