Goal: Determine Q, Ll.E and W; Wand Q depend on path; Ll.E is a state variable, independent of path 2Guiding Principle: a.. Loop: A closed conductor path dResistors in series or parall
Trang 1I
- - - - ,-
sin6 - cos6 Rule
1
Mass of electron me 9 llxI0-3 kg Mass of proton I11p 1.67x I0-27 kg , Avogadro Constant NA 6.022x I023 mol I
Elementary charge e 1.602xl0 19 C Faraday Constant F 96,4 85 C / mol
Speed of light c 3x 1 0~ m s-I Molar Gas Constant R 8.3141 mol-I K I Boltzmann Constant k 1.38x I0-23 1 K-I Gravitation Constant G 6.67x I0-11 m3 kg-I s 2
i Permeability of Space 110 4n:x1 0 7 N/A2
Permittiviy of Space Eo 8.85x10 12 F/m
MATHEMATICAL CONCEPTS
A Vector Algebra
I Vector: Denotes directional character using (x,y, z) components fig I
a V nit vectors: i along x, j along y, k along z
b Vector A = Axi + A"j + A=k
c Length of A = IAI = I'A C-:,'- +-AC-:,':-+ -A :-C:,'
2 Addition of vectors A & B, add components:
A + B = (A,+B,)i + (A,,+B, ,)j + (A=+BJk
, Sample Addition and Length Calculations: A=3i+4j-3k IAI = -/9 + 16 +9
= 134 =5.83
B=-2i+6j+5k IBI= /4+36+25
= 165 = 8.06 A+B=i+IOj+2k V4 B I= II +100 +4
= /lOS = 10.25 Note: V41 + IBI ~ IA + BI
3 Multiply A & B:
a Dot or scalar product: A - B = IAI IB lcos6 =
(A,B, ) + (A,.B y ) + (A= B=)
Note: 6 is the angle between A and B;
A-B = 0,if6 = n:/2 tig2 Sample: Scalar product:
A = 5i + 2j B = 3i + 5j
A- B = 3 x 5 + 2 5 = 15 + 10 = 25
IAI=/25 +4 =/29= 5.385
IBI=-/9 +25=134= 5.831
cos6 IA II B I 5.385 x5.83 0.796
6 = cos·I(0.796) = 37° = 0.2n: fad fig 3
Base Units
Length
Mass
Temperature
Time
Electric Current
Derived Units
Acceleration
Ang Acce!
Ang Momentum
Ang Velocity
Angle
Capacitance
Charge ,
~
Electric Field
Electric Flux
Electromotive
Force (EMF)
Energy
Entropy
Force
Frequency
Heat
Magnetic Field
Magnetic Flux
Momentum
Potential
Power
Pressure
Resistance
Torque
Symbol
/, x m,M
T
/
Symbol
a
a
e,
Q, q,
p
S, d, h
E
+"
;r
E, U,K
S
F
Iv
+m
r
Vnlt Meter - m
Kilogram - kg
Kelvin - K
Second - s
Ampere - A (Ci s)
Unit
m/s 2
radian/s2
kg m2 /s
radian/sec
radian
Farad F (CN )
Coulomb C (A s)
kg/m3
meter - m
V i m
Joule J (kg m2
s-J
Newton - N (kg m/s
Hertz - Hz (cycle/
Tesla (Wb/
Weber Wb (kg m2 1A s2)
kg m/s
jVoltage V (l/C) ' Watt - W (1/s)
Pascal - Pa (N/m2)
Ohm n (VIA)
Nm
~-= 1m)
Ener
I t Force
Volume
Pressure
ILength
sin 6 = 2:'
r
cos 6 = f tan 6 = ~ = ~~~
sinl + COS2 = 1
2 Sinand Cos waves fig (,
Unit
" (degree)
Erg
Electron Volt
Dyne
Liter
Bar
Angstrom
Description
180" = n: rad
CGS unit (g c m 2 / s 2)
I erg = 10 7 J
I cV = 1.602x 1O-19 1
CGS unit (g em/s2 = erg/em)
I dyne = 10 5 N
I L= I dm3
I Bar = 105 Pa
I A= Ix lO 10m
~
