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Unformatted text preview: (4Ā»: J04 FHN ā(a bw Let V = 10(p + 1).:2 cosĀ¢V in free space. a) Let the equipotential surface V = 20 V deļ¬ne a conductor surface. Find the equation of
the conductor surface: Set the given potential function equal to 20, to ļ¬nd: (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=āVV=āā __ ___,
6p ap p 345 alt 62 a'
2 ā p+1 9 .
= ~102 cosĀ¢ap+10 zā smgbaĀ¢ā2O(p+1)zcosĀ¢az p Then
E(.10, .27r, 1.5) = ā18.2 ap + 145 21Ā¢ ā 26.7az'V/m c) Find Ipsl at that point: Since E is at the perfectlyāconducting 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! ,  āā surface (2) Given the potential ļ¬eld V = 100:1:z/ (:1:2 44 4) V. in free space:
a) FindD at thesurface z = 0: Use ā . 8 :1; V 100:1: At 2 = 0', we use this to ļ¬nd D(z ; 0) = eoE(z = 0) = ā1006c;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 ļ¬nd the total charge on that portion
of the conductor deļ¬ned by 0 < a: < 2, 3ā< y < 0: We have ' _100ā¬oa: 
z=0 $2 + 4 C/m2 7 so: Ā§ā;V/Ā°I/22.'15Ā°ā¬5%a Ā„āĀ§('3)(i00)e lilin.A(>2+4i)j2_I;15061112ā4092110
730 z?+_4 1" of Ā° *"ā, .  Atomic hydrogen contains 5.5 X 1025 atoms/m3 at a certain temperature and pressure. When
an electric ļ¬eld of 4 kV/m is applied, each dipole formed by the electron and positive nucleus
has an eļ¬'ective length of 7.1 X 10ā19 m. :1) Find P: With all identical dipoles, we have P = qu = (5.5 x 1025x1602 >< 1019)(7.1 x 1019) = 6.26 x 10ā12 C/rn3 = 6.26 pC/māā '0) Find 51.: We use P = eoxgE, and so P 6.26 x 1013 4
= ._ = ā_ā.ā__, = 1.76 x 10
X5 60E (8.85 x 1012)(4 x 10Ā°) 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, ļ¬nd: a) D and E as ļ¬mctions of p: Use _ P __ (2/p) x 109ap 144.9
_ eo(e,  1) _ (8.85 x 10ā12)(1.56) _ p aā Wm
Then
ā9 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 ā 3
V;=ā āāād =144.91n āā =12V
b A p p (0.8) 9
Xe = e, ā 1 = 1.56, as found in part a. c) If there are 4 x 1019 molecules per cubic meter in the dielectric, ļ¬nd p(p): Use (2 x 109/p) = 5.0 x 1029 4x1019 aā Tapeā The surface 2: = separates two perfect dieleCtrics. For J: > 0, let 6,. = an = 3, While 613 = 5
where r < 0. If E; = 80ac ā 6āan ~ 30a; V/m, ļ¬nd:
a) ENL: This will be E1  az = 80 V/m. b) Em. This has components of IE; not normal to the surface, or Em = ā60a4, ā 30a: V/m.
ām c) En = Vā(60)2 + (30ā2 = 67.1V/m_. d) E1 = Ā«/(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 ā91 = E1 104.4 ā f) DNg = DNI = ErlĆ©oENl = X 10~12)(80) = 2.12nC/m2.
g) 1312 = ergeoETl = 5(8.85 x 10ā12)(67.1) = 9.97nC/m2.
