Homework_5_Solutions

Homework_5_Solutions - M"" we" c I $78M t a,v i...

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Unformatted text preview: M" "? we" c, I: $78M t a ,v i, . . E rotate oatde cottaettetrce ot a h/tfld trartetetofirt {a} The gate or;th capacitartee yer rtttrt area ot a trifle traneieter With a dich Eager oi" 1?; >~< iti— cttt thichneae can he toned ae e e e 3.9 x 8.85 x 10-12 , C : .925. 2 7’ 0 g N as ~2 ‘ 3 ex tax tax 2 X EQ_3 .. 373 X 10 F—m (b) Using L = 5 x 10‘4 cm and W : 2 x 10’3 cm, the totai gate capacitance is r: m tjng = career e: we 2r iii-3 x 5 >< rem-t X a X row?) m tyre 3: $330 Qoaxiai cagracitor with two dieieetriee. (a) This yrohierh can he soived irt two different waye. in the first method, we can take advantage of the fomruia for capacitance per unit iength of the coaxiai line, as determined in Exampie 4w27, nameiy i 27% 111(1) / a) Ccoax z where e is the permittivity of the dieiectric titling. The configuration shown in Figure 4.70 can he considered to be two ‘haltcoaxiai’ capacitors at length E connected in parade}, with capacitart ea respectiveiy of t 2775”! 1 27(6ng 0‘ =7itnte/a) and CZ=§1n(b/o} which combine additiver (capacitors iii parailei) to give the totai capacitance per unit length ot‘ c = W 1n(b/ a) The same reenit can also he ohteined rising Game’s iaw arid noting that the electric tieid has to he radiat (dare to symmetry) and most have the same vahre in hoth dielectrics, doe to the eiectroetatic 3 boundary condition {4.66} dictating the continuity of the tangentiat component Qt." the eteetiic tieid é acrose a dietectric hoahdary. Considering a eyiiridiicai Gaussian surface ct" terrgth i and radius 1 d< r < hwecanthert write 3 i t 2 7r 1 27F ~gD-ds=jji €0€1TE¢Td¢dz+ij eoeererzbdzr—pll 0 G 0 71’ Pl + E = l —> E = ——-————-—-—- 1 —> mite” e2on 1‘ Pt 1“ Mead” + 621») This eiectric field is radiaiiy outward if the inner conductor is positively charged, so that pl > O. The potential difference between the inner and outer cylinders of the coaxrai hoe is then 3 :01 057" __ .01 m E} a: a. e zfifi’o: were =~jf Edit—it Wiwa M K > a > E: r ct; W’Eoiéh- 52¢} t" nectar? +52” \a/ se shes we have C: Q = pll = W€0{€1r+€2r)l @ba ‘13:»; ENE/a) {23) From Table 4.}; We have 6194 = 2.3 fer oil and 62¢ = 5.4 fee mice, which gives N M885 x m-lzme + 5.4)(5 x 10-2) C 111(3) 2 9.75 HF (C) With (miy oii used throughout we have eh“ = 627- = 2.3 so that w 7218.85 >< 10—12)(2.3 + 2.3)(5 x 10-2) 0 , E3363}; 3 518211;? 5&5“ Eesih eeeeeiseh (a) 338ng {4.47} wish a z 63% x E05 m, the eepaeitahee e‘f she eerie can he caisuiafiecfi as 0 = 47:” x (8.85 x 10"”)(6371 x 106) 2 709 he (1)) The electric fieid on the surface of the earsh can be expressed in terms of the total charge stored on she eafih as E7. = Q 2 : ww—ELW—E =1eev-m'3 47mm 47r(8.85 >< 10‘12)(6.371 x 106) from which we find Q 2 4.52 x 105 C. The energy stored on the earth can be found as 4. 105 2 %% 21.44 x1014 J N (e) The maximum cherge shes: can be stored on the earth is determined by {he dieieetric hreakdewn eieessie fiehi Egg 2 3 whim”; Qsieg this mine? we find 69%}: m Lgfi >< mm C 5:: Ti ‘ he“ i _ @137. Ffiehes charge. We zest the vahdiEy 0f the different eefientiei hmcéiens by requiiing that; 2 82 32 a? . VQ): EEJrETyfiQ ®=0 for |yl>0(1-e-,p=0) We have V2<D1 = e“?! cosh a: + cosh :1: 6‘11 = 2 cosh 338"” # G We = ~2sin(2m)ev‘/§ + sin(2m)e‘y‘/§ = — sin(2m)e”y‘/§ ate V2634 = ~3sinmsinysinz #0 so that @19 @3 and (D4 ase not valid solutions. We now consider (Dz: V2®2 = -— cos we‘ll + 6‘?! cos ac = 0 —> valid solution {is} The ceisesgiemiiiig eieeiiic fieiii is given by Ezm, 3,1,2) = —V®2 = (e‘y sin ms + (e‘y cos x)? for y > 0 By symmetry we must heve E2(1:, ~y, z) = ~E2(m, y; 2). Also, siece $2 and E2 dc set depeiici on 2, the surface charge density p3 aiso does moi depend on 2. Thus, we moss have ,05 = ps(:1:). To find p3{$), consider a Gaussian surface as shown and let Ay -> 0. Using Qauss’s law, we have / 60E; 'd51 4-; 60312 ~ (132 =f deA S] 32 A whei'e we can igmi'e the