Physics 111 Exam #1 January 4, 011 Name Muliple hoice /16 Problem #1 /8 Problem # /8 Problem #3 /8 Toal /100
ParI:Muliple hoice:irclehebesansweroeachquesion.nyohermarks willnobegivencredi.eachmuliple choicequesionisworh4poinsoraoalo 16poins. 1. When he elecric ield is zero a all poins in any region o space a. he elecric poenial mus be zero a all poins in ha region. b. he elecric poenial mus have a consan value a all poins in ha region. c. he elecric poenial canno be ound unless we know is value a a minimum o wo poins in he region. d. he region migh be ound beween he plaes o a charged capacior.. poin charge Q is ixed in space. Suppose ha a small es charge -q is placed a a disance r rom Q. I he es charge is released rom res he oal energy o charge disribuion a. increases wih increasing disance beween Q and -q. b. remains consan. c. increases wih decreasing disance beween Q and -q. d. decreases wih decreasing disance beween Q and -q. 3. Plae a (o a parallel-plae air-illed capacior) is conneced o a spring having orce consan k, while plae b is ixed. I a charge +Q is placed on plae a and Q is placed on plae b, and i he elecric ield is or a plae o charge is given by E = Q, plae a moves ε o a. x = Q Q o he le. b. x = o he righ. kε o kε o c. x = Q o he le. kε o Q d. x = o he righ. kε o 4. Suppose ha you have wo poin charges Q 1 and Q separaed by a disance r so ha here exiss a orce o repulsion wih magniude F. I each o he magniudes o he charges were ripled and he separaion beween hem doubled, he magniude o he repulsive orce would now become a. c. 3 4 F. b. 3 F 9 F d. 9 4 F
ParII:FreeResponseProblems:Thehreeproblemsbelowareworh84poins oalandeachsubparisworh7poinseach.pleaseshowallworkinorderoreceive parialcredi.iyoursoluionsareillegibleorillogicalnocrediwillbegiven. numberwihnoworkshown(evenicorrec)willbegivennocredi.pleaseusehe backohepageinecessary,bunumberheproblemyouareworkingon. 1. Given below is a se o daa on a collecion o poin charges along wih heir locaions (in meers) in a sandard aresian coordinae sysem. a. Wha is he elecric ield a he origin due o poin charges q 1 q 4? harge (µ) Locaion (x,y) q 1-6 (0, ¾) q 3 (1, 0) q 3-5 (0, -½) q 4 1 (-¼, 0) E ne,x = E + E 4 = kq r + kq 4 r 4 E ne,y = E 3 + E 1 = kq 3 r + kq 1 3 r 1 E ne = E ne,x + E ne,y @θ = an 1 = 9 10 9 Nm = 9 109 Nm E ne,y E ne,x 3 10 6 1m + 1 10 6 ( ) ( 0.5m) 5 10 6 0.5m + 6 10 6 ( ) ( 0.75m) =1. 105 N = 8.4 104 N =1.44 10 5 N @θ = 35o below he + x - axis b. Wha is he elecrosaic orce on a -½ µ poin charge placed a he origin? F ne = qe ne = 0.5 10 6 1.44 10 5 N = 0.07N@θ =145o above he + x - axis c. Wha is he elecric poenial a he origin due o poin charges q 1 hrough q 4? V P = V 1 + V + V 3 + V 4 = kq 1 + kq + kq 3 + kq 4 r 1 r r 3 r 4 V P = 9 10 9 Nm 6 10 6 0.75m + 3 10 6 1.0m 5 10 6 0.5m + 1 10 6 = 9.9 10 4 V 0.5m d. How much work was done on he -½ µ poin charge i i was brough in rom ininiy o he origin? ( )[ 0V ( 9.9 104V )] = 0.050J = 50mJ W = qδv = 0.5 10 6
