Solution Manual For Introduction To Electrodynamics, 4th Edition

26 CHAPTER 2. ELECTROSTATICS Chapter 2 Electrostatics Problem 2.1 (a) Zero. 1 qQ , where r is the distance from center to each numeral. F points toward the missing q. 4π0 r2 Explanation: by superposition, this is equivalent to (a), with an extra −q at 6 o’clock—since the force of all twelve is zero, the net force is that of −q only. (b) F = (c) Zero. 1 qQ , pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as 4π0 r2 a cancellation in pairs of opposite charges (1 o’clock against 7 o’clock; 2 against 8, etc.), with one unpaired q doing the job, then you’ll need a different explanation for (d). (d) Problem 2.2 This time the “vertical” components cancel, leaving 1 E = 4π 2 rq 2 sin θ x̂, or 0 E= 1 qd x̂. 4π0 z 2 + d 2 3/2 2 – E AUA r θ AA z A A s As- x q −q 1 qd From far away, (z d), the field goes like E ≈ 4π 3 ẑ, which, as we shall see, is the field of a dipole. (If we 0 z set d → 0, we get E = 0, as is appropriate; to the extent that this configuration looks like a single point charge from far away, the net charge is zero, so E → 0.) c 1 Problem 2.3 CHAPTER 2. ELECTROSTATICS 27 1 Problem 2.3 Problem 2.3 z z ” !” ! θ η z dq = λdx #L !x “# $!x θ r z dq = λdx % L λ dx 1% 2 Ez =1 4π” =2 z 2 +2 x2 ; cos θ z= ηz ) L 2 cos θ; 2(η R λ ηcos 0 dx 0 λ dx 2 2 θ = η) z EzE = =4π”01 0 L θ; (η = z 2 η r =+z 2x+; xcos ; cos θ = r ) z 4π0 0 r 2 cos θ; ( % L 1 % 1 =1 4π”0 λzLR L0 (z12 +x2 )3/2 dx 1λz 1 = =4π” dxdx 2 3/2 ‘( λz0 &(z2 +x 0 2 ) 2 )3/2 (L 4π x ‘(L 10 & h0 1(z√+x 1 λ√ L . =1 4π”0 λz1 √z2 x z2 +x(2i L ( =1 4π” L zL 2 +L2 1λ 0√ λz√ . . = =4π”01λzλz z%2 12 z√2 +xx2 ( =0 =4π” 2 +L2 0 z z% z L 2 2 2 2 4π0 1% z λ dx 4π 0 1 x dx z +x +L %0λR zx dx 0 − E = −1 4π”0LR L 2 sin θ = 3/2 λ0 dx 1 4π” ηsin (x2x+z #L 1 1λ 0 dx2 ) ExEx=x =− − θ= − 4π” 2λ dx 2 +z 2 )3/2 & ‘( & ‘ ! sin θ = − λ 4π” η 0 0 0 (x 2 L 2 2 )3/2 4π10 & 0 r 4π0 1 & (x +z ! “# $ ( ‘( 1 1 1’ i L x iL h z−√ = −1 4π”0 λh √− √ = − . λ ( ( 1 1 1 1 2 2 2 2 x 4π” x1 +z = z1 +L. 1λ − √ 1λ 0 1 √√ = =− − −− − 0= 2 +z 2 ( 2 +L2 λ − λ − . 4π” 4π” z 0 0 x z 2 2 2 2 4π0 z z +L )* + * 4π0 + x, +z 0 0 λ z + * L + , 1 )* −1 + √z z x̂ + √L L ẑ . E =1 1 λ λ √√ z 2 + L2x̂ + √√ z 2 + L2ẑ ẑ. . ++ EE = = 4π% 0 z −1 −1 x̂ + 2 2 2 2 4π% z z+ L L2 z + L L2 0 z z 2+ 2+ 4π 0 1 λL For z # L you expect it to look like a pointz charge q = λL: E →1 4π” λL z 2 ẑ. It checks, for with z # L the x̂ 1 20λL For z# L Lyou expect it ittotolook like a point charge q = λL: E → ẑ. ẑ.It Itchecks, L Lthe 4π” For z you expect look like a point charge q = λL: E → checks,forforwith