e

10 cm

1cm

e

-e

CalculateEy here.

Cathode Ray Tube Conducting Paper

+10 Volts

0 Volts

A+

B+

C+

Ex

Ey

+ -

+

VOUT

VIN

-

VOUT

VIN

V=6 volts

V=8 volts

V=4 volts

V=2 volts

V=0 volts

V=-2 volts

= 1cm

E=?

6 V

5 V

4 V

a b c

d e f

g h i

3 V 3 V

3 V

A. B.

C.

3 C

4 C

3 cm

-5 C

3 C

4 C

3 cm

-5 C

q

3 C

4 C

3 cm

-5 C

q

3 V

6 V

1,000

F on q 2

by q1

kq1q2

r2ˆ r

E caused

by q1

kq1

r2ˆ r

Ufrom q1and q 2

kq1q2

r

Vfrom q1k

q1

r

Baseball Diamond Heuristicof Electrostatics Equations

E caused

by q1

kq1

r2ˆ r

Ufrom q1and q 2

kq1q2

r

Vfrom q1point charge

kq1

r

F on q 2

by q1

q2

E from q1

Won q 2 q2V

(Usually find with Gauss’s Law.)

(Remember: V, the electric potential, has units of energy per unit charge.)

F

U

xˆ x

U

yˆ y

U

zˆ z

U(r ) U0

F d

s

r 0

r

E

V

xˆ x

V

yˆ y

V

zˆ z

V (r ) V0

E d

s

r 0

r

(scalar)

(scalar)

(vector)

(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 “work-energy theorem”: K = W.)

(The potential energy stored in having 2charges at a distance r from each other.)

(The force between 2 charges at a distance r from each other.)

3 V 3 V

3 V

B.

C.

3 V1 2 3

3 V1 2 3

VR1 R2

VRDMM

VD

MM

VR1 RDMM

VD

MM

3 V1 2 3

1

100 200 100

10

100 200 100

10

V

VR1

R2

3 V1 2 3

1 2

3 V1 2 3

1 2

1

9 V R1= 1 R2= 2 R3= 3

R4=4

I4=? V4=?

I1=? V1=? I2=? V2=? I3=? V3=?IBattery=?

Reffective=?

SCOPE

t V

t

Vmotor

t

Vresistor=|Vsource|-Vmotor

T

t1 t2

Vmotor,on

Vresistor,on

on on on onoff off off offoff

on on on onoff off off offoff

OSCOPE

Voltage(0.5 volts per div)

Time(1 secondper div)

OSCOPE

Y-axis: Voltage (0.5 volts per division)

X-axis: Time(1 second

per division)0

1.5

t (sec) V (Volts)0

0.0010.0030.0050.0070.0090.0110.0130.0150.0170.0190.0210.0230.025

t (sec) V1 (Volts) V2 (Volts)

00.0030.0060.0090.0120.015

R

C

+

-

red1

bottomground

red2

RVsource (t) = VMAXsin(t)

where VMAX = 5 Volts/(2) = 1,000 Hz

VR (t) = -Vsource (t) = -VMAXsin(t)

Vsource (t) = VMAXsin(t)

where VMAX = 5 Volts/(2) = 1,000 Hz

Vamp=3 V

330

CH1 CH2

red1

red2

bottomground

x-ymode

200

100

red 1

red 2(channelinverted)

black(middleground)

+

-

200

100

red 1

black(bottomground)

red 2

+

-

R

C

Vsource (t)

SCOPE

Y-axis: Voltage (5 volts per division)

X-axis: Time(3 millisecondper division)

f VR,MAX VC,MAX XC

I

I

I

Magnet

B

Magnet

BClose is strong

BFar is ~ zero

Magnet

B

IL

I

I

I

I

I

Beginning Position 180o Rotated Position

current direction

reversed (so is

force on wire)

I

I

I

I

current direction

always the same (so

is force on wire)

DC Power Supply+ -

these wiresfixed

brushesallow goodcontact as

loop rotates

I

A.

NI

B.

SI

C.

