E 1 E 2 E 3 free fall cord extending h 1 h 2 max velocity at this position e For max velocity need Bridge A h 1 160 m e max 80 m Put in numbers to find position at which max velocity occurs e then find ID: 760225
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Slide1
Bungee jump – use conservation of energy
E1 = E2 = E3
free fall
cord
extending
h
1
h
2
max velocity at this position
e
For max velocity need
Bridge A: h
1
= 160 m
e
max
= 80 m
Put in numbers to find position at which max velocity occurs e, then find
v
max
from above equation, but to do this need values for m and k
Answers to depend upon mass and spring constant but not for free fall part of motion
Slide2spark
hair
repulsion
van de
Graaff
Slide3Slide4Slide5neutron
proton
electrons
nucleus
electron orbits
Slide6Slide7glass rod (
+
)
electrons (
- )
silk cloth ( - )
wooden rod (
-
)
electrons (
- )
woollen cloth (
+
)
Slide8A
B
A
B
+ + + +
-
-
-
-
-
+
+
+
+
+
A
B
+ + + +
-
-
-
-
-
+
+
+
+
+
A
B
-
-
-
-
-
+
+
+
+
+
(1)
(2)
(3)
(4)
Slide9- - - - -
+
+
+
+
+
-
-
-
-
-
- - - - -
+
+
+
+
+
-
-
-
-
-
electrons move to the ground through finger
+
+
+
+
+
(1)
(2)
(3)
(4)
Slide10+
0
centres of positive and negative charges coincide
+ + + +
centres of positive and negative charges don’t coincide
distorted electron cloud -
atom is electrically polarized
Slide11+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
F
attraction
F
repulsion
F
attraction
>
F
repulsion
wall: atoms become polarized
-
-
-
-
-
-
-
-
balloon sticks to wall
Slide12Slide13medium
vacuum
Slide14Force of charge B on charge A
X
Y
Force of charge A on charge B
charges are of opposite sign, so, they attract each other
Putting in the numbers
Newton’s 3
rd
Law
Slide15Slide16Slide17X
Y
C
C
X
Y
Slide18X
Y
C
30
o
Slide19W.A. salt mine
Slide20Cl
-
Na
+
Na
+
Na
+
Cl
-
Cl
-
H
2
O
+
+
Slide21+
+
Slide22q
1
q
2
Slide23sphere 1
sphere 2
E
large
E
small
Slide24electric field
+ charge
electric field
- charge
+
-
Slide25Slide26photon
E = h f
nucleus
positron e
+
electron e
-
Pair Production
Annihilation
electron e
-
positron e
+
before pair annihilation
after pair annihilation
photon
photon
Slide27strong electric field
weak electric field
strong electric field
weak electric field
q
< 0 vectors electric field and force are antiparallel
q
> 0 vectors electric field and force are parallel
Slide28metal ring hit ceiling
Slide29+
++
Slide30+ + + + + + + + + + +
-
-
-
-
-
-
-
-
-
-
-
Slide31100 V
80 V
60 V
40 V
20 V
0 V
-
-
-
-
-
-
-
+
+
+
+
+
+
+
V
100 V
0 V
x
0
d
A
Slide32+ + + + + + + + + + +
-
-
-
-
-
-
-
-
-
-
-
parabolic path
straight line
path
Y
X
screen
Slide33uniform electric field region
Slide34strong
electric field
weak
electric field
Free electrons move to the left because of the influence of the external electric field. Conductor becomes polarized, creating an internal electric field that opposes the external field.
+
Slide35+
+
+
+
+
+
copper wire
electron
Slide36cross-sectional area [ m
2
]
constant:
(0, 0)
constant:
Slide37+
+
+
+
+
charged particle density [ m
-3
]
volume
V
number of charged particles of type
c
in volume element
charge on species of type
c
[ C]
Slide38N
electrons in volume
V
average drift velocity of electrons
v
[ m.s
-1
]
Conduction in a metal is due to the drift of free electrons
(0, 0)
constant:
A, q, n
Slide39t
I
0
t
I
0
t
I
0
+
-
t
I
0
+
-
varying DC current
constant DC current
ac current
sinusoidal
ac current
Slide40uniform gravitational field
X
uniform electric field
+
Slide41Slide42Slide43Slide44recall image
other images
story line drama title
Slide45y
[ m ]
0
1
2
3
#A
y
A
= 0
Y
X
#B
y
B
= 1 m
#C
y
C
= 2 m
#D
y
D
= 3 m
Slide46+ + + + + + + + + + +
-
-
-
-
-
-
-
-
-
-
-
+
A
B
uniform
0 V
Slide470 V
earth
A + + + + + + + + +
V
A
= + 9 V
B + + + + + +
VB = + 6 VC – – VC = – 2 VD – – – – – – – VC = – 7 V
high potential
low potential
potentials w.r.t earth
The
potential differences between any two points is simply the difference in the potentials between the two points
VAB = VA – VB = +9 - (+6) V = + 3 VVBA = VB – VA = +6 - (+9) V = - 3 VVCD = VC – VD = -2 - (-7) V = + 5 VVDC = V A – VB = +9 - (+6) V = + 3 VVAC = VA – VC = +9 - (-2) V = + 11 VVCA = VC – VA = -2 - (+9) V = - 11 VVAD = VA – VD = +9 - (-7) V = + 16 VVDA = VD – VA = -7 - (+9) V = - 16 V
Note: a potential difference exists when both points have the same type of charge. What is important is the extent of the charge imbalance.
