Columbia University
Department of Physics
QUALIFYING EXAMINATION
Friday, January 13, 2017
3:00PM to 5:00PM
General Physics (Part II)
Section 6.
Two hours are permitted for the completion of this section of the examination. Choose 4 problems
out of the 6 included in this section. (You will not earn extra credit by doing an additional problem).
Apportion your time carefully.
Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s) which
question you are answering (e.g., Section 6 (General Physics), Question 2, etc.).
Do NOT write your name on your answer booklets. Instead, clearly indicate your Exam Letter
Code.
You may refer to the single handwritten note sheet on 8 12 ” × 11” paper (double-sided) you have
prepared on General Physics. The note sheet cannot leave the exam room once the exam has begun.
This note sheet must be handed in at the end of today’s exam. Please include your Exam Letter
Code on your note sheet. No other extraneous papers or books are permitted.
Simple calculators are permitted. However, the use of calculators for storing and/or recovering
formulae or constants is NOT permitted.
Questions should be directed to the proctor.
Good Luck!
Section 6
Page 1 of 7
1. A glass is filled to height h0 with a volume of water V0 with density ρ. A straw with uniform
cross sectional area A is used to drink the water by creating a pressure at the top of the straw
(Ptop ) that is less than atmospheric pressure (PAtm ). The top of the straw is a distance htop above
the bottom of the glass.
(a) Determine the pressure (Ptop ) at the top of the straw as a function of the height (h) of the
water in the glass so that the velocity of water coming out of the top of the straw remains
constant at a value of vtop .
(b) How much work is done by the person in order to drink all the water in the glass with the
constant velocity vtop ?
(c) How much time does it take to drink all the water in the glass with the constant velocity
vtop ?
Section 6
Page 2 of 7
2.
(a) Imagine one day the sun ceases to shine. The temperature on the earth will be cold such
that the atmosphere is liquefied, and the earth surface will be covered by an ocean of
liquid air. Estimate the average depth of this liquid air ocean. Compare it to the average
depth of the true ocean, which is on the order of 5 km.
(b) Estimate the largest height of a mountain on earth. Are the typical heights of mountains
on the moon higher or lower than those on earth?
Hint: Assume that rock melts at 1000 Kelvin and the average mass number of rock is 30.
The rest mass of the proton and neutron is about 1 GeV, and 1 eV corresponds to about
10,000 K of temperature. As a crude approximation, you can assume that the latent heat
of fusion is of the same order as kT m , where T m is the melting temperature.
Section 6
Page 3 of 7
3. Uniform plasma with electron number density n is in thermal equilibrium with Planck radiation of temperature T and energy density U = aT 4 , where a = π2 k4 /15c3 ~3 is the radiation constant and k is the Boltzmann constant. The electrons (mass me ) and photons interact
through Thomson scattering with cross section σT = (8π/3)re2 where re = e2 /me c2 is the
classical electron radius.
(a) Find the photon mean free path λ.
(b) Estimate the electron mean free path to photon scattering, l.
Hint: consider the time between successive scatterings from the electron point of view.
(c) Estimate the average number of scatterings experienced by a photon as it diffuses through
a distance L λ.
Section 6
Page 4 of 7
4. A two-dimensional electron gas (2DEG) is separated from a conducting metal plane by a
dielectric of known thickness. The 2DEG can be patterned into any shape and with any configuration of electrical contacts, and is cooled in a cryostat to 2K with a variable magnetic
field.
(a) Describe a transport measurement that could be made to deduce the electron density in a
channel formed by the 2DEG.
(b) The capacitance formed by the conducting metal plane and the 2DEG is typically very
small, and therefore difficult to measure directly. Describe a transport measurement that
could allow determining the relative permittivity of the dielectric between the conducting
metal plane and the 2DEG.
Note: Be sure to indicate all quantities to be measured in the experiments.
Section 6
Page 5 of 7
5. An astronaut wants to have some coffee with milk. At her disposal are some coffee (200 ml)
at 97 ◦C and some milk (200 ml) at 7 ◦C. She would like to use all the coffee and milk for the
drink. Unfortunately she has to finish some work that will take her 15 minutes before she can
begin to consume her drink. She would like her drink to be as hot as possible when she drinks
it. She is trying to find out if she should mix the coffee and milk before or after finishing her
work.
