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Unformatted text preview: (4»: J04 FHN “(a bw Let V = 10(p + 1).:2 cos¢V in free space. a) Let the equipotential surface V = 20 V deﬁne a conductor surface. Find the equation of
the conductor surface: Set the given potential function equal to 20, to ﬁnd: (p+ 1).:2 c0545 = 2 b) Find p and E at that point on the conductor surface where 915 = 0.271 and z = 1.5: At the given values of {15 and 2, we solve the equation of the surface found in part a. for p,
obtaining p = ._1_Q. Then 8V 1 8V 8V E=—VV=—— __ ___,
6p ap p 345 alt 62 a'
2 ‘ p+1 9 .
= ~102 cos¢ap+10 z“ smgba¢—2O(p+1)zcos¢az p Then
E(.10, .27r, 1.5) = —18.2 ap + 145 21¢ — 26.7az'V/m c) Find Ipsl at that point: Since E is at the perfectly—conducting surface, it will be normal
to the surface, so we may write: E ' E ... 
p5 = EOEn = so =em/EE : coy/(18.2? + (145)2 + 26.7 2 = 1.32 110 1112
IE! ,  —— surface (2) Given the potential ﬁeld V = 100:1:z/ (:1:2 44 4) V. in free space:
a) FindD at thesurface z = 0: Use ‘ . 8 :1; V 100:1: At 2 = 0', we use this to ﬁnd D(z ; 0) = eoE(z = 0) = —1006c;a:/(a:2 +4) a: C/mz. b) Show that the'z '= 0 surface is an equipotential surface: There are two reasons for this:
1) E at z = O is everywhere zdirected, and so moving a charge around on the surface
involves doing no "work; 2) When eValuatiug the given potential function at z = 0, the
resultisOforallxandy._' ' c) Assume that the z = Ousurface is a conductor and ﬁnd the total charge on that portion
of the conductor deﬁned by 0 < a: < 2, 3—< y < 0: We have ' _100€oa: 
z=0 $2 + 4 C/m2 7 so: §‘;V/°I/22.'15°€5%a ¥‘§('3)(i00)e lilin.A(>2+4i)j2_I;15061112—4092110
730 z?+_4 1" of ° *"—, .  Atomic hydrogen contains 5.5 X 1025 atoms/m3 at a certain temperature and pressure. When
an electric ﬁeld of 4 kV/m is applied, each dipole formed by the electron and positive nucleus
has an eﬁ'ective length of 7.1 X 10‘19 m. :1) Find P: With all identical dipoles, we have P = qu = (5.5 x 1025x1602 >< 1019)(7.1 x 1019) = 6.26 x 10—12 C/rn3 = 6.26 pC/m’’ '0) Find 51.: We use P = eoxgE, and so P 6.26 x 1013 4
= ._ = —_—.—__, = 1.76 x 10
X5 60E (8.85 x 1012)(4 x 10°) Then 5, = 1+ Xe = 1.000176. A coazdal conductor has radii a. = 0.8 mm and b = 3 mm and a. polystyrene dielectric for . which a, = 2.56.,If P = (2/p)ap nC/m2 in the dielectric, ﬁnd: a) D and E as ﬁmctions of p: Use _ P __ (2/p) x 109ap 144.9
_ eo(e,  1) _ (8.85 x 10—12)(1.56) _ p a” Wm
Then
—9 9 r)
D=60E+P=2><10 ap 1 +1 =3.28x10 a,J C/mz=3..8ap nC/m2
p 1.56 p p b) Find Vab and Xe: Use
0.8 I
144.9 ‘ 3
V;=— ———d =144.91n —— =12V
b A p p (0.8) 9
Xe = e, — 1 = 1.56, as found in part a. c) If there are 4 x 1019 molecules per cubic meter in the dielectric, ﬁnd p(p): Use (2 x 109/p) = 5.0 x 1029 4x1019 a” Tape“ The surface 2: = separates two perfect dieleCtrics. For J: > 0, let 6,. = an = 3, While 613 = 5
where r < 0. If E; = 80ac — 6‘an ~ 30a; V/m, ﬁnd:
a) ENL: This will be E1  az = 80 V/m. b) Em. This has components of IE; not normal to the surface, or Em = —60a4, — 30a: V/m.
