AP Physics: Volume 2
Chapters 9-17
Chapter 9: Solids and Fluids
9.1: States of Matter
3 states:
Solid- Definite volume and shape
Liquid- Definite volume; changing shape
Gases- Neither definite volume nor shape
Plasma- #4- A highly ionized substance containing equal amounts of
+ and – charges.
*It exists inside stars.
*Plasma is the most prevalent state of matter in our
universe.
*This is formed by matter being heated to ultra high temperatures
resulting in a substance with free electrically charged particles.
*Solids are either crystalline or amorphous.
*Crystalline Solids- The atoms have an ordered structure.
Ex. Salt (Pg. 257)
*Amorphus Solids- Atoms are arranged randomly
Ex. Glass
*Liquids are always at a higher temperature then the solid form of the same substance.
The high temperature causes the molecules to wander through the substance randomly.
*Gases- Molecules are always in random movement and have only weak nuclear forces
on each other. Their separation is quite large; much more so than liquids. They even
collide occasionally.
9.2: The Deformation of Solids
For this chapter, we will assume that all solids maintain their shapes, regardless of how
much force is put upon them.
*Stress- The force causing a deformation.
*Strain- The degree of deformation.
Elastic Modulus =
stress
strain
Tensile Stress- The ratio of the magnitude of the external force (F) to the cross sectional
area. The units are:
1N
=1 Pa (pascals).
m2
Tensile Strain- The ratio of ∆ length to original length
L
Lo
Youngs Modulus- Elasticity in length. Uses
stress
.
strain
 L  FLo
F
 
Y =    
 A
 Lo  AL
*This is usually used to characterize a rod or a wire under either tension or compression.
*Pg. 260: Fig. 9.4
*Exceeded the stress/strain curve goes from straight to almost a square root graph
until it reaches the breaking point
*Exceeding the elastic limit results in deformation.
Shear Modulus- Elasticity of shape
S=
shear _ stress  F   x  Fh
=     
shear _ strain  A   h  Ax
*∆x = horizontal distance the sheared face moves
*h = height of object
*F = force applied at a tangential direction
*A = area of the face being sheared
Bulk Modulus- Volume Elasticity
B=
*Since
volume _ stress
 F   V 
 V 
= 
 
  P   

volume _ strain
 A   V 
 V 
F
= Pressure, use P instead
A
*The negative sign is used to make a positive reading possible because an increase in
pressure causes a decrease in volume.
Ex. +∆P causes -∆V
*The reciprocal of bulk modulus is called the compressibility of the material.
*Notice on the chart on Pg. 259 that solids and liquids have bulk modulii. However, they
have no Youngs Modulus nor Shear Modulus; a liquid will not sustain a shearing stress
or a tensile strength because it flows.
**DO EXAMPLES ON PG. 261**
9.3: Density and Pressure
*The density of a substance of uniform composition is defined as its mass per unit
volume.
p=
m
mass
=
V
volume
p = Rho (Greek abbreviation for density)
*Pg. 262: Table 9.2
Density is
kg
g
or
3
cm 3
m
*The density of most liquid and solids varies slightly with temperature and pressure.
*The density of gases varies greatly with temperature and pressure changes.
*Under normal conditions, densities of liquids and solids are 1000 times that of gas
*This implies that the average distance between molecules is 10 times in gases since
103=1000
The specific gravity of any substance is the ratio of its density to the density of water @
4oC
water @ 4oC = p = 1*103
kg
m3
*Any force that applies stress to an object in a liquid tends to compress the object. The
force exerted on an object in a fluid is always perpendicular to the surfaces of the object.
Recall: P =
F
A
P = pressure
**DO TACK LAB ON PG. 263**
*Place between thumb and forefinger and squeeze
*Fthumb = Fforefinger per newtons 3rd law
*Why does only sharp end hurt? P =
F
A
*If force and pressure are equal, than area is the difference.
*The same goes for baseball spikes or steal pointy bottom spikes. They have a small area
on which to apply your weight onto the floor. This is why snowshoes keep you above
snow.
Ex. A waterbed is 2 m x 30 cm x 2 m
a. Find its weight.
Pwater = 1000
kg
m3
Vbed = (2*2*0.3) m3 = 1.20 m3
M = pV = (1000
kg
) (1.20m3) = 1.2*103 kg
m3
Fweight = Mg = (1.2*103 kg) (9.8
m
) = 1.18*104 N
2
s
Note: 1.2*103 kg = 2640 lbs (Check for yourself.)
b. Find the pressure the waterbed (full) exerts on the floor.
From part “a”… w = 1.18*104 N and cross-sectional area
(touching floor) = 2 m * 2 m = 4 m2
1.18 *10 4
N
 2.95 *10 3 2 = 2.95*103 Pa
P=
2
4.0m
m
c. Find the pressure if the bed is on its side.
P=
1.18 *10 4 N
= 3.9*104 Pa
2 * 0.3
d. Is the pressure the same if its stood on end as it is when its sat on its side?
Why?
9.4: Variation of Pressure with Depth
Normal atmospheric pressure at sea level is Po = 1.01*105 Pa = 14.7
lbs
and the above
in 2
equation tells us that at a depth (h) below the surface of a liquid (open to the atmosphere)
is greater than Po by pgh.
*Recall from last year Pascal’s Principle:
1. Pressure in a fluid depends on depth
2. Any increase in pressure at the surface is distributed evenly throughout the
fluid.
*Remember that P =
F1 F2

A1 A2
*This is why a hydraulic jack the size of an air pump lifts a car.
**SEE TOP OF PG. 266 – EGYPTIAN LEVEL**
9.5: Pressure Measurements
Ex. Assume body area of 2000 in2
*Find atmospheric pressure (ap) on body.
Answer is approximately 3*104 lbs
Ex. Car Lift- Compressed air lift, exerts force on:
*1st Piston (P1) Radius = 5 cm
*2nd Piston (P2) Radius = 15 cm
What force must be put on P1 to raise a car W = 1.33*103 N?
A
F1 =  1
 A2
  (5 *10 2 ) 2 

(1.33 *10 4 N ) = 1.48*103 N
 F2  
2 2 

  (15 *10 ) 
What air pressure produces this force?
P=
F1 1.48 *10 3 N
= 1.88*105 Pa

