e 10 cm 1cm e -e

Calculate E y here.

Cathode Ray Tube Conducting Paper +10 Volts 0 Volts A + B + C +

ExEx EyEy + -

+ V OUT V IN -

V OUT V IN

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 abc def ghi

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 

Baseball Diamond Heuristic of Electrostatics Equations

(Usually easier to use Gauss’s Law.) (V POINT not really useful.  V usually measured or otherwise derived.) (scalar) (vector) (Usually change in electrical potential is related to work on the particle.)

3 V B. C.

3 V 1  2  3 

V R1R1 R2R2

V R DMM V DMM V R1R1 R DMM V DMM

3 V 1  2  3  1 

100  200  100  10 

100  200  100  10  V

V R1R1 R2R2

3 V 1  2  3  1  2 

3 V 1  2  3  1  2  1 

9 V R 1 = 1  R 2 = 2  R 3 = 3  R 4 =4  I 4 =? V 4 =? I 1 =? V 1 =?I 2 =? V 2 =? I 3 =? V 3 =? I Battery =? R effective =?

SCOPESCOPE tV

t V motor t V resistor T t1t1 t2t2 V motor,on V resistor,on on off on off

OSCOPEOSCOPE Voltage (0.5 volts per div) Time (1 second per div)

R C + - red1 bottom ground red2

R V source (t) = V MAX sin(  t+  ) where V MAX = 5 Volts  /(2  ) = 1,000 Hz

V source (t) = V MAX sin(  t+  ) where V MAX = 5 Volts  /(2  ) = 1,000 Hz

V amp =3 V 330  CH1CH2 red1 red2 bottom ground x-y mode

200  100  red 1 red 2 (inverted) middle ground  100  red 1 bottom ground red 2 + -

R C V source (t) = V MAX sin(  t+  )

SCOPESCOPE

I I I Magnet B B Close is strong B Far is ~ zero Magnet B I L

I I I I I I Beginning Position180 o Rotated Position current direction reversed (so is force on wire)

I I I Alternating Contacts current direction always the same (so is force on wire) DC Power Supply + - these wires fixed brushes allow good contact as loop rotates

I A. N I B. S I C. N S S NS N N SS N

4 V 0.5  A B 1.5  2.5  4  4 V20 V

12 V 1  2  3  1  2  1  2  BATTERY A B

6  1  2  S 2  F 6 V

Direction of I ? NS V velocity

NS Direction of I ?

NS V velocity

SN Direction of I ?

V receiver, amplitude f f maximum transmission

I resistor,amplitude f drive f resonance

R = 2,000  C = 15  F V source amplitude = 15 V L = 75 mH f drive = 750 Hz

R [Ohm] C [Farad] V source L [Henry]

10  L

C L

I resistor,amplitude f drive f resonance Same L and C with lower R

L R red 1 red 2 ground C R red 1 ground red 2

C R red 1 ground red C R V source (t)=V source amp sin(  D t) + -

Pulses let through by the diode move speaker with frequency of desired audio wave. Quantum mechanical turn-on voltage of diode. Modulate Wave Transmitted by Diode to Speaker

Function Generator RF Modulator INOUT Variable Capacitor Speaker Diode

Speaker Diode Solenoid ASolenoid B

Speaker Diode 3,600

RF Modulator INGROUND Variable Capacitor Speaker Diode (This is just to provide a ground.) external antenna

I2I2 I1I1 P d1d1 d2d2 I W H D

2.0 Amp 1.0 Amp P 1.0 meter 2.0 meter

Current carrying region 2. Current carrying region 1. Non-conducting material a b c

6  1  2  S 2  6 V

I I r a

A. N B. S C. N S S NS N N SS N

D1D1 D2D2

(use more frames if necessary) Cartoon Frames

30 V Ground 1000 V 2000 V 3000 V to ground constant voltage charge separation

x V a (x)

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

EaEa E d,v E d,h - +

V d Volts 0 Volts d w v o,x x y coordinates z

V d Volts 0 Volts d w x y coordinates v o,x v f,y yy

V d,y Volts 0 Volts d w x y coordinates v o,x v f,y yy VaVa L  y’ DyDy acceleration in x-direction acceleration in y-direction while crossing deflection plates constant motion while crossing remaining distance to screen

S (magnet) B S B OR N (magnet) B N B OR Assessment #1Assessment #2

I through  V applied

I LED  V applied (many various applied V’s) (a non-Ohmic graph) V TURN ON ÷ R

VRVR  V applied (many various applied V’s)

I through R

 V applied

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 V source Q Cap (0)=0 Q Cap (∞)= Q Max C Q Cap (0)=Q o R Q Cap (∞)= 0

t V Cap (t) Delineate vertical scale:

Algebraic EquationDifferential Equation y+3 = 2 (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) (-1)+3 = 2…True! (check solution by plugging point into original algebraic equation) (check solution by plugging function into original differential equation) …True!