University of Alabama Department of Physics and Astronomy. PH126: Exam 1
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1 University of Albm Deprtment of Physics nd Astronomy PH 16 LeClir Fll 011 Instructions: PH16: Exm 1 1. Answer four of the five questions below. All problems hve equl weight.. You must show your work for full credit. 1. The circuit below is known s Whetstone Bridge, nd it is useful circuit for mesuring smll chnges in resistnce. Perhps you cn figure out why. Three of the four brnches on our bridge hve identicl resistnce, but the fourth hs slightly different resistnce, by n mount δ such tht its totl resistnce is δ. In terms of the source voltge V s, bse resistnce nd chnge in resistnce δ, wht is the reding on the voltmeter, V? You my ssume the voltmeter nd voltge source re perfect (drwing no current nd hving no internl resistnce, respectively). Vs ΔV δ Solution: Lbel the nodes on the bridge d, s shown in the figure below, nd let current I 1 flow from point d through to c, nd current I flow from d through b to c. Looking more crefully t the bridge, we notice tht it is nothing more thn two sets of series resistors, connected in prllel with ech other. This immeditely mens tht the voltge drop cross the left side of the bridge, following nodes d c, must be the sme s the voltge drop cross the right side of the bridge, following nodes d b c. Both re V dc, nd both must be the sme s the source voltge: V dc = V s. If we cn find the current in ech resistor, then with the known source potentil difference we will know the voltge t ny point in the circuit we like, nd finding V b is no problem. Let the current from the source V s be I. This current I leving the source will t node split into the seprte currents I 1 nd I ; conservtion of chrge requires I=I 1 I. At node c, the currents
2 I I1 I Vs d b c δ Lbeling notes nd currents in the Whetstone Bridge recombine into I. On the leftmost brnch of the bridge, the current I 1 cretes voltge drop I 1 cross ech resistor. Similrly, on the rightmost brnch of the bridge, the resistor hs voltge drop I nd the lower resistor hs voltge drop I ( δ). Equting the totl voltge drop on ech brnch of the bridge: V s = I 1 I 1 = I I ( δ) = I 1 = V s V s I = δ Now tht we know the currents in terms of known quntities, we cn find V b by wlking from point to point b nd summing the chnges in potentil difference. Strting t node, we move towrd node d ginst the current I 1, which mens we gin potentil difference I 1. Moving from node d to node b, we move with the current I, which mens we lose potentil difference I. Thus, the totl potentil difference between points nd b must be ( Vs V b = I 1 I = (I 1 I ) = ) ( = V s V b = V s ( 1 δ δ 4 δ V s δ ) ) If the chnge in resistnce δ is smll compred to (δ ), the term in the denomintor cn be pproximted 4 δ 4, nd we hve V b = 1 4 V s ( ) δ (δ ) Thus, for smll chnges in resistnce, the voltge mesured cross the bridge is directly proportionl to the chnge in resistnce, which is the bsic utility of this circuit: it llows one to mesure smll chnges on top of lrge bse resistnce. Fundmentlly, it is difference mesurement, mening tht one directly mesures chnges in the quntity of interest, rther thn mesuring the
3 whole thing nd trying to uncover subtle chnges. This behvior is very useful for, e.g., strin guges, temperture sensors, nd mny other devices.. Three conducting pltes re plced prllel to one nother s shown below. The outer pltes re connected by wire. The inner plte is isolted nd crries chrge mounting to 10 5 C per squre meter of plte. In wht proportion must this chrge divide itself into surfce chrge σ 1 on one fce of the inner plte nd surfce chrge σ on the other side of the sme plte? 5cm σ1 σ σtot = 10 5 = σ1 σ V1 V 8cm V1 Solution: Let s tckle the specific cse first, nd then work out the more generl cse. If we connect the two outer pltes, they re effectively prt of the sme conductor, nd thus they will be t the sme potentil. Cll tht V 1. Let the inner plte be t potentil V. Conservtion of chrge dicttes tht the totl surfce chrge on plte two is σ=σ 1 σ =10 5 C/m. In the region between the upper plte nd the middle plte, which we will cll region I, the totl electric field must be the sum of the field from the upper plte nd tht of the middle plte. The field from the middle plte in this region is tht of n infinite plte with surfce chrge σ 1, E = σ 1 /ɛ o. The upper plte, under the influence of the top side of the middle plte, will hve n induced chrge σ 1. It will contribute the sme electric field in the region between the upper nd middle pltes, i nd in totl we hve E I = σ 1 /ɛ o σ 1 /ɛ o = σ 1 /ɛ o (1) The electric potentil between the upper nd middle pltes must be V V 1, nd it must be equivlent to integrting the electric field cross the gp between the pltes. Since the electric field is independent of distnce, this is esy: V V 1 = E I d l = El = E (5 cm) = (5 cm) σ 1 /ɛ o () I i The induced chrge on the upper plte is negtive, but the field is in the opposite direction.