2
A1  (5 *10 )m
Note: Only 2 atmospheres.
Ex. Ocean pressure @ 1000 m depth
Pwater = 1*103
kg
m3
Po = 1.01*105 Pa
P = Po + Pgh
P = 1.01*105 Pa + (1*103
kg
m
) (9.8 2 ) (1.0*103 m)
3
s
m
P ~ 9.9*106 Pa
What is the force on a 30 cm window at this depth?
**ASSIGNMENT page #300 # 2, 3, 5, 18,19
9.6: Buoyant Forces- Archimedes’s Principle
*Any body partially or completely submerged in a fluid is buoyed up at a force with
magnitude equal to the weight of the fluid that the body displaces.
B = pfVg = Mg = Wf
B = buoyant force
pf = density of fluid
V = volume of cube
g = gravity
M = mass of fluid
Wf = weight of fluid displaced
Weight of submerged object is given by:
Wo = Mg = poVg
But:
Wf = pfVg
And:
Wo = poVg
*These two formulas lead us to the conclusion that if density of the object is greater then
the density of the fluid, it sinks. If the density is less than the fluid, it floats.
*If the magnitude of the buoyant force equals the weight of the object, only part of the
object will be submerged.
*Floating Object:
po VF

p F Vo
**TURN TO PG. 270 AND TALK ABOUT THE BRAIN FLOATING IN SPINAL
FLUID**
**First, do this problem:
A raft (wood) with p = 600
kg
, volume of 0.6 m3, and surface area of 5.7 m2, is
m3
placed in fresh water. When pw = 1000
kg
, how much of the raft is below the water
m3
line?
Weight of the Raft:
Wr = pr g Vr = (600
kg
m
)
(9.8
) (0.6 m3) = 3.5*103 N
2
3
s
m
Upward Force on the Raft Equal to Wr:
B = pw g Vw = pw g Ah = 3.5*103 N
*Solve for h:
Nkgm
3.5 *10 3
B
s2
h=
=
= 0.06 m
kg 
m
p w gA 
2
1000 3  9.8 2  5.7m
m 
s 



**NOW DO THE EXAMPLE ON PG. 272, #9.6**
Assign #20, 21 on page 300
*Optional from here to end of chapter*
9.7: Fluid Flow
pV1A1 = pV2A2 = Equation of Continuity
*The fluid flow is constant through any cross section of the pipe.
*So…if p = p:
V1A1 = V2A2 tells us that if cross sectional volume is reduced, than V 
*Turn on sink with and without nozzle on to show example.
Ex. A hose with a 1.0 cm radius fills a 20 liter bucket. If it takes 1.0 min. to fill, what is
Vw as it leaves the hose.
*1 liter = 103cm3
*Cross-sectional area of the hose is:
A = π r2 = π (1)2 = π cm2
Flow rate = Av
(A) Av = 20.0
liters 20 *10 3 cm 3
cm
20 *10 3 cm 3

 106
=
3
min
s
60 sec( A)
60 sec(cm )
Reduce the hose to 0.5 cm.
V = 20 liters
cm
20 *10 3 cm 3
= 424
2
s
60 sec(0.25cm )
*Why did it not only double?
-
Because 22 = 4
Bernoulli’s Equation- Word done by a fluid and work done on one end of a tube (on the
fluid) is equal and opposite to work done on the other end.
*As a fluid is moved along a pipe, part of the work is done changing KE, and part
changing PE.
P1 +
1
1
pM1V12 + pgy1 = P2 + pV22 + pgy2
2
2
OR
P+
1
pV2 + pgy = Constant
2
**DO EXAMPLE 9.19 ON PG. 277**
*Surface Tension: ‫= ﻻ‬
F
L
**DO #41,42**
*Surface tension is in
J
N Nm


2
m m2
m
*Surface Tension is the energy content of the fluid at its surface per unit surface area
‫=ﻻ‬
F
2L
*L = circumference of the wire
*F = force on spring
*‫ = ﻻ‬surface tension
Water Bug- Walks on Water
Ex. An insect with a mass of 2*10-5 kg has 6 feet, each with a radius of 1.5*10-4 m. Find
the angle θ of depression the legs cause the water to indent.
*The length (L) along which the force acts is the distance around the insect’s foot, 2πR.
(assume the pressure is equally distributed to all 6 legs).
F = ‫ﻻ‬2L = ‫ﻻ‬2πR
We want the vertical force:
Fv = ‫ﻻ‬2πRcosθ
Distributed to 6 legs  Fv =
1
w
6
*γ = surface tension of water = 0.073
N
(chart on Pg. 281)
m
1
1
w  mg   2R cos 
6
6
1
mg
6
 cos
 2R
2

1
m 

2 * 10 5 kg  9.8

6
s




N

2R 0.073 
m

 cos 
62 = θ
*Cohesive Forces- forces between like molecules (water on water).
*Adhesive Forces- forces between unlike molecules (glass on water).
*Water climbs (clings) the glass on the edge because the adhesive forces between the
class and water are greater than water on water.
*Mercury curves down at the glass because the cohesive forces between the mercury
atoms are greater than the adhesive forces between mercury and glass.
Capillary Action:
F=‫ﻻ‬L=2πR‫ﻻ‬
Vertical component of this force is:
Fv = ‫ ﻻ‬2 π R cosф
In order to be in equilibrium:
Fv = Wcylinder of water at height(h)
W = Mg = pVg = p g π R2 h
Therefore:
Fv = W
‫ ﻻ‬2 π R cosф = p g π R2 h  h =
 2 R cos 
pgR 2 h
If the angle ф = 0, then:
h=
2
pgR
Ex. Find the height the water rises up the tube with a radius of 5*10-5m.
O
ф = very small  use ф = 0.
N