C = A x B = V4 IIBlsin6 e
6 - Angle between A and B, vector e is -a perpendicular to A and B
m
i j k
A = 2i + j B = i + 3j
i j k I
A x B = 2 I 0 = I (6 - I) k = 5k
130 -IfA and B are in x-y plane, A x B is along
-6 is the angle formed by A : sin6 = IAC B
Given: IAI = IS IBI = /iO ICi = 5 sin6= 5/(1S x /iO)~51!50 = 1 /1
< AB: 6 = 45" = n:/2 radians
c The Right-Hand Rule gives the orientation of vector e flt!.t
B Trigonometry
I Basic relations for a triangle fi" 5
Values Of sin cos and tan
erad ["] sin!i cos tJ tan tJ
0 [O"]l -OOO 1.00 0.00 f-;;H,[ J(J"t 0.50 0.866 0.577 rr/4l45"J 0.707 '0.707 100 f-;;;' J1,6l fj 0.866 0.50 , 1732 "
~ O"] O.t) ~ 1.0010·00 -a
Trang 2MATHEMA11CAL CONCEPI'S (cont.)
C Geometry
Circle: Area = rrr2; Circumference =2rrr
Sphere: Volume = 4/) rrr\ Area = 4rrr2
Cylinder: Volume = hrrr
Triangle: Sum of angles = 180" IiI!
D Coordinate Systems
I One dimension (I-D): position = x lig R
• The x position is described relative to an origin
a Calculate (r, e) from (x, y):
r=./ x ' +y ' ; e = sin· 1( )
b Calculate (x, y) from (r, e), or x and y
components of a vector "r" with angle e; •
Sample: Generate x and y vector v = dx l dt ;
a = dv l dt
components, given: r = 5.0, e = ~ (30")
x = r cos( ~) = 5.0 x 0.866 = 4.33
y = , sin( ~) = 5 x 0.50 = 2.50 i \ '
2.52 + 4.332 = 6.25 t- 18.75 = 25.00 Cartesian: (x, y)
3Three Dimensions (3-0)
a Cartesian (x y, z) : The basic coordinate CD!
• Polar coordinates, with a z axis
• Calculate (r , e) from (x,
• Same process as for 2-d polar;
z: same as Cartesian
x = rcose,
x = r sinq> cosS, y = r sinq> sine,
z = r cosq>, ? = x 2 + y2 + z 2 fill II r2 = x2 + y2
'Calculate (r, e, q» from (x, y, z)
Spherical
jJ
polar coordina
E Use of Calculus in Physics
I Methods from calculus are used physics definitions, and the derivations In Ny
of equations and laws
Physical meanin s of calculus expressions: x
x = r sinq> cose
a Derivative - slope of the curve: d ~~ X) y = r sinq> sine
z = r cosq>,
r 2 = x2 + y2 + z1
b Integral -area under the curve: IF(x)d~
Samples:
• Position: x or F(x)
Velocity: vex) = dF( x )
Acceleration: a = dr(x ) and integrals
,-Fe x ) dF (x) ' IF( x) dx II
dt
E=IFd~
I x I 1 x 2
b ' ~= G dx - G ' dx IJ Il - 1 _ I_ x" ,
x " II +1
c Partial derivative:
Inx
x I x "
e Integration by parts:
-f Symbol for integration of
c os ( x ) -sin(x) sin(x) closed surface or volume:.f ·
I Vector vs scalar
a Vector: Has magnitude and direction
b Scalar: Magnitude only, no direction
2 Number and unit
a Physical data, constants and equations have numerical values and units
• Electrostatic potential energy:
q , q
U c = 1/(4rr£o) -, constant: 1/4rrEo; units are J m' C2
