Hmwk 4 Solutions

Hmwk 4 Solutions - 30-; H“) Sehiw Let a point charge Q125...

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Unformatted text preview: 30-; H“) Sehiw Let a point charge Q125 nC be located at P1(4, —2, 7) and a. charge Q2 = 60 110 be at P2(—3, 4, —2). a.) If e = 60, find E at P3(l,2,3): This field will be _ 10—9 [25R13 + 6013.23] 47r€o lR13l3 [Heals where 1511;3 = —3a¢ +4ay —4az and R23 = 4a¢ — 2a1, +5az. Also, |R13[= m and [12.23]: «45. So _ 10‘9 25 x (—3a¢ + 4ay - 4a,) + 60x (4am — Zay + 5a,) ‘ 4m (41)1-5 (45)1-5 = 4.58% — 0.15% + 5.51a, E E b) At what point on the y axis is Ex = 0? P3 is. now at (0,y,0), so R13 5 —4a¢ + (y + 2)ay —— 7az and R23 = 334; + (y — + 23.2. A130, lle = 65 + (y + 2)2 and IR23I = V13 + (y —4)2. Now the :1: component of E at the new P3 will be: E _ 10‘9 25 x (—4) + 60 x 3‘ E _ 4m [65 + (y + 2))2]1-5 [13 + (y - 4)2]1-5 To obtain EI = O, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic: 0.48312 + 13.92;, + 73.10 =0 which yields the two values: y = —6.89, —22.11 A 2 [1C point charge is located at A(4, 3, 5) in free space. Find Ep, Egg, and 1:72 at P(8, 12, 2). Have E 2 x 10—6 RAF 2 x 10‘6 [434. + 9a.y — 32., P:——-—— —-—— ———— _ = . .3 — . 4WD IRAPI3 47m) (106015 ] 65 9a¢ + 148 a1, 49 4a,. Then, at point P, p = VS2 + 1’)2 = 14.4, 45 = tan"1(12/8) = 56.3°, and z = 2. Now, E, = E, -a,, = 65.9(a¢ -a,,) + 148.3(ay - a,-,) = 65.9 cos(56.3°) + 148.3 sin(56.3°) = 159.7 and Egg = Ep - a¢ = 65.9(a1, -a¢) + 148.3(ay : 3.4,) = —65.9sin(56.3°) + 148.3 cos(56’.3°) = m Finally, E, = —49.4 V/m A 100 nC point charge is located at A(—1, '1, 3) in fiee space. 3.) Find the locus of all points P(a:, y,z) at which E, = 500 V/m: The total field at P will be: . 100' X 10—9 RAP E}: = —-——-——- 3 41reo IRAPI where RAP = (m+l)a¢+(y—1)ay+(z—-3)az, and where [RAP] = [(a:+,1)2+(y—1)2+(z—3)2]1/2. The a: component of the field will be 100 x 10‘9 (a; + 1) = — —————— = 500 V Ex 47reo. + 1)2 + (y _ 1)2 + (z __ 3)2]1.5 /m And so our condition becomes: (a: + 1) = 0.56 [(z + 1)2 + (y — 1)2 + (z — 3)2]1-5 b) Find yl if P(—2, 311,3) lies on that locus: At point P, the condition of part a becomes 3.19 = [1 + (311 — 1)”]3 from which (311 — l)2 = 0.47, or y1 '= 1.69 or 0.31 A charge Q0 located at the origin in free space produces a. field for which E2 = l kV/m at point P(—2, 1, —l). 3.) Find Q0: The field at P will be ,Since the 2 component is of value 1 kV/in, we find Q0 = —47l'€061'5 X 103 = —1.63 110. b) Find E at ill/[(1, 6, 5) in cartesian coordinates: This field will be: E _ —1.63 x 10—6 a1 + 6a” + 5a, M " 4m, [1+36+25]1'5 01' EM = —30.11a¢ —180.63a.y —150.53az. c) Find E at ZVI(1, 6,5) in cylindrical coordinates: At M, p = \/1 + 36 = 6.08, (15 = tan‘1(6/l) = 8054", and z = 5. Now ‘ Ep = EM -a,, = —30.11 cosqb — 180.63 sind) = —183.12 E), = EM -a¢ = —30.11(-— sin (15) —— 180.63 cos <15 = 0 (as expected) so that EM = —183.12ap - 150.53az. d) Find E at .M(1,6, 