e. 10 cm. e. 1cm. e. Calculate E y here. Cathode Ray Tube Conducting Paper. C. +. B. +10 Volts. +. A. +. 0 Volts. E y. E x. +. . V OUT. . +. V IN. V OUT. V IN. V=8 volts. = 1cm. V=6 volts. V=4 volts. E=?. V=2 volts. V=0 volts. V=2 volts. a. b. c. 6 V. d.
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e
10 cm
e
1cm
e
Calculate
Ey here.
Cathode Ray Tube Conducting Paper
C
+
B
+10 Volts
+
A
+
0 Volts
Ey
Ex
+

VOUT

+
VIN
VOUT
VIN
V=8 volts
= 1cm
V=6 volts
V=4 volts
E=?
V=2 volts
V=0 volts
V=2 volts
a
b
c
6 V
d
e
f
5 V
g
h
i
4 V
A.
B.
3 V
3 V
C.
3 V
3 C
5 C
4 C
3 cm
3 C
5 C
4 C
q
3 cm
3 C
5 C
4 C
q
3 cm
3 V
6 V
1,000
Baseball Diamond Heuristic
of Electrostatics Equations
RELATING IMPORTANT CONCEPTS
(vector)
(The force between 2 charges at a distance r from each other.)
(vector)
(scalar)
(The potential energy stored in having 2
charges at a distance r from each other.)
(Usually find with Gauss’s Law.)
(The change of electric potential a particle experiences moving from one position to another can be used to find the change in its kinetic energy via the “workenergy theorem”: K = W.)
(scalar)
(Remember: V, the electric potential, has units of energy per unit charge.)
Note: F, U, E, and V are all functions of position.
B.
3 V
3 V
C.
3 V
3 V
3
2
1
3 V
3
2
1
V
R1
R2
V
R1
VDMM
RDMM
V
VDMM
RDMM
3 V
3
2
1
1
100
200
100
10
V
100
200
100
10
V
R1
R2
3 V
3
2
1
1
2
3 V
3
2
1
1
1
2
Reffective=?
R4=4
I4=? V4=?
9 V
R1= 1
R2= 2
R3= 3
IBattery=?
I1=? V1=?
I2=? V2=?
I3=? V3=?
t
V
S
C
O
P
E
Vmotor
on
on
on
on
off
off
off
off
off
Vmotor,on
t
t1
t2
Vresistor=VsourceVmotor
on
on
on
on
off
off
off
off
off
Vresistor,on
t
T
Voltage
(0.5 volts per div)
O
S
C
O
P
E
Time
(1 second
per div)
Yaxis: Voltage (0.5 volts per division)
1.5
O
S
C
O
P
E
Xaxis: Time
(1 second
per division)
0
red1
R
+
red2

C
bottom
ground
R
VR (t) = Vsource (t) = VMAXsin(t)
Vsource (t) = VMAXsin(t)
where
VMAX = 5 Volts
/(2) = 1,000 Hz
Vsource (t) = VMAXsin(t)
where
VMAX = 5 Volts
/(2) = 1,000 Hz
red1
xy
mode
Vamp=3 V
red2
CH1
CH2
330
bottom
ground
red 1
red 1
100
100
+
+
red 2
(channel
inverted)
black
(bottom
ground)
black
(middle
ground)
red 2


200
200
R
Vsource (t)
C
S
C
O
P
E
Yaxis: Voltage (5 volts per division)
Xaxis: Time
(3 millisecond
per division)
I
L
I
B
I
I
B
BFar is ~ zero
BClose is strong
Magnet
Magnet
Magnet
Beginning Position
180o Rotated Position
I
I
I
I
I
I
current direction
reversed (so is
force on wire)
I
I
DC Power Supply
+

