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Unformatted text preview: ‘ CALIFORNIA STATE FOLYTECHNIC Umveaezw; POMONA
Electrical & Computer Engineering Department
L. Ferguseh I '  7 ECE $32 . ' ‘ HOMEWORK e 5'" ‘ , y ( Q . Determine the work done in carrying a 2uC charge from (2, 1, ~41) to (8, 2,
—1) in the ﬁeld E = yax + xay along (a) the parabola x = 2312, (b) the ,
hyperbolax = 8/(7 — 3y); (c) the straight line x = 6y —— 4. , charge on a circular arc, the circle centered at the origin, from x = a to xi: y = (IA/2.. A potential ﬁeld in free space is expressed as V = 20/ (xyi) V. (a) Find the
total ehergy stored {within the cube 1, < x, y, z < 2. (b) What value would be
obtained by assuming a unifoxm energy density equal to the value at the
Center of the cube? (9
(2) Given E = —xax —l_— yay, ﬁnd the work involved in moving a unit positive
(3 l @ F0”? 99in: Charges: "53511 05 Value lieea; C arekeituafted' at the vertices 14,553,, C
B (loeatedfin that sequence) of a square ef side's 1m. Find the foilowing: {ai'the
: I..'ih’t3‘t‘k required ta assemble the chatgecdistt‘ibutioti; (53} the additiona'i 'wetit re
quired to move‘the charge at i} to the ceeter of. the Square; and"{e} the additienal work. required to then metre the charge at C to. the center of. the side 5354 4 Am. “(3);” 63.93739 3; (313)192961533; ta) 36.1746555 . " * 5 Three peintcitatges ef values 1; 2, are situated at the eernete ofan equilat
eral triangle of sides 1 m. it is de'sited? te ﬁnd the work required to move these, '
charges to the corners et‘ ea equilateral triaegle ef shorter sidee was shown in Fig. 3?“ " .. ' ._...__..——_._ Ahshw‘i we: n '59 I . . “itﬁ‘eo ": ﬁg?! Etingingithree point? 4
chatgesimmjhe WES e: a target .. ., equilaterial ttfieegie ta the eeteets of; 3‘ stnaﬂeijmuﬂa‘ieml tsiengie. ' ' j :Two I'Sphericaiﬂchargem each of the ariiriémrgciius a and this same; uniform charge ~
density p0 C/mJ, are situated inﬁriiteiy' Egan, The twn Spherical charges arr} ‘ brought together and made inm aﬁsirigle spheric'él ghargeof uniform density. Find .
the following: (a) his mark required if~the radiuspf the new spherical charge is'a: f (h) the work required if ihe charge: density cf the, new spherical charge is pa; and '.= ‘(ciihe radius and the charge density. of the new Spherical charge {Dr whichihe:
WOri'i required is zero. ' K ' ‘ ' A215. .(a)‘.1.57sspsa5/sdi; (h), Maggie's/.503; ($251,025,063 ® A Viiirnnc charge'Llistrihutimi Ls given in Sphuricni cnmmiinzncs by, .='__ huh/u) {hr 1' <1:
. 1r, ii)" ‘ [hr 1‘ >u in) Find the wmi». required in rearrange the charge with uniiiirnr dunsny within
' ihu rcgiun r < in ‘ I . (hi Find the mum's (if the Sphcricui region and the charge density liarwhich nci‘ thcr additional 'wnriL is rcquirqd hur wnrir. is mth zwziiiubie in redistributing
(in: charge uniiiirml‘y within tin: region. ' .... _...—..
.._— .... qu..,..,.,._.. ,
... . .. ....._....K.. V... . .__.._.._ ._.._. .__...... ..... ... .. ., iChnrgiéisT and Q are distributed Uniiiirmly 0n. the surfaces of Lwh I Tsiihcr23"n't“rzidii‘Z! Enid“b"(>71)1'rEEﬁEEiWééf."'TiiE'Yiﬁtéi‘"§ﬁﬁéf€iiiiﬁzidé‘up at two, scparuhiu hemispheres. Find the: work rcquircd in carry me two hemisphere$ in an
1 infinite distance i'rmn tin: inncr sphtrc anti juin them to harm it sphere ui umiuriniy ilimri‘hutud charge {_). i..———" ﬂ“ ._ a ..