r in m, q] and q2 in Coulomb Units of U c = (J mlC 2 )C 2 / m = J
b A correct answer must include the correct numerical 5 Using Conversion Factors
a The # of sigfigs reflects the accuracy of experimental b SI units: MKS (m-kg-s) and CGS (cm-g-s)
b For multiplication: The # of sigfigs in the final d Conversion factors are obtained from an answer is limited by the entry with the fewest sigfigs equality of two units
c For addition: The # ofdecimal places in the final answer Sample: 100 em I m
is given by the entry with the fewest decimal places • This equality gives two conversion tactors:
·If the last digit is <5 round down
• Use the l;t factor to convert "cm" to "m"
·If digit = 5, round up if preceding digit is odd
1.245 + 0.4 = 1.6 (I decimal place)
• Use the 2nd to convert "m" to "cm" 1.345 x 2.4 = 3.2 (2 sigfigs)
m
a Two key issues:
2 Have a correct mathematical strategy
I Prepare a rough sketch of the problem
physical co
"right" equation in your notes or text
6 Pitfall : If the units ar e wrong, the answer is
3 Describe the physics using a
* Take special care if you derive the equation 4 Obtain the relevant physical constants
Do you have all the essential d ta'!
information
in the correct ov rall unit
5The hard part: Derive or obtain a Samples: The en rgy unit is Joules for kin tc,
gravitational and Coulombic energy
problem; use dimcnsional a alysis to
- "'+v ~ 6 The easy part: Plug n mbers into the:
m in kg, v in m / s Units of K = kg
I Goal: Determine position, velocity acceleration I Goal: Similar to " A " with 2 or 3 dimensions
2 Key terms: Acceleration: a = dvl d l velocity: v = ddd t 2 Key concept: Select Cartesian, polar
spherical coordinatcs, depending on the type
3 Key Equations: x = V i ' + ~ aI 2 1= V i + al
motion
with 'i 'i; how do we set up the problem?
Step I Detine x as horiLontal and)' vertical
Step 2 Detennine initial " ,i and "r , 1'" 16
" xi = V r ; cosS
Step 3 IdentifY a, -Gravitational fo e => al · = -g
2
Trang 3Step 4 Identify a" -No horizontal force => a, = 0
Step 5 Develop x-and y-equations of motion
X = Vi,,1 + 2axt-= viI
1 Goal: Examine force and acceleration
Law #2 Forces acting on a body equal the mass
action
a Law #2: F = III a or LF = III
• Surface -ti'iction: = Fj' = flF"
Sample: F, exerted on
object on a horizontal plane
F( = flF" = fl rg =fl III g
Net force = F, - F lig 17
examine Fg & F f
Fj '= fl F Il = fl III g cos8
e Law #3:
FI2 = - F21 or 1111 a l = - 1112 al
bullet fired from a rifle L - - - '_ _" ' - _ - - '
Rifle recoil = ,,(bullet) x III (bullet)
D Circular Motion fig 19
use 2-d polar coordinates: (r, 8
Key variables:
