Phys Sp 9 Exam, Wed 8 Feb, 8-9:3pm Closed Book, Calculatos allowed Each question is woth one point, answe all questions Fill in you Last Name, Middle initial, Fist Name You ID is the middle 9 digits on you ISU cad Special codes K to L ae you ecitation section, fo the Honos section please use the two-digit section numbe that you shae ecitation with Best of luck, Caig Ogilvie VERSION A Please fill in A on Q4 Acknowledgement: Some questions in this exam wee dawn fom extenal souces, old textbooks, pevious exams, and adapted fo
c=3. 8 ms - h=6.63-34 Js ε = 8.85 - F/m μ = 4π -7 H/m e=.6-9 C m e =9. -3 kg m p =.67-7 kg ev=.6-9 J qq F = ˆ point chages F E = qtest q E = ˆ point chages qencl E da = E σ = ε ε above with chage density σ f ΔU = qe dl ΔU = qδv ΔV = qq U = f i i E dl conducting suface point chages q V = point chage V V V E = (, ) x y z Q C = V Aε C = paallel plate capacito d U = QV C = KC dieletcic vacuum u = ε E enegy density dq i = = n q Avd dt J = I / A Ohm's law E = ρj σ = ρ ρl R = A ΔV = ir P = IΔV loop ΔV τ = RC = q( t) = Q ( e t / RC sin( a) + sin( b) = t / RC q( t) = Qe ( + x) n = + nx + n( n ) x /! +... sin ( a + b)cos ( a b) 3 VolumeSphee = 4π /3 sufaceaeasphee = 4π aeacicle = π cicumfeencecicle = π b ± ax + bx + c =, x = A B = ABcos( θ ) A B = ABsin( θ ) ) b 4ac a
. Two small chaged objects A and B ae sepaated by.3 mete, and epel each othe with a foce of 4. x -5 N. If we move body A an additional.3 metes away, what is the electic foce now? a. 4. x -5 N b.. x -5 N c.. x -5 N d..5 x -5 N If you double the sepaation between the chages the foce deceases by facto of 4. Fou identical conducting sphees mounted on Dy Ice pucks ae held on a smooth level table at the fou cones of a squae. Once eleased, the pucks will move without fiction. Two of them cay a chage of +q and the othe two cay a chage of -q each, as shown below. If all the pucks ae eleased at the same time, in what diection will puck 4 move? Dawing the foces on puck 4. One foce towads, one foce towads 3, and a epulsive foce away fom. Howeve the epulsive foce is a facto of smalle than the othes, since the sepaation is sqt() lage Hence the net foce is towads the cente. F F 3 F a. puck 4 will acceleate towads the cente of the squae b. puck 4 will acceleate away fom the cente of the squae c. puck 4 will acceleate towads puck d. puck 4 will acceleate towads puck 3
3. Thee equal positive chages ae located at thee cones of a squae, and a negative chage (same magnitude of chage) is located at the fouth cone. What is the diection (if any) of the electic field at the cente of the squae? a. Towad the negative chage b. No diection, since field is zeo c. Away fom the negative chage d. Pependicula to the diagonal that goes though the negative chage. + + + + - The E-field at the cente has 4 contibutions. The top ight and bottom left chages give equal and opposite E- fields at the cente, so those two cancel. The emaining positive chage (top left) and the negative chage both give E- fields towads the ve chage so the total field is towads the negative chage 4. A chage of q is at a distance fom a fixed chage q. If q is allowed to move feely away unde the epulsion of the Coulomb field, what will be its kinetic enegy at a vey lage sepaation? q a. qq b. qq c. q d. e. depends on the choice of efeence level fo the potential The kinetic enegy at infinity is equal to the potential enegy at qq and U=
5. Two identical chages q ae nailed in place on the y axis, equal distances a above and below the oigin of coodinates. Find an expession fo the electostatic potential on the x axis, as a function of distance x fom the oigin. (Note: the efeence level fo the potential is chosen so that V= at infinity.) V = q a. a + x q V = 4 πε ( a + x ) b. q V = πε ( a + x ) c. V = πε q d. a + x e. V = The potential at a point is the sum of the potential fom each chage. At each point on the x-axis, the distance to each chage is sqt(a +x ). Hence V = V + V = = πε a + x q a q + x + a q + x