The astronaut starts her analysis by assuming that her coffee and milk are exchanging heat with
their surroundings primarily through black body radiation. She assumes her coffee and milk
are suspended in air as liquid spheres due to zero gravity. The mixture would also be a sphere
suspended in air. The environment is at the room temperature (300 K). She also assumes that
the specific heat capacity and density of coffee and milk are the same as water, and that all
liquids are in thermal equilibrium throughout their volume as they cool down or warm up.
(a) Based on her assumptions, should the astronaut mix the coffee and milk before or after
her work? You can use the following approximation valid for small ∆T
Z T1 +∆T
∆T
1
dT
≈
(1)
(T 4 − T 04 )
T 14 − T 04
T1
(b) What is the maximum temperature of the coffee when she begins to drink it?
Section 6
Page 6 of 7
6. In a parallel plate capacitor, electrical forces squeeze the dielectric layer that separates the two
plates. If the dielectric material is compliant, the reduction in the distance between parallel
plates is no longer negligible. To explore potential uses of this phenomenon, we would like to
determine the changes in electrical and mechanical properties of this system in response to an
applied voltage.
(a) Write an expression for the electrical force pulling the capacitor plates in terms of plate
area A, plate separation h, dielectric permittivity d , and the applied voltage V.
(b) Assume that the dielectric layer behaves as a linear spring with a spring constant k, and
it has an initial thickness of h0 . Determine the voltage V necessary to reduce the plate
separation from h0 to h.
(c) The spring constant depends on the Youngs modulus Ed , area A, and thickness h0 by
k = Ed A/h0 . Using this relationship, express your result in (b) in terms of Ed , d , h, and
h0 .
(d) Once the plates in (b) reach their equilibrium separation h, a small external force, ∆F,
is applied to the parallel plates to compress the dielectric layer by ∆h. Determine the
effective spring constant ke f f (h) = −∆F/∆h.
(e) Evaluate ke f f (h) at h = 2h0 /3. Interpret your result.
Section 6
Page 7 of 7
Shaevitz
Section 6-1
Quals Problem Fluids
M. Shaevitz
Dec, 2016
A glass is filled to height h0 with a volume of water V0 with density ρ. A straw with
uniform cross sectional area A is used to drink the water by creating a pressure at the top
of the straw (Ptop) that is less than atmospheric pressure (PAtm). The top of the straw is a
distance htop above the bottom of the glass.
vtop
Ptop
a) Determine the pressure (Ptop) at the top of
the straw as a function of the height (h) of the
water in the glass so that the velocity of water
coming out of the top of the straw remains
constant at a value of vtop .
b) How much work is done by the person in
order to drink all the water in the glass with
the constant velocity vtop ?
htop
h0
c) How much time does it take to drink all the
water in the glass with the constant velocity
vtop?
V0
Solution:
a) Use coordinate system with +y up and zero at the bottom of the glass.
Assume velocity of the water at the top of the fill is zero.
Compare point at the top of liquid with height h to point at the top of the straw with height htop .
2
PAtm + ρ gh = Ptop + 12 ρ vtop
+ ρ ghtop
(
2
Ptop = PAtm − 12 ρ vtop
− ρ g htop − h
)
b) Areaglass = Aglass = V0 / h0
Calculate the work against gravity to raise the water to the top of the straw.
h0
h ⎞
h ⎞
⎛
⎛
Wgravity = ∫ ρ gAglass htop − h dh =ρ gAglass h0 ⎜ htop − 0 ⎟ = ρ gV0 ⎜ htop − 0 ⎟
0
⎝
⎠
⎝
2
2⎠
(
)
Calculate the work to increase the kinetic energy all the water to a velocity v top
2
WKE = 12 ρV0 vtop
h ⎞
⎛ ⎛
2 ⎞
Wtotal = ρV0 ⎜ g ⎜ htop − 0 ⎟ + 12 vtop
⎟⎠
⎝
⎠
⎝
2
(
)
(
)
Could also use: W = ∫ F dx = ∫ PAtm − Ptop A dx = ∫ PAtm − Ptop Aglass dh
=∫
h0
0
(
1
2
)
2
ρ vtop
+ ρ g ( htop − h ) Aglass dh ⇒ Which gives the same result.
c) Flow _ Rate = Avtop
⇒ ttotal =
V0
Avtop
Hughes
Section 6-2
Beloborodov
Sec 6-3
Problem:
Uniform plasma with electron number density n is in thermal equilibrium with Planck radiation of
temperature T and energy density U = aT 4 , where a = π 2 k 4 /15c3 h̄3 is the radiation constant and k is
the Boltzmann constant. The electrons (mass me ) and photons interact through Thomson scattering
with cross section σT = (8π/3)re2 where re = e2 /me c2 is the classical electron radius.