“m c) En = V’(60)2 + (30‘2 = 67.1V/m_. d) E1 = «/(80)2 + (60)2 + (30)2 = 104.4V/m. e) The angle 01 between E1 and a normal to the surface: Use Elax 80 m5 ‘91 = E1 104.4 ‘ f) DNg = DNI = ErléoENl = X 10~12)(80) = 2.12nC/m2.
g) 1312 = ergeoETl = 5(8.85 x 10‘12)(67.1) = 9.97nC/m2.
h) D3 = erleoEmax + ErgfoETl = 2.123a — 2.66% — 1.33az nC/m2. P2 = D2 — EQEQ = D2 [1 — (ll/67.2)] = = — —1.068.z nC/mz. j) the angle 02 between E2 and a normal to the surface: Use Egax=D2az E2 D2 COS 62 = = Thus 92 = cos1(.581) = 54.5°. Two perfect dielectrics have relative permittivities 5,.1 = 2 and 672 = 8. The planar interface
between them is the surface r—y+22 = 5. The origin lies in region 1. If E1 = 100a,, +200ay ——
50az V/m, ﬁnd E2: We need to ﬁnd the components of E1 that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. The normal component
will be E’Nl = E1  n. Taking f = :1: — y + 22, the unit vector that is normal to the surface is This normal will point in the direction of. increasing f, which will be away from the origin, or.
into region 2 (you can visualize a portion of the surface as a. triangle whose vertices are on the
three coordinate axes at :1: = 5, y = —5, and z = 2.5). So Em = [100 — 200 — 100] = —81.7V/m. Since the magnitude is negative, the normal component points into region 1 from
the surface. Then EM = —81.65 (%),[a¢ — a1, + 2az] = —33.33a¢ + 33.33ay — 66.67az V/m Now, the tangential component will be En = E1 — ENl = 133.3az + 166.7ay + 16.67az. Our
boundary conditions state that En = Em and EN2 = (erl/e,2)EN1 = (1/4)EN1. Thus 1
E2 = ETz + EN2 = ET1 + zENl = 133.3% + 166.7% +16.67az — 8.3% + 8.3% — 16.673z
= 125ac + 175a.y'V/m ‘ \ Let V(2:, y) = 462‘” + f(a:) — By2 in a region of free space where pu = 0. It is lmown that both
Ex and V are zero at the origin. Find f (1:) and V(x, y): Since p1, = 0, we know that V2V = 0, and 5° 32V 62V 6’ f
2 __ __ = 22 _:_ _ =
V V — —a$2 + ayz 166 + dmz 6 0
Therefore d2 d
Eg=—1652”+6 ¢ 3%: 8622+6$+01 Now 6V d Ex = — :: 8522 + —f 3 d2; and at the origin, this becomes E40) =s+ d—f dz; i=0 = 0(as given) Thus df/da: [3:0 = ~8, and so it follows that 01 = 0. Integrating again, we ﬁnd f(a:, y) =' —4e2m + 3:1;2 + Cg which at the origin becomes f(0,0) = —4+ C2. However,_ V(0,0) = 0 = 4 + f(0,0). So
f(O, 0) = —4 and 02 = 0. Finally, f(:27,y) = 4162” + 32:2, and V(m, y) = 452” — 4432‘” + 3:1;2 —
3312 = 3(22 — ' Given the potential ﬁeld V = (Ap4 + B p“) sin 4gb:
a.) Show that VQV = O: In cylindrical coordinates, q 1 3 8V) 1 62V
V‘V = —— — + ——,—
p3p< 39 P? 3¢2 >
1 8 _ . 1  .
= (p(4Ap3 — 4Bp 5)) sm4¢ — F16(Ap4 + 3,0 4)Sm4¢
= 16(Ap3 + Bp's) sin4q§ — 1—S(Ap4 + Bf“) sin4q5 = 0 b) Select A and B so that V = 100 V and El = 500 V/m at P(p = 1,¢ = 225°; 2 = 2): ’1 First, E=—VV=2Kap—lﬂa¢ 6p . p 8¢ .
