Find magnitude and direction

Charge expressed in sum of protons and electronsq = e ! Np " Ne( )

Coulomb s Law

F = kq1q2r2

k = 8.99 !109 N !m2

C 2

k = 14" #0

$0 = 8 85 !10%12 C 2

N !m2

Equilibrium Position (charge at origin and elsewhere is positive)

F1!3 = F2!3

kq1q3

x3 " x1( )2= k

q3q2x2 " x3( )2

x3 =q1x2 + q2 x1q1 + q2

Charged balls hang from ceiling with insulated ropes find massT sin! " Fe = 0 T cos! " Fg = 0

Fg = mg Fe = kq2

d 2

sin! =d / 2l

Fe = kq2

d 2= k q2

4l2 sin2!T sin!T cos!

=FeFg

m =kq2

4gl2 sin2! tan!

Force between electrons of electrostatic force and gravitational forceFeFg

=kq2eGm2

e

Bead on a wire slanted find mass of second bead

Fe = kq1q2d 2

Fg = m2gsin!

kq1q2d 2

= m2gsin!

m2 =k q1q2d 2gsin!

Fx = kq1q4d 2

+ kq2q42d( )2

cos45° = kq4d 2

q1 +q22cos45°!

"#$%&

Fy = kq2q42d( )2

sin 45° ' kq3q4d 2

=kq4d 2

q22sin 45° + q3

!"#

$%&

F = F2x + F

2y tan( =

FyFx

F =kq4d 2

q1 +q22cos45°!

"#$%&2

+q22sin 45° + q3

!"#

$%&2

( = tan'1

q22sin 45° + q3

!"#

$%&

q1 +q22cos45°!

"#$%&

Four charged objects find force on fourth charge

Chapter 21: Electrostatics

Electric Field from a point chargeEx =

F1,x + F2,x + ...+ Fn,xq0

Ey =F1,y + F2,y + ...+ Fn,y

q0Electric field do particles at center from 4 point charges

!Ecenter ,x = k

q2r2(! x) + k q3

r2(x) = k

r2q3 ! q2( )

!Ecenter ,y = k

q1r2(! y) + k q

r2(y) = k

r2q ! q1( )

Ecenter =!E2center ,x +

!E2center ,y

"kr2

q3 ! q2( )2 + q ! q1( )2 = 2ka2

q3 ! q2( )2 + q ! q1( )2

# = tan!1 q ! q1q3 ! q2

$%&

'()

Electric field everywhere on x axis (dipole)

E =q4!"0

1

x # 12d$

%&'()2 #

1

x + 12d$

%&'()2

*

+

,,,,

-

.

////

Vector electric dipole moment

!p = q

!d

Electric field far away from electric dipoleE =

p2!"0x

3

d = (10!10m)cos52 5° = 0 6 "10!10mp = 2ed = 2 "10!29Cm

Electric dipole moment of water

Electric field from three point charges

E1 = kq1b2!x

E3 = kq3a2!y

E2 =kq2 cos!a2 + b2

"#$

%&'!x +

kq2 sin!a2 + b2

"#$

%&'!y

E = E2x + E

2y

! = pE sin"Torque on electric dipole

Electric flux! = EAcos"

! =

!E "d!A"##

Guass s Law

! =

!E "d!A ="##

q$0

cyclindrical symmetryE =

!2"#0r

=2k!r

planar symmetryE =

!2"0

planar symmetry conductorE =

!"0

Charge distributed uniformly throughout sphere insideEins de =

!r13"0

Eins de =Qr1

4#"0R3 =

kQr1R3

Eoutside =kQr22

outside

Electric field from a ring of chargeEz (z) =

kdqr2! cos" = k dq

r2zr!

r = z2 + R2 dq = q!Ez (z) =

kqzz2 + R2( )3/2

ay =Fm

=qEm

y = 12at 2 L = vt

a = 2y(L / v)2

=2yv2

L2

m =qEa

Mass of inkdrop on inkjet

Chapter 23: Electric PotentialConstant electric field

Electric field

Electric field at x

!E(!x) =

!Fq0

= kqr2r

Charge in a cube one of the sides!oneface =

!6=16Q"0

Chapter 22: Electric Field

W = q!E !!d = qEd cos"

displacement same direction as e # fieldW = qEd $U = #qEddisplacement opposite direction as e # fieldW = #qEd $U = qEd

V =Uq

!V = "We

q

Electric Potential

K = !"U = ! q1V + q2 (!V )[ ]K =

12mv2 # v = 2K

m

Energy of tandem acc

!V = "We

q= 0#V is cons tan t

0 Work constant V

W = q

!E !d!s

i

f

"Work

!V = Vf "Vi = "

We

q= "

!E #d!s

i

f

$Potential Difference

PotentialV (r) = kq

r V (!x) =

!E !d!s

x

"

# V = Vii=1

n

! =kqirii=1

n

!

V =kqirii=1

3

! = k q1r1+q2r2

+q3r3

"#$

%&'= k q1

b+

q2a2 + b2

+q3a

"#$

%&'

Superposition of Electric Potential

ES = !"V"s

U = q2V1(r)

U =kq1q2r

U12 =W = q2V = q2kq1d

W13 +W23 =U13 +U23 = q3V = q3kq1d+ q3k

q2d

V (C) = q1R+q2R

+q3R

V (C) = k !eRi=1

12

" = k !12eR

E(C) = 0

Calculating Field from Potential

Electric Potential from two charge system

Electric Potential of triangle system

Electric Potential from charges at center

12 electrons circle

C =qV

=!0Ad

Farad

Capacitance of a parallel plate capacitor

Area of a parallel plate capacitor

Cylindrical capacitor Spherical capacitor

Single conducting sphere capacitor

Chapter 24: Capacitors

Circuit in parallelCircuit in series

Work required bring capac to full charge

Work stored as EP energy for capacitor

energy density

EP energy for parallel plate cap

Thundercloud find potential diff EF and total E

Energy stored in capacitors (xJ) Dielectric cons

C =! 2"#0Lln r2 / r1( )

Capacitance of a coaxial cable

Capacitor half filled with a dielectricCharge on a cylindrical capacitor

E =Eai

!=

q!"0 A

=q"A

" =!"0

C =!Cair

C =

!0 Ad

; V =q / 2C

; E =Vd

;U =12

qV

A =

dC!0

q = (C1 + C2 + ..Cn )VCeq = C1 + C2 + ..Cn

Ceq = Cii=1

n

!

