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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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10 cm

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10 cm

e

10 cm

-e

1cm

e


10 cm

Calculate

Ey here.


10 cm

Cathode Ray Tube Conducting Paper

C

+

B

+10 Volts

+

A

+

0 Volts


10 cm

Ey

Ex

+

-


10 cm

VOUT

-

+

VIN


10 cm

VOUT

VIN


10 cm

V=8 volts

= 1cm

V=6 volts

V=4 volts

E=?

V=-2 volts

V=0 volts

V=2 volts


10 cm

a

b

c

6 V

d

e

f

5 V

g

h

i

4 V


10 cm

A.

B.

3 V

3 V

C.

3 V


10 cm

3 C

-5 C

4 C

3 cm


10 cm

3 C

-5 C

4 C

q

3 cm


10 cm

3 C

-5 C

4 C

q

3 cm


10 cm

3 V

6 V

1,000 


10 cm

Baseball Diamond Heuristic

of Electrostatics Equations


10 cm

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 “work-energy 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.


10 cm

B.

3 V

3 V

C.

3 V


10 cm

3 V

3 

2 

1 


10 cm

3 V

3 

2 

1 


10 cm

V

R1

R2


10 cm

V

R1

VDMM

RDMM

V

VDMM

RDMM


10 cm

3 V

3 

2 

1 

1 


10 cm

100 

200

100 

10 


10 cm

V

100 

200

100 

10 


10 cm

V

R1

R2


10 cm

3 V

3 

2 

1 

1 

2 


10 cm

3 V

3 

2 

1 

1 

1 

2 


10 cm

Reffective=?

R4=4 

I4=? V4=?

9 V

R1= 1 

R2= 2 

R3= 3 

IBattery=?

I1=? V1=?

I2=? V2=?

I3=? V3=?


10 cm

t

V

S

C

O

P

E


10 cm

Vmotor

on

on

on

on

off

off

off

off

off

Vmotor,on

t

t1

t2

Vresistor=|Vsource|-Vmotor

on

on

on

on

off

off

off

off

off

Vresistor,on

t

T


10 cm

Voltage

(0.5 volts per div)

O

S

C

O

P

E

Time

(1 second

per div)


10 cm

Y-axis: Voltage (0.5 volts per division)

1.5

O

S

C

O

P

E

X-axis: Time

(1 second

per division)

0


10 cm

red1

R

+

red2

-

C

bottom

ground


10 cm

R

VR (t) = -Vsource (t) = -VMAXsin(t)

Vsource (t) = VMAXsin(t)

where

VMAX = 5 Volts

/(2) = 1,000 Hz


10 cm

Vsource (t) = VMAXsin(t)

where

VMAX = 5 Volts

/(2) = 1,000 Hz


10 cm

red1

x-y

mode

Vamp=3 V

red2

CH1

CH2

330 

bottom

ground


10 cm

red 1

red 1

100 

100 

+

+

red 2

(channel

inverted)

black

(bottom

ground)

black

(middle

ground)

red 2

-

-

200 

200 


10 cm

R

Vsource (t)

C


10 cm

S

C

O

P

E


10 cm

Y-axis: Voltage (5 volts per division)

X-axis: Time

(3 millisecond

per division)


10 cm

I

L

I

B

I

I

B

BFar is ~ zero

BClose is strong

Magnet

Magnet

Magnet


10 cm

Beginning Position

180o Rotated Position

I

I

I

I

I

I

current direction

reversed (so is

force on wire)


10 cm

I

I


10 cm

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)


10 cm

A.

B.

C.

S

N

S

N

S

N

I

I

I

S

N

N

N

S

S


10 cm

1.5 

A

4 V

0.5 

4 

1.5 

4 V

20 V

B

2.5 


10 cm

A

1 

2 

2 

3 

BATTERY

12 V

1 

1 

2 

B


10 cm

S

2 F

6 V

6 

2 

1 


10 cm

S

N

Vvelocity

+/-Q?


10 cm

S

N

Vvelocity

Direction of I ?


10 cm

S

N

Vvelocity

Direction of I ?