4 Proceeding similrly in region II between the lower nd middle pltes, we find V V 1 = (8 cm) σ /ɛ o (3) Dividing the lst two equtions, we find (5 cm) σ 1 /ɛ o = (8 cm) σ /ɛ o = σ 1 σ = 8 5 (4) Noting σ=σ 1 σ, we find σ 1 = 8 13 σ σ = 5 13 σ (5) Wht bout the more generl cse? Let the spcing between the upper nd middle pltes be d, nd the spcing between the upper nd lower pltes be D (nd thus the spcing between the lower nd middle pltes is D d). Proceeding s bove, we still hve E I =σ 1 /ɛ o, nd V V 1 = E I d l = El = Ed = dσ 1 /ɛ o (6) I In the region between the lower nd middle pltes, we hve E II =σ /ɛ o, nd Thus, V V 1 = E II d l = El = E (D d) = (D d) σ /ɛ o (7) II σ 1 = D d σ d (8) Agin noting σ=σ 1 σ, ( D d σ 1 = D ) σ σ = ( ) d σ (9) D We cn find the energy stored by integrting the electric field squred over ll spce. Outside ll of the pltes, E = 0. We cn brek up the integrl over the region between the two pltes into n integrl over region I nd n integrl over region II. Since the electric field is constnt in ech region, the integrls simply reduce to the volume contined in the region between the two pltes. Assume ech plte s n re A.
5 U = 1 ɛ o E dτ = 1 ɛ o E I dτ 1 ɛ o I II E II dτ = 1 ɛ o (σ 1 /ɛ o ) I dτ 1 ɛ o (σ /ɛ o ) U = σ 1 ɛ o Ad σ ɛ o A (D d) (10) II dτ Now we my substitute σ = d D σ nd σ 1 = ( ) D d d σ: [ U = σ 1 ɛ o Ad σ ɛ o A (D d) = A ɛ 0 = Aσ D ɛ 0 [ d (D d) d (D d) ( ) D d ( ) d d D σ D σ (D d)] ] = Aσ ( dd D Dd d 3 d D d 3) ɛ 0 U = Aσ D ɛ 0 ( dd d D ) = Aσ Dɛ 0 ( dd d ) (11) We wish to optimize U with respect to d: du dd = Aσ Dɛ 0 (D d) = 0 = d = D (1) We cn verify this is mximum potentil energy by noting d U/dd < 0 for ll d. A nice result: mximl energy is stored for symmetric plcement of the middle plte, just wht we might hve expected. Another wy to pproch this problem would be to notice tht this is relly just two cpcitors connected in series, which leds you to the sme result. 3. Two grphite rods re of equl length. One is cylinder of rdius. The other is conicl, tpering linerly from rdius t one end to rdius b t the other. Show tht the endtoend electricl resistnce of the conicl rod is /b times tht of the cylindricl rod. Hint: consider the rod to be mde up of thin, disklike slices, ll in series. Solution: The cylindricl conductor is trivil: if it is of rdius nd length l, nd hs resistivity ρ, then cyl = ρl π (13) Of course, we don t know the length l or resistivity ρ, but they will not mtter in the end. Wht bout the cone? Brek the cone up into mny disks of thickness dx. Stcking these disks up with incresing rdius cn build us cone: If we strt out with rdius t one end of the cone, nd the other end hs rdius b, then the rdius s function of position long the cone is esily determined. Let our origin (x = 0) be
6 b ( ) b r(x)= l x r x=0 x=l dx the end of the cone with rdius, nd ssume the cone hs totl length l, sme s the cylinder. Agin, we will not need this length in the end, but it is convenient now. The rdius t ny position long the cone is then ( ) b r(x) = x l (14) If the current is in the x direction, then ech infinitesimlly thick disk is bsiclly just tiny segment of wire with thickness dx nd crosssectionl re π [r(x)]. If we ssume the sme resistivity ρ, the resistnce of ech disk must be d cone = ρdx π [r(x)] = ρdx π ( ( b l ) x ) (15) The totl resistnce of the cone is found by integrting over ll such disks, from x=0 to the end of the cone t x=l. For convenience, let c=(b ) /l. l cone = d cyl = 0 ρdx π ( cx) = ρ [ π 1 c ( cx) ] l = ρ [ 1 0 π cb 1 ] = ρ [ ] b c πc b = ρl πb (16) Here we hve nice result: the resistnce of cone is the sme s resistnce of cylinder whose rdius is the geometric men cone s rdii. Tht is, if we substitute b in our usul formul for the resistnce of cylinder, we hve the result for cone. Anywy: the desired result now follows redily, cone cyl = b (17)