= 4 [(Ap3 — Bp'°) sin 445 an + (Ap3 + Bp's) cos 4915 ad
and at P, E; = —4(A —— B) a,,. Thus Ep = i4(A — B). Also, Vp = A+ B. Our two
equations are: 1‘ 4(A _— B) = i500 and
A+B =100 We thus have two pairs of values for A and B: A = 112.5, B = —12.5' or A = 12.5, B = 112.5 a“ Let V = (cos 2gb) / p in ﬁres space. V _
a) Find the volume charge density at point 'A(0.5, 60—",1): Use Poisson’s equation: _; 2 _:_ 13 Eli  i‘ﬂ
p”_ E°VV~ 60(pap pap +p23¢2
(1 8 <~cos2¢> 4 cosZcp) 360c0s2q’9
=—Eo — —— —— 53p p 92 9 113
So at A we ﬁnd: 3 1900
m = M = —l2eo = —106pC/m3
0.53 ——'
b) Find the surface charge density on a conductor surface passing through E (2, 30°, 1): First,
we ﬁnd E: 8V 1 6V
E: —VV =—a—pap _ cosZgb a + 2sin2¢ a,
92 p P2 4' At point B the ﬁeld becomes cos 60° a + 25in 60°
4 p 4 The surface charge density will now be E3 = a4, = 0.125 a,."+ 0.433 as ' m = $11331 = WEB] = $045150 = $0.399pC/m2 The charge is positive or negative depending on which side of the surface we are consid
ering. The problem did not provide information necessary to determine this. Coaxial conducting cylinders are located at p 0.5 cm and p = 1.2 cm. The region between
the cylinders is ﬁlled with a homogeneous perfect dielectric. If the inner cylinder is at 100V
and the outer at 0V, ﬁnd: a) the location of the 20V equipotential surface: From Eq. (16) we have 1n(012/p)
_» ln(.012/.OO_5)
We seek p at which V: 20 V, and thus we need to solve: V(p) = 100 V ln(.012/p) _ .012
ln(2.4) => p 20 = 100 b) Ep mm: We have
3V W 100 Ep=———_——= 8p dp p ln(2.4) « whose maximum value will occur at the inner cylinder, or at p = .5 cm:
4 . 100 =____=,2 04v =22.8kV
Em“ .005m(2.4) 2_8x.1 /m J c) eT if the charge per meter length on the inner cylinder is 20 nC/m: The capacitance per meter length is
216061 Q
C = m(2.4) ‘ vb We solve for 67: 
 (20 x 109)1n(2.4)
= — = 3.15
21Ij_eo'(100) I 1y Concentric conducting spheres are located at 7' = 5 mm and r = 20 mm. The reoicn between
D the s heres ' ﬁlled ‘   ‘  .
Spherlz at 0 ‘1: With a. perfect dielectric. If tne inner sphere is at 100 V and the outer 3.) Find the location of the 20 V equipotential Surface: Solving Laplace’s equation gives us V(r) = V0 where V0 = 100, a = 5 and b = 20. Setting V(7‘) = 20, and solving for 1' produces
r = 12.5 mm. b) Find Emma: Use dV VOar
E = VV = —— = ——
W m e
V
Em“ = E(r = a) = 0 — ——109——— = 26.7 V/nun = 26.7 kV/m a(1  (0/5)) _ 5(1 — (5/20)) c) Find er if the surface charge density on the inner sphere is 1.0 pC/m2: p, will be equal
in magnitude to the electric ﬂux density at r = «1. So ,05 = (2.67 X 104 V/m)ereo =
10‘6 0/1112. Thus er = 11;; ' We want in design a spherical vacuum capacitor wiih a given radius a for the outer
sphere. wnicn wul he able in sicre the greatest amount of electrical energy subieci in
the constraint that the eiecingﬁeid strength at the surface of ihe inner sphere may not exceed Ea. What radius 33' ehouid be chosen icr iheinner spherical csnductcr. and? how mucﬁjﬁggg can be stored? .. I __ _ ‘_ A __._—... . 5 Pique" I»: seesaw: swig“
_ = m" = 4Wé° “’5 ~_ “WotanL _,___jnenls___ 4' °~ ‘L‘D cvka~ 1—11 a :3 __,\_.._: _. “QM.. __ _._..__... _ ._ ._..__ _ __~ _ ... __ .— e = a . \3 e a e a.