1’.“ —" @ A sphera of "radius Rama; 5. charge: Q‘dis'tributed unjfbrmly
Omar its sillfact How large a spheretcntaing 90 percenth thegnergy
sibred'inihe alectrosiatic ﬁgirilaf ihi; chars; histnb‘utmn? Z Q w: ’28“? @ w: ~a2/z
saw—:38??? is) WZZOiPT '” 7a; ,Wz; 7T3”; 0.. j—_ Yr: a ck, _. 1°00 2
*7“ Q ‘.
’5 8 W 2 «T
U erred? L. Ferguson CALKFOHMA STATE PoLweeHme 'Umveeszw', PomoNA
Electrical s Computer Engineering Department ' eoe sea" , p ,
HQMEWORK #g' stem“: ' ( . , . , l _ , ' V ' '
© Determine the work clone in carrying a 2pC charge from (2,1,4) to, (8,2,4). in the ﬁeld E a) the parabola a; = 2y2: = yam + may along 
As a look ahead, we can show (by taking its curl) that E is conservative. We therefore expect the same answer for all three paths. The general
expression for the work is ‘  ' ' _ , ' B I 8 2
=“Q/ E'dL=q[/ ydmtf catty]
A 2 1 In the present case, a: =, 2y2, and so y = «Av/2. Substituting these and the charge, we
get . 8 2 _ A a ’ .
W,=_2x10‘5 x/x/2dz—lf 2y2dy] =—2x10—6 [3W2] +3y32 =_2s in
2 1 3 2' 3 1 b) the hyperbole a: = 8/(7 — 33;): We. ﬁnd y = 7/3 —— 8/353, and the worlr is 4 @ 8 . 2
__ 7 8 8
2 i 2 3 3:1: » w+ 1 7~3ydy _ 7 8 ' 8 .8 ~ '2 =—2 106— — ._~ _ __ _ I =_
X‘ [3(8 2) 3 ln (2) 3 ln(7 3301] 28 it}
c) the straight line a: = 63,: ——' 4: Here, 3; = :E/ﬁ + 2/3, and the work is g 8 , 2
W3 = —2 x 10—6 + dart—f (ﬁg—4) dy] = —28 [,LJ
’ 2 1 ’ Given E = —:ra,,, + gay, ﬁnd the worlr involved in moving a unit positive charge on a circular
arc, the circle centered at the origin, from at = a to :1: = y = (1/ In moving along the arc, we start at qi = O and move to 45 = 77/4. The setup is W=—q/E'dL=—— .Ead45a¢=—/V (—zazva¢+‘yaya¢)adq5
0 0 .WJv W4 —sinq5 cosqb . 77/4 1r/4 » ' _ '
'7; — f . 253 Sin qbcos qb dd) =n—— / a2 sin(2¢) d9!) = —a,2/2
a D 0 i5 where q = 1, a: = acos <25, and y = asinqS. Note that the ﬁeld is conservative, so .We would get the same result by integrating along .