I
these
wires
fixed
I
brushes
allow good
contact as
loop rotates
I
current direction
always the same (so
is force on wire)
A.
B.
C.
S
N
S
N
S
N
I
I
I
S
N
N
N
S
S
1.5
A
4 V
0.5
4
1.5
4 V
20 V
B
2.5
A
1
2
2
3
BATTERY
12 V
1
1
2
B
S
2 F
6 V
6
2
1
S
N
Vvelocity
+/Q?
S
N
Vvelocity
Direction of I ?
S
N
Vvelocity
Direction of I ?
S
N
Vvelocity
Direction of I ?
Vreceiver, amplitude
f
fmaximum
transmission
Iresistor,amplitude
fdrive
fresonance
Vsource amplitude = 15 V
fdrive = 750 Hz
R = 2,000
C = 15 F
L = 75 mH
R [Ohm]
Vsource
C [Farad]
L [Henry]
10
L
C
L
Iresistor,amplitude
Same L and C with lower R
fdrive
fresonance
R
ground
red 1
C
red 2
red 1
R
red 2
ground
L
R
+
Vsource(t)=Vsource ampsin(Dt)
C

R
ground
+
red 1
C

red 2
Modulate Wave Transmitted by Diode to Speaker
Quantum mechanical
turnon voltage of diode.
Pulses let through by the diode move speaker with
frequency of desired audio wave.
Function
Generator
RF
Modulator
Diode
OUT
IN
OUT
Speaker
Variable
Capacitor
Diode
Speaker
Solenoid A
Solenoid B
Diode
3,600
Speaker
(This is just to
provide a ground.)
RF
Modulator
external antenna
Diode
IN
GROUND
Speaker
Variable
Capacitor
W
H
D
I
P
d1
d2
I2
I1
1.0 meter
P
1.0 Amp
2.0 meter
2.0 Amp
Current carrying region 2.
Nonconducting
material
Current carrying region 1.
c
a
b
S
2
2
6 V
6
1
a
r
I
I
A.
B.
C.
S
N
S
N
S
N
S
N
N
N
S
S
D2
D1
Cartoon Frames
(use more frames if necessary)

+
constant
voltage

charge
separation


+

+
+


+
to ground
Va(x)
200
100
xi
xf
x
100
200


+
+
{outward}
+
{upward}

{accelerated}
“{upward}” and “{outward}” describe
which way the electron is deflected.

Ea
+

+
Ed,h
Ed,v
+

coordinates
y
Vd Volts
x
z
d
0 Volts
w
vf,z
Vd Volts
d
0 Volts
w
vf,y
y
vf,z
y
x
coordinates
y
z
coordinates
Vd,y Volts
y’
vf,y
Dy

+
vf,z
y
d

+
0 Volts
Va
w
L
acceleration
in zdirection
acceleration
in ydirection
while crossing
deflection plates
constant motion
while crossing
remaining distance
to screen
Assessment #1
Assessment #2
(magnet)
(magnet)
(magnet)
S
S
B
N
B
B
OR
OR
(magnet)
B
N
V
I
I
V
ILED
(a nonOhmic graph)
÷ R
Vapplied
(many various
applied V’s)
VTURN ON
VR
Vapplied
(many various
applied V’s)
R
Ithrough
Generic Plot of Energy Bands for Semiconductor
Energy (eV)
conduction band
(empty)
E is called Band Gap Energy
valence band
(filled with electrons)
Momentum
R
R
QCap(0)=Qo
C
QCap(∞)= 0
QCap(0)=0
Vsource
C
QCap(∞)= QMax
R
QCap(0)=0
Vsource
C
QCap(∞)= QMax
R
QCap(0)=Qo=2 [coul]
C
VCap(t)
Delineate vertical scale:
t
Algebraic Equation
Differential Equation
y+3 = 2
(involves coordinate y)
(involves a function y(t)
and it’s parameter t)
y = 1
(solution is a point/number)
(solution is a function of t)
…True!
(1)+3 = 2
…True!
(check solution by plugging point
into original algebraic equation)
(check solution by plugging function
into original differential equation)
red 1
red 2
R
Vsource
black
motor
R
+
Vsource(t)=Vsource ampsin(Dt)
C

R
ground
+
red 1
C

red 2