6. A small squae suface,. cm x. cm, is placed at a distance of. m fom a point chage of 3. x -9 C. What is the appoximate electic flux though just the font face of this squae if it is face on to the electic field? See figue below. a..7 N/C m b..7 N/C m c..7 N/C m d..7 N/C m e..7 N/C m Flux though one face ~EAea, since E is appoximately constant eveywhee on the suface q Φ = A = 4π 8.85 =.7N / Cm 3 9 (..) 7. A point chage of. x - C is located at the cente of a cubical Gaussian suface. What is the electic flux though each of the faces of the cube? a. insufficient info to detemine b..38 N/C m c..56 N/C m d..75 N/C m e..3 N/C m By Gauss law the total flux though the cube is q/ε. Since this flux goes though the six faces of the cube and that the chage is in the cente, the flux though each face is /6 of the total flux. Hence q Φ face = = * =.38N / Cm 6 ε 6 8.85
8. At a point P on the suface of a conducting mateial the chage density is σ. The chage caies ae static Outside the conducto, at a point vey nea P, the electic field is σ/ε outwads, nomal to the suface. Is this field due to a. the chages on the conducto nea P b. all the chages on the conducto c. the chages on the conducto nea P and any othe chaged objects outside the conducto d. all the chages on the conducto and any othe chaged objects outside the conducto The field at any point is due to sum of the E-fields fom all the chages. See fo example the pictue below. The E-field just outside the conducto is due to both the chages on the conducto, and the E- field fom the extenal plates. The field fom the extenal plates affects the magnitude of the chage density σ on the conducto, and the hence the dependence of the extenal field is captued by the expession σ/ε
9. The tube of a Geige counte consists of a thin conducting wie of adius "a" stetched along the axis of a conducting cylinde of adius "b". The wie and the cylinde have equal and opposite chages of "Q" distibuted along thei length "L". What is the fomula fo the electic field in the space between the wie and the cylinde; conside that the electic field is that of an infinitely long wie and cylinde. Q E = a. b. c. d. E = πε E = E = πε Q L Q L Q al e. E E = Q E = πε b Q E. da = ε cuvedcylinde Q da = ε Q E(πL) = ε πε Q L Since the chage distibution is cylindically symmetic we can daw a cylindical Gaussian suface at a adius fom the axis. The side view E is pependicula to the cuved face of the cylinde So thee is no flux though the two endcaps. Hence
. An odinay flashlight battey has a potential diffeence of.5 V between its positive and negative teminals. What is the change in potential enegy of an electon that moves fom the positive teminal to the negative teminal? ΔV< a..5 J b. -.5 J c..4x -9 J d. -.4x -9 J e. Fom the positive teminal to the negative teminal the potential deceases by.5 V, i.e ΔV =-.5V. ΔU=qDV, so ΔU =(-.6* -9 )*(-.5V) = +.4-9 J. The incease in potential enegy makes sense, since now the electon has moe potential enegy which is available to be dissipated as it tavels aound the cicuit.. A conducting spheical shell of adius. cm is chaged to a potential of 5 V. What is the value of the electostatic potential inside the sphee a distance 5. cm fom the cente of the sphee? Assume that the efeence level is chosen to be V= at infinity a. V b..8 V c. 5. V d. 8.7 V e. 5 V The potential inside a conducting sphee is the same as at the suface, all points on a conducto in static equilibium ae on an equipotential. This is because E=, so ΔV as the line integal of E gives ΔV=. In some egion of space, the electostatic potential is the following function of x and y, but not of z: V =x + xy whee the potential is measued in volts and the distance in metes. The units fo this ae made coect by the numeical factos. Find the x-component of the electic field at the point x =, y =. a. -8 N/C b. -4 N/C c. N/C d. 4 N/C e. 8 N/C V E x = = x y = * + * = 8N / C x
3. High-dielectic-constant thin-film capacitos ae attactive fo digital memoy applications; fo example, when baium stontium titanate (BaSTi O 6 ) is used as a 5-nm-thick dielectic mateial, a capacitance pe unit aea of.9 F/m fo a paallel plate capacito can be achieved. What is the appoximate dielectic constant of this mateial? a.. b. c. 74 d. 7 e. 5 Aε C = κc = κ d C d 5 k = =.9* A ε 8.85 9 = 5 4. A "supecapacito" has a huge capacitance of 6.8 F, but can withstand a potential diffeence of only.5 V. A powe-supply capacito has a capacitance of 8 μf and can opeate up to 4 V. Which can stoe moe enegy? a. The supecapacito b. The powe-supply capacito c. In this example, both stoe the same amount of enegy d. Insufficient infomation povided U = QV = CV Usup e = 6.8*(.5) = J 6 Usup ply = 8 *(4) = 66J