V = V1 +V2 +V3 = q 1C1

+1C2

+1C3

!"#

$%&

V =qCeq

1Ceq

=1Cii=1

n

'

Wt = dW! =12q2

C

U =12q2

C=12CV 2 =

12qV

u = UAd

=CV 2

2Ad=12!0

Vd

"#$

%&'2

=12!0E

2

U =12CeqV

2 =12nCV 2

n = 2UCV 2

C =! 2"#0Lln r2 / r1( )

q = CV = ! 2"#0Lln r2 / r1( )

$

%&'

()V

u =U

volume 1F =

1C1V

C =qV

=!L

!2"#0

ln r2 / r1( )=

2"#0 Lln r2 / r1( )

C = 4!"0

r1r2

r2 # r1

False: A negative charge placed in Zone 2 can be in equilibrium True: A positive charge placed in Zone 3 can be in equilibrium True: A negative charge placed in Zone 3 can be in equilibrium True: A negative charge placed in Zone 1 can be in equilibrium True: A positive charge placed in Zone 1 can be in equilibrium False: A positive charge placed in Zone 2 can be in equilibrium 0: charge o the hydrogen atom, consisting o a proton and an electron -1/3: charge o the down quark 0: charge o the neutron +2/3: charge o the up quark 0: charge o the helium atom, consisting o two protons, two neutrons, and two electrons -1: charge o the electron +1: charge o the proton

F = k q1q2r2

! k e2

c2* 2 * b

c

radius = atmosphere + earth

# of protons = rate* time* (4!r2 )total charge = # of protons * (1 602i10"19 )

Cosmic rays bombarding Earth find total charge

kg *1000gkg

*1molxg

* 6.023*1023atoms

1mol* #.of .electrons

1atom

Electrons on mass of water

F = kq1q2d 2

4.3F = kq1q2xd 2

x =14.3

= 0.232

new.dis = 0.232 *d 2

How far apart two charged spheres

Ratio of electrostatic/gravitationFe = k

q1q2r2;Fg = G

m1m2

r2

Fe / Fg !k q1q2r2

G m1m2

r2

Two beads on insulating string find mass of sec beadFe = Fg

Fe = kq2

d 2;Fg = mg

m =k q

2

d 2g

Fe = kq1q2d 2

Fg = m2gsin!

kq1q2d 2

= m2gsin!

d =k q1q2m2gsin!

Two beads slanted wire find force between

T sin! " Fe = 0 T cos! " Fg = 0

sin! =d / 2l

Fg = mg Fe = kq2

d 2=

q2

2l sin!( )2=

q2

4l2 sin2!T sin!T cos!

= tan! =FeFg

l = kq2

4gmsin2! tan!

Two balls on wire find length of string

kq1q3

(x3 ! x1)2 = k

q3q2(x3 ! x2 )

2

q1 (x3 ! x2 ) = q2 (x3 ! x1)

x3 =q1x2 ! q2 x1q1 ! q2

One pos charge on origin and neg charge find position of third

CAPA #1

CAPA #2Square, pos charge and neg charge diagonal, find E at other corners

wo point charges at corners o right triangle, fine E o third and direction o E

Hollow metal sphere with a solid sphere insidetotal charge surface of inner sphere

total charge on inner surface of hollow sphere

total charge surfaceof outer surface

Max torque with dipole moment measured in D

Net orce o water molecule near point charge

wo semicircular shaped insulating rods find E at center

Ex =!2kq"r2

2Ex

E needed to counteract weight o e- at Earth s sur ace

Ratio o two di erent wires, charge density lamda and radius r

CAPA #3

Speed of e- w th VK = !"U = !qV

K =12mv2

v = 2Km

#2(qV )m

Two ons w th +1 and -1 charge, find work to nc. d stW = q!V

"W = q k qdi

# k qd f

$

%&'

()

Battery w th two para e meta p ates, acc from neg to pos p ate, fine KE and speedK =U = qV K =U = qV K =

12mv2

qV =12mv2 ! v 2qV

m

K = q!V = (16qe )!VP = (16qe )!V " (rate)

Acc erator str ps eʼs at a rate, find power

V = k qr! "V = k 4q

a2

#$%

&'(2

+ b2

#$%

&'(2

Four same point charges at corners of rectangle fine potential at center

Hollow spherical conductor with surface charge

Voutside =kQR

Vinside =kQR

=Q

4!"0RFour point charges on corners of square three pos and one neg find V at certain point above centerk qr! = k q + q + q " q

a2 + a2 + c2

# k 2q2a2 + c2

Ex = !"V"x

Ey = !"V"y

Ez = !"V"z

ie(x2 + xy2 + yz)

Ex = !"V"x

= 2x + y2

Ey = !"V"y

= 2xy + z

Ez = !"V"z

= z

Find V from equation of volume with certain coordinates

nsulating sheet with charge distribution uni orm ind change in V when charge is moved rom position A to B

!0!Eid!A""" = !0 EA + EA( ) = q = #A

E =#2!0

;W = QEd cos$

%W =Wf &Wi = Q#2!0

'()

*+,dy1 &Q

#2!

')

*, dy2

V = &%WQ

Surface area of capacitor

A =Cd!0

Capacitor Defibrillator charged find capacU =

12CV 2

C =2UV 2

Capacitor with mylar between plates find work to remove mylar and potential difference one Mylar is removed!W =Wi "Wf

#12CV 2 "

12C$V$( )2

V =!V

Parallel plate capacitor find charge energy stored area and if filled withdielectric

q = VCU =12qv A =

dC!0

C =! "0Ad

C =! 2"#0Lln r2 / r1( )

Two soup cans filled with soup

Chapter 25: Direct CurrentsElectric current

i = dqdt

q = dq! = i dt0

t

!

1A =1C1s

ontophoresis

q =amount

applicationrate

q = it ! t =qi

Current Density

Dri t Velocity

Current through a wire

Resistance

Resistivity

Resistance o Wire

Convert rom AWG size to wire diameter

emp dependence o the resistivity o metals

emp dependence o the resistance o conductor

Ohm s Law

Resistors in Series

nternal Resistance o Battery example

J =

iA

i = dqdt

= nevd A

J =iA= nevd

J = di / dA

Aoute ' = !R2 " ! R / 2( )2= ! 3R2

4i = JA '

R =Vi=

ELJA

=!LA

1" =1V1A

! =EJ

units = VmA

= "m R = ! L

A

d = 0 127 !92 36" AWG( )/39 mm

! " !0 = !0# T " T0( )

R ! R0 = R0" T ! T0( )

Vemf = iR

Vemf = iR1 + iR2... = iReq

Req = Rii=1

n

!

i =V / RVt = iReq = i R + Ri( )R + Ri( ) = Vt

i

Ri =Vt

i! R

nternal Resistance o Battery

Resistance in Parallel (have same voltage)current is sum o individualPower in Electric circuits

Energy

emp Dependence o the Resistance o a Light bulb

Galvanometer as current-measuring device

Dri t velocity o e in copper wire

i =V / RVt = iReq = i R + Ri( )R + Ri( ) = Vt

i

Ri =Vt

i! R

i =Vemf

1Req

1Req

=1Rii=1

n

!