. Suppose ha you charge a 0.19nF capacior o ull charge using a 10.0V baery. The capacior plaes are circular each wih a diameer o 0cm. a. Suppose you waned o build his capacior ou o he wo meal plaes (wih dimensions above and raed capaciance) and urher you waned o insulae he plaes rom one anoher wih a piece o rubber (κ = 7), how hick would he insulaing layer need o be o creae his capacior? = κε o d ( ) d = κε o = 7 8.85 10 1 π 0.1m Nm 0.19 10 9 F = 0.010m =10mm b. How much charge is iniially sored on he capacior when conneced o he baery? Q max = V = 0.19 10 9 F 10V =1.9 10 9 =1.9n c. Suppose ha you remove he baery and decide o discharge he capacior hrough a 1000Ω resisor. How long will i ake or he capacior o dissipae 63% o is iniial sored charge? Q ( ) = Q max e R Q( = R) = 0.37Q max = Q max e R ( ) = 1000Ω 0.19 10 9 F ln( 0.37) =1.9 10 7 s = R ln 0.37 d. How much energy was iniially sored in he capacior and how much energy will remain aer one ime consan has passed? E max = 1 V = 1 0.19 10 9 F( 10V ) = 9.5 10 9 J = 9.5nJ E( ) = 1 V ( ) = 1 V maxe E( = R) = E max e R R = E max e R = 1 V maxe R = 9.5 10 9 J 7.39 =1.3 10 9 J =1.3nJ
3. Suppose ha you have a beam o elecrons ha you wan o urn a 90 o corner as shown in he igure below, by using a parallel plae capacior. ssume ha he elecron has a kineic energy KE = 3x10-17 J and moves up hrough a small hole in he boom plae o he capacior. y x a. Should he boom plae be charged posiively or negaively (relaive o he op plae) i you wan he elecron beam o urn righ? Explain you choice ully. To repel he elecron rom he op plae, i should have a negaive charge and hus he boom plae should be charged posiively, which will also arac he elecron oward he exi hole. b. Wha is he magniude o he velociy o he elecron? KE = 1 m v e e v e = KE = 3 10 17J 9.11 10 31 kg = 8.1 106 m s c. Using he coordinae sysem on he igure, wha consan magniude elecric ield would be needed i you waned he elecron o emerge 1.0cm rom where i eners raveling a a righ angle o is original direcion? Draw he elecric ield s direcion on he diagram above. (Hin: ssume ha he elecric ield bisecs he righ angle in he igure above.) v y = 0 = v iy a y rise = v iy F x v ix = v i sinθ ee x v i cosθ E = m v e i sinθ cosθ = 9.11 10 31 kg 8.1 10 6 m s ex 1.6 10 19 0.01m direced rom he posiive o he negaive plae. ( ) sin45cos45 = 3.7 10 4 N d. Wha minimum separaion,, would he capacior plaes need o have? = y = y i + v iy rise 1 a y rise = x anθ eex 4 v i cos θ = 0.01m an45 1 x = v i sinθ 1 v i cosθ e E 1.6 10 19 3.7 10 4 N ( 0.01m ) ( ) cos 45 4 9.11 10 31 kg 8.1 10 6 m s x v i cosθ = 0.008m =.8mm
τ = NIB sinθ = µb sinθ PE = µb cosθ B = µ 0I πr ε induced = N Δφ B Δ onsans g = 9.8 m s 11 Nm G = 6.67 10 kg 1e =1.6 10 19 k = 1 = 9 10 9 4πε o 1 Nm 1eV =1.6 10 19 J µ o = 4π 10 7 Tm c = 3 10 8 m s Nm = N h = 6.63 10 34 Js = 9.11 10 31 kg = 0.511MeV c m p =1.67 10 7 kg = 937.1MeV c m n =1.69 10 7 kg = 948.3MeV c 1amu =1.66 10 7 kg = 931.5MeV c N = 6.0 10 3 x + Bx + = 0 x = B ± B 4 Δ ( B cosθ ) Δ Physics111EquaionShee ElecricForces,FieldsandPoenials Elecricircuis LighasaWave F = k Q Q 1 r ˆ r F E = q E Q = k Q r ˆ PE = k Q 1Q r V ( r) = k Q r E x = ΔV B, Δx W,B = qδv,b MagneicForcesandFields F = qvb sinθ F = IlB sinθ ε o = 8.85 10 LighasaParicle&Relaiviy NuclearPhysics Geomery ircles : = πr = πd = πr Triangles : = 1 bh Spheres : = 4πr V = 4 3 πr3 Misc.Physics110 Formulae F = Δ p Δ = Δ( mv) = ma Δ F = ky F = m v R r ˆ W = ΔKE = 1 m v v i PE graviy = mgy PE spring = 1 ky x = x i + v ix + 1 a x v x = v ix + a x v vx = v ix + a x Δx ( ) = ΔPE