PH 102 Exm I Solutions 1. Three ienticl chrges of = 5.0 µc lie long circle of rius 2.0 m t ngles of 30, 150, n 270 s shown below. Wht is the resultnt electric fiel t the center of the circle? By symmetry, the fiel is zero t the center. If you rw the irection of the fiel from ech chrge t the center of the circle, there is 120 between ech of the three fiel vectors. If you brek it own by components, you hve one purely long y, n two tht re 60 incline from the x xis. Then you ll en up with something like this for the x component of the fiel: E x = k e r 2 + k e r 2 where r is the rius of the circle. Similrly, E y =0. cos 60 + k e r 2 cos 60 = 0 (1) y 270 150 30 x 2. If the electric fiel strength in ir excees 3.0 10 6 N/C, the ir becomes conuctor n current my flow (i.e., sprk occurs). Using this fct, etermine the mximum mount of chrge tht cn be crrie by metl sphere 2.0 m in rius. Hint: recll tht from Guss lw tht for points outsie sphericlly-symmetric chrge istribution, the electric fiel is ienticl to point chrge t the center of the sphere. Outsie the sphere, Guss lw sys tht the electric fiel is just tht of point chrge, s we iscusse in clss. Tht mens for rius r>2 m from the center, the fiel is E = k e r 2 (2) where is the chrge on the sphere. Since we wish for E to excee 3.0 10 6 N/C, E = k e r 2 3.0 10 6 N/C (3) Er2 k e 1.3 10 3 C (4)
Now think bck to the vn e Grff emo... 3. Four point chrges ech hving chrge Q re locte t the corners of sure hving sies of length. Wht is the totl potentil energy of this system? Using the principle of superposition, we know tht the potentil energy of system of chrges is just the sum of the potentil energies for ll the uniue pirs of chrges. The problem is then reuce to figuring out how mny ifferent possible pirings of chrges there re, n wht the energy of ech piring is. The potentil energy for single pir of chrges, both of mgnitue, seprte by istnce is just: PE pir = k e 2 We nee figure out how mny pirs there re, n for ech pir, how fr prt the chrges re. Once we ve one tht, we nee to figure out the two ifferent rrngements of chrges n run the numbers. In this cse, there re not mny possibilities. Lbel the upper left chrge in ech igrm 1 n number the rest clockwise. The possible pirings re then only 1 2, 1 3, 1 4 2 3, 2 4 3 4 Since there re the sme number of possibilities for either crystl, the totl potentil energy in either cse is just ing ll of these pirs contributions together. Except for pirs 2 4 n 1 3, which re seprte by istnce 2, ll others re seprte by istnce. Thus, PE = k e 1 2 + k e 1 3 2 + k e 1 4 + k e 2 3 + k e 2 4 2 + k e 3 4 All we nee to o now is plug in for ll the chrges, since they re ll the sme: PE = k e 2 e 2 e e e 2 + k e 2 = k e 2 ( 4 + ) 2 (5) (6) 4. In Rutherfor s fmous scttering experiments tht le to the plnetry moel of the tom, lph prticles (hving chrge +2e n msses of 6.64 10 27 kg) were fire towr gol nucleus with chrge +79e. An lph prticle, initilly very fr from the gol nucleus, is fire t spee of v i =2.00 10 7 m/s irectly towr the nucleus, s shown below. How close oes the lph prticle get to the gol nucleus before turning roun? Assume the gol nucleus remins sttionry, n
tht energy is conserve. +2e +2e +79e vi vf = 0 This one is just conservtion of energy. The lph prticle (α) strts with some potentil energy, n it cn only work ginst its repulsion from the gol nucleus until it hs converte ll tht kinetic energy into potentil energy. At this point it comes to rest for n instnt, n the repulsive force pushes it bck to where it cme from. If the prticle strts from n infinite istnce wy with purely kinetic energy, n stops with no kinetic energy t some istnce r f from the gol (Au) nucleus: KE = PE (7) 1 2 mv2 i = k e α Au r f (8) r f = 2k e α Au mv 2 i 2.74 10 14 m (9)
5. A prllel plte cpcitor hs cpcitnce C when there is vcuum between the pltes. The gp between the pltes is hlf fille with ielectric with ielectric constnt κ in two ifferent wys, s shown below. Clculte the effective cpcitnce, in terms of C n κ, for both situtions. Hint: try breking ech sitution up into combintion of two seprte cpcitors. () (b) κ κ () Dielectric prllel to the pltes: C eff = 2K 1+K C. It is esiest to think of this s two cpcitors in series, both with hlf the plte spcing - one fille with ielectric, one with nothing. First, without ny ielectric, we will sy tht the originl cpcitor hs plte spcing n plte re A. The cpcitnce is then: C 0 = ɛ 0A The upper hlf cpcitor with ielectric then hs cpcitnce: (10) The hlf cpcitor without then hs Now we just the two like cpcitors in series: C = Kɛ 0A /2 = 2Kɛ 0A = 2KC 0 (11) C none = ɛ 0A /2 = 2ɛ 0A = 2C 0 (12) 1 C eff = 1 + 1 2KC 0 2C0 (13) C eff = = 4KC 2 0 2KC 0 + 2C 0 (14) 2K 1 + K C 0 (15) (b) Dielectric perpeniculr to the pltes: C eff = K+1 2 C.