2 0.073 
m

h=
= 0.29 m
m

3 kg 
5
1 *10 3  9.8 2  5 *10 m
m 
s 



**DO # 43, 45-47 ON PG. 302**
Chapter 10: Thermodynamics
10.1-10.3
Recall- the exchange of energy between two objects that produces a change in
temperature is called heat, i.e. Thermal Energy
F = Fahrenheit
C = Celsius
γ = 2α (usually)
γ = avg. coeff. of area expansion.
Lo = original length
α = alpha = coeff. of linear expansion
β = beta = coeff. of volume expansion
Vo = original volume
β = 3α (usually)
Chart on Pg. 319
K = Kelvin
Conversions:
K = C + 273.15
F = 9/5 C + 32
C = 5/9 (F – 32)
10.3
ΔL= α LO ΔT
Therefore: new length
For area of square objects use:
L = L O + ά LO ΔT
A = L2 = (Lo + α Lo ΔT)(Lo + α Lo ΔT)
= Lo2 + 2 α Lo2ΔT + α2 Lo2ΔT2
But squaring the coefficient of linear expansion creates such a small value we can ignore
the last term
Therefore:
A = L2 = Lo2 + 2 α Lo2ΔT
= Ao + 2 α Ao ΔT
ΔV = β Vo ΔT
New: V = Vo + β Vo ΔT
Ex.
A cross sectional area of a hole in a piece of steel is 100cm2 at 20˚C. What is the
area of the hole if heated to 100˚C?
ΔA = γ Ao ΔT = ( 22x10-6 C-1)(100cm2)(80˚C) = .18cm2
Therefore: new hole = 100cm2 + .18cm2 = 100.18cm2
Now try the above using the whole equation for A = L2 including the last term with the
coefficient of linear expansion squared to show why we disregard this term.
Look at Pg. 322.
It talks about the behavior of water from 0˚C → 4˚C
Read pgs. 322-323. Do the thinking physics 2
**Assign Pg. 335 # 3,8,12,18,19
10.4
PV= nRT ←ideal gas equation
Also
PV= nRT can be used as
Pf Vf
Tf
=
R = universal Gas constant
n = number of moles
P = pressure
V = volume
T = Temperature in Kelvins
R = 8.31 J/(mol)(K)
= 0.0821 (L)(atm)/(mol)(K) ← in atmospheres
Pi Vi
Ti
Do number 25 on the board
** Assign Page 336 # 21, 22, 24, 28
10.5-10.6
PV= nRT
can be rewritten to include kb called Boltzmann’s constant
Kb = R
Na
Na = 6.02 x 1023 molecules/mole (Avogadro’s
number)
N =N
PV = N kb T
Na
R = 8.31 J/(mol)(K
**SEE PAGE 328 FOR HOW
AVOGADRO’S
kb = 1.38
x 10-23 J/K NUMBER WAS FOUND
Kinetic energy of a gas molecule is directly proportional to the absolute temperature of
the gas. So we can say:
KE = ½mv2 = 3/2 kb T
And
Vrms of a molecule of gas is
______
______
V = √3 kb T = √3 RT
m
M
** Assign Page 336 # 33, 36, 38, 41
Chapter 11: Heat
11.1: Mechanical Equivalent of Heat
Recall – heat flows from hot to cold until equilibrium is reached
-
Heat gained or lost is equal to the work done on or by the system
-
1 J (Joule) is equal to the amount of work required to move one
Newton one meter
-
1 cal (calorie) is equal to the amount of energy (heat) needed to raise 1
gram of water to 1˚C (if the calorie is written as a capital C is
understood as 1000cal or 1 kilocalorie (this is one calorie in food))
-
Now we will use Joules in heat and the conversion is:
1 J = .239 cal or 1 cal = 4.186 J
This is known as the mechanical equivalent of heat. *BTU (British thermal unit) = heat
needed to raise 1 lb of water 1˚F.
Ex.
If Colin consumes 2000 calories of candy and Pluta finds out he makes him work
it off lifting 50kg masses in the weight room. How many times must he lift the mass 2m?
Remember 1C = 1000 cal
So 2000 food calories is 2000C
Converting to cal gives
(2000C) (1000cal) = 2 x 106 cal
1C
W = (2 x 106 cal)(4.186 cal) = 8.37 x 106J
J
Because W = mgh
Total work will use N for the number of times he has to lift the weight therefore
W = nmgh
Solving for N
n = W = 8.37 x 106J
= 8.54 x 103 times
mgh (50kg)(9.8 m/s2)(2m)
Moral of the story, Colin don’t eat candy.
Question: How many hours would this take if Colin lifts the weight once every 10
seconds.
Answer: apx. 24 hours
11.2: Specific Heat
Q = mcΔT
You should recall this from last year.
11.3: Conservation of energy (Calorimentry) table of “C’s” on Pg. 342
Q`w + Q`m = Q``w + Q``m
Heat of water plus heat of material before hand is equal to heat of the two substances
after.
This assumes NO heat loss to surroundings. (yes we are going to do a calorimeter lab)
mwCwTiw + mmCmTim = mwCwTfw + mmCmTfm
Ex.
A blonde dear hunter fires a 2g silver bullet @ 200m/s from her 30-06 misses a
10pt buck and hits the pine walls of the outhouse. (Not to worry, her brother was all that
was in the outhouse at the time.) Assuming that the bullet retains all the energy (heat) it
generates on impact, find its temperature change.
KE = ½mv2 = ½ (2 x 10-3kg)(200m/s) = 40 J = Q
Q = mCΔT → ΔT = Q = 40 J
mC (2 x 10-3kg)(234J/kg C)
All this ΔT is due to friction
= 85.5˚C
Q: What would the ΔT be if Wendy used lead bullets instead of silver?
A: 157˚C thus lead would cause twice as much friction due to its different specific heat.
Assign Pg. 363 # 1, 2, 5, 8, 11, 15, 18
11.4: Latent Heat & Phase changes
This book uses L (a capital L for heat of fusion and heat of vaporization rather than H as
last years book did) i.e. L = latent means heat constant
Thus Lf Latent heat of fusion
Lv = Latent heat of Vaporization
We do the problems the same way
i.e. if ice @ -30˚C to steam @ 120˚C
warm ice
Q = mCΔT
fuse ice
Q= Lf m
warm water
Q = mCΔT
vaporize water
Q = Lv m
warm steam
Q = mCΔT
Heat total = Σ (sum) of the heats
Try this for 1g of ice
A: 3110 J
Cooling just yields a negative answer because it gave off heat.
Ex. 11.5 Pg. 349
Liquid helium with a boiling point of 4.2 K and a Lv = 2.09 x 104 J/kg, and is supplied a
constant power of 10w (watts) (1w = 1 J/s) from an immersed electric heater. At this rate,
how long would it take to boil away 1kg of the liquid helium?
Since Lv = 2.09 x 104 J/kg this is how much thermal energy it will take to boil
away 1kg and since P = w/t and the power is 10w = 10 J/s
T = w = 2.09 x 104 J = 2090 seconds
P 10 J/s
Try this for liquid nitrogen (same 1kg) and for water @ 100˚C (1kg) how long would it
take?
For nitrogen: apx. 5.6 hours
For water: apx. 2.617 days
Assign Pg. 363 20, 22, 26, 29a
11.5: Heat Transfer By Conduction (touching)
The heat transfer rate (heat of current) is H (capital H)
H= Q
ΔT
Note Q = Joules T = seconds
so heat H is J/s watts
For heat flow from one side of a piece of material to another use:
H = k A (T2 – T1)
L
k = Constant of thermal Conductivity
L = Thickness of material
A = Cross-Sectional Area
T2 = Higher Temperature
T1 = Lower Temperature
k = Large for metals lower for non heat conduction
H = Heat Transfer Rate
See chart on Pg. 351
Ex.
For a concrete basement wall 2m high 3.65m long and 20cm thick find the
amount of heat transferred through the wall if the inside temperature is 20˚C and the
outside temperature is 41˚F in one hour.
41˚F = 8˚C
T2 = 20˚C
T1 = 8˚C
k = 1.3 J/s · m· C˚
A = (2m)(3.65m) = 7.3m2
T = 1 hour = 3600 seconds
Solving for Q
Q = k A ΔT (T2 – T1) =
L
= (1.3 J/s · m · C˚)(7.3m2)(15˚C) = 2.6 x 106J
