Physics Exam II Chapters 25-29

Transcription

1 Physcs Exam II Chaptes 5-9 Answe 8 of the followng 9 questons o poblems. Each one s weghted equally. Clealy mak on you blue book whch numbe you do not want gaded. If you ae not sue whch one you do not want gaded you may put down two choces (not moe than two) and the one that s most ncoect wll dopped. 1. Detemne the ntal decton of the deflecton of chaged patcles as they ente the magnetc felds as shown below. Consde a smple paallel-plate capacto whose vey lage plates ae gven equal and opposte chages and ae sepaated by a small dstance d. If we then ncease the sepaaton between the plates but stll have the dstance between the plates beng much smalle than the aea of the plates, what happens to each of the followng? Neglect fnge effects at the edges of the capacto. a) the feld between the plates b) the feld outsde the plates c) the potental dffeence between the plates d) the enegy stoed n the system e) the chage on the plates f) the capactance of the system 3. In the followng DC ccuts, a. Chage flows though a lght bulb. Suppose a we s connected acoss the bulb as shown. When the we s connected, what happened to the bghtness of the bulb? Explan.

2 b. The ccut below conssts of two dentcal lght bulbs bunng wth equal bghtness and a sngle 1 V battey. When the swtch s closed, what happens to bghtness of bulb A? Explan. 4. You connect a battey to a essto, poducng a potental dffeence V acoss t and causng a cuent to flow though the essto. Next, the essto s emoved fom the ccut and cut n half cosswse (so ts length s halved). One of the halves s placed back nto the ccut, wth the battey connected to t. (a) What happens to the potental dffeence acoss the essto? (b) What happens to the cuent acoss the essto? 5. Consde the system of capactos shown below wth C1 = 3. µf and C = 4. µf. a. Fnd the equvalent capactance of the system. b. Fnd the potental dffeence acoss each capacto. c. Fnd the chage on each capacto. d. Fnd the total enegy stoed by the goup. 6. Suppose that you wsh to fabcate a unfom we out of 1. g of coppe (esstvty = 1.7 x 1 8 Ω-m). The we s to have a esstance of R =.5, and all of the coppe s to be used. a. What s the length of ths we?

3 b. What s the damete of ths we? 7. If R = 1. k and ε = 5 V n the Fgue below, detemne (a) whch esstos ae n sees and whch ae n paallel. (b) Reduce the ccut to a smple one combnng and sees o paallel esstos. Clealy daw and name all cuents n the ccut, namng dentcal cuents wth the same name. (c) Deve at least one equaton elatng these cuents usng Kchoff s uncton ule. Clealy mak on you dawng of the ccut whch unctons you ae lookng at. (d) Deve moe equatons usng Kchoff s loop ule so that you have as many unknowns as equatons. Clealy ndcate whch loops you ae analyzng. You do not need to solve the equatons (that mght take too long). 8. A we.8 m n length caes a cuent of 5. A n a egon whee a unfom magnetc feld has a magntude of.39 T. Calculate the magntude of the magnetc foce on the we fo the followng angles between the magnetc feld and the cuent, (a) 6. o (b) 9. o, and (c) 1. o. 9. A unfom electc feld of magntude 35 V/m s dected n the negatve y decton as shown below. The coodnates of pont A ae (-., -.3) m, and those of pont B ae (.4,.5) m. (a) Calculate the potental dffeence V B - V A, usng the vetcal and hozontal paths. (b) Calculate the potental dffeence V B - V A, usng the dagonal path.

4 Possbly Useful Infomaton. 1 q1 q F = ε = 885. X 1-1 ( C / N m ) e = 1.6 X 1-19 C E = F q q E = εφ= ε EdA. = q enc x = x - x 1, t = t - t 1 v = x / t s = (total dstance) / t v = dx/dt a = v / t a = dv/dt = d x/dt v = v o + at g = 9.8 m/s x-x o = v o t + (½)at = x $ + y $ + zk $ v = v o + a(x-x o ) = - 1 x-x o = ½( v o + v)t = (x - x1) + (y - y1) + (z - z1) k x-x o = vt -1/at v= / t, v=d/dt a = dv / dt a= v/ t U = U f - U = -W U=-W V = V f - V = -W/q = U/q V = -W /q V V f = E f. ds V= E. ds f 1 q n 1 q V = V = V = = 1 = 1 1 dq V = V E s = V E s x E V y E V x = ; y = ; z = z V E = 1 q1q U = W = s 1 Q = CV A C = ε d l ab C = πε C = ln( b / a) b a C = R C = C (paallel) eq n

5 1 1 C eq C Q U = = 1 CV C u = 1 ε E C = κ C I= dq/dt ρ = 1 σ L R = ρ A V = IR P = IV P = I R=V /R I = ε ( R + ) P emf = Iε R = R (sees) 1 1 R = eq R (paallel) q(t)= Q(1-e -t/rc ) I = (ε/r)e -t/rc q(t) = Qe t/rc I = (Q/RC)e -t/rc, I = (Q/RC) F= qv x B F = IL x B d F = Ids x B τ = µ x B = NI A µ µ db = Ids x 4π 3, µ = 4π x 1-7 T. m/a B = µ I/ π B = µ ni (solenod) F/l = (µ I 1 I )/πa B.ds = µ Ienc I = ε B = (µ IN)/(π) (tood) d dφ E dt eq Φ B = BdA.