P = iV = i2 R =

V 2

Rwithi = V

RandV = Ri

1kWh = 1000W i3600s = 3 6i106 joules

Power whenlighted

P =V 2 / R ! R =V 2 / P

R0 =R

1+" T # T0( )

V = i nt Ri;V = imax Req

!V = i nt Ri = imax Req

1Req

=1Rii=1

n

" =1Ri

+1Rs

!imax

iint Ri

=1

Req

=1Ri

+1Rs

Rs = Ri

iint

imax # iint

x = vdt;iA= nevd

n =# conductionelectrons

volume

n =1electron

atom6.02i1023 atoms

Xgdensity g

cm3

1.0i106 cm3

m3

vd =i

neA

x = vdt =i

neA!"#

$%&

t

Chapter 26: CircuitsKircho s Junction Rule

Kircho s Loop Rule

Charging a batteryWheatstone bridge

Voltmeter in a simple circuit

Capacitor ully charged

Capacitor discharging

ime to charge a capacitor

i1 = i2 + i3he sum o voltage drops around a complete circuit loop must

sum to zero

!iRi !V +Ve = 0Ve = iRi +V

loop adb :!i3R3 + ig Rg + i1R1 = 0

loopcbd :+iu Ru ! ig Rg ! iv Rv = 0

junctionsi1 = ig + iui3 + ig = ivWhenbridgebalancedi1R1 = i3R3;iu Ru = iv Rv

i1 = iu ;i3 = ivFinallyiu Ru

i1R1

=iv Rv

i3R3

" Ru =R1

R3

Rv

Before

i =VR

Connected

1Req

=1R+

1Rv

! Req =RRV

R + RV

i =VReq

qtot = CVemf

q = q0e!

tRC

"#$

%&'

i = dqdt

=q0

RC"

#$%

&'e

!t

RC"#$

%&'

q(t) = q0 1! e!

tRC

"

#$%

&'

t = !RC ln 1!q(t)q0

"

#$%

&'ime constant o circuit

Rate o energy storage in a capacitor

q = q0 1! e!

tRC

"#$

%&'

"

#$$

%

&''

whereq0 = CVemf ;( = RC ;V (t) = q(t) / C

V (t) =Vemf 1! e!

tRC

"#$

%&'

"

#$$

%

&''=Vemf !Vemf e

!t

RC"#$

%&'

RC = ( =!t

ln! V (t) !Vemf( )

Vemf

"

#$$

%

&''

q = CV 1! e! t / RC( )U =

12

q2

CdUdt

=qC

dqdt

=qC

i

"CV 1! e! t / RC( )

CVR

#$%

&'(

e! t / RC =V 2

Re! t / RC 1! e! t / RC( )

Chapter 27: MagnetismLorentz orce

Magnetic orce on moving chargeesla

Guass

Proton in B ield (find orce)

Cathode Ray ube (calc vel and acc)

Particle Orbits in Uni orm B

Mass spec , find mass

Momentum o a track (find momen and speed)

Period, req, and ang req o particle in orbitDeuteron in Cyclotron

Magnetic field (Hall e ect)

Hall probe (find mag o magnetic field)

orce on a current carrying wire

orque on a current carrying loop (square and circle)

Magnetic Dipole Momentum

Magnetic potential energy in external magnetic field

riangle Shaped Loop

!F = q!v !

!B( !x)

FB = qvBsin!

1T = 1

NsCm

= 1NAm

1G = 10!4T

K =12

mv2 ! v =2Km

F = qvBsin"

!K = "!U = qV

12

mv2 = eV # v =2eV

m

F = ma = qvB # a =qvBm

m vr= qB !

pr= qB

!K = 1 / 2mv2;!U = "qV

!K + !U = 0 #1 / 2mv2 " qV = 0

v =2qV

m

r =mvqB

# m =B2qx2

8V

p = mv = erB

vc=

mvmc

T =2!r

v=

2!mqB

f =1T=

qB2!m

" = 2! f =qBm

B =

2!mfq

B =

VH

dv=

VH dhnedi

=VH hne

i

n =# electrons

volume

n =1electron

atom6.02i1023 atoms

Xgdensity g

cm3

1.0i106 cm3

m3

B =VH hne

i

q = ti =Lv

i;(v is drift velocity)

F = qvBsin! =Lv

i"#$

%&'

vBsin!

( F = iLBsin!

!! = !r "

!F

! = iaB( ) a2

#$%

&'(

sin) + iaB( ) a2

#$%

&'(

sin)

* ia2 Bsin) = iABsin)

µ = NiA

U = !µBcos"

hypotenuseF = i

!L !!B

i!L ||"B " F = 0

x # sideBx = #Bcos$ = 0

By = Bsin$

$ = tan#1(Opp / Adj)F = iLx By

current is constant

Vemf =V1 +V2

voltage is constant

i = i1 + i2

Chapter 28: Mag Fields of Moving ChargesMagnetic ield

B ield rom a Long, Straight Wire

B ield rom a Loop

Parallel Current Carrying Wires

B orce on wire rom two other wires

orce on square LoopB ield inside long wire

B ield inside ideal solenoid

B ield inside toroid

ind vel where net orce on E and B is 0 (pos charge)

B ield rom Current Distribution

B field rom Current distribution (3 circuits)

B field o Long wires, find point where B field is 0

Circular motion, find speed and radius taken

Atoms as magnets

dB =µ0

4!idssin"

r 2

µ0 = 4! #10$7 TmA

B r!( ) = µ0i

2"r!

B =

µ0i2R

F12 = i2 L

µ0i2!d

"

#$%

&'=µ0i1i2 L2!d

Bb =µ0ib2!d

; Bc =µ0ic

2! 2d( )Bbc =

µ0ib2!d

"µ0ic4!d

=µ0i

4!d

Fabc = ia LBbc =µ0i

2 L4!d

Fdown =µ0i1i2 L

2! 0.100m( )Fup =

µ0i1i2 L2! 0.100m + 0.250m( )

F = Fup " Fdown

Ampere s Law

!B ! d!s"" = µ0ienc

B r!( ) = µ0i

2"R2

#

$%&

'(r!

B =

µ0 Ni2!r

q!E ! !v !