.2m
Heat loss through walls
Q = A (T2 – T1)
ΔT
ΣR
ΣR = Sum of the R values of the materials used to construct the walls
values on page 352
Assign 30-33
11.6: Conduction – Heat transferred
By the movement of a heated substance, i.e. forced air heat, hot water heat.
11.7: Radiation
Warming w/o contact nor conduction, i.e. warming hands by a fire. No contact and
conduction would be upward not outward (heat rises it does not move horizontally
through conduction unless forced)
Stefan’s Law
P = θ AeT
4
Pnet = θ Ae (T4 – To4)
P = Power (in watts)
e = Constant of emissivity
A = Surface area (in meters squared)
θ = Constant = 5.6696 x 10-8 w/m2 · K4
T = Objects Temperature in Kelvins
To = Temperature of Surroundings
Anything that absorbs all the energy incident upon it is called an ideal absorber. Its
emissivity is said to be equal to unity. It is also known as a Blake Body.
By contrast anything with an emissivity of zero, absorbs no energy it, reflects all back
off, is known as a perfect reflector.
A person with surface area of 1.5m2 has a temperature of 98.6˚F in a 20˚C room. If they
removed all clothing how much heat would be lost in 10mins.
Pnet = θ Ae (T4 – To4) =
= (5.67 x 10-8 w/m2 · K4)(1.5m2)(.9)(310K4 – 293K4)
= 1.43 x 102 J/s
In 10 mins.
Qtotal = (Pnet)(Time) = (1.43 x 102 J/s)(600s) = 8.6 x 104 J
.Assign 37-39
Test on Chapters 9, 10, 11 in 2 days
eskin = .9
Chapter 12: The Laws of Thermodynamics
12.1 & 12.2: Heat and Internal Energy & Work and Heat
*Internal Energy- All energy in a system that is stationary- includes heat, nuclear,
chemical, and strain (stretched / compressed spring) energy.
*Thermal Energy- The portion of internal energy that changes when temperature of
system changes.
*Heat Transfer- Caused by a temperature difference between the system and
surroundings- Glass, Ice.
*Recall the thermal energy of a monatomic gas is associated with the motion of its atoms.
This thermal energy is just kinetic energy on a microscopic scale.
Temperature ↑  Ice ↑  Thermal Energy ↑
*But… thermal energy also includes rotational, vibrational, and potential energy on the
molecular level.
**GO OVER * PARAGRAPH ON GAS/WORK ON PG. 374**
W = F∆y = PA∆y
*F = force
*∆y = change in height
*P = pressure
*A = cross sectional area
*W = work
Remember if the piston moves:
∆V = A∆y
*∆V = change in volume
*Positive work is done by the system.
*Negative work is done on the system.
*Graphically, the work done from Vo to Vf is the area under the curve on a PV graph.
*W = P (Vf – Vo)
*If it’s a curved line, you must find the equation of the line and integrate from a = Vo to
b = Vf
Integral*dx
*Adiabatic Free Expansion- NO heat is transferred through the insulated walls of a
system. Thus the temperature of the system remains constant.
 No heat transfer = No work done.
*Heat transfer and work depends on the path between the initial and final states.
12.3: The First Law of Thermodynamics
*∆U = change in internal energy of a system
*Isolated systems have no change in internal energy because they allow for no heat
transfer, thus Q = W = O and ∆U = O  Uo = Uf.
*In a cyclic process (engine), the system begins and ends at the same state. Internal
energy doesn’t change so +heat = work during ∆U = O but Q = W.
Thermal Dynamics  ∆U = Uf – Uo = Q – W
*U = internal energy function
*+Q = heat transferred into system
*+W = work done by system
*The First Law of Thermodynamics is essentially the principle of conservation of energy
generalized to include heat as a mode of energy transfer.
*Isobaric Process- Pressure remains constant and work done/heat transferred are both ≠0.
Ex. A gas in a cylinder with a piston has a cross-sectional area of 0.10 m2, and a
constant pressure of 8000 Pa. It has heat slowly added to it, resulting in the piston
being pushed up 4.0 cm. If 42 J of heat is added to the system, what is the change
in internal energy of the system?
Work done by gas:
W = P∆V = (8000 Pa) (0.10 m2) (4*10-2 m) = 32 J
*According to the First Law of Thermodynamics, the change in internal energy is
∆U = Q – W = 42 J – 32 J = 10 J
*If you added 42 J to a system without allowing the piston to move, then you would have
W = 0 and ∆U = 42 – 0 = 42 J
**LOOK AT EXAMPLES ON PG. 380-381 AND DO #2, 7, 11, 15, 20**
12.4: Second Law of Thermodynamics
It is impossible to construct a heat engine that, operation in a cycle, produces no other
effect than the absorption of heat from a reservoir and the performance of an equal
amount of work.
W = Qh – Qc = Qnet
*Qh = heat from a hot reservoir
*Qc = heat from a cold reservoir
Qh > 0
Qc > 0
*Because it cycles ∆U = 0, therefore the net work done by a heat engine equals the net
heat flowing into it.
*Efficiency = e
e=1–
Qc
Qh
**DO EXAMPLE 12.6 ON PG. 383**
Second part is e = 1 – Qc / Qh
0.20 = 1 –
3000 J
Qh
Qh (-0.80) =
 3000 J
 0.80
Qh = 3750
W = Qh – Qc = 3750 – 3000 = 750 J
**GET OUT THERMO COIL AND PUT INTO CUP OF HOT WATER TO
DESMONSTRATE CONVERSION OF HEAT (THERMAL ENGERGY) TO KE**
**DO # 24, 26, 28, 36, ON PG. 398-399**
Chapter 13: Vibrations and Waves
13.1: Hooke’s Law
Fs   Kx
*It is negative because the force exerted by the spring is always directed opposite the
displacement of the mass.
*The direction of the spring’s force (restoring) is such that the mass is either being
pushed or pulled towards the equilibrium position.
*Simple harmonic motion occurs when the net force along the direction of a motion is a
Hooke’s law type of force. (Ex. Net force is proportional to displacement and in the
opposite direction).
*Amplitude (A)- max distance traveled by an object from its equilibrium position
*Period (T)- time (seconds) it takes to execute one complete cycle of motion
*Frequency (F)- # of cycles per unit of time (or # of vibrations)
Fs  Kd  mg  K 
mg
 We will use in lab to measure spring constants
d
*Acceleration can be used instead of g for use on a horizontal surface.
F   Kx  ma  a  
K
x
m
13.2: Elastic Potential Energy
*Elastic Potential Energy- energy stored in a stretched or compressed spring.
(Ex. Cannon)
PE 
1 2
Kx
2
*Because we always square the distance x, the PE’s must always be positive.
KE  PE
g
 PEs i  KE  PEg  PEs F
*If friction is present, then final energy does not equal initial energy, then work (heat)
was done- this work is the result of the final energy minus the initial energy.
NC- Non Conservative Forces:
WNC  KE  PEg  PEs F  KE  PEg  PEs i
13.3: Velocity as a Function of Position
*At maximum extension = A, the total energy of a spring system is
*Anywhere between A and equilibrium
Total Energy = KE  PE
1
1
1
KA 2  mv 2  Kx 2
2
2
2
*x = present position
*A = max position
1
KA 2
2
Solving for V:
V 