withz # z thex̂ x̂ 1 λL 0 z 2 4π term → 0, and the ẑ term →1 4π” 0 z λ L z ẑ. 1 0λzẑ. L term → 0, and the ẑ term → term → 0, and the ẑ term →4π”4π ẑ. 0 z z 0 z z Problem 2.4 Problem 2.4 -q 2 . /. a /2 Problem 2.4 2.1, with L → a and z → z +a 222 (distance from center of edge to P ), field of one edge is: From Ex. 2 z→ from center of of edge toto P ), field of of one is:is: z 2z+ From Ex. 2.1, with LL →→a2 aand 2 +2 a (distance From Ex. 2.2, with and z → (distance from center edge P ), field oneedge edge 2 2 1 λa λa. E =1 1 λa -0q 2 .a2. E1E=1= 4π% a2q 2 a2 + z + z + 2 2 2 1 4π%0 a 4 2 a 24 a 24 4π0 z 2z+ 2 +4 a2 z z+ 2 +4 a+ 4a 4 4 + 4 z There are 4 sides, and we want vertical components only, so multiply by 4 cos θ =q4 q z 2 a2 : There are 4 sides, and we want vertical components only, so multiply by 4 cos θ = 4 z 2+ :4 z q There are 4 sides, and we want vertical components only, so multiply by 4 cos θ = 4 z2 + a a2 : 4 z2 + 4 4λaz 1 4λaz E =1 1 . ẑ. / 4λaz 2 2 EE = = 4π% . 0 z 2 a+2 /aq 2 2 aẑ. ẑ. 4π% 0 z2 + 4 2 z a+ 2 4π 0 z 2 +4 a2 z z+ 2 +2 a2 4 Problem 2.5 Problem 2.5 Problem 2.5 ! ! θ z θ z r r r 2 0% 1 1 θ o ẑ. 10%nR λdl cos “Horizontal” components cancel, leaving: E =1 4π” 2 1 0 λdlλdl θ θẑ. ẑ. “Horizontal” components cancel, leaving: = =4π”4π “Horizontal” components cancel, leaving:E E 2 % ηcos 2 cos 0 0 % η Rrdl = 2πr. So Here,2 η2 2 =2 r2 2 +2 z2 2 , cos θ z= zηz (both constants), while Here, z z, cos θ θ==η (both constants), while So So Here, ηr ==r r++ , cos constants), whiledl =dl2πr. = 2πr. r (both 1 λ(2πr)z λ(2πr)z E =1 1 λ(2πr)z ẑ. 3/2 EE = = 4π ẑ. ẑ. 4π%00(r(r22++zz3/2 22 3/2 2 + z 2 ) )) 4π%0 (r Problem 2.6 Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the field of each ring. Total charge of a ring is σ · 2πr · dr = λ · 2πr, so λ = σdr is the “line charge” of each ring. 1 (σdr)2πrz 1 Ering = ; Edisk = 2πσz 3/2 2 2 4π0 (r + z ) 4π0 1 1 1 Z R r 0 3/2 (r2 + z 2 ) c “2005 Pearson Education, Inc., Upper Saddle River, rights reserved. Edisk = NJ. All2πσz − √ This material ẑ. is 2 + z 2 may be protected under all copyright laws as they currently exist. 4π0No portionz of thisRmaterial reproduced, in any form or by any means, without permission in writing from the publisher. c c “2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. dr. 1 CHAPTER 2. ELECTROSTATICS 28 1 For R z the second term → 0, so Eplane = 4π 2πσẑ = 0 σ ẑ. 20 −1/2 2 2 1 1 R2 R2 For z R, √R21+z2 = z1 1 + R ≈ 1 − , so [ ] ≈ z1 − z1 + 12 R 2 2 z z 2 z z 3 = 2z 3 , 2 1 2πR σ 1 Q 2 and E = 4π 2z 2 = 4π0 z 2 , where Q = πR σ. 0 Contents X z # Problem 2.7 E is clearly in the z direction. From the diagram, dq = σda = σR2 sin θ dθ dφ, r 2 = R2 + z 2 − 2Rz cos θ, cos ψ = z−Rrcos θ . z ψ r So θ R ” y φ Z R 1 σR2 sin θ dθ dφ(z − R cos θ) Ez = . dφ = 2π. 4π0 (R2 + z 2 − 2Rz cos θ)3/2 ! Z π x 1 (z − R cos θ) sin θ θ = 0 ⇒ u = +1 = (2πR2 σ) dθ. Let u = cos θ; du = − sin θ dθ; . 