N

S

S

N S

N

N

S S

N

4 V 0.5

A

B

1.5

1.5

2.5

4

4 V 20 V

12 V

1 2 3

1

2

1

2

BA

TT

ER

YA

B

6

1

2

S

2 F6 V

+/-Q?

NS

Vvelocity

NS

Vvelocity

Direction of I ?

Direction of I ?

NS

Vvelocity

NS

Vvelocity

Direction of I ?

Vreceiver, amplitude

ffmaximum

transmission

Iresistor,amplitude

fdrive

fresonance

R = 2,000

C = 15 F

Vsource amplitude = 15 V

L = 75 mH

fdrive = 750 Hz

R [Ohm]

C [Farad]Vsource

L [Henry]

10

L

C

L

Iresistor,amplitude

fdrive

fresonance

Same L and C with lower R

L

R

red 1

red 2 ground

C

R

red 1ground

red 2

C

R

red 1ground

red 2

+

-

C

R

Vsource(t)=Vsource ampsin(Dt)

+

Pulses let through by the diode move speaker withfrequency of desired audio wave.

Quantum mechanical turn-on voltage of diode.

Modulate Wave Transmitted by Diode to Speaker

FunctionGenerator

RFModulator

IN OUT

OUT

VariableCapacitor

SpeakerDio

de

SpeakerDio

de

Solenoid A Solenoid B

SpeakerDio

de

3,600

RFModulator

IN GROUND

VariableCapacitor

SpeakerDio

de

(This is just to provide a ground.)external antenna

I2I1

P

d1 d2

I

W

H

D

2.0 Amp

1.0 Amp

P1.0 meter

2.0 meter

Current carrying region 2.

Current carrying region 1.Non-conducting

material

ab

c

6

1

2

S

2 6 V

II

r

a

A.

N

B.

S

C.

N

S

S

N S

N

N

S S

N

D1

D2

(use more frames if necessary)

Cartoon Frames

• 30 V• Ground• 1000 V• 2000 V• 3000 V

to ground

constantvoltage

chargeseparation

+++++- ---

- -

-

x

Va(x)

-200

100

-100

200

xi xf

+

-

-

+{upward}

{outward}

“{upward}” and “{outward}” describewhich way the electron is deflected.

- +

{accelerated}

+

-

-

+

Ea

Ed,v

Ed,h

- +

Vd Volts

0 Volts

d

w

vf,z

x

ycoordinates

z

x

y

coordinates

Vd Volts

0 Volts

d

w

vf,z

vf,y

y

Vd,y Volts

0 Volts

d

w

z

y

coordinates

vf,z

vf,y

y

-

-

+

+

Va L

y’Dy

accelerationin z-direction

accelerationin y-direction

while crossingdeflection plates

constant motionwhile crossing

remaining distanceto screen

S

(magnet)

B

S

(magnet)

BOR

N

(magnet)

B

N

(magnet)

B

OR

Assessment #1 Assessment #2

I

V

I

V

ILED

Vapplied(many variousapplied V’s)

(a non-Ohmic graph)

VTURN ON

÷ R

VR

Vapplied(many variousapplied V’s)

Ithrough

R

Energy (eV)

Momentum

Generic Plot of Energy Bands for Semiconductor

conduction band(empty)

valence band(filled with electrons)

E is called Band Gap Energy

C

R

VsourceQCap(0)=0

QCap(∞)= QMax

C QCap(0)=Qo

R

QCap(∞)= 0

C

R

VsourceQCap(0)=0

QCap(∞)= QMax

C QCap(0)=Qo=2 [coul]

R

t

VCap(t)

Del

inea

te v

ertic

al s

cale

:

Algebraic Equation Differential Equation

y+3 = 2

dy(t)

dt2y(t)

(involves a function y(t)and it’s parameter t)

(involves coordinate y)

y = -1(solution is a point/number) (solution is a function of t)

y(t) e2t

(-1)+3 = 2 …True!

(check solution by plugging point into original algebraic equation)

(check solution by plugging function into original differential equation)

d e2t dt

2e2t …True!

R

Vsource motor

red 1

red 2

black

C

R

red 1ground

red 2

+

-

C

R

Vsource(t)=Vsource ampsin(Dt)

+