Slide48+
high potential
low potential
current
positive charges
negative charges
Slide4930 V
10 V
10 V
10 V
0 V
+30 V
10 V
20 V
30 V
-10 V
-20 V
-30 V
0 V
-30 V
potential
differences
positive
potential
negative
potential
Slide50Y
X
+
A B
high potential
low potential
Slide51+
+
+
+
+
+
Slide52Mauna Loa Volcano Hawaii
height in feet
Slide53Y
X
A(-0.5, 0)
O
+(0.5, 0)
x [m]
y [m]
S(0, 0.5)
T(0.25, 0.5)
Slide54Two charges of equal magnitude and same sign
-
q
-
q
Two charges of unequal
magnitude and opposite signs
-
q
2
+q1
Two charges of unequal magnitude and same sign
+
q
1
+
q2
[2D] quadrupole – four charges of equal magnitude and alternating sign lying on the corners of a square
+
q
+
q
-
q
-
q
[1D] quadrupole of 3 charges in a straight lines
+2
q
-
q
-
q
Four equal charges at the corners of a square
+
q
+
q
+
q
+
q
A short capacitor model
Slide55S
T
T
Slide56Slide57Y
X
S
Y
X
S
negative
positive
repulsion
attraction
Slide58Y
X
+
-
The
negatively
charged object is attracted to the positive plate.
Slide59+
+
+
+
net attractive force between the two spheres A and B
A
B
Slide60+
+
+
A
B
C
net force on A
Slide61+
+
?
0.10 m
0.10 m
+
C
Slide62Y
X
electron
Direction of electric field is in the direction of the force that would act upon a positive charge
alpha particle
Slide63At each point, the direction of the
force
on a
positive charge
is in the direction of the tangent of the electric field line.
Y
X
Slide64Y
X
uniform
electric field
parabolic path of negative charge
constant force on negative charge
Y
X
uniform
electric field
constant force on negative charge
Slide65Negative charge moves in a straight line, parallel to the electric field direction. The negative charge falls with decreasing speed then stops and then rises with increasing speed due to its acceleration caused by the electrostatic force acting on it.
Y
X
uniform
electric field
constant force on negative charge
time
velocity
vectors
Slide66potential
V
potential energy
U
Slide67Slide68equipotential
A
B
Slide69large slope implies large electric field
small slope implies small electric field
equipotential lines close together implies a strong electric field
Electric field and electric potential for a positive point charge
equipotential lines across the electric field lines are right angles
Slide70hot filament
Slide71+
-
uniform electric field between the plates
+30.0 V
0.0 V
Y
X
V
[V]
y
[m]
(0,0)
30
-0.12
constant slope
constant electric field
Slide72strong
electric field
weak
electric field
Earth neutral 0V
+30 V
V
is constant throughout the conductor
Net charge resides on surface of conductor.
Electric field lines must be at right angle to surface of conductor.
Slide73electrostatic force
electrostatic
potential energy
electric field
electric
potential
V
Slide741000 V
800 V
600 V
200 V
400 V
0 V
F
E
D
C
B
A
G
Slide751000 V
800 V
600 V
200 V
400 V
0 V
F
E
D
C
B
A
+
q
=+2 C
G
Slide76Y
X
electric field
initial velocity
charge
q
< 0
electrostatic force
displacment
final velocity
The electrons moves from a lower potential to a higher
potential point
Slide77Slide78uniform electric field
electron
proton
Newton’s Second Law
Displacement after time
t
Kinetic energy
V
e
=
V
p
= V
Slide79NO
NO
YES
Slide80