K 2
A  x2
m

**GO OVER EXAMPLES ON PG. 410-413**
**DO # 1, 3, 5, 6, 9, 12, 14, 16 ON PG. 434**
13.4: Comparing Simple Harmonic Motion with Uniform Circular Motion
T  2
L
m
 2
g
K
V  C A2  x 2
*V = velocity
*C = constant
**SEE PG. 414, FIGURE 13.9**
A2  x 2
*A ball moving at a constant velocity (Vo), tangent to the circular path at this instant
(Vball) in the x-direction is:
V  Vo sin   sin  
sin  
V

Vo
V
Vo
A2  x 2
A
A2  x 2
A
Therefore:
V
Period:
Vo
A
A 2  x 2  C A2  x 2 
V0
C
A
2A
T
Vo 
T
2A
Vo
1
1
2
KA 2  mVo
2
2
A

Vo
T  2
f 
m
K
m
K
1
1

T 2
K
m
*Where units of frequency are s-1=Hertz angular frequency is omega (  )
  2f 
K
m
**DO EXAMPLE 13.6 ON PG. 417**
**DO # 14, 16, 17, 19, 20, 25, 28 ON PG. 436**
13.5: Position, Velocity, and Acceleration as a Function of Time
x  A cos
Because:
cos  
x
A
*Since we will use a constant rotational velocity, we can say that   t
Therefore:
x  A cos(t )
*In one revolution the ball rotates 2π radians in the period (T).
Therefore:

 2

 2f
T
T
*f = frequency of rotational motion
*But the angular speed of rotation on the reference circle = angular frequency of simple
harmonic motion.
Therefore:
x  A cos( 2 ft)
*We will use this equation to represent the position of an object moving with simple
harmonic motion as a function of time. The curve of this graph is a trig. function
(sinusoidal).
*In each case below, the object was released from rest at its maximum distance from
equilibrium.
*x = displacement
*V = velocity
*a = acceleration
Pg. 436, #25
PROBLEM:
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of
7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is
attached to the free end of the spring. The particle is pulled horizontally so that it
stretches the spring 5.00 cm and is then released from rest at t = 0.
(a.) What is the force constant of the spring?
(b.) What are the period, frequency, and angular frequency (  ) of the motion?
(c.) What is the total energy of the system?
(d.) What is the amplitude of the motion?
(e.) What are the maximum velocity and the maximum acceleration of the particle?
(f.) Determine the displacement, x, of the particle from the equilibrium position at
t = 0.500 s.
SOLUTION:
(a.) By Hooke’s Law:
F  Kx  K 
F 7 .5 N

 250 N/m
x .03m
(b.) Angular Frequency  (omega) is:
K
m

*K = spring constant
*m = mass attached
250 N
= 22.4 Rad/s
0.5kg
 
And Period is:
T
2


2
22.4 Rad
= 0.281 sec.
s
And Frequency is:
f 
1
1

= 3.56 Hz
T 0.281
(c.) Total Energy of System:
E  Eo 
E  0
1
1
2
2
mVo  Kxo
2
2


1
250 N 0.05m 2
m
2

E = 0.313 J
(d.) At maximum displacement from equilibrium, x = A = Amplitude. Plus, it is at
rest (for and instant) implying all the energy is elastic potential energy.
Therefore:
1
E  0  KA2  A 
2


2(0.313 J ) 
2E
 
K
 250 N 
m 

E = 0.05 m
(e.) Vmax is at equilibrium (x = 0) where all the energy is kinetic
Therefore:
E
1
2E
2

mVmax  0  Vmax 
2
m
2(0.313 J )
 1.12 m/s
0.5kg
*Note that from Newton’s Second Law, the maximum acceleration of the particle occurs
when the spring exerts maximum forces on it (when spring is stretched or compressed to
its maximum). x  A
*Therefore: Qmax
(f.) Since  


N
Fmax
KA 250 m 0.05m 



= 25 m/s2
m
m
0.5kg
K
and our object starts at rest where x  A and T  0 , the
m
displacement at any time is:

K

x  A cos(T )  A cos T

m


*So, at T  0.5 sec onds :
 
 250 N  
 

m  = 0.92 cm
x  0.05m  cos 0.5 sec
 
 0.5kg  
 

 
 
*Done in radians.
**DO # 23, 24, 26 ON PG. 436**
13.6: Pendulums
If   15 o , pendulums exhibit simple harmonic motion and the resultant force acting on
the mass is equal to the component of the weight tangent to the circle. Its magnitude is
mg sin  . Since this force is always directed towards   0 , it is a restoring force.
Ft  mg sin   restoring force
*If we use radians with angles less than 15o, the radian measure is approximately equal to
sin  .
*Therefore, for all   15 o , use:
FT  mg
Hooke’s Law F  KS , we can also say that S  L   
FT 
 mgS
L
Also use last year’s formulas for pendulums:
T  2
K
 spring
m
T  2
L
 pendulums
g
**LOOK AT EXAMPLE 13.8 ON PG. 421**
S
L
**DO PROBLEM # 31 ON PG. 437**
a.) Period of pendulum:
L
gt 2
T  2
L
g
4 2
Earth: L 
(9.8)(1) 2
 0.25m
4 2
(3.7)(1) 2
Mars: L 
 0.094m
4 2
b.) Period of vibration of mass on spring:
T  2
m
KT 2
m
K
4 2

10 N 1sec 
m

2
4 2
 0.25kg
Since this formula does not use gravity, the mass on Earth = mass on Mars.
**DO #27-30 ON PG. 437**
13.7, 13.8, & 13.9: Damped Oscillations, Wave Motion, and Types of Waves
*Damped Oscillations- This occurs in real systems where friction is present. They reduce
amplitude, acceleration, velocity, period, frequency, etc.
*Wave Motion- To produce mechanical waves, you must have:

An elastic medium to disturb.

An energy source to provide the disturbance.

A physical mechanism that allows adjacent portions on the
medium to influence on another (transfer the wave- slinky).
*The 3 Parameters Important in Characterizing Waves:
1. Wave Length
2. Frequency
3. Wave Velocity
*3 Types of Waves:
1. Transverse- Particles of disturbed medium move perpendicular to
direction of wave velocity (slinky shook side to side).
2. Longitudinal- Particle’s displacement is parallel to direction of wave
motion (slinky compressed (section) and released).
3. Surface Waves- A combination of both- circular motion.
*There is a 4th type of wave which the book does not consider:
4. Electromagnetic- Never cross; always in same direction- more about these
in February.
1310 & 13.11: Frequency, Amplitude, and Wavelength & The Speed of Waves on
Strings
*Wavelength is Lambda: 
*Amplitude: A
*Hertz 
1
sec ond
*Period (1 vibration): T
V 
x 
  f
t T
**DO EXAMPLE PROBLEM #13.9 FROM PG.427**
*Wave length of 0.4m  
*Amplitude of 0.15m  A
*Frequency of 8.0Hz
*Find velocity
1 1
  0.13 sec.
f 8
Period:
T
Speed:
 1 
V  f  (8 Hz )(0.4m)  
(0.4m)  3.2 m/s
 8 sec 
**DO #32-37 ON PG. 437**
Also V 
F