2 2 3/2 θ = π ⇒ u = −1 4π0 0 (R + z − 2Rz cos θ) Z 1 1 !2 σR21sin θ dθ dφ(z z − Ru ” − R cos θ)du. Integral = can be done by partial fractions—or look it up. 2 + z 2 − 2Rzu)3/2 . dφ = 2π. Ez4π = 0 (2πR σ) −1 (R 2 2 3/2 4π%0 (R + z − 2Rz cos θ) 1 ! # $ 1 1 2 1 π zu(z−−RR cos θ) sin θ 1 2πR2 σ (z − R) (−z − R) θ = 0 ⇒ u = +1 2 √ = = (2πR σ) = − . dθ. z 2 Let |zu − = R| cos θ; du =R| − sin θ dθ; . 2 −22Rzu 3/2 0 4π04π%0 (2πR σ) z 2 0R2(R |z + +2z+ θ = π ⇒ u = −1 z − 2Rz−1 cos θ)4π ! 1 1 z − Ru = (2πR2 σ) Integral can 1 be q done by partial fractions—or look it up. q 1 2 4πR2 σ 1 du. 2 4π%0the sphere),−1E(R For z > R (outside = 3/2 z0 −z2Rzu) 2 z =+ 4π 4π0 z 2 , so E = 4π z 2 ẑ. 0 &1 % # $ 1 2πR2 σ (z − R) (−z − R) 1 1 zu − R 2 For z R (outside thecenter, sphere), Ez all = exterior 2 , so E wereFor concentrated at the while contribute nothing. Therefore: 4π”0 z 2 shells 2 0 z 4π% z 0 For z < R (inside), Ez = 0, so E = 0. E(r) = 1 Qint r̂, 4π0 r2 Problem 2.8 to Prob. 2.7,interior all shells to Outside the pointthe (i.e. at smaller r) charge contribute as though where QAccording charge to interior the point. sphere, all the is interior, so their charge int is the total were concentrated at the center, while all exterior shells contribute nothing. Therefore: 1 Q E= 1 r̂.Qint r̂, E(r)4π =0 r 2 4π%0 r2 Inside the Q sphere, only that fraction of the total which is interior to the point counts: where int is the total charge interior to the point. Outside the sphere, all the charge is interior, so 4 πr3 r3 1 r3 1 1 Q Q 2 r̂ = r. Qint = 43 3 Q = 3 Q, so EE== 1 Q r̂. 3 3 R 4π R r 4π πR 2 0 0 R 4π%0 r 3 Inside 2.9 the sphere, only that fraction of the total which is interior to the point counts: Problem ∂ (a) ρ = 0 ∇·E = 0 r12 ∂r r2 · kr3 = 0 r12 k(5r4 ) = 50 kr2 . c "2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by anycmeans, without permission in writing from the publisher. 4 πr43 3 r3 3 1 r3 31 1 Q r so E = 1 Qr r̂ 1= 1 r.Q Qint = 43 33πr Q = 3 Q, 3 2 3 Qint = Q = Q, so E = Q r̂ = r. R 4π" R r 4π" 0 0R 3 3 2 3 3 πR4 πR R r3 4π"10 Rr3 r 1 4π"10 R3Q 4 3 πr3 Qint = 43 3 Q = 3 Q, so E = Q r̂ = r. Problem 2.9 R 4π"0 R3 r2 4π"0 R3 ! Problem 2.9 1 ∂ 2 ! 3 " 3"πR1 2 (a) ρ = "0 ∇·E = "0 r2 ∂r1 r∂ · kr = " k(5r4 ) = 5" kr . (a) ρ = "0 ∇·E = "0 r2 ∂r r2 · kr3 0=r2"0 r12 k(5r4 ) = 0 5"0 kr2 . Problem 2.9 !