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Unformatted text preview: cauEQtuua 57%th hereafterastute numeesw‘e ears/ices _
 ﬂeaneat e. Cementee Engineering Sepertrnent
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I Let V = lO(p + 1):2 cos 45 V in free space. (a) Let the equipotentiul surface
V = 20 V deﬁne a conductor surface. Find the equation of the conductor
surface. (1)) Find p and E at that point on the conductor surface where qb :
O.er an a z 1.5. (c) Find lpgl at that point. Given the potential ﬁeld V = lOO.r;/(.r2 + 4) V in free space: ((1) Find D at
. the surface = 0. (b) Show that the : = 0 surface is an equipotential surface.
(c) Assume that the a = 0 surface is a conductor and ﬁnd the total charge on that portion of the conductor deﬁned by 0 < r < 2, ——3 < y < 0. ® Atomic hydrogen contains 5.5 x if)25 atoms/ni3at 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 effective length of 7.1 x 10‘” m.
((1) Find P. (b) Find 6,. ' ® A coaxial conductor has radii a = 0.8 mm and «£7 = 3 mm and a polystyrene
dielectric for which er 2 2.56. if P = (2/io)ap nC/m2 in the dielectric, ﬁnd:
(a) D and E as functions of p; ([2) Va], and Xe. (c) If there are 4 x 10'19
molecules per cubic meter in the dielectric. ﬁnd p(p). The surface .r = O separates two perfect dielectrics. For .1: > 0 let e,. 5.1 = 3, while 5,; = 5 where x < 0. If E1 = 80213. — 60a. ~ 303: Wm. ﬁnd: C l= rt ((0 EN}; ([1) ET}; (c) E;; (d) the angle 9, between E and a normal to the c; «,4 .9 surface; (e) DNZ; (f) 073; (g) 1);; (/1) P2; (1') the angle (92 between E; and a (9 $61 3 4’ h normal to the surface. 8 4, (e iw *5 t
G A t *5 Two perfect dielectrics have relative permittivities 6,, = 2 and 63 = 8. The
planar interface between them is the surface .r — y + 22 = 5. The origin lies in region 1. le; : lOOax + 2003), — 503E V/m, find 5;. (given the potential ﬁeld V : (Ap‘l + Bp—4)Sin 495; (a) Show that VZV _ O
(7') Select A and B so that V = 100 V and E : __ _ '
(ﬁr—22.59.32”. , l 500 V/m at P(p— i, Let V(r. y) = 463" + ftx) — 3323 in a regiOn of free space Where pp 2: 0. _It
is known that both EA. and V are zero at the origin. Find f(.t') and V(.r, 32). Let V = (cos 2q§)/,0 in free space. ((1) Find the volume charge density at
point AtOj. 60: . l). (b) Find the surface charge density on a conductor
v surface passing through the point 30.. 30“. i). Coaxial conducting cylinders are located at ,0 = 0.5 cm and p = l .2 cm. The region between the cylinders is filled with a homogeneous perfect dielectric.
If the inner cylinder is at lOO V and the outer at O V. ﬁnd: {(1) the location of
the 20 V equipotential surface; (1)) 15,, "m; (c) 6,. if the charge per meter
length on the inner cylinder is 20 nC/ni. Concentric conducting spheres arelocztted at r = 5 mm and'r = 20 mm. The
region between the spheres is ﬁlled with a perfect dielectric. If the inner sphere is at lOO V and the outer sphere is at O V: (a) Find the location of the 20 V equipotential surface. ([2) Find Erlnm. (c) Find 6,. if the surface charge
density on the inner sphere is 1.0 uC/mz. ' @ we went in design a smartest yarnsunthdatiacitdr with a siren fail‘55.? “liaiiiiﬁgﬁuiar‘
sphere. whiten wit! he atria tn atnra thegreatest amount or electrical energy $135933 in
the ennetratnt that the electrinﬁald strength at the euriaca’ at the inner sphere may not
exceed Ea. What radius h' ehnuid he. chosen. tar theinner spherical centimeter. and i
hora much energy can be stirred? The potential ﬁiﬁ'erence A's? between the A * L1i( V) l
. plates of a spherical capacitor isltept' constant: MN“ ‘ A b
Show that then the electric ﬁeld at the surface of i
the inner sphere will be a a = ﬁr.