5) in spherical coordinates: At JV[, 1' = \/1 + 36 + 2 = 7.87, (I) = 80.54° (as before), and 9 = cos—1(5/7.87) = 5058". Now, since the charge is at the origin, we expect to obtain only a radial component of EM. This will be: Er = EM ' a, = —30.115in9cos¢ — 180.63sin95inc15 —— 150.53cos9 = ~237.1 A uniform volume charge density of 0.2 ,uC /1:n3 is present throughout the spherical shell extendin from 'r = 3 cm to r = 5 cm. If p1, = 0 elsewhere: 3.) find the total charge present throughout the shell: This will be Q /27r 1r .05 2 T3 .05 = 0.21' sinfl drdfld': 4 02 _. = _ —5 = 0 f0 [03 <1: [ 1r( ) 3103 8 21 x 10 to 82.1 pC b) find 1'1 if half the total charge is located in the region 3cm '< 7' < 1'1: If the integral over 7' in part a is taken to r1, we would obtain [Mo-2s; =‘4.105 x 10-5 Thus [3 x 4.105 x 10-5 1-1 — .—____ 1/3 _ a _ 0'2 x 4” + (.03) ] _ 4.24 cm 3 A uniform line charge of 16 nC/rn is located along the line defined by y = —2, z = 5. If e = 60: a) Find E at P(l,2, 3): This will be where R; = (1, 2, 3) — (l, —2, 5) = (0, 4, —2), and [Rp]2 = 20. So 16 X 10—9 4ay — 2a, ' ; = .__ __ ='7.5 —28.8 ,v EP 2m, [ 20 l a ay 8‘ /m b) Find E at that point in the z = 0 plane where the direction of E is given by (1 / 3)ay — (2/ 3)az: With 2 = 0, the general field will be z=0 = L (y + may - 521;] 27reo (y + 2)2 + 25 We require [E,[ = —l2Eyl, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes: Ez=o = —"’ [‘2'5"? ‘ 5'3“) = 23:21., - 46a: An infinite uniform line charge pL = 2 nC/m lies along the :1: axis in free space, while point charges of 8 n0 each are located at (0,0,1) and (0,0,-1). 3.) Find E at (2,3,4). The net electric field from the line charge, the point charge at z = l, and the point charge at z = —1 will be (in that order): E 1 2pL(3ay—4az)+q(2a¢+3ay—5az)+q(2a¢+3ay—3az) = ._ x E \ tat 47reo 25 (38ls/2 (22)3/2 Then, with the given values of p1, and q, the field evaluates as Em = 2.0aI + 7.3ay — 9.43z V/rn M b) To what value should pL be changed to cause E to be zero at (0,0,3)? In this case, we only need scalar addition to find the net field: pL q q o = = E( ’0’ 3) 2150(3) + 47reo(2)2 + 47reo(4)2 0 l 1 1 2 1 : q[_+_]=_fl : pL=_— =—0.47q=—3.75nC/m i Spherical surfaces at 7' = 2, 4, and6 rn carry uniform surface charge densities of 20 nC/mz, —4 nC/mz, and p30, respectively. a) Find D at r = 1, 3and5 m: Noting that the charges are spherically-symmetric, we ascertain that D will be radially-directed and will vary only with radius. Thus, we apply Gauss’ law to spherical shells in the following regions: 7' < 2: Here, no charge is enclosed, and so D, = 0. 