rotation center
e l rad
a ' s
I
angular
l ex rad/52
motion arc;
s I m
the center fig 19
V, = roo = 6.378 X 106 m x 7.3 x 10.5 rad/s v, =465 m/s
E Energy and Work
with forces acting on an object
d Woet = K tinal - K initial ; K is converted to work
a 50 kg box 10 m; given: a = g = 9.8 m/s2
Equations: F = III g => W = III g d
gravitation ( V = mgh),
b E = K + U Conservative system: No external force
Sample: Examine K & V for a launched rocket
D
Next, resolve into x and y components: K y i & Kl'i
the flight
ground: V = 0, K = Ki 40 0 a fig 20
a Types of collisions:
b Relative motion and frames of reference: A body moves with velocity v in frame S; in frame S', the
relative to S, then V = J{,+ v'
d Conserve K & p for conservative system (no
external forces):
Ll2 m v· I 2 = Ll2 l1lv/ 2 Lm Vi = LmVj
Conservation of momentum:
ml Vii + m2 V2i = (1111 + m2)vr
before collision
"
"'1 "'2 ~ vf
" "
X Lm = Em, Y ,."nI = Lm , Ze -Em ;
& a 2 kg ball connected by a 1.00 m bar ball I: XI = 0.00, 1111 = I kg; 1111 XI =0.00 kg m
ball 2: Xl = 1.00 III 2 = 2 kg; 1111 x 2 = 2.00 kg III
Lilli xi = 1111 XI + 1112 x2 = 0.00 + 2.00 = 2.00 kg III
= LIIl ,Xi = 2.00kg III = 0
O.66m
,··· ···1 O.33m
center
3
b Moment of inertia:
1= Lilli I}, with ri about the center of mass along a
specific axis
~
I =l.mR2
Rotating sphere of radius,
1=1.mR2
5
Sample: Determine th r for a spherical Earth
assume uniform M ;
Data: M = 6 x I 024 kg, r = 6.4 x 106 m
r= 1M r 2 = 1 x 6xl024 kg X (6.4xI06 m)2
5
= 9.8 X 1037 kg m
2 Key variables and equations
L = /oo = rxp = frxvdlll F
note: vector cross product lig
L force = 0 & Ltorq ue = 0
component; any net
moves the object,
fig
2 Case 2 Examine deformation of a solid body
For equilibrium:
y =
L11/ I
1_/o-IMI-Note: Force F is longiludinal
co F,/A
•• Axlh
fig 27b
B= F"IA L1V/V , _ L v
,,-!:.:'
,
F"
Trang 4MECHANICS (continued)
K, Universal Gravitation r ·_·
I Goal: Examine gravitational energy and force fig 28 MI··· ·M ~
2 Case 1: Bodies of mass MI & M2 separated by,
I
a GravitatIOnal Energy: U, = ; I
b Uravltatlona ,orce: ,=~ I
I
I
c Acceleration due to gravity: g = G
For objects on the Earth's surface g =
T
~
Sample: Verify "g" at the Earth's
Equation: g = G M(earth)
Given: M = 6xl024 k, r = 6.4x
Calculation - (6.4 X I 0" In) ' - 9.8 ms
HOOKE'S
4 Case 2A body interacts with the Earth fig 29
LAW
5 Key Equation:
a Gravitational potential energy: Ug = 11/ g h; object
the Earth's surface, h 0; U =
b Weight = gravitational force; Fg = III g
Sample: Calculate escape velocity, I'esc' for an
P Hint: K = Ug at point ofescape; r = h + r (earth)
2'11 vesc - - r - ' t erelore, vesc - -,-.