Find this minimum value or” E. lg —— ————— 5 l3) A quindrteai. capaettnr has neuter artdinner sundael} anyway 3/1
tiara whose are in the ratio or? sir/n =é/l. 'l"rt'e'=i inner ennduetnr ' is to he replaced by assire entree; .
I radius is one—haﬁ the radius of the originalinner ennduetnr. E? teltat factor should the length he in—_ creased. in. order tn obtain snapanimus to that . n? the original capacitor? ‘ ' ' Currentcarrying components in highvoltage power equipment must be cooled to carry
away the heat caused by ohmic losses. A meansof pumping is based on the force transmit— ted to the cooling ﬂuid by charges in an electric ﬁeld. The electrohydrodynamic (EHD)
pumping is modeled in Figure 6.1. The region between the electrodes contains a uniform
charge p0, which is generated at the left electrode and collected at the right electrode. Ca1
culate the pressure of the pump if p0 = 25 mC/m3 and V0 = 22 kV.' rigtre 6.1 An electrohydrodynamic pump
Example 6.1. Argo2m:
E = 550 'M/ml l Area S / ' A point charge Q is. located a distancez‘r from 3. Conducting plane. Deter mine the locuscf paints in the plane at" which the magnitude of the surface mmgdmmyaogmt AMWﬂJ ;{cigggk ED} 4 A. posztive poth charge {2 is ﬁxed '10 cm above a horizontal
CUHdUCHng plane. An equal negative charge Q is to be located some Awsz: 1when: along the perpendicular dro
’ t . V pped from Q4 to the lane. Where a. . am
can —Q be placed sothat the total force on it will he 23m? 1 ‘3 ‘ 3.0 g C l6 a.) In the ﬁeld of the point charge over the plane (Fig. H i, if you
genesis ﬁeld line that starts out from the point charge in a horizontal .1 '  _ FT" S
Irectlon, that is, parallel. to the plane, where does it meet the surface AM i R a l 3 n
of the conductor? (You’ll. need Gauss’s law. and a simple integration.) HGURE 3 Some .ﬁeld lines for the charge above
_ 7 V the lane.
field strength at the surface, p The dete ‘
the surface charge density Pf . rmmes b) A charge Q is located 12 cm above a conducting plane, (1;; £310) Asked to predict the amount of work that would haveto
be done to move this charge out to inﬁnite distance from the plane,
one student says that it is the same as the workrequired to separate
to inﬁnite distance two charges Q and  Q which are initially 211 cm
apart, hence W =KQ2/2h. Another student calculates the force that
acts on 'the'charge as it is being moved and integrates F dx, but gets
a different answer. What did the second strident get, and who is right? "AN: W 2 lV’tragthZ/‘thl ® Given the current density .l =  lO4[sir1(2:t)e‘_3—"aI —t— cos(2x)e‘3"a\.l l(.A../‘11‘lzi
((1) Find the total current crossing the plane y = l in the a}. direction in the
region 0 < x < l, 0 < z, < 2. (1’)) Find the total current leaving the region
0 < x, y < l, 2 < z < '3 by integrating E  515 over the surface of the cube.