#2 " 1 ∂ 1 3 4 2 ) = 4π" (b) By law: =Q"enc )(4πR (a)Gauss’s ρ = "0 ∇·E · #kr3 =="0"(kR ) = 5" kr2 . kR5 . 0 r 2=∂r"0 r E·da 0 r 2 k(5r (b) By Gauss’s law: Qenc = "0$ E·da = "0 (kR3 )(4πR2 )0 = 04π"0 kR5 .$ $ R CHAPTER 2. ELECTROSTATICS 2 $ = R (5" $ R0 kr2 )(4πr By direct integration: Qenc = ρ#dτ dr) = 20π"0 k 5 0 r$4Rdr 429 = 4π" kR5 .! 0 3kr2 )(4πr 2 2 dr) = 20π" integration: ρ dτ = (5" k 0 r dr = 04π"0 kR5 .! 0 0 (b)ByBydirect Gauss’s law: QencQenc = "= E·da = " (kR )(4πR ) = 4π" kR . 0 0 0 $ $ $R 4 Problem 2.10 2 2 5 Problem 2.10integration: QencH = ρ dτ = R (5" By direct kr )(4πr dr) = 20π" k r dr = 4π" .!the surface 5 0 0 0 kRup 3 2 0 0 Think this cube as one of the charge. of0the 24 which make (b)Think ByofGauss’s = 80ofsurrounding = 0 (kR )(4πR )Each = Each 4π kR . squares enc one of thislaw: cubeQas 8E·da surrounding the charge. of the 24 squares which make up the surface Rother RR 4 R Problem 2.10 of this larger cube gets the same fluxRas every one, 2 so: 2 5 By larger direct integration: ρ dτ (5 kr )(4πr dr) = 20π k r dr = 4π kR .X of this cube gets theQsame as=every other one, so: enc = flux 0 0 0 0 0 squares which make up the surface Think of this cube as one of 8 surrounding the charge. Each of the 24 Problem 2.10cube gets the same flux as of this larger every other one, so: % % % the 1charge.% Each of the 24 squares which make up the surface Think of this cube as one of 8 surrounding 1 E·da. E·da = =one, so:% E·da. of this larger cube gets the same flux as every other 24 %E·da 24 one whole 1 Z Z face one largewhole face E·da = 1 large E·da. cube E·da = 24cube E·da. one %face 24 whole large % one q whole q .cube The latter is "10 q, 1by Gauss’s law. Therefore faceE·da = large = 0cube . The latter is "0 q, by Gauss’s law. Therefore %E·da24" 24" one q0 face one Z . The latter is "10 q, by Gauss’s law. Therefore face E·da = q 1 The latter is 10 q, by Gauss’s law. Therefore oneE·da = 24"0. 240 face one Problem 2.11 face Problem 2.11 1 # # = E(4πr2 ) =2 1 Qenc Gaussian surface: Inside: E·da Problem 2.11 1 = 0 ⇒ E = 0. " 0 Gaussian surface: Inside: E·da = E(4πr ) = "0 Qenc = 0 ⇒ E = 0. Problem"2.11 " (As in Prob. 2.7.) 2 # (As in Prob. 2.7.) ! 2 2= 0. 1 2 1 Qenc = 0σR ⇒ σR Gaussiansurface: surface:Outside: Inside: E(4πr E·da2= E(4πr E )= r̂. r ! ! Gaussian 1 ) = ")0 ⇒ 2 E= 2 "0 (σ4πR ! 2 Gaussian surface: Outside: E(4πr ) = "0 (σ4πR ) ⇒ E"= r̂. r r " 0 "0σR r2 2 (As in Prob. 2.7.) ! 