h) D3 = erleoEmax + ErgfoETl = 2.123a ā 2.66% ā 1.33az nC/m2. P2 = D2 ā EQEQ = D2 [1 ā (ll/67.2)] = = ā ā1.068.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) = 54.5Ā°. Two perfect dielectrics have relative permittivities 5,.1 = 2 and 672 = 8. The planar interface
between them is the surface rāy+22 = 5. The origin lies in region 1. If E1 = 100a,, +200ay āā
50az V/m, ļ¬nd E2: We need to ļ¬nd the components of E1 that are normal and tangent to the boundary, and then apply the appropriate boundary conditions. The normal component
will be EāNl = E1  n. Taking f = :1: ā 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 = ā5, and z = 2.5). So Em = [100 ā 200 ā 100] = ā81.7V/m. Since the magnitude is negative, the normal component points into region 1 from
the surface. Then EM = ā81.65 (%),[aĀ¢ ā a1, + 2az] = ā33.33aĀ¢ + 33.33ay ā 66.67az V/m Now, the tangential component will be En = E1 ā 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 ā 8.3% + 8.3% ā 16.673z
= 125ac + 175a.y'V/m ā \ Let V(2:, y) = 462āā + f(a:) ā 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 5Ā° 32V 62V 6ā f
2 __ __ = 22 _:_ _ =
V V ā āa$2 + ayz 166 + dmz 6 0
Therefore d2 d
Eg=ā1652ā+6 Ā¢ 3%: 8622+6$+01 Now 6V d Ex = ā :: 8522 + āf 3 d2; and at the origin, this becomes E40) =s+ dāf dz; i=0 = 0(as given) Thus df/da: [3:0 = ~8, and so it follows that 01 = 0. Integrating again, we ļ¬nd f(a:, y) =' ā4e2m + 3:1;2 + Cg which at the origin becomes f(0,0) = ā4+ C2. However,_ V(0,0) = 0 = 4 + f(0,0). So
f(O, 0) = ā4 and 02 = 0. Finally, f(:27,y) = 4162ā + 32:2, and V(m, y) = 452ā ā 4432āā + 3:1;2 ā
3312 = 3(22 ā ' Given the potential ļ¬eld V = (Ap4 + B pā) sin 4gb:
a.) Show that VQV = O: In cylindrical coordinates, q 1 3 8V) 1 62V
VāV = āā ā + āā,ā
p3p< 39 P? 3Ā¢2 >
1 8 _ . 1  .
= (p(4Ap3 ā 4Bp 5)) sm4Ā¢ ā F16(Ap4 + 3,0 4)Sm4Ā¢
= 16(Ap3 + Bp's) sin4qĀ§ ā 1āS(Ap4 + Bfā) sin4q5 = 0 b) Select A and B so that V = 100 V and El = 500 V/m at P(p = 1,Ā¢ = 225Ā°; 2 = 2): ā1 First, E=āVV=2Kapālļ¬aĀ¢ 6p . p 8Ā¢ .
= 4 [(Ap3 ā Bp'Ā°) sin 445 an + (Ap3 + Bp's) cos 4915 ad
and at P, E; = ā4(A āā B) a,,. Thus Ep = i4(A ā B). Also, Vp = A+ B. Our two
equations are: 1ā 4(A _ā B) = i500 and
A+B =100 We thus have two pairs of values for A and B: A = 112.5, B = ā12.5' or A = 12.5, B = 112.5 aā Let V = (cos 2gb) / p in ļ¬res space. V _
a) Find the volume charge density at point 'A(0.5, 60ā",1): Use Poissonās equation: _; 2 _:_ 13 Eli  iāļ¬
pā_ EĀ°VV~ 60(pap pap +p23Ā¢2
(1 8 <~cos2Ā¢> 4 cosZcp) 360c0s2qā9
=āEo ā āā āā 53p p 92 9 113
So at A we ļ¬nd: 3 1900
m = M = āl2eo = ā106pC/m3
0.53 āā'
b) Find the surface charge density on a conductor surface passing through E (2, 30Ā°, 1): First,
we ļ¬nd E: 8V 1 6V
E: āVV =āaāpap _ cosZgb a + 2sin2Ā¢ a,
92 p P2 4' At point B the ļ¬eld becomes cos 60Ā° a + 25in 60Ā°
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 ļ¬lled with a homogeneous perfect dielectric. If the inner cylinder is at 100V
and the outer at 0V, ļ¬nd: a) the location of the 20V equipotential surface: From Eq. (16) we have 1n(012/p)
_Ā» 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=āāā_āā= 8p dp p ln(2.4) Ā« whose maximum value will occur at the inner cylinder, or at p = .5 cm:
4 . 100 =____=,2 04v =22.8kV
Emā .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) ā vb We solve for 67: 
 (20 x 109)1n(2.4)
= ā = 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 ' ļ¬lled ā   ā  .
