Physics 114 Exam 2 Fall Name:

Transcription

1 Physcs 114 Exam Fall 015 Name: For gradg purposes (do ot wrte here): Questo Problem Aswer each of the followg questos. Pots for each questo are dcated red. Uless otherwse dcated, the amout beg spread amog parts (a,b,c etc) are equal. Be sure to show all your work. Use the back of the pages f ecessary.

2 Questo 1. (10 pots) Descrbe the moto of a proto after t s released from rest a uform electrc feld: (a) How does ts ketcs eergy chage (crease, decrease or stay the same)? (b) How does the potetal eergy of the proto-feld chage f at all? (c) Descrbe the dfferece electrc potetal (Vf V) betwee the tal ad fal postos of the proto (s t zero, postve, or egatve)? How do the aswers to all of these questos chage f a electro s released stead of a proto (d) Ketc eergy, (e) potetal eergy, ad (f) electrc potetal? Soluto (a) It s ketc eergy creases due to the accelerato from the force of the electrc feld. (b) The potetal eergy must decrease due to coservato of mechacal eergy. Ths s lke droppg a mass the earth s gravtatoal feld. (c) The dfferece s electrc potetal s egatve. The accelerato s the drecto of the electrc feld sce the force o the proto s ths drecto. The electrc feld pots drecto gog from hgh to low potetal. Also, f V E. ds E. s. The agle betwee the dsplacemet vector ad the electrc feld s zero degrees so Vf V s egatve. (d) The electro just goes the opposte drecto. Ketc eergy stll creases. (e) The potetal eergy decreases. (f) Now the dsplacemet s the opposte drecto of the electrc feld. Vf V s postve.

3 Questo. (10 pots) Two capactors wth capactaces C1 = F ad C = 4 F are coected parallel to a 1 V battery. (a) What s the rato of the voltages o each capactor, V1/V? (b) What s the rato of charges o each capactor, Q1/Q? (c) What s the rato of the eergy stored each capactor, U1/U? Now the capactors are coected seres to the 1 V battery. (d) What s the rato of the voltages o each capactor, V1/V? (e) What s the rato of charges o each capactor, Q1/Q? (f) What s the rato of the eergy stored each capactor, U1/U? Soluto. (a) The rato s 1. Parallel devces have the same voltages across them. (b) Q1/Q = C1V/CV = C1/C = 0.5. (c) U1/U = (½C1V )/ (½CV ) = C1/C = 0.5. (d) V1/V = (Q/C1)/(Q/C) = C/C1 =. (e) Q1/Q = 1. Seres capactors have the same charge o them. (f) U1/U = (Q /C1)/(Q /C) = C/C1 =.

4 Questo 3. (10 pots) Cosder the crcut below whch there are 3 detcal lght bulbs, labeled A, B, C. I aswerg each questo, clude a explaato. A C B (a) How does the brghtess of B compare to that of C? Brghtess goes as the power whch we ca wrte as I R. Both bulbs have the same resstace ad the same curret sce they are seres, so the brghtess s the same. (b) How does the brghtess of A compare to that of B? We ca also express the power as V /R. The etre emf s across bulb A but ths same emf s splt betwee B ad C. Thus, A s brghter tha B. (c) Whe A s uscrewed, what happes to the brghtess of bulbs B, C? Nothg happes to the brghtess of B ad C as these are parallel wth bulb A (lke your house), so bulbs B ad C stll have the same voltage across them. (d) Whe B s uscrewed, what happes to the brghtess of bulbs A ad C? Nothg happes to the brghtess of A as t s parallel. Bulb C goes out as the resstace that brach of the crcut goes to fty (e) If a capactor were seres wth bulb A (see fgure below), how does the brghtess of A mmedately whe the swtch s closed compare to whe t has bee closed for a very log tme? Whe the swtch s closed, there s o charge yet the capactor ad curret flow through t as f t were a wre. After a certa amout of tme, the capactor s fully charged ad o curret flows through t. Thus, Bulb A s brghtest mmedately after the swtch s closed ad the dms to othg as tme goes o. (f) I ths same crcut, how does the brghtess of B mmedately whe the swtch s closed compare to whe t has bee closed for a very log tme (assume a tugste flamet wth a postve temperature coeffcet of resstace, )? The voltage across B s costat oce the swtch s closed. Its resstace creases over tme due to heatg. Thus, the P=V /R decreases over tme, ad t s brghtest mmedately upo closg the swtch. C A B