1 2 2 ! Gaussian surface: Outside: E(4πr ) = "0 (σ4πR ) ⇒ E = r̂. r "0 r 2 Contents Contents Gaussian surface Problem 2.12 Problem 2.12 Problem 2.12 Problem 2.12 }} } # 1 4 1 34 #H = E · 4πr2 = 221 Qenc E·da πr ρ. 3 So E·da = =E E ·· 4πr 4πr "0= = 110Q Q= enc"0=3 1 4 πr3 ρ. So E·da "0 enc = "00 33πr ρ. So # Gaussian surface 21 1 E·da = E ·E4πr = ρrr̂. 1Qenc = "10 43 πr3 ρ. So r " "1 =E = 0 # r# " ρrr̂. E = 3"0 30ρrr̂. Gaussian surface 3"01 r " Q = 1 ρrr̂. # Q in Prob. 2.8). Problem 3E= πR42πR ρ, 2E r̂ Q(as Since QtotQ=tot43 = 31 3" Since ρ, E4π" E r̂ in (asProb. in Prob. R 2.13 0 R14π 0 ρ, == r̂3 (as 2.8).2.8). Since Qtot = 433πR 0 3R 4π"0 R $R % $ Gaussian surface Q 4 E 2· 2πs · l = 1!100 Q = !1in λl. So 2.8). Prob. Since QE·da tot = = 3 r̂ (as R 3 πR ρ, E = 4π" Renc 0 Problem 2.13 Problem 2.13 Problem 2.13 " $ Problem 2.13 ! s # %H Gaussian surface Gaussian surface λ= E · 2πs · l1= 1 1 Qenc1= 1 1 λl. So #E·da E·da Gaussian surface Problem 2.13 E·da =E ·E2πs ·l= = So So ŝ· 2πs E = (same 2.1). == · l =asQEx. = λl.10λl. 10Qenc enc ! "# $ E·da 2π#0 sE · 2πs · l"0= !0"0 Qenc"0= !0"0 λl. So & " % sl ! s% # & Gaussian surface s λ E · 2πs · l = 1 Q = 1 λl. So E·da enc E =λ λλ ˆ s (same as Eq. 2.9). "0 "2.1). 0 E= as Ex. ŝ=s(same E = 2π E = (sameas asEx. Ex.2.1). 2.1). 0 ŝŝ (same & & ! '( "# ) % 2π"2π# $ 0s 0s s & '( ) 2π"0λs l l l E= ŝ (same as Ex. 2.1). Problem 2.14 Problem 2.14 2π"0 s Problem Problem2.14 2.14& '( ) l H R R This2 material is Problem 2.14 c !2005 Pearson Education, Inc., Upper2 Saddle 1 River, NJ. 1 All rights reserved. =E · 4πr = currently QencRiver, = ρ portion dτ = 1of (kr̄)(r̄ This sin θmaterial dr̄ dθ dφ) c !2005 Pearson Education, Inc., All rights reserved. is protected under all E·da copyright laws as Upper they exist. No Saddle 0 NJ. 0 0 this material may be Rasr they protected all copyright currently Nowriting portion of this may be 4 Problem 2.14 reproduced, in under any form or by1anylaws means, without permission in from the material publisher. 4πk rexist. πk 3 4 = or kInc., 4π r̄ dr̄Saddle = 0River, rinrights . writing in anyEducation, form any Upper means, without permission from the publisher. Gaussian reproduced, surface c !2005 Pearson NJ.0All reserved. This material is 0 by 4 = 0 protected under all copyright laws as they currently exist. No portion of this material may be r# $ reproduced, in any form means, without permission in writing from the publisher. 