*F= Ftension on string
*  = linear density of string (mass per unit length)
**DO EXAMPLE PROLEM #13.12 ON PG. 429**
**DO #38-40 ON PG. 437**
Chapter 14: Sound
14.1 & 14.2: Producing a Sound Wave & Characterizing a Sound Wave
*Sound waves are longitudinal waves traveling through a medium, usually air, the motion
of the wave particles is back and forth along the wave.
*High Density- compression of condensation
*Low Density- rare faction
*Categories:
-
infrasonic waves < 20 Hz (long)
-
audible waves  20  20,000 Hz (long)
-
ultrasound waves > 20,000 Hz (long)
*Ultrasound- Transforming electrical energy into mechanical energy is called the
Piezoelectric effect.
2
   t 
 *100
PR   i
 i  t 
*  i = density of fluid
*  t = density of object (kid)
14.3: Speed of Sound
*Speed of sound in air @ 0oC at sea level is 331 m/s (humidity can affect it too).
*The sensation of sound is approximately logarithmic in ear.
*The relative intensity of a sound is the intensity level called decibel level.
*Threshold of human hearing is 1 * 10 12 w/m2
*Threshold of pain for humans is 1 w/m2
*Recall from Chapter 9 the formula for Bulk Modulus.
stress
P
B

V
strain
V
* P = change in pressure
* V = volume change
* V = original volume
*The speed of all mechanical waves is given by:
V
elastic
( properties )
inertial
*In a solid rod, the speed of a longitudinal wave (go in the direction) of the particle
(along rod).
Solids:
V 
y

* y = Young’s Modulus of a solid
*  = density of material
Liquids & Gases:
B
V 

* B = Bulk Modulus

V  331m
T
s 1  273
* T = temp. in Celsius
* V = velocity in air
V  f
*Recall that:
**DO #3 ON PG. 476**
*Range of human hearing is from 20  20,000 Hz. Find the wavelengths at 37oC.
*First find V:
V @ 27 oC

V  331 m
27
s 1  273
 347 m/s
*Now find wavelength:
347 m
V
s  0.01735m  17 cm
s  
f 2 * 10 4 Hz
L 
m
V 347 s

 17 m
f
20 Hz
**DO PROBLEMS #1-7 ON PG. 476**
14.4: Energy and Intensity of Sound Waves
*The intensity of a spherical wave produced by a point source is proportional to the
average power emitted and inversely proportional to the square of the distance from the
source.
I
P
A
* P = power
* A = area (d2)
* I = intensity
 I 

I
 o
  10 Log 
*  = intensity level (in dB= decibels)
* I = 1*10-12 w/m2 & reference & intensity level
*Recall that 1*10-12 = min. for human hearing, and that 1 w/m2 = threshold for pain.
Ex. A noisy grinder in a factory produced sound at 1*10-5 w/m2. Find the decimal level
of this machine. Then calculate what would happen if a second machine was added.
 1 *10 5 w 2 

m   10 log 10 6  70 dB
  10 log 

 1 * 10 12 w 2 
m 

 
 2 * 10 5 w 2 

m   73 dB
  10 log 

12 w
 1 * 10

m2 

**DO PROBLEMS #8,10,11 ON PG. 476**
14.5: Spherical and Plane Waves
*When a small spherical object oscillates so that its radius changes periodically with
time, it will produce a spherical sound wave. This wave moves outward from its source
at a constant velocity (like dropping a stone in a pond) only in 3 dimensions.
*
PAV
 surface area of a sphere
4R 2
*I = intensity (this assumes no absorption into the medium)
I
P
averagepower PAV

 AV 2
area
A
4R
*Since the intensity of the wave will decrease the further it gets from the source (called a
point source), the intensity will be greater at 1 then at 2. A, B, and C are called wave
fronts for spherical waves. The distance between A and B is the same as between B and
C, or the point source and A. This distance is  (lambda) waves.
*The further the waves get from the point source, a small portion cut from the wave will
have rays almost parallel to one another. This portion is called a plane wave.
**LOOK AT EXAMPLE #13 ON PG. 476**
*Train horn heard at 50 dB, 10 km away. Find:
a) Average power of the horn
b) Intensity level of horn’s sound heard by a person
50 m away
*Treat horn as a point source with no absorption of the sound by the air.
a)   10 log
I
Io
From 14.4 on Pg. 449:
*Io = 1*10-12 w/m2
*  = 50 dB (given)
 I 
50dB  10 log  
 Io 
 I 
5  log  
 Io 
I  I o 10 5

I  1 * 10 12 w
10 
m
5
2
I  1*10-7 w/m2
*The train horn is a point source, so A  4R 2 (spherical surface area). Average
power emitted is P  IA .
*Use I and A as same distance from source (R).
R  10km  1.0 *10 4 m
I  1.0 * 10 7 w
m2
Therefore:


P  4 1 *10 4 m 1 * 10 7 w
m2
  1.3*10 w
2
b) at R  50m
I
P 1.3 *10 2 w

 4.1*10-3 w/m2
A 4 50m 2
*Also at 50 m decibel level is
 4 *10 3 w 2 
I

m 
  10 log  10 log 

I0
 1.0 *10 12 w 2 
m 

  10 log 4.1*10 9 
  96.13 dB
**DO PROBLEMS #12-14 ON PG. 476**
14.6: Doppler Effect
*To have a Doppler effect, the needs to be a relative motion between the sound wave
source and the observer. Its effect is obvious, as it approaches you the sounds intense
increases and vise versa.
*Two cases:
1st- Sound stationary observer moving because you are moving through
waves, you are hearing additional wave fronts.
VoT

*V0 = V of observer
*T = time
*  = wave length
*Because we are looking for frequency we can solve this for f.
f '  f VBSmoving  f 

Vo

V
f
*V = velocity of sound
*Vo = velocity of observer
* f  = frequency with observer
*f = frequency of sound wave
*Also sound stationary and observer moving away.
 V  Vo 
f f

 V 
*2nd case: source moving- observer stationary
Source moving away from person:
 V 

f   f 
 V  Vs 
Source moving toward observer:
 V 

f   f 
V

V
s 

*Vs = velocity of source
*V = velocity of sound
* f  = frequency heard
*f = frequency of sound in air
**SEE SKETCH ON PG. 454**
**SEE EXAMPLE FROM PG. 455-456**
14.7: Interference of Sound Waves
*Interference of sound waves occurs when waves produced by two different sources
overlap.
*Constructive Interference- two or more sound waves that combine (waves are timed
correctly) to produce maximum results.
*Destructive Interference- when waves are ½ off, they will cancel each other out.
*To find destructive interference, simply solve for  
V
and move one source 1 
2
f
away from the observer to attain minimum sound intensity.
14.8: Standing Waves
*When a string is vibrated so it oscillates up and down at a constant speed and one end is
attached to a stationary object, the wave will invert and return to the oscillating source.
*If the waves at one point are exactly opposite, they will cancel (destructive interference)
and the line will be flat, a node occurs where two waves have equal, but opposite
magnitude.
Ex. Net displacement of this spot is zero.
*Although there exists no movement at the nodes, halfway between nodes, there is a
point where the maximum amplitude occurs. This point is called an antinode.
**SEE FIGURE 14.16 ON PG. 459**
**DO PROBLEMS #16-19, 24,26 ON PG. 477**
*Hint on # 19: If both observer and source are moving, use:

f   

 V  Vo
f 
 V  Vs

 