!B( ) = 0

qE = qvBsin 90°( )v =

EB

1 dB1 =µ0

4!idssin 0°( )

r 2 = 0 " B1 = 0

2 dB2 =µ0

4!idssin 180°( )

r 2 = 0 " B2 = 0

3 B3 = dB# =µ0

4!iRd$R2 =

0

! / 2

#µ0i2R

%

&'(

)*

f om fieldat cente ofloop

=µ0i ! / 2( )

4!R=µ0i8R

Upper ! half

" = # R = 3r Buppe =µ0i

12Ra and b

BR= "=# =µ0i4r

c

BR= "=# / 2 =µ0i8r

BR=2 "=# / 2 =µ0i16r

Final

Ba =µ0i

12R+µ0i4r

=µ0i3r

Bb =µ0i

12R!µ0i4r

= !µ0i6r

Bc =µ0i

12R+µ0i16r

+µ0i8r

µ0i12!r

=µ0i2

2! d " r( )i1r=

i2d " r( ) # r =

i1 d " r( )i2

12

mev2 = eV ! v =

2eVme

Fcent =mev

2

r; Fmag = vBe

vBe =mev

2

r! r =

meveB

Current

i =eT=

e2!r( ) / v

=ve

2!r

Magnetic moment

µo b = iA =ve

2!r!r 2 =

ver2

Orbital angular momentumof e"

Lo b = rp = rmv

# Lo b = rm2µo b

er$

%&'

()=

2mµo b

e

Magnetic Dipole Moment!µo b = "

e2m!Lo b

B = µ0 in;wherenis# turns per unit length

!B =!B0 + µ0

!M ;!H =

!B / µ0!

B = µ0

!H +

!M( ) = µ0 1+ !m( ) !H

Re lative permeability" m = 1+ !m

Magnetic permeabilityµ = 1+ !m( )µ0 =" mµ0

Magnetic Prop o matter

Chapter 29: Electromagnetic InductionMagnetic flux

araday s Law o nduction

nduction in a flat loop

Special cases or flat loop

Changing B ield, current inc according to current, find induced voltage at loop

!B =!Bid!A"" = BAcos#

1Wb = 1Tm2

Vemf = !

d"B

dt

Vemf = !Acos" dB

dt! Bcos" dA

dt+# ABsin"

A,! cons tan t Vemf = "Acos! dBdt

B,! cons tan t Vemf = "Bcos! dAdt

A, Bcons tan t Vemf =# ABsin!

A = N!R2

B(t) = B0 1+ 2.4s"2t2( )A,# cons tan t Vemf = "Acos# dB

dt

Vemf = "Acos# ddt

B0 1+ 2.4s"2t2( )( )= "Acos# B0 2.4s"2 2t( )

Motion Voltage, find induced voltage as a unction o time

Lenz s Law - our Cases

nduced Voltage on a Moving Wire in B ield

nduced em , field o a solenoid, find induced em due to changing field

nduced em

nductance

RL Circuit, switch connected

A(t) = w ! d(t) = w ! d " vt( );t f = d / v

Vemf = "Bcos# dAdt

= "Bcos# ddt

w ! d " vt( )( )Vemf = wvBcos# from0tot f

a) An increasing magnetic field pointing to the right induces a current that creates a magnetic field to the leftb) An increasing magnetic field pointing to the left induces a current that creates a magnetic field to the rightc) A decreasing magnetic field pointing to the right induces a current that creates a magnetic field to the rightd) A decreasing magnetic field pointing to the left induces a current that creates a magnetic field to the left

V = vLB

B ! field of outer solenoidB = µ0in = µ0iN / L;i =V / R

voltagetimedependV =V0 sin(t)

Induced emf

Vemf = !Acos" dBdt

find derivative

!E ! d!s = "

d#B

dt"$

N!B = Li

L"# $% =!B"# $%i"# $%

&1H =1Tm2

1A

µ0 = 4' (10)7 H / m

Vemf L = !L didt

Vemf ! iR ! L didt

= 0

i(t) =Vemf

R1! e! t / " L( )( );" L = L / R

nductance o a solenoid

Sel induction

RL Circuit, em suddenly removed

L =

N!B

i=

nl( ) µ0in( ) Ai

= µ0n2lA

Vemf L = !

d N"B( )dt

= !d Li( )

dt= !L di

dt

i(t) = i0e

! t /" L ;i0 =Vemf / R

RL Circuit solenoid, how long current to reach hal

L didt

+ iR =Vemf i(t) =Vemf

R1! e! t / L / R( )( )

12

Vemf

R=

Vemf

R1! e! t0 / L / R( )( )

12= 1! e! t0 / L / R( ) "! ln2 = !t0 / L / R( )

t0 =LR

ln2

nstantaneous power provided by em source

Energy stored in B field o the inductor

P =Vemf i = L di

dt!"#

$%&

i

U B = P dt

0

t

! = Li 'di '0

i

! =12

Li2

Chapter 30: Electromagnetic Oscillation and Currents

chargeas a functionof time

q = qmax cos !0t + "( )angular frequen

!0 =1

LCcurrent

i = #imax sin !0t + "( )imax =!0qmax

electrical energy

U E =qmax

2

2Ccos2 !0t + "( )

magneticenergy

U B =L2

i2max sin2 !0t + "( ) = qmax

2

2Csin2 !0t + "( )

energy incircuitU =U E +U B

=qmax

2

2Ccos2 !0t + "( ) + qmax

2

2Csin2 !0t + "( )

=qma

2

2Csin2 !0t + "( ) + cos2 !0t + "( )( ) = qmax

2

2C

LC Circuit

RLC Circuit

rateof energy lossdUdt

= !i2 R

chargeas a functionof time

q = qmaxe!

Rt2 L cos "t + #( )

angular freq

" = " 20 !

R2L

$%&

'()

2

where "0 =1

LCenergy incircuit

U E =q2

max2

2Ce!

RtL cos2 "t + #( )

oscillation freq

!0 =1

LCenergy stored

U =qmax

2

2C=

CVemf( )2

2Cchargeof capac at giventime

q = qmax cos !0t + "( );q = qmax when" = 0

q = qmax cos !0t( )

LC Example

sinusoidal voltageas functionof timevemf =Vmax sin!t

induced current

i = I sin !t "#( )voltageand current amp.acrossresistorVR = IR R

capacitivereac tan ce

XC =1

!Ccurrent incircuit

iC =VC

XC

cos!t = IC sin !t + 90°( )voltageand current amp,acrosscapac.VC = IC XC

inductivereac tan ceX L =!L

curent in inductor

vL = iL X L $ iL = "vL

X L

cos!t = IL sin !t " 90°( )voltageand current amp,across inductorVL = IL X L

current incircuit

I =Vmax

R2 + X L " XC( )2

impedance

Z = R2 + X L " XC( )2

phasecons tan t

! = tan"1 VL "VC

VR

#

$%&

'(= tan"1 X L " XC

R#

$%&

'(

For XL > XC, φ is positive, and the current in the circuit wi l lag behind the voltage in the circu tFor XL < XC, φ is negative and the current in the circuit will

lead the vo tage in the circuitFor XL = XC, φ is zero, and the current in the circuit will be

in phase with the voltage in the circuit

rateenergy is dissipated intheresistor (Power)

P = i2 R = I sin !t "#( )( )2R = I 2 Rsin2 !t "#( )

average power

P =12

I 2 R =I

2

$%&

'()

2

R; I ms =I

2

* P = I ms2 R = I msVmax ms

RZ= I msVmax ms cos#

max when# = 0socos# is power factormisc.