1 or by any 2 ∴ E = πkr r̂. Gaussian surface 4π0 r $ # Problem 2.15 c (i) Qenc = 0, so E = 0. Problem % 2.15 & & (ii) E·da = E(4πr2 ) = !10 Qenc = !10 ρ dτ = !10 r̄k2 r̄2 sin θ dr̄ dθ dφ (i) Qenc = 0, so E = 0. ' ( &r k r−a 4πk 4πk &r̂.k 2 =% !0 a dr̄ = !02 (r −1a) ∴ E =1 & 2 1 |E| (ii) E·da = E(4πr ) = !0 Qenc = !0 #0ρ dτ r= dφ !0 r̄ 2 r̄ sin θ dr̄ dθ ! ' ( & k r−a & 4πk 2r 4πk4πkb(r − a)4πk r̂. ∴ E = so 2 (iii)=E(4πr == !0 a) dr̄ |E| !0 !0a dr̄ = !0 (b − a), # r 0 ! ' ( k b−a & r̂. E= 2 b 2 dr̄ = 4πk (iii) E(4πr #0 ) =r4πk !0 !0 (b − a), so a ' ( % Gaussian surface r $ # 111 Contents 30 CHAPTER 2. ELECTROSTATICS Problem 2.15 Qenc = 0, so E = 0. Problem(i)2.15 & & (i) Qenc =%0, so E = 0. 2 ) = !10 QencR= !10 ρ dτR= !10 r̄k2 r̄2 sin θ dr̄ dθ dφ H (ii) E·da = E(4πr 1 1 2 2 '1 k ( Problem 2.13 (ii) E·da = E(4πr 0 Qenc = 0 ρ dτ = & r ) =4πk k 0r −r̄2ar̄ sin θ dr̄ dθ dφ 4πk Problem r̂. dr̄ = !0 (r − a) ∴ kE =r − a 2 = R r !0 a2.16 |E| #0 rr̂. 4πk ! = 4πk dr̄ = (r − a) ∴ E = 0 0 a 2 0 r & 4πk 4πk b dr̄ = (b − a), so (iii) E(4πr2 ) = R !0 a b !0 ' a dr̄ ( (iii) E(4πr2 ) = 4πk = 4πk 0 0 (b − a), so k b−a ! E=b − a 2 r̂. Problem Problem 2.13 ! E·da = E · 2πs · l = !10 Qenc = !10 ρπs2 l; Problem2.13 2.13k(i) Gaussian surface #0 r E= r̂. 2 % Problem 2.16 Problem Problem 2.16 2.16 0 r ρs l r b ŝ. a E= 2#0 Problem 2.16 2.16 Problem H ! E·da = E · 2πs · l = 1 Q = 10 ρπs2 l; Gaussian c !2005 Pearson Education, Inc., Upper NJ. All! reserved. This material is0 enc !rights !surface " Saddle River, 222 under all copyright! laws they currently exist. No portion of may (i) Q l; E·da E 2πs s asGaussian (i) Q = ρπs l;l; ! E·da = EE·!·ρs 2πs = (i) Qenc = !!110!010ρπs ρπs ! E·da= =this ·material 2πs···lll= = !!110!010be (i) protected(ii) 1 1 enc 2 enc= surface Gaussian surface Gaussian surface E·da =E · 2πs ·l = E= ˆ s.publisher. reproduced, in any form or by any means, without permission in writing from the !0 Qenc = !0 ρπa l; ρs ρs ρs 2 0 lll 2 ŝ. E ŝ. E = ŝ. E= = 2# 2# 2#000 E = ρa ŝ. 2#0 s l Contents Contents (ii) (ii) (ii) (ii) " " " sss (iii) lll ! H ! ! "Gaussian Gaussian surface Gaussiansurface surface E·da = E · 2πs ·1l = 10 Qenc = 10 ρπa2 l; ! Gaussian!!!surface s 222 E·da ·!··2πs l; E·da = EEρa 2πs = Q = ρπa l;l; E·da= =E 2πs = !!110!010Q Qenc = !!110!010ρπa ρπa 2 ···lll= enc enc= 1 Q E·da = E · 2πs · l = E = 222 ˆ s. !0 enc = 0; ρa ρa ρa 20 s ŝ. E ŝ.ŝ.E = 0. E = E= = 2# 2# 2#000sss l ! ! ! Gaussian Gaussian surface Gaussiansurface surface " " " sss !!!H 1 = E·da ···ll·l= E·da= ·2πs 2πs l==!!110!01Q Q = 0; E·da = EEE···2πs 2πs = QQ ==0; 0;0; E·da ==E enc enc enc enc 00 E = 0. E = 0. EE == 0.0. (iii) (iii) (iii) (iii) Contents |E| " lll Problem 2.17 # |E| |E| |E|a" On the x z plane E = 0 by symmetry. Set up Gaussian “pillbox” with and the a one face b in this plane s " " other at y. c2.17 Pearson Upper River, NJ. Allup rights reserved. This