*Convert km/h to m/s first.
**DO REVIEW PROBLEMS #49-51,53,54,59,60,63 OF CH. 14 ON PG. 479-480**
**QUIZ TOMORROW**
Chapter 15: Electric Forces and Electric Fields
15.1, 15.2, & 15.3: Properties of Electric Charges, Insulators and Conductors, &
Coulomb’s Law
*Recall that like charges repel and unlike charges attract.
*All charges originate inside the atom.
*Positive charges carried by protons and atomic particles bond permanently inside the
nucleus with the neutrons. To change from positive to negative charge, an atom must
gain electrons. Particles that revolve around the nucleus in paths called valence shells.
The charge on a proton is equal and opposite that of an electron.
*Electric charge is always conserved.
Ex. Electric charges are transferred, not created.
*Robert Millikan in 1909 found that charges were always in full units- not fractional.
These units are said to be quanitized as “e” ±e, ±2e, ±3e, etc.
*The value of “e” is 1.60219*10-19C.
*C is a unit of electrical charge called a Coulomb.
*Ways to change stuff:
*Conduction- when two objects come in contact with each other and they did not
have the same charge.
*Induction- no contact- usually just passing close…Ex. Electric Vandegraff
Generator.
*The Earth is an infinite supply (reservoir) of electrons.
*3 Classes:
*Semiconductors- controlled magnitude of electrons.
*Conductors- electrons move freely
*Insulators- not chargeable- no electron movement
*If there exists more positive charges on one side of a molecule, that molecule is said to
be polarized. This happens because the negative electrons move to the other side due to
an object passing near, or a grounding cable being attached.
*1 Coulomb= current of one amp. amount of charge that flows through a
wire point in one second.
*Coulombs Law:
F  Ke
q1 q 2
R2
*Ke = 8.99*109 Nm2/c2
*q1 = charge one
*q2 = charge two
*R2 = distance between charges
e=
charge
mass
electron
-1.6*10-19
9.11*10-31
proton
1.6*10-19
1.67*10-27
neutron
0
1.67*10-27
1
 6.3 * 1018 protons  1C
e
+ proton
e  1.6 *10 19 
- electron
e  1.6 * 10 19 
1
 6.3 * 1018 electrons  1C
e
* 10 6 C  1C  mu coulomb
*In 1cm3 of copper, there exists 1023 electrons.
*F21 = force of q2 on q1
*F12 = force of q1 on q2
*Coulomb’s law is a vector quantity.
* F21   F12 in this case- Newton’s 3rd Law
Ex. In a hydrogen atom, an electron and a proton are separated by 5.3*10-11 m. What
forces do they exert on each other?
*Electromagnetic Force:
Fe  K e
e
R
2

2
 8.99 *10 Nm
9
C2
15..63**1010
19
11

m
C
2
2
 8.2*10-8 N
*Gravitational Force:
Fg  G



m * m2
9.11*10 31 kg 1.67 *10 27 kg
11 Nm 2

6
.
67
*
10
 3.6*10-47 N
2
2
2

11
k
R
5.3 *10 m


*Notice that electromagnetic force is 2*1039 greater, thus the gravitational force is
negligible when compared to electromagnetic force- for an electron and a proton.
**DO PROBLEMS #1-5 ON PG. 514**
*Now for the superposition principle: finding the electromagnetic force on a charge
when more than two charges are present.
Ex. Three charges (points) at the corners of a triangle.
F12  8.99 *10 9
5 *10 C 6 *10 C   3*10-6 N
9
9
0.3m2
*Q1 and Q2 are positive, therefore Q1 goes left.
F13  8.99 *10
5 *10 3 *10   1.35*10-5 N

9
9
9
0.1m2
*Q1 = positive, Q3 = negative, thus charge goes down towards Q3
**FIND THE RESULTANT— YAY VECTORS!!!**
FR 
tan 1
F12 2  F13 2
 1.38 *10 5 N
F13
 77.5o  259.5o
F12
FR = 1.38*10-5 N @ 257.5o
**LOOK AT PROBLEM 15.3 ON PG. 493**
Ex. Two positive and one negative change with all three charges along a line (on the axis)
*Place Q2 at the origin.
*Find where Q3 (a negative charge) must be placed so there exists a negative force of
zero.
q1  15C
q 2  6C
F31  K e
F32  K e
q3 1.5 *10 5 C 
2  x 2
q3 6 *10 6 C 
x2
*Since the magnitude of the resultant on Q3 = 0, F31 = F3C, so set them equal to each
other and cancel out the Ke and Q3.
1.5 *10 C   6 *10 C 
5
2  x 2
6
x2
*Also multiply both sides by 106
15
2  x 
2

