Transcript Physics 30 – Unit 2 Forces and Fields – Part 2

Physics 30 – Unit 2 Forces and
Fields – Part 2
To accompany
Pearson Physics
Electric Fields
• QuickLab 11.1 Shielding Cellular Phones
• Watch electric cable inspector video
• Ancient Greeks: “violent” and “natural” forces
• Effluvium Theory
• Field Theory developed to explain forces at a
distance: gravity, electrostatic, and magnetic forces
• Quantum Theory necessary to understand “how”
these forces work
Electric Fields
• Because electric force direction will vary depending on
the type of charge on the object, it is necessary to
define electric field direction
• It is defined as the direction of force on a positive test
charge
• As the diagram shows, field direction is away from the
(+) charge and towards the (-) charge
Electric Fields
• The symbol for electric field is
E
• The symbol for electric field strength is
E
• Note the difference between this and the symbol for
energy which is a scalar! If you get them confused,
the consequences can be disastrous
Electric Fields
Fe
E
q
where q is the charge of the charge in the electric
field (the test charge) not the field source
• Review Example 11.1, page 548 and try Practice
Problem 1, page 548
Electric Fields
• Practice Problem 1, page 548
E 
Fe
q
1.00  10 N / C 
3
Fe
1.60  10 19 C
Fe  1.00  103 N / C 1.60  10 19 C  1.60  10 16 N
• This question asked for the magnitude of the field
therefore the absolute value signs were used
Electric Fields
• If q1 represents the field source and q2 the test
charge, the electric field due to q1 is
k q1 q2
Fe
2
k q1
kq
r
E

 2 commonly written as E  2
q2
q2
r
r
• Review Example 11.2, page 549
• Try Practice Problem 1, page 549
Electric Fields
• Practice Problem 1, page 549
E
kq
r2
N m2
8.99  10
q
2
C
40.0 N / C 
2
0.0200
m


9
40.0 N / C   0.0200 m 
q
 1.78  10 12 C
2
Nm
8.99  10 9
C2
2
• Because the question states that the field is
directed away from the charge, the charge must be
(+)
Electric Fields
• Review Example 11.3, page 550
• Try Practice Problem 1, page 550
Electric Fields
• Practice Problem 1, page 550
x
.
2.10 x 10-2 m
• What is the field at point X?
• Find the field due to A and the field due to B and
add them together
• They are both in the same direction – to the left
Electric Fields
2
N
m
6
8.99  10 9