7 4. Three protons nd three electrons re to be plced t the vertices of regulr octhedron of edge length. We wnt to find the potentil energy of the system, or the work required to ssemble it strting with the prticles infinitely fr prt. There re essentilly two different rrngements possible. Wht is the energy of ech? Symbolic nswer, plese. Figure 1: An octhedron. It hs eight fces nd six vertices. Solution: 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 unique pirs of chrges. The problem is then reduced to figuring out how mny different possible pirings of chrges there re, nd wht the energy of ech piring is. The potentil energy for single pir of chrges, both of mgnitude q, seprted by distnce d is just: PE pir = k eq d First, we need figure out how mny pirs there re for chrges rrnged on the vertices of n octhedron, nd for ech pir, how fr prt the chrges re. Once we ve done tht, we need to figure out the two different rrngements of chrges nd run the numbers. How mny unique pirs of chrges re there? There re not so mny tht we couldn t just list them by brute force which we will do nywy to clculte the energy but we cn lso clculte how mny there re. In both distinct configurtions, we hve 6 chrges, nd we wnt to choose ll possible groups of chrges tht re not repetitions. So fr s potentil energy is concerned, the pir (, 1) is the sme s (1, ). Pirings like this re known s combintions, s opposed to permuttions where (1, ) nd (, 1) re not the sme. Clculting the number of possible combintions is done like this: ii wys of choosing pirs from six chrges = ( ) 6 = 6 C = 6!! (6 )! = = 15 ii A nice discussion of combintions nd permuttions is here: permuttionscombintions.htm
8 We cn verify this by simply enumerting ll the possible pirings. Lbel the chrges t ech vertex in some fshion, such s this: q q1 q6 q5 q3 q4 We hve six chrges t six vertices, nd thus 6 C = 6!!4! =15 unique pirings of chrges. Nmely, q 1 q, q 1 q 3, q 1 q 4, q 1 q 5, q 1 q 6 q q 3, q q 4, q q 5, q q 6 q 3 q 4, q 3 q 5, q 3 q 6 q 4 q 5, q 4 q 6 q 5 q 6 Here ll the q i hve the sme mgnitude, the lbels re just to keep things stright. At given vertex, ll four nerestneighbor vertices re t distnce, while the single nextnerest neighbor is t distnce. This mens tht there re three pirs chrges which re seprted by distnce, nd the other twelve pirings re t distnce. We hve highlighted the pirings bove. How cn we find two different rrngements? Since there re n odd number of nextnerest neighbor pirings, the first suspicion is tht the difference between the two rrngements will be in nextnerest neighbor pirings. If you experiment for while, the two different rrngements re these: A B Now we need only dd up the potentil energies of ll possible pirs of chrges. All the nerest
9 neighbor pirs will hve the sme energy, viz., U nn = kq (18) All the nextnerest neighbor pirs will hve U nnn = kq (19) For the first rrngement we hve 1 nerestneighbor pirs: eight of them re pirings, nd four of them re or pirs. We hve three nextnerest neighbor pirs, two or, nd one. Thus, the totl energy must be [ ] kq U A = 8 4 [ kq ] [ ] kq 1 [ kq ] = kq [ ] [ ] = 4 U nn 3.9 U nn (0) For the second rrngement, of the 1 nerestneighbor pirs we hve six pirs nd six or pirs, nd thus the totl energy of nerestneighbor pirs will be zero. We re left with only the nextnerest neighbor terms, nd for this rrngement, ll three re pirs. Thus, U B = 3 kq = 3 U nn.1 U nn (1) Thus, U A < U B, nd the first lttice is more stble, owing to its lower nerestneighbor energy. Though the second lttice hs smller nextnerest neighbor energy, there re fewer nextnerest neighbor pirs, nd their energy is smller thn the nerest neighbor pirs. Usully, minimizing the nerestneighbor energy gives the most stble crystl, simply becuse the potentil is decresing with distnce. 5. Show tht the expression Q /C is the energy stored in sphericl cpcitor (two concentric hollow metl spheres) by integrting the energy density u = 1 ɛ oe over the region between the spheres. Use the volume between two spheres or rdius r nd rdr s volume element.
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