Note: "esc varies with altitude, but not rocket mass
t
L Oscillatory Motion
I Goal: Study motion & energy of oscillating body
e
a Force: F = - kl'!.x (Hooke'S Law)
Surface
b Poten(ial Energy: Uk = :!-kl'! cJ
Liquid
c Frequency = 2~ /! fig 30
3 Simple Pendulum
a Period' T = 2IT Vg fI VWIAA"", Surface
b Potential energy: U = m g h [ ] p Liquid
4 For both cases:
a Kinetic energy: K = 1 m ,,2
2
bConservation of Energy: E = U + K
M Forces in Solids and Liquids
I Goal I: Examine properties of solids & liquids
a Density of a solid or liquid: p = mluss
1'0 ume
- " Sample: A piece of metal, 1.5 cm x 2.5 em x 4.0 em, has a mass of 105.0 g;
detennine
Equation: p =
Data: m = 105.0 g, V = 1.5 x 2.5 x 4.0 em3 = 15 em3
Calculate: p = 105.0115.0 g/ em 3 = 7.0 g/cm
b Pressure exerted by a tluid: P = area
c Pascals's Law: For an enclosed tluid, pressure is equal at all points in the vessel
Sample: Hydraulic press: F = P I A for enclosed liquid; A is the sUlfaee area
of the piston inserted into the tluid
Equation: A IFI = A 2 F2 ; cylinder area detennines force fig 32
d A column of water generates pressure, P increases with
Equation: P2 = PI + pgh
eArchimedes' Principle: Buoyant force, Fb' on a object of volume V submerged
in liquid of density p: Fb = pVg fig 34
2 Goal 2: Examine fluid motion & fluid dynamics
a Properties of an Ideal fluid: Non-vi cous, in om pre sible, steady Bow, no turbulence
At any point in the tlow, the product of area and velocity is constant: A IVI = A2v2
b Variable density: p I A I VI = P2 A2 v2; illustrations: gas flow through a smokestack
water flow through a hose fig 35
c Bernoulli's Equation: For any point y in the tluid tlow, P + :!-pv 2 + P g )' = constant
-Special case: Fluid at rest PI - P2 = P g h
A Descriptive Variables
I Types: Transverse, longitudinal, traveling, standing harmonic
a General form for transverse travding wavc: y = f(x - I'/)
right) or y I(x + 1'1) (to
b General fornl ofharmonic wave: y = Asin(kx -WI) orv =A cos(kx - w()
c Standing wave: Integral multiples of.1 fit the length of '11 I
d General wave equatIOn: dx ' =pI dl '
c Superposition Principle: Overlapping waves interact => constructive
'Peri<;d T (sec) Time to travel one -;::
I Frequency /(Hz) ' = 1 T Angular Frequency w (rad/s) W= 271 T = 2n/
Wave Amplitude A i Height 0 wave Speed 11 (m /s) 11 = AI Wave number k(m-I) I k = 2;
' - ~
~-2 Sample: Determine the velocity and period of a wave with
A= 5.2 m andf =
Equations' I' = A} ' T = f
Data: A= 5.20 mJ =
Calculations: v = AI= 5.20 m x 50.0 = 260 m /
T = t = 510 Hz = 0.02
B Sound Waves
I Wave nature of sound: Compression
2 General speed of sound: v = fIt;
note: B = Bulk Modulus (measure of volume compressibility)
For a gas: v = / -V ; note: y = c (ratIO 0 gas heat capacilles) Sample: Calculate speed of sound in Helium at 273
Helium: Ideal gas, y= 1.66; M =
= ( 1.66 X 8.314 kg m' Is ' X 273K
= ; "" 9 -c4-c-1,9OO m ,s'-'-' -'-,-""'·, = 971 mls note: r applies to the unils
3 Loudness as intensity and relative intensity
a Absolute Intensity (I = Powerl Area) is an inconvenient meas
b Relative loudness: Decibel scale (dB): ~ = 10 log +;
threshold of hearing; ~(l0) = 0
c Samples: let plane: 150 dB; Conversation: 50 dB; a represents a 10-fold increase in ,
4 Doppler effect: The sound frequency shifts 7
of source and listener;
"0 - listener speed; "s -source speed; v - speed of so
f' v +\'0 II \' - 'u