material with is Problem!2005 On theEducation, x z planeInc., E= 0 bySaddle symmetry. Set a Gaussian “pillbox” one face in this plane protected under all copyright laws as they currently exist. No portion of this material may be and the other at y. reproduced, in any form or by any means, without permission in writing from the publisher. Gaussian pillbox # # 1 1 # E·da = aa E a · A = 0bbbQenc = 0ssAyρ; s ρ E= y ŷisisis(for |y| d). “0 E$ ρd “0 −d d ! y Problem 2.17 On the x z plane E = 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the other at y. ! Gaussian pillbox # CHAPTER 2. ELECTROSTATICS E·da = E · A = !10 Qenc = !10 Ayρ; ρ E = y ŷ (for |y| d). Qenc = 0 Adρ ⇒ E = “d0 ŷ (for y > d). 0 E$ ρd “0 −d d ! y Problem 2.18 Problem 2.18 From Prob. 2.12, the field inside the positive sphere is E+ = 3ρ0ρr+ , where r+ is the vector from the positive From Prob. 2.12, the field inside the positive sphere is E+ = 0 r+ , where rρ+ is the vector from the positive center to the point in question. Likewise, the field of the negative3!sphere is − 30ρr− . So the total field is center to the point in question. Likewise, the field of the negative sphere is − 3!0 r− . So the total field is ρ EE== ρ(r(r r−r) ) + −− “r−r− r + − 33″ *&−− 0 0 % r+r+ d ρρ d r But (see diagram) r − r = d. So E = d. + − But (see diagram) r+ − r− = d. So E =30 d. ++ 3″0 Problem Problem2.19 2.19 Z” r̂η̂ ρ dτ = 11 Z” #∇× $ η̂r̂ %& ρ dτ (since ρ depends on “r0 , not r) 11 ∇×E = ∇× ∇× r 22 ρ dτ = ∇× r2 2 ρ dτ (since ρ depends on r , not r) ∇×E = 4π 4π 0 0 4π” η 4π”00 η $ % r̂ η̂ ==0 0 (since 1.63). (since∇× ∇× 22 = 0, from Prob. 1.62). rη Problem Problem2.20 2.20 ‘ ‘ ‘ x̂ ŷ ŷ ẑ ẑ ‘ x̂ ‘∂ ∂ ∂ ∂ ∂ ∂ ‘ ‘ ‘ =k k[x̂(0 (1) ==k k∂x (1)∇×E ∇×E [x̂(0−−2y) 2y)++ŷ(0 ŷ(0−−3z) 3z)++ẑ(0 ẑ(0−−x)] x)]6=#=0,0, 1 1 ‘ ∂x ∂y∂y ∂z∂z = ‘ ‘ xy2yz xy 2yz3zx 3zx’ sosoEE ananimpossible impossibleelectrostatic electrostaticfield. field. 1 1is is ‘ ‘ ‘ x̂ x̂ ŷ ŷ ẑ ẑ ” ‘∂ ∂ ∂∂ ∂∂ ‘ ‘ [x̂(2z−−2z) 2z)++ŷ(0 ŷ(0−−0)0)++ẑ(2y ẑ(2y−−2y)] 2y)]==0,0, (2)∇×E ∇×E (2) ==k k∂x [x̂(2z 2 2 ‘ =k k ‘ ∂x ∂y∂y ∂z∂z = 2 2 ‘ 2y 2xy + 2z 2yz ‘ y 2xy + z 2yz possibleelectrostatic electrostaticfield. field. sosoEE a apossible 2 2is is z 6 Let’s go by the indicated path: c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be any form or by any means, without permission in writing from the publisher. 