10 cm

S

N

Vvelocity

Direction of I ?


10 cm

Vreceiver, amplitude

f

fmaximum

transmission


10 cm

Iresistor,amplitude

fdrive

fresonance


10 cm

Vsource amplitude = 15 V

fdrive = 750 Hz

R = 2,000 

C = 15 F

L = 75 mH


10 cm

R [Ohm]

Vsource

C [Farad]

L [Henry]


10 cm

10 

L


10 cm

C

L


10 cm

Iresistor,amplitude

Same L and C with lower R

fdrive

fresonance


10 cm

R

ground

red 1

C

red 2

red 1

R

red 2

ground

L


10 cm

R

+

Vsource(t)=Vsource ampsin(Dt)

C

-

R

ground

+

red 1

C

-

red 2


10 cm

Modulate Wave Transmitted by Diode to Speaker

Quantum mechanical

turn-on voltage of diode.

Pulses let through by the diode move speaker with

frequency of desired audio wave.


10 cm

Function

Generator

RF

Modulator

Diode

OUT

IN

OUT

Speaker

Variable

Capacitor


10 cm

Diode

Speaker

Solenoid A

Solenoid B


10 cm

Diode

3,600

Speaker


10 cm

(This is just to

provide a ground.)

RF

Modulator

external antenna

Diode

IN

GROUND

Speaker

Variable

Capacitor


10 cm

W

H

D

I

P

d1

d2

I2

I1


10 cm

1.0 meter

P

1.0 Amp

2.0 meter

2.0 Amp


10 cm

Current carrying region 2.

Non-conducting

material

Current carrying region 1.

c

a

b


10 cm

S

2 

2 

6 V

6 

1 


10 cm

a

r

I

I


10 cm

A.

B.

C.

S

N

S

N

S

N

S

N

N

N

S

S


10 cm

D2

D1


10 cm

Cartoon Frames

(use more frames if necessary)


10 cm

-

+

constant

voltage

-

charge

separation

-

-

+

-

+

+

-

-

+

  • 30 V

  • Ground

  • 1000 V

  • 2000 V

  • 3000 V

to ground


10 cm

Va(x)

200

100

xi

xf

x

-100

-200


10 cm

-

-

+

+

{outward}

+

{upward}

-

{accelerated}

“{upward}” and “{outward}” describe

which way the electron is deflected.


10 cm

-

Ea

+

-

+

Ed,h

Ed,v

+

-


10 cm

coordinates

y

Vd Volts

x

z

d

0 Volts

w

vf,z


10 cm

Vd Volts

d

0 Volts

w

vf,y

y

vf,z

y

x

coordinates


10 cm

y

z

coordinates

Vd,y Volts

y’

vf,y

Dy

-

+

vf,z

y

d

-

+

0 Volts

Va

w

L

acceleration

in z-direction

acceleration

in y-direction

while crossing

deflection plates

constant motion

while crossing

remaining distance

to screen


10 cm

Assessment #1

Assessment #2

(magnet)

(magnet)

(magnet)

S

S

B

N

B

B

OR

OR

(magnet)

B

N


10 cm

V

I


10 cm

I

V


10 cm

ILED

(a non-Ohmic graph)

÷ R

Vapplied

(many various

applied V’s)

VTURN ON


10 cm

VR

Vapplied

(many various

applied V’s)


10 cm

R

Ithrough


10 cm

Generic Plot of Energy Bands for Semiconductor

Energy (eV)

conduction band

(empty)

E is called Band Gap Energy

valence band

(filled with electrons)

Momentum


10 cm

R

R

QCap(0)=Qo

C

QCap(∞)= 0

QCap(0)=0

Vsource

C

QCap(∞)= QMax


10 cm

R

QCap(0)=0

Vsource

C

QCap(∞)= QMax

R

QCap(0)=Qo=2 [coul]

C


10 cm

VCap(t)

Delineate vertical scale:

t


10 cm

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)


10 cm

red 1

red 2

R

Vsource

black

motor


10 cm

R

+

Vsource(t)=Vsource ampsin(Dt)

C

-

R

ground

+

red 1

C

-

red 2


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