~
Key: Identify relative speed of source and listener """""
THERMODYNAMICS
A Goal: Study of work, heat and energy of a system tig ,6
-IHeat: Q +Q added to the system
lWork: W + W done by the system
i Energy: E Systcm internal E ~
Entropy: S Thermal disorder
- - -
Temperature: T Measure of thermal E
Pressure: P Force exerted by a gas
Volume: V Space occupied
Trang 5~ Ai", too.~ilIl4 Uii,
Types of Processes E The Kinetic Theory of Gases A Electric Fields and Electric Charge
Isothermal Ll.T= 0
P V = constant
Adiabatic I Q = O
Isobaric: Ll.P = O
Iochoric Ll.V =O
' - -
B Temperature & Thermal Energy
I Goal: Temperature is in Kelvin, absolute
temperature: T(K) = T("C) + 273.15
Note: T(K) is always positive; lab temperature
must be converted from °C to Kelvin (K)
" Sample: Convert 35" C to Kelvin:
T(K) = T(°C) + 273.15 = 35 + 273.15 =
308.15 K
2 Thermal Expansion of Solid, Liquid or Gas
a Goal: Determine the change in the length
(L) or volume (V) as a function of
temperature
b Solid: Ll L = uLl.T
L
c Liquid: LlV =
d Gas' Ll.V = (7; -T.)nR
3 Heat capacity: C = iT or Q= CLl.T
a Special cases: Cp constant P; Cv-;;onstant V
bCarnot's Law: For ideal gas: Cp - Cv = R
oLl.E=C Ll.TLl.H=C-Ll.T
oExact for monatomiC gas, modify for
'"
~ molecular gast!s
~
=
'I CPre s sure (Pa) JL2 Temperature (K)
C Ideal Gas Law; PV= nRT lig ,7
I Goal: Simple equation of state for a gas
2 Key Variables: P(Pa), V(m3 ), T(K), n moles of
gas (mol); gas constant R = 8.314 J mol-I K-I
LG Pitfall: Common errors in T, P or V units
3 Key Applications:
a P a -&' T fixed: Boyle's Law
b P a T, Vfixed
c VaT, P fixed: Charles' Law
d Derive thermodynamic relationships
D Enthalpy and 1st Law of Thermodynamics
I Goal: Determine Q, Ll.E and W; Wand Q
depend on path; Ll.E is a state variable,
independent of path
2Guiding Principle:
a 1st Law of Thermodynamics: Ll.E = Q - W
oKey idea: Conservation of Energy
b Examine the T, P, W & Qfor the problem
3 Enthalpy: H = E + PV; Ll.H = Ll.E + PLl.V
b Variable temperature: Ll.H = JCpdT
c For constant Ci Ll.H = CpLl.T
4 Work: W = JPdV
a W depends on the path or process
b Ideal Gas, Reversible, Isothermal :
W= nRTln II,
V
c.ldeal Gas, Isobaric: W = PLl.V
I Goal: Examine kinetic energy ofgas molecules
2Key Equations: E = IMv 2 and E = lRT
a Speed, vrms = j 3::
Sample: Calculate the speed of Helium at 273 K
Helium: M = 0.004 kg/mole
vrms = ~=
/ 3 x8.314 kg m'ls' X 273 K
V rms = ! l 702, 292 mls = 1305 mls
b Kinetic energy for Ideal Gas: K = lRT
2
c For real gas: Add terms for vibrations and rotations
F Entropy & 2nd Law of Thermodynamics
I Goal: Examine the driving force for a process
2 Key Variables:
a Entropy: S, thermal disorder; dS = dF
b S(univ) = S(system) + S(thermal bath)
3 Guiding Principle: 2nd Law of Thermodynamics:
For any process, Ll.Suniv > 0; one exception: Ll.Suniv = 0 for a reversible process
4 Examples:
a Natural heat flow: Qflows from Thot to fcold fig 3R
Ll.Suniv = Ll.Shot + Ll.Scold =
j) hint: Ll.Suniv > 0 for a natural process
Q(phase change)
b Phase Change: Ll.S = T (phme chang e1 c.ldeal Gas S(T): Ll.S = nCpln¥,
dIdeal Gas: S( V): Ll.S= nRln~
G Heat Engines
I Goal: Examine and W
2Thermal Engine
- - "
transfers Q from a
hot to a cold reservoir, and produces W fig 39
3 Efficiency of engine: ~ = QW = I _ QQ' '''
I "" h " 1
4 Idealized heat engine: Carnot Cycle fig 40
Carnot Cycle
P
A
Ll.T=O
Tcold
C
~ -V