(0) Repeat part (b), but use the divergence theorem. Let 3 = 400 sin 19 /(r2 + 4) ar A/mZ. (a) Find the total current ﬂowing
through that portion of the spherical surface r = 0.8, bounded by 0.17: < 9 < 0.3%, O < (15 < 272'. (In) Find the average value of} over the
deﬁned area. ‘ Let 3 = 25/pa,, —— Ell/(,02 + 0.0!) a: Almz. (c1) Find the total current
crossing the plane .: = 0.2 in the a: direction for p < 0.4. ([1) Calculate
apt/lit. (c) Find theoutward current crossing the closed surface deﬁnedby
p = 0.01, ,0 = 0.4, 2 = 0, and z :4. 0.2. ((1) Show that the divergence
theorem is satisiﬁed for .i and the surface speciﬁed in part (c). 69:) Two perfectly conducting cylindrical surfaces of length E are located at,
p = 3 and p :: 5 cm. The total current passing radially outward through the
medium between the cylinders is 3 A dc. ((1) Find the voltage and resistance
between the cylinders, and E in the region between the cylinders, if a
conducting material having or 4— 0.05 S/m is present for 3 < p < 5 cm.
(17) Show that integrating the power dissipated per unit volume over the a. volume gives thetotal dissipated power. ﬁns 3 (a) E = [(9'55ypmap Wm, V = _ R = (1.63)/ Z S2, wherel is the cyﬁnder_lengt
A“ wag ; (not given) (19) 14.64/l W '. 1.
Mfg...“ (c210: “1221:0311: = 2 .. _, .
(12);: = 0.10,E(.10,.27r, 1.5) = —1s.2
145%. — 26.7az V/m (c) 1.32 ire/m? (a) D01 = 0) = —(100«sox)/<x2 +49? (a) 12.5 mm (b) 26.7 kV/m
(c) 4.23 (with given p5 = 1_0,,LC/m2) ’ (LO0.92119 me) long: Gaga
(a) 6.26'pC/m2 (12) 1.000176 0 ' " " " ' ' ’
L1 ((2)133 = [(l44.9)/,0]ap V/m, D : ' _ 21 i h "t 3
(3.283,:VP nC/m2 (b) Vab =192 V, Xe 21,56 ‘3) Wmm‘i’i_ 511(H‘TC’9 be a" (C) x 10—29)/p]ap C h m 7 . ., i ,. , ',,.._,___ ® (a) 80 Win (/7) ~60a 1 — 303s V/m (c) 67 l V/rn ‘ t r
I .1 l.  a »—1.23
(d) l04l.4 V/m (a) 40.0“ (f) 2.12 nC/m2 (g) 2.97 @( ) M (b) 0 (C) 0’ as expecmd‘ 7 . 2
, nC/m; __ 2.663)! _ nC/mz (a) 53.032' A1111
(‘1') 1.70211~ — 2.132}. — 1.063: nC/m2 (j) 545° ! (a) "—178'OA (b) O (C) O @ 125.2. + l7Sa_\, V/m @ ;f(.1‘.‘.\=) 2 ~49?‘ + 3x2. V(_x, y) = 3(x2 — y?) 6‘) (b) A = 112.5, B = —12.5 or
A = —12.5. s = 112.5 (a) — 106 pC/m3 (b) :l:0.399 pC/m2 (depending on
which side of the surface is considered) ® (a) 1.01 cm ()2) 22.8 kV/rn (c) 3.15 302 HM e 7 558% Let V = 10(p'+ 1)z2 cos 45V in free space. I; 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: (,0 + l)z2 00345 = 2 b); Find p and E at that point on the conductor surface Where gt = 0.271" and z = 1.5: At the given values of cf) and z, we solve the equation of‘the surface found in part a. for p,