5 Problem 1. (15 pots) Cosder the system of two charges +q alog the y-axs show the Fgure below (gore the presece of the charge Q for ow). Let q = C ad let d = 8 m. (a) Calculate the electrc potetal at a pot o the x-axs a dstace x = 3 m from the org, due to the two C charges. (b) Calculate the electrc potetal at the org. (c) Now let the magtude of the egatve charge Q = 3mC whch has a mass of Kg be released from rest from the pot alog the x-axs at x = 3 m. What would the speed of the charge be at the org? Solutos 1 q (a) V V = (/5 + /5) V r = V. The dstace of 5 m was obtaed usg the pythogorea theorem. 1 q (b) V V = (/4 + /4) V r = V (c) Uf + Kf = U +K. K = 0 sce t s released from rest. So we have U = -Kf qv = -½ mv (-0.003)( ) = - v v = m/s

6 Problem. (15 pots) Cosder the crcut below, (a) What s the equvalet capactace of the four capactors? (b) Fd the charge o each of the four capactors. (c) Fd the voltage across each of the four capactors. Soluto

7 Problem 3. (15 pots) Cosder the crcut below. (a) Determe f there are ay resstors seres or parallel. (b) Use Krchoff s frst law appled at pot c to obta oe equato relatg the three currets, I1, I, ad I3. (c) Apply Krchoff s loop rule to the top loop to obta a equato relatg I1 ad I. (d) Apply Krchoff s loop rule to the top loop to obta a equato relatg I1 ad I3. (e) Gve that I1 = A, fd the other two currets. Solutos (a) Noe of the resstors here are seres or parallel. (b) I1 + I = I3. (c) -14+6*I1-10-4*I = *I1 4*I = 0 (d) 10-6*I1-*I3 = 0. (e) Takg the last equato we got, we have 10-1= *I3 I3 = -1 A Now takg the jucto rule equato, we have I = I3 I1 = -1- = -3A So I ad I3 are actually gog the opposte drecto as orgally guessed.

8 Possbly Useful Iformato 1 q1 q F 885. X 10-1 ( C / Nm ) r e = 1.6 X C E F q 0 q E, E = / 00EdA. q ec 4 0r x = x - x1, t = t - t1 v = x / t s = (total dstace) / t v = dx/dt a = v / t a = dv/dt = d x/dt v = vo + at g = 9.8 m/s x-xo = vot + (½)at r = x + yj + zk v = vo + a(x-xo) r=r -r 1 x-xo = ½( vo+ v)t r = (x - x1) + (y - y1) j + (z - z1) k x-xo = vt -1/at v = r / t, v=dr/dt a = dv / dt a = v / t U = Uf - U = -W U=-W V = Vf - V = -W/q0 = U/q0 V = -W /q0 V V f E f. ds V E. ds f 1 q 1 q V V V 40 r r Uf + Kf = U +K 1 dq V 4 r K = ½ mv 0 V Es V E s x E V y E V x ; y ; z z V E 1 q1q U W s 40 r1 Q = CV A C 0 d l C 0 C l( b / a) 4 0 ab b a

9 C 4 0R Ceq Cj (parallel) 1 1 C eq C Q U 1 j CV C u 0E C = C0 I= dq/dt 1 L R A V = IR P = IV P = I R=V /R I ( R r) Pemf = I R R (seres) 1 1 R R (parallel) eq j I = (R)e -t/rc I = (QRC)e -t/rc, I0 = (QRC) E = o eq q(t)= Q(1-e -t/rc ) j q(t) = Qe t/rc = Q/L, = Q/A, = Q/V