a Four steps in the cycle: two isothermal, two adiabatic; for overall cycle: Ll.E = 0 and Ll.S= 0
bEfficiency = I - ~ "
1 ",
5
I Goal: Examine the nature of the field gt!nerated by an electric charge, and forces between charges
2 Key Variables and Equations
a Coulomb C: "ampere sl!c" of charge
b e - charge on an electron; 1.6022 x 10 19 C
c Coulomb' Law - electrostatic force: F = 4it: q; ? ,e
oVector direction defined bye "
d Electric Field: E = %
j) Hint: Calculation shortcut:
r(!n )
Note: q in Coulombs and r in meters
3 Superposition Principle: Forces and tields are composites of contributions from each charge
F = 1: Fi , E c ~ 1: Ei ; j) Hint: Forces and electric fields are vectors
B Gauss's Law
I Goal: Define electric flux, +
2Key Variables and Equations:
a Gauss's Law: +e = • ' IE x dA = ~ ~I\
bThe electric flux, +e' depends on the total charge in
the closed region of interest
C Electric Potential & Coulombic Energy
I Goal: Determine Coulombic potential energy
2 Key Variables and Equations
P t t • I U - I q , q
a 0 en la energy - 47ft: " r
b Potential: V(q/) = 1L = I !l!.
l/ 47ft: " r
Note: The potential is scalar depending on 1/' 1
c For an array of charges qi' V total = I.V i
dShortcut to V(r): U = 9 x 10 J r(m-)
3 Continuous charge distributions: V = 4it: J d;, I" 41
Radius R, Charge Q
V = -4 I Q for r < R
7ft: " R
V = _ 1- Q for r > R
V 47f t: u r kQ H
4Dielectric elTect: V& Fdepend on
the dielectric constant, K; replac
I ! -+r
Eo with KtO for the material;
F(lC) = IF(va c uum)
K
D Capacitance and Dielectrics
I Goal: Study capacitors, plates
charge Q
dielectric material fi~ ·
2 Key Equations:
a Capacitance, C = ~ , V is the measured voltage
b Parallel plate capacitor, vacuum, with area A,
spacing d: C = EOA; E = Q A
d t:u
c Parallel plate capacitor, dielectric lC with area A,
spacing d: C = Kto1 \AI ;: J C V l
.'r Z
3 Capacitors in series: t = L t 1(
4Capacitors in parallel: Ctot = I.C i fI'l43
Cz tot I 2 I Z
Two Capacitors in Parallel
lCJc;l - cl+C Z
Trang 6o
ELECTRICITY
MAGNETISM
~ E Current and Resistance
I Goal: Examine the current, I , quantity of charge, Q,
resistance, R; determine the voltage and power dissipated
2 Key Equations:
a Total charge, Q = It
b V = IR , or R = T V
R t o = ~Ri rt)!
d Resistors in Parallel:
Two Resistors in Parallel
lR.J!Q I I 1
e Power = IV = 12R tlJR to; R;+Tz
F Direct Current Circuit
I
containing
2 Key Equations and Concepts:
V h = 1 r, r internal battery resistance
b Junction: Connection of 3 or more conductors
c Loop: A closed conductor path
dResistors in series or parallel => replace with Rtot
e Capacitors in series or parallel => replace with Ctot
3 Kirchoff's Circui~ Rules
a ForanyLoop:~V = ~/R;
P Hint: Conserve energy
b For any Junction: ~I = 0;
Z
.tIIIII PHint: Conserve charge;
'liliiii define "+" flow Ii~ 4"
G Magnetic Field: B
I Key concepts: •
~ a Moving ch<\fge => Magnetic Field B
~ c Force on charge, q and v, moving in B:
= F= qv x B = qvBsinll ; v parallel to B => F = 0; v
perpendicular to B => F = qvB
~ dMagnetic Moment of a Loop: M = 1 A
e Torque on a loop: 't = M x B
f Magnetic Potential Energy: U = -M • B
g Lorentz Force: Charge interacts with E and B;
F=qE+qvxB
H Faraday's Law and Electromagnetic Induction
Key Equations:
I Faraday's Law: Induced EMF: if= f E ds = -d J m l dt
2Biot-Savart Law: Conductor induces B; current 1,
).ill r
length dL : dB = 47f1dL x ?
