Pof. Anchodoqui Poblems set # 12 Physics 169 My 12, 2015 1. Two concentic conducting sphees of inne nd oute dii nd b, espectively, cy chges ±Q. The empty spce between the sphees is hlf-filled by hemispheicl shell of dielectic (of dielectic constnt ɛ/ɛ 0 ), s shown in Fig. 1. (i) Find the electic field eveywhee between the sphees. (ii) Clculte the sufce-chge distibution on the inne sphee. (iii) Clculte the poliztion-chge density induced on the sufce of the dielectic t =. 2. A coxil cpcito of length l = 6 cm uses n insulting dielectic mteil with ɛ/ɛ 0 = 9, see Fig. 2. The dii of the cylindicl conductos e 0.5 cm nd 1 cm. If the voltge pplied coss the cpcito is V (t) = 50 sin(120πt) wht is the displcement cuent? 3. The pllel-plte cpcito shown in Fig. 3 is filled with lossy dielectic mteil of eltive pemittivity κ nd conductivity σ. The seption between the pltes is d nd ech plte is of e A. The cpcito is connected to time-vying voltge souce V (t). (i) Obtin n expession fo I c, the conduction cuent flowing between the pltes inside the cpcito, in tems of the given quntities. (ii) Obtin n expession fo I d, the displcement cuent flowing inside the cpcito. (iii) Bsed on you expessions fo pts (i) nd (ii), give n equivlent-cicuit epesenttion fo the cpcito. (iv) Evlute the vlues of the cicuit elements fo A = 4 cm 2, d = 0.5 cm, κ = 4, σ = 2.5 (S/m), nd V (t) = 10 cos(3π10 3 t) V. [Hint: 1 S = 1Ω 1, (S stnds fo siemens)] 4. Figue 4 shows plne electomgnetic sinusoidl wve popgting in the x-diection. Suppose tht the wvelength is 50 m, nd the electic field vibtes in the xy plne with n mplitude of 22 V/m. Clculte (i) the fequency of the wve nd (ii) the mgnitude nd diection of the mgnetic field when the electic field hs its mximum vlue in the negtive y-diection. (iii) Wite n expession fo the mgnetic field with the coect unit vecto, with numeicl vlues fo B mx, k, nd ω, nd its mgnitude in the fom B = B mx cos(kx ωt). 5. Some science fiction wites hve descibed sol sils tht could popel intestell spceships. Imgine gint sil on spcecft subjected to dition pessue fom ou Sun. (i) Explin why this ngement woks bette if the sil is highly eflective the thn highly bsoptive. (ii) If the sil is ssumed highly eflective, show tht the foce exeted by the sunlight on the spcecft s sil is given by F d = P A 2π 2 c, whee P is the powe output of the Sun (3.8 10 26 W), A is the sufce e of the sil, is the distnce fom the Sun, nd c is the speed of light. (Assume tht the e of the sil is much lge thn the e of the spcecft so tht ll the foce is due to dition pessue on the sil, only. (iii) Using esonble vlue fo A, compute the foce on the spcecft due to the dition pessue nd the foce on the spcecft due to the gvittionl foce of the Sun on the spcecft. Does this esult imply tht such system will wok? Explin you nswe. 6. A pulsed lse fies 1000 MW pulse tht hs 200 ns dution t smll object tht hs mss equl to 10.0 mg nd is suspended by fine fibe tht is 4.00 cm long. If the dition is completely bsobed by the object, wht is the mximum ngle of deflection of this pendulum? [Hint: Think of the system s bllistic pendulum nd ssume the smll object ws hnging veticlly befoe the dition hit it.] 7. An electomgnetic wve hs fequency of 100 MHz nd is tveling in vcuum. The
mgnetic field is given by B(z, t) = 1.00 10 8 cos(kz ωt)î. (i) Find the wvelength nd the diection of popgtion of this wve. (ii) Find the electic field vecto, E(z, t). (iii) Detemine the Poynting vecto, nd use it to find the intensity of the wve. 8. A dish ntenn hving dimete of 20 m eceives (t noml incidence) dio signl fom distnt souce s shown in Fig. 5. The dio signl is continuous sinusoidl wve with mplitude E m = 0.2µV/m. Assume the ntenn bsobs ll the dition tht flls on the dish. (i) Wht is the mplitude of the mgnetic field in this wve? (ii) Wht is the intensity of the dition eceived by the ntenn? (iii) Wht is the powe eceived by the ntenn? (iv) Wht foce is exeted by the dio wves on the ntenn? 9. Show tht ny function of the fom y(x, t) = f(x ct)+g(x+ct) stisfies the one-dimensionl wve eqution fo light, 2 y 1 2 y = 0. x 2 c 2 t 2 10. Suppose tht we hve cylindicl cpcito, s seen in the Fig. 6. Suppose futhe tht we put n AC cuent coss the pltes, stting t low fequency, ω. As the voltge ltentes, the positive chge on the top plte is tke off nd negtive chge is put on. While tht is hppening, the electic field disppes nd then builds up in the opposite diection. As the chge sloshes bck nd foth slowly, the electic field follows. At ech instnt the electic field is unifom, s shown in the figue, except fo some edge effects which we e going to disegd. We cn wite the electic field s E = E 0 cos(ωt), whee E 0 = Q 0 /ɛ 0 A is constnt, nd A = π 2 is the e of the plte. Now will this continue to be ight s the fequency goes up? No, becuse s the electic field is going up nd down, thee is flux of electic field though ny cicul loop, sy Γ, of dius inside the cpcito. And, s you know, chnging electic field cts to poduce mgnetic field. Fom Mxwell s equtions, the mgnetic field is given by c 2 B d l = d dt E d A c 2 B2π = d dt (Eπ2 ), o B = ω 2c 2 E 0 sin(ωt). So, the chnging electic field hs poduced mgnetic field ciculting ound inside the cpcito, nd oscillting t the sme fequency s the electic field. Now, e we done? No! This mgnetic field lso oscilltes, which poduces new electic field! The unifom field, E 1 = E 0 cos(ωt), is only the fist tem! The chnging mgnetic field poduces new electic field, E 2, such tht the totl field is E = E 1 + E 2. Now, in genel, E 2 is lso oscillting! This mens tht thee will be new mgnetic field fom E 2, which will be oscillting, which will cete new electic field, E 3, which will cete new mgnetic field... You tsk is to clculte the fist fou tems of the seies, enough to get the ptten, nd wite the totl electic field, tking the field t the cente of the cpcito to be exctly E 0 cos(ωt), (i.e., thee is no coection t the cente). Then, compe you esult with Bessel functions, nd see if you cn wite the full electic field you find in tems of one of the Bessel functions, in closed fom. Cn you find n exct expession in tems of one of the Bessel functions fo the mgnetic field?