_ <<<< °~ _¥:T________.________  _._ ____._____ _
if“: 2 Q (J'Reb  <7 _ . _ __ ___ ._ M _._.__._ .. _€____.___,E=E. .L.Hﬁ_e6¥’:_ezlﬁj_ _ ﬂF—ﬂwu ' MN""_~—ﬁE§'Z:\—}e5— swk.) k5 _
u a I _. _ Q ~ “I j k —. ‘IIH ﬁlm—1
I 01H 01H ' I ' The potential diﬁ‘erence AV bemeenfthem
plates of a spherical capacitor iser conchan’r~ E
Show that then the elecen'e ﬁeld'at'the surface of ‘.
the inner sphere will be a minimum if a = 1515. I _ Find this minimum value of E. b
a Q dz: Q [l_}_]
OV(r) " LMEO? 41mm?» 1: V
l 1} = 4! —
Q L!“ 43L; I V . a.  h I ;
Smeea<0 for a<i, andgg>o fora);
I' L‘K—E/iindﬁfcaleaoa—citor has outer and inner conduc tors whoseradii are in the ratio of b/a = 4/1; The?
inner conductor is to be replaced by a wire whosezi
radius is onehalf the radius of the original inneri;
conductor. By what factor should the length be creased in order to obtain a eapacitance equal to that ' of the original capacitor? ' V3.0: lb wherea:=% Ili=_b_=§
, 2kln(;) a: ¢1/2 6:
__. ————l—;'”' “ " ““l2_ ,.__ “Hm—~N‘
2kC = = ——
' 1n(;‘; 111%) . 1
1, _ meg) _ ln(2 5;)_ 1n2+1n(;¢;) _ 1.1 _ ‘ Emu"; “175? he,» ' 1am ‘ _ .——___._"=_'1'.WL. ._._.. . i. s :. i I.
i . E . E
l
! an » USS @
Currentcarrying components in highvoltage power equipment must be cooled to carry away the heat caused by ohmic losses. A means of pumping is based on the force n‘ansmit ‘
ted to the cooling ﬂuid by charges in an electric ﬁeld. The electrohydrodynamic (El1D) pumping is modeled in Figure 6. The region between the electrodes contains a uniform charge p0, which is generated at the left electrode and collected at the right electrode. Cal culate the pressure of the pump if p0 = 25 rnC/m3 and V0 = 22 kV. Figure 6 An electrohydrodynarnic pump; Area S Since pv .—’ O, we apply Poisson’s equation '~ ' ' Viv = ——
8 The boundary conditions V(., = 0) = V0 and V(z = d) = 0 show that Vdepends only on 3
(there is no p or (b dependence). Hence sz _ p°
d:2 8
Integrating once gives
dV poz
_ = ' A
dz 8 T
Integrating again yields ,
,2
= —£9”— + A: + B
a 28 where A and B are integration constants to be determined by applying the boundary condi—
tions. When z = O, V = V0, Vo=—O+O+B—>B=Vo
Whenz=d,V=0, 2
0——p°d +Ad+vo
28 '
01‘
_&é_ﬁ
A— s d The electric ﬁeld is given by The net force is d .
F=fvadv=p°JdSJ Edz _ VOZLE 2— :ld .
—poS[d :28(z dz) Dag
F=PaSVoaz The force per unit area or pressure is p = g = pOVO = 25 x 103 x 22 x 103 =_ 550 N/m2 _ I _ . ,. ‘h 4””.  '5.“ ' ‘1 ‘ ‘ _ V ._ _ M W“ Whig“? l .. . ‘ .. . .~ ., ._. _. _.._ __  _ ___. .:’N_'.'.___ em / Farce reguired +a move +his‘ Q T@ charge upward =35Q2/(2xf
' ‘r (’3 '* .
h l T The second studen‘i‘
WW7 77777777777777? CGICUICH'ES C15 ‘FOHOWS:
I .
i I 00 1' . z
I , _ _ KG a 3.,
l H...) I] WOFK —f(2x)2  Inmal state ." h ' .a This is the correct answer.