1.50

10
C
2
kq
7
C
E  2 

3.06

10
N / C left
2
A
r
 0.0210 m 
N m2
6
8.99  10

2.00

10
C
2
kq
6
C
E  2 

6.17

10
N / C left
2
B
r
 0.0210 m  0.0330 m 
9
E
at X
 3.06  107 N /C  6.17  106 N /C  3.67  107 N /C left
Electric Fields
• Do Check and Reflect, page 553, questions
1, 2, 4, 5 and
• SNAP p. 81, questions 2, 3, 5,
7-9, 13
Electric Field Lines
• Drawing electric field lines:
Text gives rules and rationale on page 554
• Light particles sprinkled in oil will line up if an
electric field is set up within the oil
• Read pages 555 – 559
• At the simplest level the field lines show the path of
movement of a (+) charge if placed along a field line
Electric Fields
• Diagrams of electric fields can be drawn using the
principles given
• Example:
+
-
• Try the following: a single positive charge, 2
negative charges, a (+)ly charge hollow sphere, a
(+)ly charged plate with a (-)ly charged plate
• Discuss
Electric Potential
• Electrical potential energy is similar to gravitational
potential energy
• Gravitational potential energy is easier to
understand
• We’ll use a comparison between the two to help you
understand electrical
Electric Potential
Gravitational
Electrical
work done in lifting an object is stored
as gravitational potential energy
work done in moving a charge with
respect to another charge is stored as
electrical potential energy
W   E p  F d
W   E p  F d
Reference point → surface of earth
Reference point → charges in contact
∆Ep between surface of earth and final
point
∆Ep between 2 charges in contact and
in the final position
∆Ep for macroscopic objects easy to
measure and sensible
∆Ep for sub-microscopic objects easy to
measure but not sensible for individual
particles
No analogous concept for gravitational
Electric potential (voltage) defined as
V
E p
q
1 Volt = 1 J/C
Electric Potential
• Electric Potential Difference (commonly called
voltage) is the difference in electric potential
between any two points
V  Vfinal  Vinitial
• For example the potential difference between the
(-) and (+) electrodes of an alkaline dry cell is
1.5 V
Electric Potential
V
E p
q
can be rewritten as
E p  qV
e-
• Basis for a non-SI unit used for
energy of subatomic particles
• If one electron was accelerated
across a potential difference of 1 V it would
have:
Ep  qV 1e 1V 1 eV energy
• 1 eV = 1.60 x 10-19 J (page 2 of Data Sheets)
1V
Electric Potential
• Review Example 11.8, page 566
• Do Practice Problem 2, page 566
Electric Potential
• Practice Problem 2, page 566
Ep  qV 1e  4.00  104 V  4.00  104 eV
4.00  10 4 eV 1.60  10 19 J / eV  6.40  10 15 J
Electric Field between Parallel
Plates
•
kq
E 2
r
is not valid for electric field between
parallel plates
• Recall that
W  E p  F d
for both electric and gravitational fields
• Also recall that
Fe  q E
• Put the 2 together and you get a new formula for
electric field between parallel plates:
Electric Field between Parallel
Plates
 E p  q E d
 Ep
q
 E d
V  E d
V
E
d
Field between parallel plates is constant
everywhere between the plates
Electric Field between Parallel
Plates
• Electric field between parallel plates has units V/m
• Earlier you used N/C for electric field units
• Your book shows on page 568, that these are really
the same units
• You should be capable of showing this
Electric Field between Parallel
Plates
• Review Example 11.9, page 568
• Do Practice Problem 2, page 568
Electric Field between Parallel
Plates
• Practice Problem 2, page 568
V
E
d
V
5.00  10 3 m
V  3.00  106 V / m  5.00  10 3 m  1.50  10 4 V
3.00  106 V / m 
Electric Field between Parallel
Plates
• Do Check and Reflect, p. 569
• Questions 10a, 11, 12
Conservation of Energy and
Electric Charges
• Review Example 11.10, page 571
• Do Practice Problem 2, page 571
Conservation of Energy and
Electric Charges
• Practice Problem 2, page 571
smaller charge initially
at rest
larger charge
q = -2.00 μC
m = 1.70 x 10-3 kg
E p i  Ek i  E p f  Ek f
E p i  0  0  1.70  10  5.20  10 m / s 
1
2
E p i  2.30  106 J
3
4
2
Conservation of Energy and
Electric Charges
• Concept Check, page 572
initial motion perpendicular to plates
initial motion parallel to plates
or
Conservation of Energy and
Electric Charges
• Review Example 11.11, page 572
• Do Practice Problem 1, page 573
Conservation of Energy and
Electric Charges
• Practice Problem 1, page 573
E p i  Ek i  E p f  Ek f
qV  0  0  21  m  v 2
3.20  10 19 C  4.00  10 4 V  0  0  21  6.65  10 27 kg  v 2
19
4
3.20

10
C

4.00

10
V 2
12
2
2
v2 

3.85

10
m
/
s
6.65  10 27 kg
m2
6
v  3.85  10

1.96

10
m/ s
2
s
12
Conservation of Energy and
Electric Charges
• Review Example 11.12, page 574
F q E
F  2.6  10 12 C 1.7  105 V / m  4.4  10 7 N
F
4.4  10 7 N
8
2
a 

1.5

10
m
/
s
m 3.0  10 15 kg
Note: at this acceleration, it would be
travelling at half the speed of light in
1 s! Why is this not possible?
d  v i t  21 at 2
d  0  21 1.5  10 8 m / s 2   6.0  10 6 s 
d  2.6  10 3 m
2
Conservation of Energy and
Electric Charges
• Do Check and Reflect, page 575
• Questions 1, 2, 3, 7
• SNAP p. 90 1, 3, 4, 6, 7, 8, 10, 12, 14, 16
Conservation of Energy and
Electric Charges