Spherlz at 0 ā1: 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 Laplaceās equation gives us V(r) = V0 where V0 = 100, a = 5 and b = 20. Setting V(7ā) = 20, and solving for 1' produces
r = 12.5 mm. b) Find Emma: Use dV VOar
E = VV = āā = āā
W m e
V
Emā = E(r = a) = 0 ā āā109āāā = 26.7 V/nun = 26.7 kV/m a(1  (0/5)) _ 5(1 ā (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 ļ¬ux density at r = Ā«1. So ,05 = (2.67 X 104 V/m)ereo =
10ā6 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 eiecingļ¬eid strength at the surface of ihe inner sphere may not exceed Ea. What radius 33' ehouid be chosen icr iheinner spherical csnductcr. and? how mucļ¬jļ¬ggg can be stored? .. I __ _ ā_ A __._ā... . 5 Pique" IĀ»: seesaw: swigā
_ = m" = 4WĆ©Ā° āā5 ~_ āWotanL _,___jnenls___ 4' Ā°~ āLāD cvka~ 1ā11 a :3 __,\_.._: _. āQM.. __ _._..__... _ ._ ._..__ _ __~ _ ... __ .ā e = a . \3 e a e a.
_ <<<< Ā°~ _Ā„:T________.________  _._ ____._____ _
ifā: 2 Q (J'Reb  <7 _ . _ __ ___ ._ M _._.__._ .. _ā¬____.___,E=E. .L.Hļ¬_e6Ā„ā:_ezlļ¬j_ _ ļ¬Fāļ¬wu ' MN""_~āļ¬EĀ§'Z:\ā}e5ā swk.) k5 _
u a I _. _ Q ~ āI j k ā. āIIH ļ¬lmā1
I 01H 01H ' I ' The potential diļ¬āerence AV bemeenfthem
plates of a spherical capacitor iser conchanār~ E
Show that then the elecen'e ļ¬eld'at'the surface of ā.
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?Ā» 1: V
l 1} = 4! ā
Q L!ā 43L; I V . a.  h I ;
Smeea<0 for a<i, andgg>o fora);
I' LāKāE/iindļ¬fcaleaoaācitor 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_=Ā§
, 2kln(;) a: Ā¢1/2 6:
__. āāāālā;'ā' ā " āāl2_ ,.__ āHmā~Nā
2kC = = āā
' 1n(;ā; 111%) . 1
1, _ meg) _ ln(2 5;)_ 1n2+1n(;Ā¢;) _ 1.1 _ ā Emu"; ā175? he,Ā» ' 1am ā _ .āā___._"=_'1'.WL. ._._.. . i. s :. i I.
i . E . E
l
! an Ā» 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 nāansmit ā
ted to the cooling ļ¬uid by charges in an electric ļ¬eld. 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 .āā O, we apply Poissonās equation '~ ' ' Viv = āā
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 _ pĀ°
d:2 8
Integrating once gives
dV poz
_ = ' A
dz 8 T
Integrating again yields ,
,2
= āĀ£9āā + A: + B
a 28 where A and B are integration constants to be determined by applying the boundary condiā
tions. When z = O, V = V0, Vo=āO+O+Bā>B=Vo
Whenz=d,V=0, 2
0āāpĀ°d +Ad+vo
28 '
01ā
_&Ć©_ļ¬
Aā s d The electric ļ¬eld is given by The net force is d .
F=fvadv=pĀ°JdSJ Edz _ VOZLE 2ā :ld .