In this cse, we think of the hlf-fille cpcitor s two cpcitors in prllel, one fille with ielectric, one with nothing. Now ech hlf cpcitor hs hlf the plte re, but the sme spcing. The upper hlf cpcitor with ielectric then hs cpcitnce: The hlf cpcitor without then hs C = Kɛ 0 1 2 A = Kɛ 0A 2 = 1 2 KC 0 (16) Now we just our prllel cpcitors: C none = ɛ 0 1 2 A = ɛ 0A 2 = 1 2 C 0 (17) C eff = 1 2 KC 0 + 1 2 C 0 (18) = 1 2 (K + 1) C 0 (19) = K + 1 C 0 (20) 2 6. Fin () the euivlent cpcitnce of this circuit, n (b) the totl chrge store in this circuit. Note µ=10 6. 1μF 3μF 3μF 6μF 9V Combine the upper two cpcitors tht re purely in prllel to mke single one of 4 µf. Combine the lower two purely in series to mke single one of 2 µf. These two euivlent cpcitors re then in prllel, n to 6 µf. If the totl voltge is 9 V on the euivlent cpcitnce of 6 µf, the totl chrge is Q=C V =54 µc.
7. In time intervl of 7.00 s, the mount of chrge tht psses through light bulb is 2.51 C. () Wht is the current in the bulb? (b) How mny electrons pss through the bulb in 5.00 sec? Current is chrge per unit time, so I=2.51 C/7 s 0.359 A. In 5 s, the number of chrges pssing through is Q = I t 1.79 C. One electron is 1.6 10 19 C, so the number of electrons is N=1.79/1.6 10 19 =1.1 10 19. 8. A 0.05 kg smple of conucting mteril is ll tht is vilble. The resistivity of the mteril is mesure to be 1.1 10 7 Ω m, n its ensity is 7860 kg/m 3. The mteril is to be shpe into soli cylinricl wire tht hs totl resistnce of 1.5 Ω. Wht length n imeter of wire re reuire? Not yet vilble... but problem like this will not be on the first exm in Summer 2011.
Constnts: k e 1/4πɛ o = 8.98755 10 9 N m 2 C 2 ɛ o = 8.85 10 12 C 2 /N m 2 e = 1.60218 10 19 C m e = 9.10938 10 31 kg Qurtic formul: 0 = x 2 + bx 2 + c = x = b ± b 2 4c 2 Bsic Eutions: F net = p t = m F centr = mv2 ˆr Centripetl r Newton s Secon Lw Vectors: [ ] F = F 2 x + F 2 y mgnitue θ = tn 1 Fy F x irection Electric Force & Fiel F e,12 = E 12 = k e 1 2 r 2 ˆr 12 12 E = k e r 2 Cpcitors: Φ E = E A cos θ EA = Q insie ɛ 0 Guss PE = W = E x cos θ = E x x constnt E fiel Q cpcitor = C V C prllel plte = ɛ 0A E cpcitor = 1 Q2 Q V = 2 2C C e, pr = C 1 + C 2 C e, series = C 1 C 2 C 1 + C 2 C with ielectric = κc without Current: I = Q t = nav v = eτ E τ = scttering time m m ρ = ne 2 τ V = ρl A I = RI R = V I = ρl A P = E t = I V = I 2 R = [ V]2 R power Unit Symbol euivlent to newton N kg m/s 2 joule J kg m 2 /s 2 = N m wtt W J/s=m 2 kg/s3 coulomb C A s mp A C/s volt V W/A = m2 kg/ s3 A fr F C/V = A 2 s4 /m 2 kg ohm Ω V/A = m2 kg/s3 A2-1 N/C 1 V/m Ohm: V = IR P = E t = I V = I 2 R = [ V]2 R Electric Potentil: V = V B V A = PE power constnt E fiel V point chrge = k e r PE pir of point chrges = k e 1 2 r 12 PE system = sum uniue pirs = W = PE = (V B V A ) pirs ij k e i j r ij Power Prefix Abbrevition 10 12 pico p 10 9 nno n 10 6 micro µ 10 3 milli m 10 2 centi c 10 3 kilo k PE = V = E x cos θ = E x x10 meg M 10 9 gig G 10 12 ter T 7