80 x 10-9 2 < 1' < 4: 47rr2Dr = 47r(2)2(20 x 10—9) => D, = 2 7‘ C/m2 So D,(r = 3) = 8.9 x 10-9 C/m2. _ _ 16 x 10‘9 4 < r < 6 : 4WD, = 41r(2)2(20 x 10—9) +47r(4)2(——4 x 10-9) => Dr T2 So Dr(r = 5) = 6.4 x 10‘10 C/mz. b) Determine p50 such that D = 0 at 1' = 7 In. Since fields will decrease as 1/73, the question could be re—phrased to ask for p,o such that D = 0 at all points where r > 6 m. In this ; region, the total field will be l 16 1-9 62 i DT(T>6)=_)i+/’50() . Requiring this to be zero, we find p30 = —(4/9) x 10‘9 C/mz. l 1.2 1.2 G , Volume charge density is located as follows: pv = 0 for p < 1 mm and for p > 2 mm, 3 pu=4puC/m3forl<p<2rrun. a) Calculate the total charge in the region 0 < p < pl, 0 < z < L, Where 1 < p1 < 2 m: We find . L 27r a; L Q=f / / 4ppdpd¢dz= §"—[p§—10‘Qluc 0 o .001 _3______._ where p1 is in meters. b) Use Gauss’ law to determine DP at p = p1: Gauss’ law states that 27rp1LDp = Q, where Q is the result of part a. Thus 40)? - 10‘9) C m2 3m # / DP(P1) = where p1 is 'in meters. 0) Evaluate DP at p = 0.8 mm, 1.6 mm, and 2.4mm: At p = 0.8 mm, no charge is enclosed by a cylindrical gaussian surface of that radius, so Dp(0.8n1m) = Q. At p = 1.6 m, we evaluate the part b result at p1 = 1.6 to obtain: Dp(1.6mm) = 340016) = 3.6 x 10‘6 pC/m’ At p = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss’ law is written as 27rprp = %[(.002)2 — (.001)'*’] p0 ' from which Dp(2.4rnm) = 3.9 x 10‘6 uC/m2. .‘A\ s Let D = 5.007321r mC/m2 for r _<_ 0.08 m and D = 0.205 aT/T'2 uC/m2 for 7‘ _>_ 0.08 m (note error in problem statement). a) Find p1, for r = 0.06 m: This radius lies within the first region, and so 1 d 1 d U=V'D=—— 2 r =—— . 4 =2 3 p T2 dr(7‘ D ) r2 dr(5 001' ) 0r mC/m which when evaluated at 7‘ = 0.06 yields p1,(7‘ = .06) = 1.20 mC/ms. b) Find p1, for r = 0.1 m: This is in the region where the second field expression is valid. The 1/7'2 dependence of this field yields a zero divergence (shown in Problem 3.23), and so the volume charge density is 1312 at 0.1 m. c) What surface charge density could be located at r = 0.08 m to cause D = 0 for 7' > 0.08 m? The total surface charge should be equal and opposite to the total volume charge. The latter is 27r 1r .08 Q = / / / 20r(mC/m3) 1'2 sin6’ dr d0 dq’) = 2.5.7 x 10-3 m0 = 2.57;“: 0 0 0 So now _ _ 2.57 _ _ 2 p’ " [47r(.os)2l ‘—32"C/m In the region of free space that includes the volume 2 < 1', 31,2 < 3, l. N I M“ I D — 2 2 C 2 —;W%+m%—zwA/m i I, a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In cartesian, we find V - D = 8:1;y/ Z3. The volume integral side is now l 1 3 3 38x1; 0 1 1 ; V-de=/ / / ——dmd dz: 9—4 9—4 (-——)=3.47C 3 A01 2 2 2 23 y ( ) 4 9 ' l b. Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at z = 3 and a: = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the rightand left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces. To evaluate the surface integral side, we integrate D - 11 over all six surfaces and sum the " results. Note that since the 3 component of D does not vary with 1;, the outward fluxes E from the front and back__surfaces will cancel each other. The same is true for the left ‘1 and right surfaces, since Dy does not vary with y. This leaves only the top and bottom surfaces, where the