V ms =V

2

Vmax ms =Vmax

2current

I ms =Vmax ms

Z=

Vmax ms

R2 + !L " 1!C

$%&

'()

2

Energy and Power

Vemf =Vmax sin!t

Vemf = "Nd#B

dtrelationshipVP

N P

=VS

NS

powerPP = IPVP = PS = ISVS

secondary current

$ IS = IP

VP

VS

= IP

N P

NS

primary current

IP =NS

N P

IS =NS

N P

VS

R=

NS

N P

VP

NS

N P

%

&'(

)*1R

=NS

N P

%

&'(

)*

2VP

R

Pr imary resis tance

RP =VP

I=VP

N P

NS

%

&'(

)*

2R

VP

=N P

NS

%

&'(

)*

2

R

rans ormers

wo di erent wires, find ratios o current densities and dri t velocities

ind max resistance o rectangular wa er

ubular resistor, find resistivity and % change in resistance

ind Current at R3

J = iA;A = ! d

2"#$

%&'2

JCu =i

! dCu / 2( )2; JAl =

i! dAL / 2( )2

JCu / JAl

J = i / A = nevd

vd =Jne

vd ,Alvd ,Cu

=JAl / nAlJCu / nCu

Current Density Drift Velocities

R = ! LA

arrangeinwaythat dividesby smallest area possible

! =RAL;A = "route

2 # "rinne2

! = R"route

2 # "rinne( )L

R ! R0 = R0" T ! T0( )Rnew = R0" T ! T0( ) ! R0Percentage Increase = Rnew / R0 !100%

Resistivity Percentage Increase

Equivalent Resistance

V = iR

i = VR=

V2R3

R12 =1R1

+1R2

!"#

$%&

'1

Rop= 2R12 + R3 = Rbo tom

Req =1R

op

+1

Rbo tom

!

"#

$

%&

'1

Battery, internal resistance ind resistance to extract certain power rom R = resis tanceoutsidebatteryr = int ernal

P =V 2RR + r( )2

SolveR

3 resistorsind R to produce certain currents i1 and i2

Vemf = i120! + i2 R

orVemf = i2 R

What Vem will produce currents i1 and i2

ind currents with resistance values and Vem values

Vemf 1 ! i1R1 ! i3R3 = 0

Vemf 2 ! i2 R2 ! i3R3 = 0

i1 + i2 = i3Re arrange for matrixi1R1 + 0 ! i3R3 = !Vemf 1

0 ! i2 R2 ! i3R3 = !Vemf 2

i1 + i2 ! i3first matrix

R1 0 R3

0 R2 R3

1 1 !1

"

#

$$$$

%

&

''''

second matrix

!Vemf 1

!Vemf 2

0

"

#

$$$$

%

&

''''

R1 0 R3

0 R2 R3

1 1 !1

"

#

$$$$

%

&

''''

!1

i

!Vemf 1

!Vemf 2

0

"

#

$$$$

%

&

''''

ind currents with resistance values and Vem values

RX =

1m ! L( )L

iR1

Resistance needed to produce peak power output

P =V 2 / RR = P / V 2

q = q0e!

1RC

"#$

%&'

t = !RC ln(1! % change)

ime to charge capacitor to percentage

Current through 4 ohm resistor

Potential di erence across R1

Potential di erence across R2

Potential di erence across capacitor

Vemf ! iR1 ! iR2 ! i" = 0

i =Vemf

R1 + R2 + "i = 0

i =V

R1 + R2

Potential Difference = iR1

i =V

R1 + R2

Potential Difference = iR2

sameas R2

Charge on C1 when switch is closed

Charge on C2 when switch closed

Charge on C1 and C2 when switch is open

iR !qC

= 0;i = VR1 + R2

q =V

R1 + R2

"

#$%

&'RC

iR !qC

= 0;i = VR1 + R2

q =V

R1 + R2

"

#$%

&'RC

q =V

C1C2

C1 + C2

Wire placed at angle with magnets at ends. Find F

! = 90 " angle givenF = iLBsin!Proton entering two-plate region, find B field

Calculate radius of trajectory of proton in B field

Calculate the period of motion in plate, freq, and pitch

Value of current of wire2 that is half of wire1

Square ammeter clamped on DC current, find current

B field inside solenoid

B field inside tungsten wire with current a distance from its central axis

Aircraft, find potential diff. between wings

Loop expanding, find ind. current

Motor, single loop find max angular speed

How long will it take circuit to reach certain current

Wire moving in xy plane, find vel to induce certain pot. diff

Fully charged cap. connected to inductor, find max current and freqAC power on RLC Circuit

Transformer

V = Ed; E = vBV = dvBa

V = !Bcos" dAdt

; A = #r 2 = # r0 + vt( )2

V = !Bcos" ddt

# r0 + vt( )2( )V = !B# cos" 2r0v + 2v2t( )V = iR $ i =V / R

Vemf =! ABsin";sin90°

! =Vemf

ABsin"

i(t) = i01! e! t / L / R( );i0 =Vemf / R

t = ! lni(t)i0

"LR

F = qvBsin! with! = 90°

E = vB =Vd" v =

VdB

imax = w0qmax

where w0 =1

LCand qmax =VC

imax =1

LC

!"#

$%&

VC( )

f =

!0

2";!0 =

1LC

X L =!L

Z = R2 + X L ! XC( )2

I =Vmax

R2 + X L ! XC( )2

VR

R= IR sin!t

VR = IR R

VC = IC XC where IC is imax

VL = IL X L where IL is imax

VP

N P

=VS

NS

!VS =VP

NS

N P

IS = IP

VP

VS

= IP

N P

NS

Resistor and Capacitor, what freq and current where potential drop across cap. equal across resistor

XC = R =1

!C"! =

1RC

f =!2#

=1 RC

2#

I =Vmax

R2 + X L ! XC( )2

=Vmax

R2 + XC( )2=

Vmax

2R22

XC =

1!C

=1

2" fC

!0 =1LC

= 2" f

!02 =

1LC

C =1

L!02 =

1

L 2" f( )2

VR = IR R;VC = IC XC ; IR = IC

VR

R=

VC

XC

R =VR

XC

VC

=VR

1 / !C( )VC

= VR

1 / 2" fC( )VC

Design own RLC find C and R

Maxwell ampereʼs law

Displacement current

Displacement current with wire going through

Wave equations

Wave solution

Speed of light

Rate of energy by an electromagnetic wave

Instantaneous power per unit area

Intensity

Energy density in electromagnetic wave

Intensity and force of radiation

Pressure due to electromagnetic radiation

Radiation pressure from laser pointer, find force

Intensity of light passing through polarizer (law of malus)