(a) Find the photon mean free path λ.
(b) Estimate the electron mean free path to photon scattering, l.
[Hint: consider the time between successive scatterings from the electron point of view.]
(c) Estimate the average number of scatterings experienced by a photon as it diffuses through a distance
L λ.
Solution:
(a)
λ=
1
.
nσT
(b) Photons bombard electrons with the speed of light. From the electron point of view, the characteristic time between successive scatterings is
tsc ∼
1
,
nγ σ T c
where nγ is the number density of photons. The mean energy of a blackbody photon is Ē ≈ 3kT , and
hence
aT 4
aT 3
nγ ∼
=
.
3kT
3k
Electrons move with thermal speeds v ∼ (kT /me )1/2 , and their mean free path to scattering is l ∼ vtsc ,
(kT /me )1/2
3k(kT /me c2 )1/2
h̄
∼
∼ α−2
l∼
nγ σ T c
σT aT 3
me c
kT
me c2
−5/2
,
where α = e2 /h̄c ≈ 1/137.
(c) Photon diffusion is a random walk with steps ∼ λ on the timescale τ ∼ λ/c. The diffusion length
L is related to time t by
L2 = Dt,
where D ∼ λ2 /τ = cλ (more precisely D = cλ/3, where 3 is associated with the dimension of space).
The number of scatterings is
2
ct
L
N∼
∼
.
λ
λ
Dean
Section 6-4
Dean
Section 6-4
Dean
Section 6-4
Dean
Section 6-4
Pasupathy
Section 6-5
Anastronautwantstohavesomecoffeewithmilk.Atherdisposalaresomecoffee
(200ml)at97°Candsomemilk(200ml)at7°C.Shewouldliketouseallthecoffee
andmilkforthedrink.Unfortunatelyshehastofinishsomeworkthatwilltakeher
15minutesbeforeshecanbegintoconsumeherdrink.Shewouldlikeherdrinkto
beashotaspossiblewhenshedrinksit.Sheistryingtofindoutifsheshouldmix
thecoffeeandmilkbeforeorafterfinishingherwork.
Theastronautstartsheranalysisbyassumingthathercoffeeandmilkare
exchangingheatwiththeirsurroundingsprimarilythroughblackbodyradiation.
Sheassumeshercoffeeandmilkaresuspendedinairasliquidspheresduetozero
gravity.Themixturewouldalsobeaspheresuspendedinair.Theenvironmentisat
theroomtemperature(300K).Shealsoassumesthatthespecificheatcapacityand
densityofcoffeeandmilkarethesameaswater,andthatallliquidsareinthermal
equilibriumthroughouttheirvolumeastheycooldownorwarmup.
(a)Basedonherassumptions,shouldtheastronautmixthecoffeeandmilk
beforeorafterherwork?Youcanusethefollowingapproximationvalidfor
small∆𝑇
)* +∆)
1
∆𝑇
$ 𝑑𝑇 ≈ $
$
𝑇 − 𝑇&
𝑇- − 𝑇&$
)*
(b)Whatisthemaximumtemperatureofthecoffeewhenshebeginstodrinkit?
Solution:
(a)Thegeneralideaisthatheatislost/gainedfromtheenvironmentbyradiation.
Letusconsiderthepost-mixingandpre-mixingscenariosseparately.