fluxes are: 3 3 3 3 ~4xy —4zy 1 1 , = _ = _4 9—4 ——— =3.47C yiD (18 f? A. 32 dzdy /2 A: 22 dxdy (9 )(4 9) \—,—/ \_W___-z _ . top bottom l “A uniform surface charge density of 20 nC/Im2 is present on the spherical surface 7- = 0.6 cm in free space. a) Find the absolute potential at P(r = 1 cm, 9 = 25°, ()5 = 50°): Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential function for r > 0.6 cm will be that of a. point'charge of Q = 47ra2p3, or 47r(0.6 x 10-2)2(20 x 10-9) _ 0.081 V with r in meters 41reo7' 7' ‘ V(7') = At 7' = 1 cm, this becomes V(r = 1cm) = 8.14 V b) Find VAB given points A(7' = 2cm, 6 = 30°, 05 = 60°) and B(7' = 3cm, 9 = 45°, 45 = 90°): & 1 Again, the angles do not matter because of the spheriCal symmetry. We'use the part a result to obtain - _J “\~~\.s_._\\ VAB = VA — VB = 0.081 [$ - 03%] = 1.36 V ,/ » - M_mwnl_n_mm Let a uniform surface ch ge-density of 5 nC/Iu2 be present at the z = 0 plane, a uniform line charge denslty of 8nC/m be located at :1; = 0, z = 4, and a point charge of 2nC be present __ Pl Pl A W(p)— f2weopdp+01=--—oln(p)+01 For the sheet charge, we have Q Pt p V = —~ _ _ _ _’_ 47reo1' 27re Imp) 2 02+0 The terms in this expression are not referenced to a common origin, since the charges are at 1different positions. The parameters 7', p, and z are scalar distances from the charges, and will e treated as such here. To evaluate the constant, C', we first look at point M where V - O T — . At M,r.= v22+5?=v29,p=1,andz=5.Wethus have 2 2x10.“6 8><10‘9 ' 5x10“9 - 0: _“1111 ._ =_ 3 Mew/2.9 2m () 260 5+0 .1 c. 1.9ex10 V At point N,r=\/1+4+9=x/fi, p=\/§, anda=3._Thepotential atNisthus _ 2 x 10-6 5 x 10-9 hem/.11 ‘ 27reo - 260 VN (3) 41.93 x 103 = 1.98 x 103V = 1.98 kV w In spherical coordinates, E = 2r/(r2'+ a2)2 a, V/m. Find the potential at any point, using w the reference - . a.) V = O at infinity: We write in general ‘ 1 V(T)=-/£dT—O+C= +0 (7'2 + 03)~ 7'2 -‘r- a2 With-a zero reference at 7' ——) 00, C' = 0 and therefore V(r) = l/(r2 + (1"). b) V = 0 at 7' = 0: Using the general expression, we find Therefore c) V = 100V at r = a: Here, we find 1 . , 1 _ _ V(a)=fi+C=100 => cam—W Therefore ~ - V 1 ' '1 100— “242 +100 (T) — 1'2 +112 — .235 + _ 2a2(r?+a2) Two uniform line charges, 8 nC/m each, are—located at a: = 1, z = 2, and at a: = —1, y = 2 in free space. If the potential at the origin is 100 V, find V at P(4, 1,3): The net potential function for‘the two charges would in general be: _ _L _ ._Pl_ V _ 2m ln(R1) 2m b.1022) + 0 At the origin, R1 = R2 = M5, and 4V = 100 V. Thus, with p; = 8 x 10-9, -9 ., . ' ' 100 = —2(8—X1—0_) 1n(«/§) + 0 => 0 = 331.6 V 271'60 At P(4, 1,3), R1 = |(4,1,3)—(1,1,2)| = mam a, ; [(4,1,3)—(—1,2,3)l = «2—6. Therefore VP = _(8_>2‘;1::__9) [mg/m) + 111N273] '+ 331.6 = —68.4 n @ Uniform surface charge densities of 6 and 2 nC/m2 are present at p = 2 and 6 cm respectively, in free space. Assume V = 0 at p = 4 cm, and calculate V