3 Sample: Long conducting wire: B(r) = f; f
I Electromagnetic Waves- Key Equations and Concepts:
I Transverse Band E fi elds; £ = c
2.c = _I
,j o e
3 Electromagnetic Wave: c =
)'
x
J Maxwell's Equations:
I Gauss's Law for Electrostatics: ¢E • dA = ~ ;
key: Charge gives rise to E ' "
2 Gauss's Law for Magnetism: rf B • dA = 0;
key: Absence of magnetic charge
3 Ampere-Maxwell Law:
~B • ds = flo! + flo£o
key: Current + change in electric flux => B
4Faraday's Law: 1E • dS = _ d: m ;
key: Change in magnetic flux => E
BEHAVIOR OF LIGHT
A Basic Properties of Light Reflection and Refraction
Incident
I Goal: Examine I ight and its interaction with matter
Ra)
2 Key variables:
a c: speed of light in a vacuum b.lndex of refraction: 1/; f, = speed of light in medium Renectcd
Ru)'
c Light as electromagnetic wave: A{=
Light characterized by "color" or wavelcngth
dLight as particle: e = hI; energy of photon
o
3 Reflection and Refraction of Light fig 47
a Law of Reflection: III = Ilr fig 48 ; ,, 1 , /
b Refraction: Bending of light ray as it passes fi'om 1/ I to 11
·Snell's Law: IIlsinlll = 1l2sinIl2.1lJ, two materials fig 172 ~
c Internal Reflectance: sinll, = !!2
II
Light passing from material of higher n to
4 Polarized light: E field is not spherically symmetric
a Examples: Planellinear polarized, circularly polarized
b Polarization by reflection from a dielectric surface at angle Il,
Brewster's Law: tanll, = _'!.!
II I
B Lenses and Optical Instruments
I Goal: Lenses and mirrors generale images of objects
2 Key concepts and variables
Lens and Mirror Properties
a Radius of curvature: R = 2I
b Optic axis: Line from base of object Parameters + sign -through center of lens or mirror f - - - - f - - - j - - -
converging lens diverging lens
I focal length
c Magnification: M = f concave mirror convex mirror
d Laws of Geometric Optics: s object dis! real object virtual object
1+ 1 =.1 2.=_l! s' image dist real image virtual image
S .I' f' s' h'
h object size erect invcrt.:d
e Combination of 2 thin lenses:
I I I {J,J; h' image size erect inverted
7 =7, +]; or = J, +};
3 ~ Sample Guidelines for ray tracing:
a Rays that parallel optic axis pass through " j'''
b Rays pass through center of the lens unchanged Sample ray tracings: fig 50,
:,
' ,: "
C Interference of Light Waves
I Goal: Examine constructive and destructive interference of light waves
2 Key Variables and Concepts:
a Constructive interference: fig 51
b Destructive interference: fig 52
c Huygens' Principle: Each portion of wave front acts as a _ _ _ _-*' _ _ ">j
3 DifIraction of light, from a grating with spacing d,
an interference pattern; dsinll = inA; (m = 0, 1,2,3,
4 Single-slit experiment, slit width a; destructive interference for sinll = I7~A; (111 = 0, ±J, ±2 )
5 X-ray diffraction from a crystal with atomic spacing d; 2dsinll = 111"-; (m = 0, 1,2,3, )
/ / / : .
~r_ -t-i ~ -r - ;- x -+ ~ ~ ~ ~ ~ x
" ~~
6
Spherical Concave Mirror