obtaining p = 4Q. Then ' ‘ V
E: 471/: $273.9 lﬁagb alaz —1022eos gs ap + 10 Then
E(.10, .27r, 1.5) = ~18.2 ap + 145 at — 26.7 az'V/m c) Find {pal at that point: Since E is at the perfectlyconducting surface, it will be normal
to the surface, so we may write: p5 = EOEn M = eoﬁfﬁf) i. ecu/E  Egy’ (18.2)2 + (145)2 + (26.7) = 1.32 nC/xn2
S ace  I m "Z . Given“ the footential ﬁeld V = 10032 / (a:2 4? 4) V. in free space: a) Find’D at thesurface z 4— 0: Use *  . a a; ' 100:5
Ef‘VV ‘100245;($2+4) are—“armzméévm At A: 0‘, we use this to ﬁnd D(z = O) = 60E(z = 0) = —1005;J:z:/(x2 +4) a2 C/Inz. b) Show that the'z L—— 0 surface is an equipotential surface: There are two reasons for this: I
1) E at z = O is everywhere z—directed, and so moving a charge around on the surface
involves doing no "work; 2) When eValuéfting the given potential function at z = O, the
result is O for all :2: and y. ' c) Assume that the z =1 Oilsurface is a conductor and ﬁnd the total charge on that portion
of the conductor deﬁned by 0 < a: < 2, —3<: y < 0: We have ' £00603. ' z=O " £62 + 4 p; = D  az C/In2 160$ " . " 1 ' *2’ ' 2' .  .j
932 +4 dz: dy. (3)(l0f))eo <2) +4)0 15060111 _ n. I ~ g. . Atomic hydrogen contains 5.5 x 1025 atoms /m3: at a certain temperature and pressure. When
an electric ﬁeld of 4. kV / In is applied7 each dipole formed by the electron and positive nucleus
has an effective length of 7.1 x 10—19 m. a) Find P: With all identical dipoles, we have b) Find er: We use P = eoXeE, and so P 6.26 x 1012 e=__.=____.____——_=.6 *4
X 50E (8.85x10—12)(4x103) 17 X10 Then er = 1 + X5 = 1.000176. A coaxial conductor has radii a = 0.8 mm and b = 3 111m and a polystyrene dielectric for
V which 6,. = 2.56. .If P = (2/p)ap nC/rn2 in the dielectric, ﬁnd: 
a) D and E as functions of ,0: Use P (2/p) X 10‘9a 144.9
E : ——————— = ———————————p—— :
60(5. — 1) (8.85 x 1012)(1.56) p 3” Wm
Then
2 x 109a 1 3.28 x 109a 3 28
D: E+P=—————!’ ——— =____._P 2:  3p 2
60 p [1.56 + l] p C/rn p nC/m W b) Find Val, and X5: Use 0.8
144.9 
Vab = _/ T dp=144.9111(—3) = 192V
3 X5 = er — 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 P (2 >'< 109/p) _ 5.0 x 1029 P=N=Wap p apc‘m P = qu = (5.5 x 1025)(1.602 >< 10‘19)(7.1x10_19) = 6.26 x 10~12 C/m2 = 6.26 pC/m2 fr“ : «t / \wwﬂ The surface a: = 0 separates two perfect dielectrics. For a: > 0, let er = 671 = 3, while 6,2 = 5
where :c < 0. If E1 = 80395 — €3an — 30a; V/m, ﬁnd:
a) Ele This will be E1 a1 = SUV/m. b) Em. This has components of E1 not normal to the surface, or E111 = ~60ay —— 30a; V / m. c) ET1 = «(60)? + (30)2 = 67.1 V/m.