Intensity of light before polarizer

Electric field after polarizer

Three polarizers

Solar stationary satellite, find area of solar sail

Multiple polarizers, find fraction of incident intensity

Laser powered sailing find time to reach % of speed of light from rest

id = !0

d"E

dt

id = !0 A dE

dt

!E = Emax sin kx !"t( ) y!B = Bmax sin kx !"t( ) z

k = 2# / $," = 2# f

!E ! d

!A"

!E # d!A = 0

!B ! d

!A"

!B # d!A = 0

Emax

Bmax

=$k= c " E

B= c

EB=

1µ0%0c

c = 1

µ0!0c"

1µ0!0

= 2.99 #108 m / s

c = ! f

!S =

1µ0

!E !!B

S =!S =

powerarea

!"#

$%& nstanteous

I = 1

cµ0

E2ms where E ms = Emax / 2( )

uE =12!0 cB( )2

, uB =1

2µ0

B2

uE =uB inelectromagnetic wave

F =!p!t

=IAc

I = !U / !tA

=c!pA!t

p =

Ic

and reflected : p =2Ic

I =powerarea

,then p =forcearea

=2Ic

! force = area "2Ic

I = 1

2cµ0

E2 =1

2cµ0

E0 cos!( )2= I0 cos2!

I0 =

1cµ0

E ms2 =

12cµ0

E02

E = E0 cos!

I1 =12

I0 , I2 = I1 cos2 45! 0( ) = 12

I0 cos2 45( )I3 = I2 cos2 90 ! 45( ) = 1

2I0 cos4 45( ) = I0 / 8

Fg = F p

F p = p A; p =2Ic

!"#

$%&

; Fg = GmmSun

d 2

2Ic

!"#

$%&= G

mmSun

d 2 ' A = GcmmSun

2Id 2

I X

I0

= cos2 !"( )( )X

2Ic

=FA!

FA=

2 P / A( )c

F =2Pc

= ma ! a =2PmC

v = at = %ic ! t = % " c2PmC

Chapter 31: Electromag. Waves

Chapter 32:Geometric OpticsLaw of rays Reflection l

Mirror image Full-length mirrorrequired length half the height o the person

Concave spherical mirrorMirror equation

Magnification Concave mirror

Always virtual, upright, and reduced image

Focal length of parabolic mirror

rd=

Rd + D !r = !i

do = di and h0 = hi

heightpe son !

heightpe son ! disteyes f omtop of head

2!

disteyes f omtop f head

2

C = R

f =R2

Case ype Direction Magnification

d <f Virtual Upright Enlarged

d=f Real Upright mage at infinity

f<d <2f Real Inverted Enlarged

d =2f Real Inverted Same size

d <2f Real Inverted Reduced

1d0

+1di

=1f

m = !

di

d0

=hi

h0

di =d0 f

d0 ! fd0 = +; f = !;di = !

f =

14a

where y(x) = ax2

ndex of refraction

law of refraction (Snellʼs Law

n =

cv

n1 sin!1 = n2 sin!2

nmedium =

sin!ai

sin!medium

where !ai > !medium

internal reflection

chromatic dispersionfirst rainbow appears 42 deg then 50 deg for second one outside first one

sin!c =n2

n1

(n2 " n1)

sin!c =1n

whenentering air

shadow of a ball

d + D =dRr

d 1!Rr

"#$

%&'= !D ( d =

DrR ! r

dligh bulb = d + D =Dr

R ! r+ D

imaged formed by a converging mirror

1d0

+1di

=1f

image produced : di =d0 f

d0 ! f

magnification : m =hi

h0

=!h0

di

d0

h0

displacement of light rays in transparent material

d = Lsin!d =t

cos!med um

sin !a "!med um( ) a

where !med um = sin"1 sin!a

n#

$%&

'(

opitcal fiber, max angle of incidence

!ai = sin"1 ncos sin"1 1

n#$%

&'(

#

$%&

'(#

$%

&

'(

Chap. 33: Optical Instr.

!B ! d!s = µ0"0

d#E

dt"$ + µ0ienc = µ0 id + ienc( )Lens-maker formula Lens equation

Convex lens Concave lensalways virtual, upright, reduced in sizeMagnification for lenses

Power of lens

myopia (near) image produced in front of retinahypermetropia (far) imaged produced behind the retinaCorrective lenses, find power

Microscopeobject to be magnified is placed just outside of foc length of the objective lensMicroscope magnification

telescope magnification

two positions of a convergin lens, find dist. between two positions

image produced by two lenses, find image produced by second lens with respect to sec. lens

image from the moon, find radius of the image of the moon on the screen

image produced by a lens and mirror, find mag.

1f= n !1( ) 1

R1

!1R2

"

#$%

&'

1f=

2 n !1( )R

1d0

+1di

=1f

Case ype Direction Magnification

f<d <2f Real Inverted Enlarged

d =2f Real Inverted Same Size

d >2f Real Inverted Reduced m = !

di

do

=hi

h0

D =

1mf

1!+

1"di

=1f= ____ diopeters

m = !

0 25Lfo fe

m! = "

!e

!o

= "fo

fe

1d0

+1di

=1f

;d0 = d ! di

1d ! di( ) +

1di

=1f

di + d ! di( ) = di d ! di( )f

df = did ! di2 " di

2 ! did + df = 0

x =d ± d 2 ! 4df

2

di 1 =d0 1 f1

d0 1 ! f1

di 2 =d0 2 f2

d0 2 ! f2

;d0 2 = d ! di 1

" di 2 =d ! di 1( ) f2

d ! di 1( ) ! f2

" di 2 =

d !d0 1 f1

d0 1 ! f1

#

$%

&

'(

#

$%

&

'( f2

d !d0 1 f1

d0 1 ! f1

#

$%

&

'(

#

$%

&

'( ! f2

di =d0 f

d0 ! f"

d0 fd0

= f

m = !di

d0

=hi

h0

# hi = !h0

di

d0

hi = !R fd

di 1 =d0 1 f1

d0 1 ! f1

di 2 =d0 2 f2

d0 2 ! f2

;d0 2 = d ! di 1

m1 =f1

f1 ! d0 1

;m2 =f2

f2 ! d0 2

m = m1m2 =f1

f1 ! d0 1

"

#$

%

&'

f2

f2 ! d0 2

"

#$

%

&'

( m =f1

f1 ! d0 1

"

#$

%

&'

f2

f2 ! d !d0 1 f1

d0 1 ! f1

"

#$

%

&'

"

#$

%

&'

"

#

$$$$$

%

&

'''''