Post-mixing:
Initially,thetemperaturesofthecoffeeandmilkare𝑇./ and𝑇0/ respectively.The
coffeelosesheatandthemilkgainsheatfromtheenvironment.Aftertime∆𝑡,the
2
2
coffeeandmilkreachfinaltemperature𝑇. and𝑇0 respectively.Giventhatthe
quantityandcharacteristicsofthecoffeeandmilkareidentical,thefinal
2
6
) +)
6
temperaturereachedwillbe𝑇3 = 5 7
8
Tocalculatethefinaltemperaturesofcoffeeandmilk,wecalculatetheamountof
heatlostintimedt,andconvertthatnumbertoachangeintemperaturedTas
follows:
Heatlostbycoffeeintime𝑑𝑡:
Heatgainedbymilkintime𝑑𝑡:
𝑑𝑄 = 𝜎𝐴 (𝑇.$ − 𝑇&$ )𝑑𝑡
𝑑𝑄 = 𝜎𝐴 (𝑇&$ − 𝑇0$ )𝑑𝑡
Changeintemperatureofcoffeeintime 𝑑𝑡:
Changeintemperatureofmilkintime 𝑑𝑡:
𝑑𝑄 = 𝑚𝐶@ 𝑑𝑇. 𝑑𝑄 = 𝑚𝐶@ 𝑑𝑇0 Pasupathy
Section 6-5
Differentialequationfortemperatureofcoffeeandmilkasafunctionoftime:
𝑚𝐶@ 𝑑𝑇. = −𝜎𝐴 𝑇.$ − 𝑇&$ 𝑑𝑡
𝑚𝐶@ 𝑑𝑇0 = 𝜎𝐴 𝑇&$ − 𝑇0$ 𝑑𝑡
Thesecanbeintegratedtogetthefinaltemperaturesofmilkandcoffeebefore
mixing:
6
)7
)7A
1
𝜎𝐴
𝑑𝑇
=
−
Δ𝑡
.
$
𝑚𝐶@
𝑇.$ − 𝑇&
∆𝑇. = −
𝜎𝐴
Δ𝑡 (𝑇./ )$ − 𝑇&$
𝑚𝐶@
Putinnumbers:
massm=0.2kg
Radiusofcoffeesphere=3.63cm
Areaofcoffeesphere=165cm2
Heatcapacity=4.2J/g
Initialtemperatureofcoffee=370K
Time=15min=900seconds
Gives:
∆𝑇. = −10.6 𝐾
Sofinaltemperatureofcoffeeis86.4C
Similarlyformilk:
∆𝑇0 = 2.0 𝐾
Sofinaltemperatureofmilkis9C
Aftermixingthetwotogether,finaldrinktemperature=47.7C
Pre-mixing:
Initially,thetemperaturesofthecoffeeandmilkare𝑇./ and𝑇0/ respectively.On
) A +) A
mixing,thedrinkhasatemperature𝑇3/ = 5 7 =52C.Subsequently,thedrinkcools
8
downbyradiationasabove.Additionalpointstonotearethatthetotalmassofthe
drinkisnow2𝑚,andthesurfaceareaisnow28/I 𝐴.Followingthestepsabove,one
obtainsthedifferentialequationforthecoolingofthedrinkwithtemperature:
2𝑚𝐶@ 𝑑𝑇3 = −28/I 𝜎𝐴 𝑇3$ − 𝑇&$ 𝑑𝑡
Whichcanalsobeintegratedtogetthefinaltemperatureofthedrink.
𝜎𝐴
∆𝑇3 = − -/I
Δ𝑡 (𝑇3/ )$ − 𝑇&$
2 𝑚𝐶@
Pluginnumberstoget:
∆𝑇3 = −2.4 𝐾
Sofinaltemperatureofthedrinkis49.6C
-----
Part(a)canalsobeansweredwithoutexplicitcalculation:
Pasupathy
Section 6-5
Thekeyobservationinthisproblemisthattherateofchangeoftemperatureis
proportionaltothefourthpoweroftemperatures(asoneexpectsforradiation).
Withthisinmind,wecanusethefollowinglogictoevaluatethepost-mixingand
pre-mixingscenarioswithoutdoingexplicitcalculations:
(a)post-mixing–theinitialtemperatureofmilkisclosetoroomtemperature
already,sothetemperatureofmilkdoesnotincreasemuch.Theinitialtemperature
ofcoffeeishigh,soitwillloseheatquicklytotheenvironment.
(b)pre-mixing–theinitialtemperatureaftermixingismidwaybetweencoffeeand
milk–ie,muchclosertoroomtemperaturethantheinitialtemperatureofcoffee.