d) E1 = «(80? + (60)2 + (30)2 = 104.4V/m. e) The angle 01 between E1 and a normal to the surface: Use f) Dm = 13m = ErleoENl = 3(8.85 >< 1012)(80) = 2.12 no/mz.‘ g) 1312 = meoETl = 5(8.85 >< 10—12)(67.1) = 2.97nC/m2. V h) D2 = ErleoENlaa, + ErzéoETl = 2.12am — 2.66% — 1.33512 nC/mZ. i) P2 = D2 — 60132 = D2 [1 — (1/52)] = (4/5)D2 = 1.70am — 2.13ay — 1.06az nC/mz. j) the angle 92 between 1E2 and a normal to the surface: Use ' '9 =—’—=———l—=———————'—————=.581
'COS 2' E2 D2 (2.12)2 = (2.66)2 +0.33)2 Thus 92 = cos“1(.581) =_ 54.5°. ' Two perfect dielectrics have relative permittivities 67.1 = 2 and 5T2 = 8. The planar interface
MM" between them is the surface a: —y+22 = 5.7 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 cenditions. The normal component
will be EN1 2 E1  11. Taking f = a: — y + 22, the unit vector that is normal to the surface is ‘ _E__1_ _
“WWI—«Elam “2&4 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 z = 5, y = —5, and. z = 2.5). So Em = (l/x/EMlOO — 200 — 100] =
—81.7 V/ m. Since the magnitude is negative, the normal component points into region 1 from
the surface. Then Em = ——81.65 — a1, + 2az] = —33.33a,, + 33.33ay — 66.67az V/In Now, the tangential component will be ET1 = E1 —— ENI = 133.3aI + 166.7113, + 16.67az. Our
boundary conditions state that Em = Em and ENz = (EH/6,2)EN1 = (1/4)EN1. Thus 1 E2 ; ETZ + ENg = ETl + zENl = 133333 + 156.731, + 16673.2  8.38% 8.38.24 — 16673.2 = 125ac + 175.3y V/m Let V(:r, y) = 462$1+ f '~— 3y2 in a regionofﬁee space where p1, = 0. It is known that both}
Em and V are zero at the origin. Find f and V(a:, y): Since p1, = O, we know that VZV = 0,
and so ' ‘ 62V  82V d2 f
2 = __ V____ = 23 __ __ =
V V. 81:2 + 83/2 165 + (£032 6
Therefore d2 d I
_];=1652”+6 => i=~862m+6z+01
da; d1; ‘
NOW av ‘ df
_ __ __ 2m __
Ea: _ 8:5 86 + d2;
and at the origin, this becomes
df 1 . —— 8 + a; z=o — 0(as given) Thus df/da: [3:43 = —8, and so it follows that 01 = 0. Integrating again, we f(a:, y) =' —482‘” + 39:2 + 02
which at the origin becomes f(0,0) = —4+ 02. However, V(0,0) = 0 = 4 + f(0,0). So f(0, 0) = —4 and 02 = 0. Finally, f(:z;,y) = —4eg‘” + 33:2, and V(a;, y) = 4&2” —— 4e233 + 332 —
3312 = 3(ch — yz). _ ' Given'the potential ﬁeld V = (Ap4 + B p4) sin 4¢i
a) Show that V2V = O: In cylindrical coordinates, 15 8V5 182V
2 __,____ ____I __
vv‘paplaplﬁfé‘aw
13 _ . 1 . l . ‘
_ pap (in/)3 4Bp 5))51n4qb— Fla/1,24 +Bp 4)sm4¢ 16 .  16 _ .
= 7(Ap3 +Bp—5) sméqﬁ —.;2(Ap4 +Bp 4) 5111445 = O b) Select, A and B so that V = 100 V and = 500 V/m at P(p = 1,45 = 22.5°;z = 2): .
