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E 10 cm 1cm e -e. Calculate E y here. Cathode Ray Tube Conducting Paper +10 Volts 0 Volts A + B + C +

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Presentation on theme: "E 10 cm 1cm e -e. Calculate E y here. Cathode Ray Tube Conducting Paper +10 Volts 0 Volts A + B + C +"— Presentation transcript:

1 e 10 cm 1cm e -e

2 Calculate E y here.

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

4 ExEx EyEy + -

5 + V OUT V IN -

6 V OUT V IN

7 V=6 volts V=8 volts V=4 volts V=2 volts V=0 volts V=-2 volts = 1cm E=?

8 6 V 5 V 4 V abc def ghi

9 3 V A.B. C.

10 3 C 4 C 3 cm -5 C

11 3 C 4 C 3 cm -5 C q

12 3 C 4 C 3 cm -5 C q

13 3 V 6 V 1,000 

14 Baseball Diamond Heuristic of Electrostatics Equations

15 (Usually find with Gauss’s Law.) (Remember: V, the electric potential, has units of energy per unit charge.) (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 2 charges at a distance r from each other.) (The force between 2 charges at a distance r from each other.) Note: F, U, E, and V are all functions of position. RELATING IMPORTANT CONCEPTS

16

17 3 V B. C.

18 3 V 1  2  3 

19 3 V 1  2  3 

20 V R1R1 R2R2

21 V R DMM V DMM V R1R1 R DMM V DMM

22 3 V 1  2  3  1 

23 100  200  100  10 

24 100  200  100  10  V

25 V R1R1 R2R2

26 3 V 1  2  3  1  2 

27 3 V 1  2  3  1  2  1 

28 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 =?

29 SCOPESCOPE tV

30 t V motor t V resistor =|V source |-V motor T t1t1 t2t2 V motor,on V resistor,on on off on off

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

32 OSCOPEOSCOPE Y-axis: Voltage (0.5 volts per division) X-axis: Time (1 second per division) 0 1.5

33

34 R C + - red1 bottom ground red2

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

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

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

38 200  100  red 1 red 2 (channel inverted) black (middle ground) + - 200  100  red 1 black (bottom ground) red 2 + -

39 R C V source (t)

40 SCOPESCOPE

41 Y-axis: Voltage (5 volts per division) X-axis: Time (3 millisecond per division)

42

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

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

45 I I

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

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

48 4 V 0.5  A B 1.5  2.5  4  4 V20 V

49 12 V 1  2  3  1  2  1  2  BATTERY A B

50 6  1  2  S 2  F 6 V

51 +/-Q? NS V velocity

52 NS Direction of I ?

53 NS V velocity

54 NS Direction of I ?

55 V receiver, amplitude f f maximum transmission

56 I resistor,amplitude f drive f resonance

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

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

59 10  L

60 C L

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

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

63 C R red 1 ground red 2 + - C R V source (t)=V source amp sin(  D t) + -

64 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

65 Function Generator RF Modulator INOUT Variable Capacitor Speaker Diode

66 Speaker Diode Solenoid ASolenoid B

67 Speaker Diode 3,600

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

69 I2I2 I1I1 P d1d1 d2d2 I W H D

70 2.0 Amp 1.0 Amp P 1.0 meter 2.0 meter

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

72 6  1  2  S 2  6 V

73 I I r a

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

75

76 D1D1 D2D2

77 (use more frames if necessary) Cartoon Frames

78 30 V Ground 1000 V 2000 V 3000 V to ground constant voltage charge separation + + + + +- - - - - - -

79 x V a (x) -200 100 -100 200 xixi xfxf

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

81 + - - + EaEa E d,v E d,h - +

82 V d Volts 0 Volts d w v f,z x y coordinates z

83 x y V d Volts 0 Volts d w v f,z v f,y yy

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

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

86 I V

87 I V

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

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

90 I through R

91 Energy (eV) Momentum Generic Plot of Energy Bands for Semiconductor conduction band (empty) valence band (filled with electrons)  E is called Band Gap Energy

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

93 C R V source Q Cap (0)=0 Q Cap (∞)= Q Max C Q Cap (0)=Q o =2 [coul] R

94 t V Cap (t) Delineate vertical scale:

95 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!

96 R V source motor red 1 red 2 black

97 C R red 1 ground red 2 + - C R V source (t)=V source amp sin(  D t) + -


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