No‘tetho'i' iF two real charges Q and Q were being
pulled apart symmetrically, the tow work done "
would beKQZ/Zh, but the agency moving Q wo'ulci
Supply only halF 0F i+. , Given the current density J = —10“[si.n(2z:)e‘f"yam + cos(2:t)e"2yay] kA/m2: 3.) Find the total current crossing the plane y = 1 in the a.” direction in the region 0 < :L' < 1,
0 < z < 2: This is found through 2 1 2 1
I=//Jnl daé/ / Jay, dmdz=f / 104cos(2z)'e_2da:clz
5' S o 0 Fl 0 0 
. 1 = 404(2); sin(2a:)loe'2 = —1.23 MA b) Find the total current leaving the region 0 < 1:, a: < 1, 2 < z < 3 by integrating J ~dS over
the surface of the cube: Note ﬁrst that current through the top and bottom surfaces will
not exist, since J has no 2 component. Also note that there Will be no current through the m = 0 plane, since Jz = 0 there. Current will pass through the three remaining surfaces,
and Will be found through I=/23/(;1J.(—ay)ly=oda;dz+A3A1J(ay)ly=ldxdz+A3AIJ(ae) 3 1 3 1
= 104/ / [cos(2a;)e"° — cos(2:r)e_2] da; dz — 104/ / sin(2)e"2y dy dz
2 o 2 o . = 104 sin(2:z:)i:(3 — 2) [1 — eE] + 104 (l) amen214 z=1 dy dz 1
0(32)=Q 2 c) Repeat part b, but use the divergence theorem: We ﬁnd the net outward current through
the surface of the cube by integrating the divergence of J over the cube volume. We have
' _ 6.7., 8],, . __ ___.___ ‘4 2'y__2’ —Zy =
V J aw + By 10 [2cos(2a:)e cos(22)e ] _0_ asexpected ‘5 .Lqu _ .
v I J=4ogsmearA/mg @
7' +4 ' ' a) Find the total current ﬂowing through that portion of the spherical surface 7' = 0.8,
bounded by 0.17r < 9 < 0371', 0 < 43 < 2w: This will be 2" 3" 40051119 2 4OO(.8)227r f3" . 9 t
= . = .s sm9d9d¢=—— sm‘dﬁ
I //J H‘s da ./0 [in (3)‘+4( 464 .l7r .31r 
= 346.5 f g [1 — cos(26‘)] d9 = 77.4..x
.lvr'  b) Find the average value of J over the deﬁned area. The areais Zn .311' 9
Area = f (.8)2 sin 9 d6 dqb = 1.46 m‘
0 .l1r The average current density is thus J Mg = (774/146) 21., = 53.0 a, A/mz. (19' Let '2. 20 J=— — .
pa” p2+0.01 a, A/mz a) Find the total current crossing the plane 2 = 0.2 in the az direction for p < 0.4: Use
21r .4 I
—20
I: J. d = . d d
. nz=.2 a A ./o'P2+.01p p 45
1 _ _ (E) 201n(.01 + p2)l:(27r) = —207rln(17) = l78.0A b) Calculate apu/Bt: This is found using the equation of continuity: 3”” = —VJ = 1.3.03.5) + 6‘72. = 13(25) ' a < ‘20 )=Q c) Find the outward current crossing the closed surface deﬁned by p = 0.01, p = 0.4, 2 = 0,
andz=0.2: Thiswillbe ~  .2 271' '2 ’ 2'” I=/ / —ap  (—a..><01)d¢dz+ / —a,.  (amem
0 0 .01  o o .4 2" '4 20 d d 27r .4 _20
+_/c; A p2+_01az‘(—az)p p ‘/O‘ maz'(az)Pdpd¢;Q since the integrals will cancel each other. d) Show that the divergence theorem is satisﬁed for J
' In part c, the net outward ﬂux was found to be zero was found to be zero (as will be its volume integral).
is satisﬁed. and the surface speciﬁed in part b.
, and in part b, the divergence ofJ
Therefore, the divergence theorem Then the electric ﬁeld is found by dividing this result by 0: ‘ 3 9.55
— 27ra’pl a” — 7a” V/m The voltage between cylinders is now: 3' a — — .
' 9.3:) 9.5:; a 4.88 7apapdp—Tln(§>—TV Now, the resistance will be b) Show that integrating the power disSipeted unit volume over the volume gives the
total dissipated power: We calculate 4. .4
P=VI=§(3)=—146 W ‘ l l which is in agreement with the power' density integration. W ...
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This note was uploaded on 05/26/2008 for the course ECE 302 taught by Professor Ferguson during the Spring '08 term at Cal Poly Pomona.
 Spring '08
 Ferguson
 Electromagnet

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