. First, V 8V » 5V
1
E—~VV————B—p—ap—;E¢—a¢ I I
= ——4 [(Ap3 — BpE’) sin4q5 ap —{—'(Ap3 + Bp's) cosélqbaﬂ
and at P, Ep = —4(A — B) ap. Thus 1131:} = i4(A — B). Also, Vp = A+ B. Our two
equations are: a . *
4(A f 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 ﬂ @ 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 Let V = (cos 2¢) / p iiifree space. : 3.) Find the volume charge density at‘point ‘A(O.5, 655, 1): Use Poisson’s equation: . 2
p” = ;Eov2V =._ED 1 a pap 5—p— 35—332
‘(1 B (—coqub) 4 cos2¢> 360c052q§
:: ~50 —— — ——2 = pap p p p 9
So at A we ﬁnd: 3 1200
pw; = W = —1250 = —106pC/m3 0.53 b) Find the surface charge density on a conductor surface passing through (2, 30°, 1): First, We ﬁnd E: ‘ 7 8V 1 6V
L E=~VV=———a ————a
6p P p aqb t
cos 2d 25111245
= a + a
p2 p #2 ¢
At point B the ﬁeld becomes
EB = 905460 ap + 25mm an = 0.125 a,;+ 0.433 as, ' 4
I The surface charge density will now be pa); = 11133; = ieglngl = 4—0.45150 = £0.399pC/m2 The charge is positive or negative depending on which side of the surface we are consid—
ering. Theproblem did not provide information necessary to determine this. and the outer at 0V, ﬁnd: V
a) the location of the 20V equipotential surface: From Eq. (16) we have __ ln(.012/p)
W”) T" 100111(.012/.005) V We seek p at which V_= 20 V, and thus we need to solve: ln(.012/p) .012 I
= ——————————— = = 1.01
20‘ 100 1n(2.4) é p (24)” cm
1)) Ep mm: We have
E _ £2 _ AZ _ 100
P.“ 6p " dp ‘ was whose maximum value will occur at the inner cylinder, or at p = .5 cm: s
E _ 100_
Pm _ .0051n(2.4) c) 5,. if the charge per meter length on the inner cylinder is 20 110/111: The‘capacitance per meter length is
C _ 27reoer Q — ln(2.4) V0 ‘ We solve for 51.:  h
a (20 x 109)1n(2.4)
= —————_————— = 3.15
2t§o(100) 57' ‘ = 2.28 x 104 V/m = 22.8 kV/m @ .ic cenductmg spheres are loeated at 1' == 5 mm and 7' = 90 mm. The regien between  ietre: ‘1; ﬁlled With a. perfect dielectric. If the inner sphere is at 100 V and the outer
a a. :  Find the location of the 20 V equipetential surface: Solving Laplace’s equation gives us V(7') = V0 Where V0 = 100, a = 5 and b = 90. Setting V(7‘) = 20, and solving for 1' produees
== 12.5mm. b) Find Em”: Use dV V034—
E— vv_—H7ai,_rg(%_% V 100
Er,ma.m = E0" = a) = "ﬂ _ E’a/m = 5(1 — (5/20)) = 26.7 V/mm = 96.7 kV/m c) Find eq if the surface charge density on the inner sphere is 1.0 [LC/13121 p5 will be equal
in magnitude to the electric ﬂux density at 7' = Q. So p5 = (2.67 X 104 V/m)ereo =
10‘6 0/132. Thus 6, = $2.; ' We wem is design a sphericai eewum'eaeecéteé' with a given fadiue e ier'ihe cuter.
eehere. which win he able to share the greatest amount ei eiectricei energy subject in
the cansiraini that the eieeiqiaﬁeie avengea at the euréace of the inner genera may not
exceed E9. What radius h' etmuid he eheeen ier iheinner eeheﬁcai centimeter. and
haw much energy can he steree? V, I ——~~~~~~~MW"""‘*‘“"" H N ,. 5 z y I ‘ . t) The Potentialzdiﬁ‘erence AV betweenw‘themw w VVVV ' plates of a spherical capaeitor iskepr constant. ‘ Show that then the electrid ﬁeld'at’the surface of ‘L the inner sphere will be a minimum if a = 117. I Find this minimum value of E. ' 2 Q 1; a
twaQ‘E r ‘3 . dE. ~ 19‘ x H ’
5ma<o for a<§. andgoo fora>§ . a2 2 [ b1 _ ‘ ¥ ‘ ' 4% I
“ a ‘ E— . item is ndnimum E . = _b_ ‘ ‘ A 6fﬁ;dﬁgai capaﬁit‘o’r her enter and inner canduég tors whose‘radii are in the ratio of b/a = 4/1; T’nei
: ' V inner conductor is to he replaced by a wire whosei‘i radius is onehalf the radius of the originai irmeriE canductor. By what factor should the length be 3111—};
creased in order t0 obtain a capacitance equal to that
’ of the original capacitor? ' ' 'wherea2=g21 '5; 1:14:— at Current—carrying components in highvoltage power equipment must be cooled to carry ‘y away the heat caused by ohmic losses. A means of pumping is based on the force transmit— v
ted to the cooling ﬂuid by charges in an electric ﬁeld. The electrohydrodynarnic (ElID) .
