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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 1012 ,
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 iii3 x 5 >< remt 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
~gDds=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 mlzme + 5.4)(5 x 102) 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 102) 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 =1eevm'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(1e,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 Ewewire iiiie such es ihei ceii—
sidei‘ed in Example 429, 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);  313 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 insther ewnv teeth
line A than the inference point (in, the origin), at which the potential is taken to he seen. 453%. Field nnner highvoltage line: (a) From Example 429, 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% Vft”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 MVrn‘l this value is higher than the dielectric breakdown field oi dry air which is 3 MV—in‘l. ...
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 Summer '08
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 Electrostatics, Electric charge, charge density, totai gate capacitance, charge density p3

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