Themixturewillcoolmuchmoreslowlythanthecaseofthecoffeealone.Asmall
additionaladvantageinthiscaseisthattheratioofsurfaceareatomassissmaller
thaninthepreviouscase.
Therefore,itisclearthatpre-mixingthedrinkleadstoahigherfinaltemperature.
Sahin
Section 6-6
GeneralPhysicsIIcapacitorproblemandsolution
Inaparallelplatecapacitor,electricalforces
squeezethedielectriclayerthatseparatesthe
twoplates.Ifthedielectricmaterialiscompliant,
thereductioninthedistancebetweenparallel
platesisnolongernegligible.Toexplore
potentialusesofthisphenomenon,wewould
liketodeterminethechangesinelectricaland
mechanicalpropertiesofthissysteminresponse
toanappliedvoltage.
a) Writeanexpressionfortheelectricalforcepullingthecapacitorplatesintermsof
plateareaA,plateseparationℎ,dielectricpermittivity 𝜀# ,andtheappliedvoltageV.
b) Assumethatthedielectriclayerbehavesasalinearspringwithaspringconstantk,
andithasaninitialthicknessofℎ$ .DeterminethevoltageVnecessarytoreducethe
plateseparationfromℎ$ toℎ.
c) Thespringconstant𝑘dependsontheYoung’smodulusEd,areaA,andthicknessℎ$
by𝑘 = 𝐸# 𝐴/ℎ$ .Usingthisrelationship,expressyourresultin(b)intermsofEd, 𝜀# ,
ℎ,andℎ$ .
d) Oncetheplatesin(b)reachtheirequilibriumseparationℎ,asmallexternalforce,
𝛥𝐹,isappliedtotheparallelplatestocompressthedielectriclayerby𝛥ℎ.
Determinetheeffectivespringconstant𝑘.// ℎ = −𝛥𝐹/𝛥ℎ.
e) Evaluate𝑘.// ℎ atℎ = 2ℎ$ /3.Interpretyourresult.
Sahin
Section 6-6
Solution
a) Theforcecanbecalculatedfromtheenergydensityoftheelectricfieldas
1
𝐹 = − 𝜀# 𝐸 5 ×𝐴
2
thenegativesignindicatesthattheforceisinthepullingdirection.Substituting𝐸 =
𝑉/ℎ,weobtainthefollowingresult:
8
9:
5
;:
𝐹 = − 𝜀#
𝐴
therearevariousotherderivationsofthisfinalformula(e.g.Differentiatingtotal
energy,includingthebattery,withrespecttoh).
b) Springforceisbalancedbytheelectricalforce:
8
9:
5
;:
𝐹<=<>? = −𝑘 ℎ − ℎ$ − 𝜀#
𝐴 = 0
Rearrangingthetermsgive:
𝑉 5 = 2𝑘
ℎ$ − ℎ 5
ℎ 𝜀# 𝐴
c) Substituting𝑘 = 𝐸# 𝐴/ℎ$ gives:
𝑉5 = 2
𝐸# 𝐴 ℎ$ − ℎ 5 2𝐸# 5
ℎ
ℎ =
ℎ 1−
ℎ$ 𝜀# 𝐴
𝜀#
ℎ$
Notethattheresultisindependentofthesurfacearea,A.
d) Usinglinearapproximation,wecanwrite
𝛥𝐹/𝛥ℎ ≈ 𝑑𝐹/𝑑ℎ
Usingtheexpressionfor𝐹<=<>? in(a),wecanobtainthederivativeasfollows
𝑑𝐹
𝑉5
= −𝑘 + 𝜀# D 𝐴
𝑑ℎ
ℎ
substituting𝑉 5 gives:
𝑑𝐹
ℎ$
= −𝑘 3 − 2
𝑑ℎ
ℎ
Sahin
Section 6-6
notethat𝑘.// ℎ = −𝛥𝐹/𝛥ℎ,therefore
𝑘.// = 𝑘 3 − 2
ℎ$
,
ℎ
alternatively 𝑘.// =
𝐸# 𝐴
ℎ$
3−2
ℎ$
ℎ
e) Forℎ = 2ℎ$ /3
𝑘.// = 0
Effectivespringconstantvanishes.Thismeanstheparallelplatesystemisunstable.
Asmalldisplacementofoneplatetowardstheotherwillcauseittocollapseonto
theotherplate.