pumping is modeled in Figure 6. The region between the electrodes contains a uniform
charge pm which is generated at the left electrode and collected at the right electrode. Cal—
culate the pressure of the pump if ,00 i 25 rnC/rn3 and V0 = 22 kV. Figure 6: An electrohydrodynaniic pump; Area 5' Since pt, 7‘5 0, We apply Poisson’s equation _ VZV = ~&
8
The boundary conditions VL = O) = V0 and V(z = d) = 0 show that V depends only On 2 (there is no ,0 or qb dependence), Hence dZV : ‘po
dz2 8
Integrating once gives
dV — z
— = p“ + A
dz 8
Integrating again yields ’
_2
V = 1”” + A: + B tions.Whenz = O, V: V0,
Vo=—O«lO+B——>B=V0 Whenz=d,V=O, podz
o = — + Ad + V0
28
01‘
_ pod _ Y3
A 28 d where A and B are integration constants to be determined by applying the boundary condi f The electric ﬁeld is given by The net force is z=0 .‘
Voz _,_ po 2 ] d A‘
= — I _ d ..
poS[ d .2 (z Z) ’03;
F=poSVoaz The force per unit area or pressure is p = E = poVo = 25 x 103 x 22 x 103 =, 550 mm:2
29 A» § ff._;_'..‘>i L" 3%“2‘”? " 16'?“ ‘ “i . .4...— 171 J
u
_/
.4. _.__ , ._.H_. E " ' Force req’uired +0 move +his‘
‘ c @ Charge ‘upword =E€Q2/(27L)Z g T . The second Student
W 777W calCUiG—ies as Fol iows :
l .
i I °° 2
I ' KG Q a!
a, 1‘ Work ~de'x. ifqu x = as initial state C." . h
' ’ This is the correct answer.
Noreihcrt H3 two real charges Q and Q were being
pulled apart Symmetricoliy, the +0173! work done "
‘Wouid beKQZ/Zh, but the agency moving Q would}
Supply only he}? OF H‘; _ 4 Given the current density J = «104[sin(2:z:)e—:2'yam + cos(2$)e‘gyay] kA/mzz a) 'Find the total current crossing the plane y = l in the as, direction in the region 0 < a: < 1,
0 < z < 2: This is found through ' 2 1 2 1 ;
I=//Jn] 'dat/ / Jayl drdz=/ / —1O4cos(2a:)_e_2dxdz
5' 3 o o y=1 o o . v l
= ~104(2)%sin(2$)‘oe_2 e —1.23MA b) Find the total current leaving the region 0 < :17, a: < 1, 2 < z < 3 by integrating J ~dS over
the surface of the cube: Note ﬁrst that current through the top and bottom surfaces will
not exist, since J has no 2 component. Also note that there Will be no current through the
:1: = 0 plane, since Jz = 0 there. Current Will pass through the three remaining surfaces, and will be found through I=/:/01J.(—ay)lﬁ=od:z:dz+1231):!(ay)y=1dmdz+L3AIJ(am) 3 1 3 1
= 104 f / [cos(2m)e"0 — cos(2m)e“21 dz: dz — 104 f / sin(2)e'2y dy dz
2 o 2 o , I I I = 104 sin(2w)‘:(3 — 2) [1 — eZ] + 104 Im(2)e—2yl:(3 — 2) =_ g dy dz z=l c) Repeat part b, but use the divergence theorem: We ﬁnd the net outward current through
the surface of the cube by integrating the divergence of J over the cube volume. We have 65:: + gag—y = —10_4 [2 cos(2:r)e“2y — 2Acos(2$)e_2y] = Q as expected I VJ: ’ Lét' ~__.4DOsin0 J 721—4 a7. A/rn2 ' a) Find the total current ﬂowing through that portion of the spherical surface 7' = 0.8,
bounded by 0.11r < 6 < 0.37r, O < 4) < 271': This will be 27r .31r éoosina 2 400(.8)227r_/.31r I 9 ‘
I“//J “lad” /O [1,. (.8)2+4( Hm ‘15 4.64 1,, g
371' 1 I = 34.6.5 ' 5n — cos(29)] d9 = 77.4A .176 b) Find the average value of J over the deﬁned area. The areais
211' 371' 9
Area = f f (.8)2 sinﬁ d9 dd) = 1.46 III."