āpoS[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 _ . ,. āh 4āā.  '5.ā ' ā1 ā ā _ V ._ _ M Wā Whigā? l .. . ā .. . .~ ., ._. _. _.._ __  _ ___. .:āN_'.'.___ em / Farce reguired +a move +hisā Q [email protected] charge upward =35Q2/(2xf
' ār (ā3 '* .
h l T The second studenāiā
WW7 77777777777777? CGICUICH'ES C15 āFOHOWS:
I .
i I 00 1' . z
I , _ _ KG a 3.,
l H...) I] WOFK āf(2x)2  Inmal state ." h ' .a This is the correct answer.
Noātetho'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 = ā10ā[si.n(2z:)eāf"yam + cos(2:t)e"2yay] kA/m2: 3.) Find the total current crossing the plane y = 1 in the a.ā direction in the region 0 < :L' < 1,
0 < z < 2: This is found through 2 1 2 1
I=//Jnl daĆ©/ / Jay, dmdz=f / 104cos(2z)'e_2da:clz
5' S o 0 Fl 0 0 
. 1 = 404(2); sin(2a:)loe'2 = ā1.23 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 ļ¬rst 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.(āay)ly=oda;dz+A3A1J(ay)ly=ldxdz+A3AIJ(ae) 3 1 3 1
= 104/ / [cos(2a;)e"Ā° ā cos(2:r)e_2] da; dz ā 104/ / sin(2)e"2y dy dz
2 o 2 o . = 104 sin(2:z:)i:(3 ā 2) [1 ā eE] + 104 (l) amen214 z=1 dy dz 1
0(32)=Q 2 c) Repeat part b, but use the divergence theorem: We ļ¬nd 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],, . __ ___.___ ā4 2'y__2ā āZy =
V J aw + By 10 [2cos(2a:)e cos(22)e ] _0_ asexpected ā5 .Lqu _ .
v I J=4ogsmearA/mg @
7' +4 ' ' a) Find the total current ļ¬owing 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 sm9d9dĀ¢=āā smādļ¬
I //J Hās da ./0 [in (3)ā+4( 464 .l7r .31r 
= 346.5 f g [1 ā cos(26ā)] d9 = 77.4..x
.lvr'  b) Find the average value of J over the deļ¬ned area. The areais Zn .311' 9
Area = f (.8)2 sin 9 d6 dqb = 1.46 mā
0 .l1r The average current density is thus J Mg = (774/146) 21., = 53.0 a, A/mz. (19' Let '2. 20 J=ā ā .
paā 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
ā20
I: J. d = . d d
. nz=.2 a A ./o'P2+.01p p 45
1 _ _ (E) 201n(.01 + p2)l:(27r) = ā207rln(17) = l78.0A b) Calculate apu/Bt: This is found using the equation of continuity: 3āā = āVJ = 1.3.03.5) + 6ā72. = 13(25) ' a < ā20 )=Q c) Find the outward current crossing the closed surface deļ¬ned by p = 0.01, p = 0.4, 2 = 0,
andz=0.2: Thiswillbe ~  .2 271' '2 ā 2'ā I=/ / āap  (āa..><01)dĀ¢dz+ / āa,.  (amem
0 0 .01  o o .4 2" '4 20 d d 27r .4 _20
+_/c; A p2+_01azā(āaz)p p ā/Oā maz'(az)PdpdĀ¢;Q since the integrals will cancel each other. d) Show that the divergence theorem is satisļ¬ed for J
' In part c, the net outward ļ¬ux was found to be zero was found to be zero (as will be its volume integral).
is satisļ¬ed. and the surface speciļ¬ed in part b.
, and in part b, the divergence ofJ
Therefore, the divergence theorem Then the electric ļ¬eld is found by dividing this result by 0: ā 3 9.55
ā 27raāpl aā ā 7aā V/m The voltage between cylinders is now: 3' a ā ā .
' 9.3:) 9.5:; a 4.88 7apapdpāTln(Ā§>āTV 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=Ā§(3)=ā146 W ā l l which is in agreement with the power' density integration. W ...
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 Spring '08
 Ferguson
 Electromagnet, Electric charge, Surface, charge density, current density, continuity equation

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