Thin lens, find radius of curvature and thickness of edge of lens and how much thinner if made with high-index plastic

Chapt. 34: Wave OpticsSnellʼs Law for Hugyenʼs construction

Light traveling in an optical medium

placing object underw ater does not change color b/c freq is the same in both

constructive interference (in phase, difference is 2pi)

destructive interference (difference is pi)

double slit interference

Order of fringe (m)for constructive interferencem= give us angle of the first order of bright fringem=2 second order fringefor destructivem=0 gives us angle of the first order dark fringem= second order of fringefirst order is closest to central maximum

Constructive interference with small angle approx.

dist of bright fringes from central max.

dist of dark fringesdouble slit intensity on screen

single slit diffraction dark fringe

single slit intensity

thin film interference

color seen by thin film interference is wavelength that is interfering constructively

if n <n2, phase of the reflected wave will be changed by half a wavelengthif n >n2 then there w ll be no phase changeeven # of phase change same as no phase changeodd # of phase change is same as one phase change

min thickness that will produce const. interference

lens coating destructive interference

interferometer

can find thickness if we know its index of refractioncan find index of refraction if we know thickness

diffraction by a circular opening

intensity of interference pattern from double slits

two slit diffraction patern example, find interference fringe in central peak of diffr. envelope

diffraction grating

wavelength of monochromatic light

dispersion (diffraction grating)

resolving power

diffraction grating ex., find angle where first order max, and find angular sep. between wavelengths

diffraction on a crystal

width of central max.

CD as a diffraction grating

spy satellite, find minimum diameter of the lens

air wedge, find how thick is the film

n1 sin!1 = n2 sin!2

sin!1

sin!2

=v1

v2

=c / n1

c / n2

=n2

n1

!1

v1

=!2

v2

!n = ! v

c=!n

fn =v!n

=c / n! / n

=c!= f

!x = m" (m = 0,±1,±2...)

!x = m +

12

"#$

%&'( (m = 0,±1,±2...)

constructive !x = d sin" = m# (m = 0,±1,±2...)

destructive !x = d sin" = m +12

$%&

'()# (m = 0,±1,±2...)

when m = 0!" = 0!#x = 0bright fringe whichiscentral max.

d sin! = d y

L= m" (m = 0,±1,±2...)

y =

m!Ld

(m = 0,±1,±2...)

y =

m + 12

!"#

$%&'L

d (m = 0,±1,±2...)

E1 = Emax sin !t( )E2 = Emax sin !t + "( )E = 2Emax cos " / 2( )

IImax

=E2

E2max

I = 4Imax cos2 " / 2( )" =

#x$

2%( ) = 2%d$

sin&

I = 4Imax cos2 %dy$L

'()

*+,

asin! = m" (m = 1,2...)

position y =m"L

awherea = width

I = Imax

sin!!

"#$

%&'

2

where! =(a)

sin* =(ay)L

!x = m +12

"#$

%&'( = m +

12

"#$

%&'(a

n= 2t

m = 0,±1,±2...( ) where( =(a

n

tmin =

!ai

4n

m +

12

!"#

$%&

'ai

ncoat ng

= 2t m = 0,±1,±2...( )

Nmate ial ! Nai =

2tn"

!2t"

=2t"

n !1( )

sin! = 1.22"d#!R = sin$1 1.22"

d%&'

()*

where !R is the min. observable angular sep.

and d is the diameter of the lens

I = Imax cos2 ! sin""

#$%

&'(

2

where" =)a*

sin+ and ! =)d*

sin+

large distance " =)ay*L

and ! =)dy*L

1) asin! = " (m1 = 1)2) d sin! = m2"1)2)

=asin!d sin!

="

m2"

# m2 =da

N =Wd

d = distance, W=width, N= number of slits

d =1n

n = number of slits per unit length

d sin! = m" m = 0,1,2( )

! = sin"1 m#

d$%&

'()

m = 0,1,2...( )

di =d0 f

d0 ! f"

d0 fd0

= f

m = !di

d0

=hi

h0

# hi = !h0

di

d0

hi = !R fd

R =

!ave

"!= Nm where!ave # ! + ! + "!( )( ) / 2

d =WN

! = sin"1 m#d

$%&

'()

D =m

d cos!*! = D*#

2asin! = m" m = 0,1,2...( )

w = 2!y =

2"La

! = sin"1 #ai

d$

%&'

()

y =L

tan!

sin!1 1.22"d

#$%

&'(= tan!1 )x

h#$%

&'(

d =1.22"

sin tan!1 )xh

#$%

&'(

#

$%&

'(

2t = m +12

!"#

$%&' = m +

12+

12

!"#

$%&

m = # of bright fringes - 1extra 1/2because of dark fringe

Determining earth circumference, find how far North

Intensity of 2 polarizers, find angle between two polarizers

Throwing shadows, how far lightbulb away from wall

Partial linear polarization, find the fraction of light that is polarized

Solar sail problem, find max acc. of spacecraft and find velocity after certain time

Frequency of x-rays

Industrial laser, find RMS of E field and B field

Sirius star find distance in light years

Power of the sun radiation

Find total power incident and radiation pressure on roof

Convex mirror, find position and height of image

Neon laser in alcohol find speed, freq, and wavelength

d1

!1

=d2

!2

" d2 = !2( ) d1

!1

I = I0 cos2! =I0

2cos2!

! = cos"1 2II0

#

$%

&

'(

I0 goes down factor 2after1st polarizer

rd=

Rd + D

d 1!Rr

"#$

%&'= !D

( d =Dr

R ! r

d gh bu b = d + D =Dr

R ! r+ D

p =2Ic

=FA

FA=

2 P / A( )c

! a =2Pmc

v = at

f =

c!

RMSElect cField = cµ0 I

B =

Ec

d1light year

;1light year = 9.46i1015 m

Power = IAwhere A = 4!d 2

P = I 4!d 2( )

P = IA = I lw( )

p adiat on =

Ic

Area( ) = Ic

lw( )

1d0

!1di

= !1f

di = !1

1d0

+ 1f

;neg b / cbehind mirror

h1

d1

=h2

d2

! h2 =h1

d1

d2

nmedium =!o iginal

!medium

"!medium =!o iginal

nmed um

v = f !medium = f!o iginal

nmedium

#

$%

&

'(

nmed um =!o ginal

!medium

" !medium =!o g nal

nmed um

f =

c!