5. A non-equilibium E-field is established inside a metal conducto, pat of which is shown below. The est of the cicuit is not shown. Which location is at a highe potential? a. the left side is at a highe potential than the ight b. the ight side is at a highe potential than the left c. the top is at a highe potential than bottom d. the bottom is at a highe potential than top ΔV = E. dl Hence the potential deceases fo a path fom left to ight, hence the potential is highe on the left. 6. A table lamp is connected to an electic outlet by a coppe wie of diamete. cm and length. m. Assume that the cuent though the lamp is.5 A, and that this cuent is steady. How long does it take an electon to tavel fom the outlet to the lamp? The fee-chage density in coppe is appoximately 8.5x 8 electons/m 3. a. 6.7* -9 s b..* -3 s c. 6.4 s d. 5 s e. 57, s i=nqav_d i=nqal/t t=nqla/i i=8.5*^8*.6*^-9**pi*(./)^/.5 i=57s
7. Fo the cicuit shown below what happens to the bightness of the bulbs when the switch is closed. Assume that the battey has negligible intenal esistance. a. all bulbs incease in bightness b. all bulbs decease in bightness c. top two bulbs decease in bightness, bottom bulb inceases in bightness d. top two bulbs incease in bightness, bottom bulb deceases in bightness e. all bulbs emain at thei oiginal bightness Since the thee banches of the cicuit ae in paallel, each banch has the same potential dop ΔV acoss them. Closing the switch does not change the potential dop, it just adds anothe path fo cuent. If ΔV is unchanged then the bulbs bightness will be unchanged. 8. When two esistos ae connected in seies, the equivalent esistance is 8Ω. When they ae connected in paallel, the equivalent esistance is 5Ω. What ae the values of the two esistos? a. 4Ω, 4Ω b. 3Ω, 5Ω c. Ω, 6Ω d. Ω, 7Ω e. 5Ω, 75Ω In seies R eq =R +R, all the options give 8Ω. In paallel /R eq = /R +!/R, R eq =5Ω when R =, R =6
9. What is the equivalent esistance of the combination of fou esistos shown below? Each of the esistos has a esistance of 3.Ω. a..8 Ω b.. Ω c..5 Ω d. 3. Ω e. 4. Ω The top ight pai of esistos ae in paallel, /R topright = /R+/R R topright =3/ Ω. This is in seies with one esisto, thee R top =3+3/=4.5Ω. This is now in paallel with the bottom esisto, /R eq =/4.5+/3, Req=.8Ω.. Fou esistos ae connected in paallel to a voltage souce as shown in the figue below. Which of the below is a coect statement a. Each esisto has the same cuent though it b. The 5 Ω esisto has the most cuent though it compaed to the othe esistos c. The 6 Ω esisto has the most cuent though it compaed to the othe esistos The esistos ae in paallel hence ΔV acoss them is the same, since DV=-iR, the banch with the smallest esisto will have the most cuent though it.
. While canking the engine, the state moto of an automobile daws 8 A at V though a esistance of.5 Ω fo a time inteval of.5 s. What is the electic enegy used in the given time inteval? a. 4 J b. 96 J c. 384 J d. 56 J e. 8 J Powe=iΔV Enegy=powe*Δt=8**.5=4J. A esisto R, connected to an ideal powe supply of voltage V, is dissipating a powe P. The voltage is inceased to a value 5% lage. How should the esistance be changed to keep the powe dissipated the same? a. inceased by 5% b. inceased by 56% c. deceased by 5% d. deceased by 56% e. kept the same (ΔV ) P = iδv = R Hence if ΔV inceases by 5%, (ΔV) will incease by 56%. To keep the powe dissipated the same you need to incease the denominato by the same amount, i.e R must incease by 56%
3. In the cicuit below a V ideal battey, 5Ω esisto, a witch and an initially unchaged. Faad capacito ae connected in seies. The switch is closed, and afte 4s the voltage acoss the capacito is appoximately? a. V b. 8 V c. 6 V d. 4 V e. V q( t) = Q ( e f => V ( t) = V ( e => V ( t = 4s) = ( e => V ( t = 4s) = 4V f t / RC ) t / RC andv = q / C ) 4 / 5*. 4. Please fill in A on Q4 )