a twosegment path over a; and y as shown: 'f W=_/E.dL=_ [f/ﬂrmn/Wﬂdy] Lag/2 0 A potential ﬁeld in free space is expressed as V = 20/ 3.) Find the total energy stored Within'the cube1 < 1:,y, z‘< 2. We integrate the energy
density over the cube volume, Where My = (1/2) 50E  E, and Where ’E_=;VV=20[ 1 am+ mZyz The energy is now b) What value would be obtained by assuming a. uniform energy density equal to the value
at the center of the cube? At C(1.5,1.5,1.5) the energy density is we = 20060‘(3) = 2.07. x 1010 J/m3 ‘ This, multiplied by a. cube volume of 1, produces an energy value of 207 pl # ) I Four point charges, each of .value 441455 Crete situated at the verti‘ces A, 'B, C, and;
D (located in_that sequence) of a square of sides lm. Find the following: (a) thei
._'W0.rk required to assemble the charge distribution; (b) the additional work 1":j
‘qured t0 move the charge at D torthe cenier of. the Square; and (c) the additional? waif rggpiged g0 {hgggzgygthe Charge at'C to the center. of the side AB. 3 A: 68.03780 1
@—
(b) Energy stored in the new charge distribution =%[4zso(1+7%+«f§)+4zso(1+l+ﬁ) '
+4ﬂ$o(\715+1+J§)+4n£0(ﬁ+ﬁ+~/§)] I = 37.333130 I
. Additidnal work reqmred A » i = 37.333150 — 63.03720 = 19.296150 J .(c) Energy stored in the new charge distribution = «[2 X 87.333180 =.123.5077£g J, since the geometry is proportional to that in (b) by the factor 1/45. Additional wofk requii'ed _ =123.5077ao— 37.3331ao= 364796503 ' g i /) Three point charges of values l, and 3 C are situated at the corners of an equilat eral triangle of sides 1 in. it is desired to ﬁnd the work required to move these.
charges to the corners of an equilateral‘ triangle of shorter sides i an as shown in Fig. l ‘ «,1
ll ‘ l Figure l Bringing three point charges from the corners of a larger :3
equilaterial triangle to the corners as smaller} Waters! triangle. , l __ "the notential energy stored'in the system the larger equilateral triangle is given by 1‘ 1(‘243) (r 3) (1' — iv =— l + + + + +
2 Q J 2i: 41750”. 47780 2 471'50 47780 3 41750 41750 _l[5+8.+9]_ 11 Nm 1 _ v, 2 477750 ' a i L I 47T 50 The potential energy stored in the system of three charges at the corners Of the smaller equilateral triangle is equal to twice this value since all distances are halved. The increase in potential energy of the system in going from the larger to the
smaller equilateral triangle is equal to l 1/411'30 Nm. Obviously, this increase in en i
ergy must be supplied by an external agent, and hence the work required to move a
the charges to the corners of the equilateral triangle of sides i m from the corners of g
the equilateraltriangle of sides 1 m is equal to ll/47rao Nm. % Two spherical charges, each of the same radius a and the same uniform chargei
density p0 C/m“, are situated inﬁnitely apart. The two spherical charges are "'
brought together and made into a single spherical charge, of uniform density. Find
the following: (a) the work required if the radiusof the new spherical charge is a:
(b) the work required if the charge density of the new spherical charge is p0; and (c)r'the radius and the charge density of the new 'spherical charge for which the
work required is‘zero. ‘ v— ..___...,__......._._........._.._ _....».—n‘.. “cw—"us ,.__..,, _. _. . _ V 5 Energy stored in the initial charge distribution = 2 x 4‘93“ = ____3"p%“ (a) 2 x % m3po = m3p’, where p’ is the new charge density. pf: 2pc . 2 5
4 ‘ 162: a
Energy stored = Ts—g(ZpQFaS = 75%)—
lt’mpgas Expects .. pﬁa5 Work reqmred =‘ 1580 — 1580 = 1.6753 50 (h) 2 x g 7ra3po =§ 75(b)3po, where b is the new radius. b = 213a
f ' I 2 5
_ 42: 2 1/3 5 _ 12.69927L'p0a
Energy stored — ——1580 po (2 g a) — 1580
. 12.69927: 2a5 87: 2a5 Work required = ——15—8(%)—— — lggo
2 5
=o.9s42 319—
so
27*” (c) If. the required radius and charge density are he: and kzpo, respectively, then;
4 3 4 3 3
2><§zra p0 = Emma) kzpo or, k1k2=2 2 x 4;:p3a5 = 47:(k2p0)2(k1a)5
1560 1550 ' or, = 2 Solving, we get k1 = 2 and k2 = 114.