other sides of the cube since we ici Ag —+ 0. Since dsl = —ds2, and E205, y) = -E2(m, y) we have If EGEZ ’ 015; = f 6032 - dsz S; 32 and thus / p3(x)dxdz = 260 Egycc, y = (3)63de A 3: {p365} — 260 cos fldxdz = 0 (since Egym, y z 0) :2 cos a: 31 —> psm) = 260 coszc Fig. 4.133 Figure fee Pmbiem $38., enggi Psmiiefi eewee iiiies. (a) The power iii‘ies A see B constiime e Ewe-wire iiiie such es ihei ceii— sidei‘ed in Example 4-29, iiie capacitance pep;- uiiii length of which is given by C = wco/ him]. /a). Thus she charge induced per unit iengih of the Rimes is given by @Afl’ifég __ {449 V)(7reg} __________ N _ _1 Mali/a} " ism iii/(0,92 m); - 3-13 W m Pl: Since a is much smalier iiieii eli the other dimensions invoiveei, we can treat the iiiie charge densities as ii" they are at the center of the power iines. The total eiectiie fieid at point i is the supeigoosition of that due to lines A and B, nameiy Noting that and for the given values ofh = 60 cm, (31 = 1 m, and d2 = 15 em, we haVe R — A 2 73.5 cm and R3 2 831 cm. Substituting in values we find $1 213.56)? + 91y V—m—i Due to symmetry, the fieid at point 2 must be of the same magnitude but with an :3 component of opposite sign, i.e., e2 2». ~13.5672+9i§ V—m“‘ (b) It is convenient to take the origin 0 as the reference point of zero potential. We can use the result of Example 4—8, which specifies the difference in potential between any two points at two different radial distances from a line charge. According to thisi the potential of point i is _ _ Pl (til/2) _* Pl (all/2) @1—®A+®B—27reoln[ RA] 27r€om[ RB due to line A due to line B and similarly fOi‘ point 2 we have o12=e2—e1=~fl—21n fl - :11 n 5‘3 2:43.77 V 27750 RE Note that it tneltes sense that the potential at. point i is negative, shine noint l is inst-her ewnv teeth line A than the inference point (in, the origin), at which the potential is taken to he seen. 453%. Field nnner high-voltage line: (a) From Example 4-29, the peak electric field at the ground level {which we assnme to he tint and petfectly conducting) due to a line charge of charge density p; at a height of h z: 12 m can be written as Pl E z m neoh where a: is the direction which is perpendicular to the ground plane. To find Em, we need the charge density pg which can he expressed in terms of the voltage % of the high—voltage with respect to ground potential as 27m) 2h : 0 ~ V = Pl Single 0 in E where we used £4.50] for the per unit length capacitance of the single wire transmission line of wire radius a < h. Assuming a wire tatlins ot‘ a = 25 cm and using V0 2 765 lcV and h = 12 in yields ,0; 2 9.32 nC from which the peak electric tield is evaluated as E 2 27.9 kV—m‘l. (h) Assuming the electric field to stay uniform over the height of the person, the peak. potential difference betWeen the head and the feet of the 6 ft tall person can approximately be calculated as e = Emhp 2 (27.9)(6 x 0.3%) 2 51.1 iv (c) The flourescent bulb requires 11% V-ft”1 in order to be ignited, which corresponds to ~33€l hint—i a much smaller than ’27};3 iiV—t’i‘twi, Thais, the hnlh will definitely he ignited. satin 'l‘hnnnereionti fields (a) The capacitance of the thundercloud with it) in 2 vertical (listence of 5 ion is given by n arm swammd by 3: 6014 (8.85 x l0"12)(10 x 106) C = ~—--— 2 M N d 5 x 103 _ 17.7 11F Using the capacitance value along with the charge Q - 300 C storeci in the n _ — 1 t capacitor, we find the stored electrostatic energy to he pper p a 6 Of the 1 12 1 3002 We=—CV2=—Q—~ -—~(—)——:2.54x10121 2 20 —§17.7><10'9 (h) The potential difference hetween the top and the bottom plates can he found as 300 N l6.9 GV V —— ___ l7.7 x to"9 QiN 0.. (c) The average electric field within the cloud can be calculated as V ~ ins x We E:—_ d x te3 : 3.39 MV-rn‘l this value is higher than the dielectric breakdown field oi dry air which is 3 MV—in‘l. ...
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Homework_5_Solutions - M"" we" c I $78M t a,v i...

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