6
2
 15 x 2  62  x 
2
x
15 x  6 2  x 
15 x  2 6  6 x
21x  2 6
x  0.77 m
**DO PROBLEMS #6-8,10,12,13 ON PG. 515**
*A piece of aluminum foil with a mass of 5*10-2 kg is suspended in an electric field
directly vertically. If the charge on the foil is 3C , find the strength of a field that will
result in FT on the string of zero.
ft
m fe = qE
fg = mg
 fy = T + fe – fg = 0
 T = fg - fe = mg
therefore… T = 0
mg = qE
E = mg/q
E = mg/q = (5x10-2)(9.8m/s2)/(3x10-6c) = 1.6x105 N/c
Assignment:15, 17, 20, 22, 24
15.5. Electrical Field Lines
*Relationship of the field and the field lines…
1) The electric field vector E is tangent to the electric field lines at each point
2) The number of lines per unit area through a surface perpendicular to the lines is
proportionate to the strength of the electric field in as given region. Therefore, E is larger
when the field lines are close together and small when they are far apart.
Drawing field lines!
1) Lines must begin at the positive (+) charge (or infinity) and must terminate on negative
charges or, in the case of excess positive charges, they end at infinity.
2) The number of lines drawn from one charge (+) to another(-) is proportionate to the
magnitude of the charge.
3) No two field lines can ever cross one another!!
+
-
Ex. #29 on page 517
a) Sketch the field lines around an isolated point charge q>0 (+).
b) Now sketch around an isolated negative field when 2q<0.
a)
b)
+
-
Notice…
1) The field lines in both cases exhibit radial symmetry.
2) Lines originate at positive or infinity and terminate at negative or infinity.
3) The number of lines in ‘a’ is ½ compared to ‘b’. Therefore, they are proportional for q
and –2q.
Look at pages 498-499 for change line pictures.
15.6. Conductors in Electrostatic Equilibrium
For conductors in electrostatic equilibrium there are four rules…
1) The electric field is zero everywhere inside the conductor.
2) Any excess change resides entirely on its surface.
3) The electric field just outside the conductor is always perpendicular to its surface.
4) On the irregularly shaped conductor, the electrical charge per unit area is always
greatest at the location where the curvature of its surface is greatest. (i.e. sharp points)
take a look (
) at the figures on page 501
Assignment: pg. 517 #27, 28, 30-37 all
Give chapter 15 test in groups of three (all 15 problems from supplemental problems
book)
Chapter 16
16.1. Electric Potential and Potential Difference
When we did W= Fd we learned that the distance was all that counted, not the path taken.
This was because of the conservation of energy we learned about later. In the last chapter,
we applied this to Coulomb’s Law. As you’ve probably guessed, electrostatic force is
also conservative.
Work done by a charge by the electric force…
W = Fd = qEd
q = charge
d = distance -a  b
E = electric field
The following is only valid for a uniform electric field…
 PE = -w = -qEd
…because the change in potential energy is equal to the negative of work.
E = uniform electric field from a  b
The potential difference between the two points a and b (Va – Vb ) is defined as the
change in potential energy (  PE) divided by the charge (q) that is moved through the
field (E) from a  b.
 V = Va – Vb =  PE/q
Because PE is a scalar (direction doesn’t count) electric potential is also a scalar.
The units (which you must use!!!) are joules over Coulombs, and together they are a volt.
1 V = 1 J/C
In other words, 1J of work must be done to move a 1C charge through a potential
difference of 1V. When this happens, the 1C charge gains or loses 1J of energy.
so…  PE/q = Va – Vb = -Ed
Remember that these formulas are only valid for a uniform electric field!
For a positive charge q: It would want to move in the direction of the field so we must
apply a positive (upward) external force on the charge to move it from a to b. Therefore,
work is done on the charge so it gains electrical potential energy when moved in the
direction opposite the electric field. (This is like a boulder gaining PE when raised up
from the floor.) Likewise, if the positive charge is released at a and goes towards b, it
accelerates downward and loses PE, while gaining equal amounts of kinetic energy.
Lastly, as the positive charge moves from a to b, its electric potential decreases.
Conversely, when the charge q is negative, it moves opposite the field direction from a
point of low to high PE and, in the process, undergoes a decrease in electrical potential
energy and an increase in kinetic energy. Therefore, they act the same, but go in opposite
directions in the field.
Assignment: 1-9 all
16.2. Electric Potential and Potential Energy due to Point Charges
In electric circuits, a point of zero electric potential is defined by grounding some point in
the circuit. In the case of a point charge, the point of zero potential is taken to be an
infinite distance from the charge.
The potential at a given point in space depends on the quantity of charge on the object
setting up the potential and the distance (R) from the object to the specific point.
V = Ke(q/R)
A potential can exist in space at a point whether or not there is a charge at that point.
PE = q2V1 = Ke(q1q2/R)
Superposition principle: a scalar (not a vector) for a group of charges.
The total energy in the system is the sum of the energy between
each pair.
Ex. A 5  c point charge is located at the origin and a -2  c point charge is at (3,0).
a) Find the electric potential at point P.
b) If the electric potential is taken to be zero at infinity, Find the total electric potential at
point P, WITH COORDINATES (0,4).
(0,4) P
R1
R2
5 c +
q1 (0,0)
-
-2  c
(3,0) q2
a) V1 = Ke(q1/R1) = (8.99x109 Nm2/c2 )(5x10-6 c/4m) = 1.12x104 V
V2 = Ke(q2/R2) = (8.99x109 Nm2/c2 )(-2x10-6 c/5m) = -.36x104 V
Vp = V1 + V2 = 7.6x103 V
How much work is done bringing a 4 micro coulomb charge from infinity to point P?
W = q3Vp = (4x10-6 c)(7.6x103 V) = 3.1x10-2 J
Assignment: pg 550 #10-13,15
16.3. Potentials and Charged Conductors
16.4. Equipotential Surfaces
We know that w = -  PE and  PE = q  V
Thus W= -q  V=-q(V1 -V2)
But if V1 =V2 (both charges have the same potential then W=0)
*All points on the surface of a charged conductor in electrostatic equilibrium are at the
same potential.
*The electric potential is constant everywhere inside a conductor. Because the field
inside a conductor is zero, it takes work to move a charge from one spot to another in the
conductor. The potential inside is therefore equal to its value on the conductor’s surface.
Electron volt = the energy that an electron/proton gains when accelerated through a
potential difference of 1 volt
1Ev =1.6x10-19J
recall V= Ke(q/R)
So a change in voltage between two points is defined by the equations
V1 = Ke(q1/R1)
V2 = Ke(q2/R2)
You saw these yesterday!
Assignment: problems #16 & #17
16.6. Capacitance
C = capacitance
Q = magnitude of charge
 V = potential difference
Q = CV
C = Q/  V = coulombs/volts = farads (F)
A capacitor is a device consisting of a pair of conductors separated by an insulating
metal. A charged capacitor stores a charge (energy) that I can use later for a specific
need.
16.7. Combination of Capacitors
parallel: The potential difference across each capacitor in a parallel combination of
capacitors is the same across each one. This potential difference is always larger than any
one individual capacitor.
series: The magnitude of the charge is the same on all plates. In general, the potential
difference across capacitors in series ( or any other circuit elements) is equal to the sum
of potential differences across the individual capacitors.
Equivalent capacitor = Ceq = 1/Ceq = 1/C1 + 1/C2 + 1/C3 +… add them as reciprocals
Ex.
3 F
6  F 12  F 24  F
+
all in a series so…
1/Ceq = 1/3 + 1/6 + 1/12 + 1/24 = 5/8
inverse of 5/8 = 1.6μF
18v
Find the charge on a 12  F capacitor.
Q = Ceq  V = (1.6x10-6F)(18v) = 29  C
Ex. Find the charge from a to b.
parallel so 1+3=4
1F
4F
3F
a
b
6F
8F
2F
parallel so 6+2=8
4
4
¼+¼=½
a
inverse = 2
b
8
8
1/8 + 1/8 = ¼ inverse = 4
2
a
b
4
a
6
b
Therefore, the equivalent capacitor of the circuit is 6  F
Now if the charge was a 12V circuit the charge on the 8μF capacitor would be
Q=CeqΔV = (6 X 10-6 F )(12V) = 72μC
Assignment page 552 29-33 all
END OF CHAPTER 16 ASSIGN THE SUPPLEMENTAL PROBLEMS FROM
OTHER BOOK AS A TEST, IN GROUPS OF THREE.