Thus, radius of new charge = 2a Charge density of new charge = 0.25 po nhnne charge distribution is given in spherical Ctml'kiillillCS by r <u
1‘ >11 Muir/u) tiir ) = 
’v in ifm (at) Find the work required in rearrange the charge with unitiirm density within the reginn r < u. (it) Find the radius (il' the spherical region and the charge density for which net i
ther additional wuri; is required nor work is made available in redistributing '
the charge uniliirmly within the region. ‘ 2 Wommuimdgﬁﬂjfmigeﬁ
‘ 2080 I  780 14060 ._,_~...._____ _ . / __ _ U ‘
'r 2: 2:: . r _ e I J. I (pg EJrz sin 6dr d6d¢ forr<a i i eoLiTW'ﬁ'Ef ferr<a ' a 1: 22:” r : 4804 i
I j I (mzjrzsmedrdedtp forr>a . r E, = . E
r=o 9:0 ¢=o » , , 3 K
“‘7 for r > a i
80" mgr“la for r < a m W w N I > ‘ > WE: ‘ forr>al i (“3) Energy stored in the electric ﬁeld of the charge distribution a n: 2:: ' ‘_ ' ‘ i =j j j §eo BEETrZsin adrdedq) ‘ V r=o a=o ¢=o .4eoa e _ 4 6° 1: 21! 3 2 '_ I +j j j %£o Elia—J r2sinedrded¢ r=a a=o ¢=o 450;? I xégs . When the charge is redistributed with uniform densrty, sayi p , then :3 :ta3p = i
3 P’ =zPo >. From ﬁnd: 37 , energy stored in the new Charge distribution 3“ term 3 6’} yo E
2223 _ )2_3"_P%a_5 . i 7  —' 1550 p0 2050 A i
i kid {he radius and the Charge density of the new spherical charge.
Then gemklaﬁkzpo = zp0a3 5
ji (k10)5(kzpo)2 = FEE
15:0 720 01', 3
W52 =3; 15 Solving, we obtain 9 28 21
k1=ﬁxi§=§6 3 ‘ o 2000‘
kzzxé—x =ﬁfé‘7 _
‘ . . ' 0 . ~21 ' 2000 A
Thus the radlus and the charge denmty of the new charge are—20a and 3087 130, respectively. ChurgesQ and Q are distributed uniformly on the surfaces of two concentric spheres of rzidii a and b (> a). respectiveiy. The outer sphere is made up of two separable hemispheres. Find the work required to carry the two hemispheres to an "
intinite distance from the inner sphere and join them to form a sphere of uniformly
distributed charge (_). For the two concentric spheres, as shown in the ﬁgure, 0 Vforr<a 4,325,: Q/éafora<r<b O forr>b
0 forr<a
0 ‘
Er —=—4W2 fora<r<b
O forr>ib
b 7: 2111 Q i
W =  r23ix16drd9d¢
‘ L=aL=OL=02£O(4W2)Z
_ 2 (1.1)
31.580 a b With the two spheres inﬁnitely apart, v 'oo 7: 2:: . ' I
1 Q . 4
We 5}; L=of¢=oiao(4 zjrzsmedrdedqb +r fit I” éeo [—LT r2.sin 6dr d9d¢ 17 9:0 ¢=o 4W2
=§ie+s
mo
2 1 1 2 1 1
Work reqmmd ~8%E—O(E+E)—%(Z—E) (a I A sphere of radius R has a charge Q distributed unifcrmly i1
oyer its surface. How large a sphere contains 90 percent of the energy
stored in the electrostatic ﬁeld of this charge distribution? ...
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This note was uploaded on 08/12/2010 for the course ECE ECE 302 taught by Professor Ferguson during the Spring '10 term at Cal Poly Pomona.
 Spring '10
 Ferguson
 Electromagnet

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