Iunpo = 2Im n

I po = Imax ! I ow

I o a = Iunpo + I po

polarized =I po

I o a

Eyeglass diopter, find if near or far-sighted and strength of of eyeglasses

Confining light in fiber, find max angle of incidenceTwo positions on converging lens, find distance

Power for reading newspaper

Focal length of magnifying glass given magnification

Farthest point w/o glasses given diopter

Find object distance and magnification to form certain image distance on right side of lens

Focal length of magnifying glass given object/image size

Three converging lens, find image

Near-sightedness

1d0

+1di

=1f= D

1d0

!1di

= D

D = n !1( ) 2

r+

n !1( )hnr 2

"

#$$

%

&''solve for r

tedge = center thickness + 2 rcu ve ! r 2cu ve !

wd ame e

2"

#$%

&'

2"

#

$$

%

&

''

D = n !1( ) 2rcu ve

+n !1( )h

nrcu ve2

"

#$$

%

&''solve for rcu ve

tedge new = center thickness + 2 rcu ve ! r 2cu ve !

wd ame e

2(

)*+

,-

2(

)

**

,

.tedge o d ! tedge new

tedge o d

sin!c =1n

!c = sin"1 1n

#$%

&'(

1d0

+1di

=1f

d0 + di = dbulb and sc een

di2 ! dbulb and sc eendi + dbu b and c een f = 0

solve using quadratic equationthen d i ! d0

1d0

+1d

=1f= D

both on same side, d is negassumed d0 is0.25m

1!+

1di

= D

di =1D

m! =dnea

f=

0.25mf

" f =0.25m

m!

1d0

+1di

=1f

1d0

=1f!

1di

" d0 =1f!

1d

#

$%&

'(

!1

d0 = 1!1d

#

$%&

'(

!1

wheredi iscoeff of _ f

m = !

di

d0

m! =hi

h0

m! =dnea

f

" f =dnea

hi

h0

1d0 1

!1

di 1

=1f" di 1 =

11f! 1

d0 1

di 2 =1

1f! 1

D ! di 1( )whered0 2 = D ! di 1

di 3 =1

1f! 1

D ! di 2( )whered0 3 = D ! di 2

False:  he final image is on the left side of the third lens False:  he final image is virtual False:  he final image is upright True:  he final image is inverted True:  he final image is real True:  he final image is on the right side of the

Separation of adjacent interference maximum

Minimum thickness of soap film

Number of lines in diffraction grating

Highest Order of diffraction grating

Distance you can resolve two objects

Cd diffraction, find spacing between tracks

Great wall from space, find angular size of the wall

Min angle human eye can resolve

Wavelength of laser beam in water of fish tank

double slit, determine the separation of the bright fringes

infrared filter, find minimum thickness

double slit, find size of individual slits and separation when given m

sep f int e fe ence max =!Ld

where L = screendistd = split sep

tmin thickness =

m + 12

!"#

$%&'

2n

sin! =m"d

# d =sin!m"

grating =1d=

m"sin!

a =m!

sin"#

1N

=m!

sin"

m =1

N!

slit width =

1.22 length( )!diam.

1.)! = arctandist.betweencd and screen

first dist.abovetable"#$

%&

2.)spacing =2(

sin!

sin!1 1.22"

diameter#$%

&'

arctan

widthaltitude

!"#

$%&

1 ) f =c

!in ai

2 ) !in wate =c

nf

d = w + s wherew = widthof slits, s = dist betweenslits

separation =m!L

d=

m!Lw + s

tmin =

!ai

4n

slit size =

!ilengthmimax defraction

separation =

!ilengthmax defraction

Enterprise, find length

Two ships in the night, find speed in c

Vulcan, find value of Lorentz factor beta.

Speed, how many times faster than fighter jet

Momentum, find KE

gold nuclei, find diameter when they reach certain total energy

Voyager, find how much pop of Earth aged during trip # of light years

Same Velocity, find KE to match proton beams

Transformation, find time in seconds

Redshift, find velocity of galaxy moving way

Black Hole, find Schwarzshild radius

Burger after the show, find usable energy

L = L0 1!vc

"#$

%&'

2

where v / c( ) is fractionof c

vsum =

v1 + v2

1+ v1v2

v =2!rT

" =vc=

2!r / Tc

1) v = !cms

"#$

%&'(

miles1609.34m

(3600s

hr

2) times faster =v

speed of jet

1) E = p2 + 0 9382

2) KE = E ! 0 938

gamma =given GeV

massof nucleon(ie 0 9315)

length =Lo

gamma

KE =

given MeVmassof proton

!massof electron

! = coeff of V

gamma =1

1" ! 2

t = gamma t "!xc

#$%

&'(

V =

c z2 + 2zz2 + 2z + 2

1000

r =2Gm

c2

whereG = gravitational cons tan t

m = 2 !1030 ! solar mass

E = mc2

Chapt. 35: RelativityPostulate 1 Law o physics are the same in each intertial reference frame, independent of the motion of this reference frame.Postualte 2 Speed of light is the same in every intertial reference frame. The speed of light (in vaccuum) is the ultimate speed.Time Dilation

Usually beta is very close to 0, and gamma is only slightly larger than Length contraction Frequency Shift

Upper signs are for observer and emitted signal moving away from each other, lower for moving toward each otherWavelength shift

Red shifted object moves away from usBlue shifted objects move toward us

Red shift

Relativistic velocity addition

Relativistic momentum

Energy contained in mass of particle at rest

Particle in motion

Kinetic energy

Momentum and energy

Speed, energy, momentum

Find total energy, KE, and momentum of electron with 99% of speed of c

Schwarzschild Raidus

Same velocity example, find required KE for electron beam with given proton beam KE

v / c = 0 9995

!t =!t0

1" v / c( )2= #!t0

! =vc=

pcE

" =1

1# ! 2=

1

1# v / c( )2

L =

v!t0

"=

L0

" f = f0

c ! vc ± v

! = !0

c ± vc ! v

z = !""0

=" # "0

"0

=c ± vc ! v

#1

for small v z ~ $

Lorentz

x ' = ! x " vt( )y ' = yz ' = z

t ' = ! t " vx / c2( )

Inverse

x = ! x '+ vt '( )y = y 'z = z 't = ! t '" vx '/ c2( )

objects move in opp.

v1+2 =v1 + v2

1+v1v2

c2

objects move in same

v1!2 =v1 ! v2

1!v1v2

c2

!p = ! m!v

E0 = mc2

E = ! E0 = ! mc2

K = E ! E0 = " !1( )E0 = " !1( )mc2

E = p2c2 + m2cAntimatter

E = ! p2c2 + m2cCase of zero mass (photons)E = pc

p =

!Ec

or E =pc!

E0 = mc2

! =1

1" # 2=

1

1" 0.992

E = ! E0

K = ! "1( )E0

p =#Ec

Rs =

2GMc2

Ke + mec2

mec2 =

K p + mpc2

mpc2

Ke = mec2

K p + mpc2

mpc2

!

"#

$

%& ' mec

2

( Ke = 0.511MeV( ) K p + 938MeV938MeV

!

"#

$

%& ' 0.511MeV( )