4.10 Two concentic conducting sphees of inne nd oute dii nd b, espectiv chges ±Q. The empty spce between the sphees is hlf-filled by hemi shell of dielectic (of dielectic constnt / 0 0, s shown in the figue. +Q Q b Poblem 6.15 A coxil cpcito of length l = 6 cm uses n insulting dielectic mteil with ε ) Find the electic = 9. The dii of the cylindicl conductos e 0.5 cm nd 1 cm. If field eveywhee between the sphees. the voltge pplied coss the cpcito is Figue 1: Poblem 1. V (t)=50sin(120πt) (V) This is somewht is the displcement cuent? cuious poblem. It should be obvious tht wit Solution: dielectic the electic field between the sphees would be dil l + ~E = Q V(t) 2 2b ˆ - Poblem 6.16 The pllel-plte cpcito shown in Fig. 4 P6.16 0 is 2 filled with lossy dielectic mteil of eltive pemittivity Figue P6.15: ε nd conductivity σ. The seption We cnnot expect To find the displcement this cuent, towe need beto know unmodified E in the dielectic spce between between the pltes is d nd ech plte Figueis2: of Poblem e A. 2. The cpcito by the dielectic. is connectedhoweve, to we the cylindicl conductos. Fom Eqs. (4.114) nd (4.115), the dil electic field is tngentil Q time-vying voltge souce V (t). to the intefce between the diele E = ˆ 2πεl empty egion. Thus the tngentil, mtching condition E k 1 = V = Q ( ) b Ek 2 is uto 2πεl ln. stisfied. At Hence, the sme time I thee is no pependicul component to the V E = ˆ so thee is nothing to woy bout fo the D 1? = D 2? ln ( ) = ˆ 50sin(120πt) = ˆ 72.1 sin(120πt) (V/m), b mtching conditi ln2 D = εe A suggests tht we = ε ε 0 Eguess solution of the dil fom V(t) = ˆ9 8.85 10 12 72.1 sin(120πt) 5.75 10 9 = ˆ sin(120πt) (C/m 2 ). The displcement cuent flows between the conductos though n imginy cylindicl sufce of length l nd dius. The cuent flowing fom the oute conducto to the inne conducto long ˆ cosses sufce S whee S = ˆ2πl. ~E = A ˆ 2 whee A is constnt to be detemined. This guess is pehps not co obvious becuse one my hve expected the field lines to bend into o o dielectic egion. Howeve, we could lso ecll tht pllel fields do not coss the dielectic intefce. We my use the integl fom of Guss lw in medium to detemine t constnt A I Figue 3: Poblem 3. 0 A ~D ˆn d = Q ) 2 (2 2 )+ A 2 (2 2 )=Q () Obtin n expession fo I c, the conduction cuent flowing between the pltes inside the cpcito, in tems of the given quntities. Figue P6.16: Pllel-plte cpcito contining lossy dielectic mteil (Poblem 6.16). I d ε, σ d
nd (b) the mgnitude nd diection of the mgnetic field when the electic field hs its mximum vlue in the negtive y-diection. (c) Wite n expession fo the mgnetic field with the coect unit vecto, with numeicl vlues fo B mx, k, nd, nd its mgnitude in the fom B = B mx cos (kx-t). y E E m dy E c z B B dx dz x ELF bnd 8.0 - ) Solution 1 (foml) Wve phse speed is detemined by the eltion between these two quntities. It moves with such wy tht the phse emins constnt vlue kx Fom which c t t const. dx dt Figue 4: Poblem 4. k t on ulte ge etenetic th s f the nd egy, Figue P34.51 Figue 5: Poblem 8. 52. One gol of the Russin spce pogm is to illuminte 1
t cedit! l cpcito, s e tht, insted oss the pltes, As the voltge the top plte put on. While disppes nd ection. As the ly, the electic electic field is xcept fo some disegd. Figue 6: Poblem 10. E B E = E 0 cos (!t), nd A = 2 is the e of the plte. Now will this cy goes up? No, becuse s the electic field is going ctic field though ny cicul loop, sy, of dius ou know, chnging electic field cts to poduce utions, the mgnetic field is given by ~E d ~ A ) c 2 B (2 ) = d E 2,