0 .17r . The average current densityis thus J Mg ; 4/146) 3., = 53.0 ar A/in2. @g Lat V 25' ' 2:01,. a
. J=——'a —————az A/In2 p p— 102440.01 a) Find the total current crossing the plane z .= 0.2 in the a; direction for p < 0.4: Use ‘ 271' .4 _20
I: Jn d = . d d
. z=.2 a ‘/D j; ‘p2+.01p p gs I _ — 20m(.01 + ,0?) .4
0 (271') = 207rln(17) = —178.0A b) Calculate BpU/at: This is found using the equation of continiﬁty: 8p”. 1 8 8.1,, 1 8 8 —20
6t p8p('0 ’0)+ 62 .pap(25)+az (p2+.01)‘ “Q c) Find the outward current crossing the closed surface deﬁned by p = 0.01, p = 0.4, z = O,
and z = 0.2: This will be ' ' .2 271' 25 2 ‘37? 25 .
V I=/0I/O TO—lap(——a,,)(.01‘)d915clz+fO f0. Zap(ap)(.4)d¢dz 2" ‘4. —20 2w .4 _20
z‘ — Cl d " _ ﬂ =
+./o f0 p2+.01a < 8‘2” p ¢+fo f0 pumaz (a.)pdpd¢ Q since the integrals will cancel each other. d) Show that the divergence theorem is satisﬁed for J and the surface Speciﬁed in part b. In part c, the net outward ﬂux was found to be zero, and in part b, the divergence of J l was found to be zero (as will be its Volume integral). Therefore, the divergence theorem
is satisﬁed. ' / Two perfectly—conducting cylindrical surfaces of length l are located at p = 3 and p = 5 cm.
The total current passing radially outward through the medium between the cylinders is 3 A
dc. a) Find the voltage and resistance between the cylinders, and E in the region between the
cylinders, if a conducting material having or = 0.05 S /1n is present for 3 < p < 5 cm:
Given the current, and knowing that it is radially—directed, we ﬁnd the current density
by dividing it by the area of a cylinder of radius p and length l: 3 ___ __ 2
Jr 27ml ap A/rn :‘ rillhen ﬁeld is found by dividing this result by a": V 3 9.55
an = 721p E = 27ro'pl The voltage between cylinders is now: 3‘ 5
' 9.55 9.55 5 4.88 Now, the resistance will be , V 4 88 1 63
RT; 7 .= w“ = T 9 ' b) Shaw that integrating the power per unit volume over the volume gives the
total dissipated power: We calculate ’ ~ z 21 .05 I 2 ' . 2 ‘
3 3 5 14.64.
P: EJd = ——————— d d d = 1 _ :—
fv ” f0 jog (2w>2p2<.05)zzp p i z 27r<.05>zﬂ(3) z W We also ﬁnd the power by taking the product of voltage and current A. which is in agreement with the power; density integration. ...
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 Spring '10
 Ferguson
 Electromagnet, Electric charge, Surface, surface charge density, equipotentiul surface

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