at: a) p = 5 cm: Since V = O at 4 cm, the potential at 5 cm will be the potential difference between points 5 and 4: ‘ 5 ' .5 V -9 t ‘ V5=_ E.dL=_/ (1”5“d,o=—('0)(6)<10 )ln(§)=-3.026V b) p = 7 cm: Here we integrate piecewise from p = 4to p = 7: V7 = _ [5 upm d)” /7 (apm+bp,b) dp 4 , a EOP ‘ With the giten values, this becomeseép I - p . a W = _ [(.02)(6 x 10-9)] In (9) _ [(112) (6 x 10-9)'+ (.06)(§ x 104)] 111(7) 50 4 60 E = 29.6%8 v __._._r @ A certain potential field is given 'in spherical-coordinates by V = Vo(r/ a) sin 9. Find the total charge contained Within the region r < q: We first find the electric field through . 87‘ T69 —E-[s1n6’ar+cos0a9] The requested charge is now the net outward flux of D = 50E through the spherical shell of radius a. (with outward normal a,): ' _ 2m- 11' l I Q = / D - d5 =/ / 50E - a,,.a2 sin9d9 dgb = -21raVoeo /wsin2 9d9 = —7r2a.eoVE, C S o o o i The same result can be found (as expected) by taking the divergence of D and integrating over the spherical volume: v.D-_i3 (Hi/Esme) 1 a (flcoshinfi) =—59b-[2sin6+°°§(26)] — 1331‘ a _'rsin9% a m 5mg V - — V =— 6°.° [2sin29+1—2sin26]= 63° ..p,, msmfl mst Now r 27rI-1r a _ ' _ 2 a. ' Q=/ / f 60V" T251119 drded¢=Mf rdr=-,rzae,,v;, c 0 0 0 0 'rasin9 a Within the cylinder =>fi2, z < 1, the potential is given by V = 100 + 50p + 1,5'0p sin ¢>'V. 3.) Find V, E, D, and p1, at P(1,60°,O.5) in free space: First, substituting the given pomt, we find Vp = 279.9V. Then, E = —\7V = 331a,, — 1—3151, = — [50 + 150mm,, — [150cos ¢1a, 3/3 p 8¢> Evaluate the above at P to find Ep = —179.9ap —- 75.0% V/m Now D = 50E, so I); = —1.59ap 4 .664a¢ 110/1112. Then " 1 d 16D,, [1 . 1 . _ 50 = - = - — ————= —- 50+150sm +—150am¢ eo-——-_-eoC '0” V D (I?) dp(pD")+p at p( - (b) p p At P, this is pup = —443 pC/m3. b) How much charge lies the cylinder? We will integrate p1, 1 2W 2 5060 Q = f f / "TWP “‘45 dz .= .-27r(50)eo(2) = -5..56nc - 0 0 o . V y . over the volume to obtain: @ A spherical charge distribution is given by x ‘ . a - ‘3'} GWSS, \W ,2 ' ‘ _ _ 5 _J o . Aw 02>. r a m -= <st 0, r > a ' _— L 5:0 (3) FindEandeorraa. E . Awru . (b) FdeandeorrSa. r ‘ " Q9“. . h (c) Show that the maximum value of E is at r = 0.7450. __. . 6'0 (d) Find where V and calculate that value: ' \SeorL . ' "5 \/=- M:ch =_&‘_Ci +C‘ lSE.r' RV)?" \/<r9"‘\ :0) mm Q‘:Q 5 .. 7% V- ‘A—D‘O‘ j r 7/q \seuv l K9 “W Y<°°, iwev {vi-Shh TIN: s‘flnui r . ’L - ‘ QM ‘ j; PaCl ‘%~\4WL°‘V : 4W» “Br; '3 EY'AFSYF ‘ _ Isa” ' ‘ISt. 9‘ch —£: i—EC +C- [sea 6. no 6 “*8 c1: + 0% kc“ V 15‘ miwmu-M {W f'_<_ok wkla'e dv Q 3. J} :‘0 'Ac' — 31.30 100* ' b ‘H; 'EVLQ) mg r:. " = 17% V (=0 '°Or V( mm . QQVKU . - 1.: I U — 2 “W 1‘ 61‘! .= -:mv‘ ‘ +.—_'13._= [we r constant a imvz "—- .- o ' 2 ' - h“. _ ‘ 'o PE- 2 . ...
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Hmwk 4 Solutions - 30-; H“) Sehiw Let a point charge Q125...

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