2 r(x0 , y0 , z0 ) E·dl = (y 2 dx + (2xy + zreproduced, )dy + 2yzindz)k 6 Step I: y = z = 0; dy = dz = 0. E·dl = ky 2 dx = 0. III Step II: x = x0 , y : 0 → y0 , z = 0. dx = dz = 0. E·dl = k(2xy + Rz 2 )dy = 2kx0 y dy. R y E·dl = 2kx0 0 0 y dy = kx0 y02 . II Step III : x = x0 , y = y0 , z : 0 → z0 ; dx = dy = 0. E·dl = 2kyz dz = 2ky0 z dz. c x I = = II – -y 32 R III CHAPTER 2. ELECTROSTATICS Rz E·dl = 2y0 k 0 0 z dz = ky0 z02 . (x0 ,y R0 ,z0 ) E·dl = −k(x0 y02 + y0 z02 ), or V (x, y, z) = −k(xy 2 + yz 2 ). V (x0 , y0 , z0 ) = − 0 ∂ ∂ ∂ Check : −∇V =k[ ∂x (xy 2 +yz 2 ) x̂+ ∂y (xy 2 +yz 2 ) ŷ+ ∂z (xy 2 +yz 2 ) ẑ]=k[y 2 x̂+(2xy+z 2 ) ŷ+2yz ẑ]=E. X Problem 2.21 Rr V (r) = − ∞ E·dl.  1 q Outside the sphere (r > R) : E = 4π0 r2 r̂. q 1 E = 4π 3 rr̂. 0 R  Inside the sphere (r R: V (r) = − ∞ 4π dr̄ = 4π q 2 0 r̄ 0 1 r̄ r ∞ = q 1 , 4π0 r h 2 2 i RR 1 q Rr 1 q q 1 1 r −R and for r R, ∇V = 4π 0 ∂r q q 1 1 r̂ = − 4π 2 r̂, so E = −∇V = 4π r 2 r̂. X 0 r 0 2 q q q q 1 ∂ 1 2r r 1 When r < R, ∇V = 4π 3 − Rr 2 r̂ = 4π −R r̂ = − 4π 2 3 r̂; so E = −∇V = 4π R3 rr̂.X 0 2R ∂r 0 2R 0 R 0 1 r V(r) 1.6 1.4 1.2 1 (In the figure, r is in units of R, and V (r) is in units of q/4π0 R.) 0.8 0.6 0.4 0.2 0.5 1 1.5 2 2.5 3 r Problem 2.22 1 2λ E = 4π ŝ (Prob. 2.13). In this case we cannot set the reference point at ∞, since the charge itself 0 s extends to ∞. Let’s set it at s = a. Then s R s 1 2λ 1 V (s) = − a 4π ds̄ = − 2λ ln . 0 s̄ 4π0 a (In this form it is clear why a = ∞ would be no good—likewise the other “natural” point, a = 0.) 1 ∂ 1 ∇V = − 4π 2λ ∂s ln as ŝ = − 4π 2λ 1s ˆ s = −E. X 0 0 Problem 2.23 R0 Rb Ra R0 V (0) = − ∞ E·dl = − ∞ k0 (b−a) dr − b k0 (r−a) dr − a (0)dr = k0 (b−a) − k0 ln r2 r2 b k b = k0 1 − ab − ln ab − 1 + ab = ln . 0 a Problem 2.24 Using Eq. 2.22 and the fields from Prob. 2.16: Rb Ra Rb Ra 2 Rb 1 V (b) − V (0) = − 0 E·dl = − 0 E·dl − a E·dl = − 2ρ0 0 s ds − ρa 20 a s ds c a b +a 1 1 a − b Contents CHAPTER 2. ELECTROSTATICS =− ρ 20 s2 2 a 0 2 b + ρa 20 ln s|a = − 33 ρa2 40 1 + 2 ln b . a Problem Problem2.25 2.25 11 2q 2q q .. ! 4π 4π"00 z 22+ "dd#22 z + 22 √ R L L 1 λ √z 2 + x2 ) %L $ L √ λ2λdx (b) V = 4π 10 −L dx2 = 4πλ0 ln(x + 2 + x2 )−L % z +x √ (b) V = 4π" z = ln(x + 4π"0 −L z 2 +x2 −L 0 " # ! √ √ ' & √ 2 2 λ L + z 2+ L 2 λ z2 + ( L+ ) L2 √ λ L + z + L √ = ln = λ ln L+ z2 +L2 . = 4π0 ln −L + √z 2 + L2 . = 2π 2π"00 ln z z 4π"0 −L + z 2 + L2 (a) (a) VV == z %R √ $ σ * 2 1 Z R σ 2πr dr 1 2p 2 )% = R (c) V = 4π" r + z = 2πσ ( 1 0 0 R√σr22πr dr 1 σ R p+ z22 − z2 . 2 4π"0 √ +z (c) V = = 2πσ ( r2 +0z 2 ) 2" =0 R +z −z . 4π0 0 4π0 20 0 r2 + z 2 ∂V In each case, by symmetry ∂V ∴ E = − ∂V ∂y = ∂x = 0. ∂z ẑ. ∂V In each case, by symmetry ∂V = = 0. ∴ E = − ∂V ∂y ∂x ∂z ẑ. ( ) r x " # 2qz 1 1 2z ”3/2 ẑ = (a) E = − 4π" 2q − 21 “ " #3/2 ẑ (agrees with Prob. 2.2a). 0 d 2 2 4π"0 z12 + " d #22qz 1 1 z +( 2 ) 2z 2 (a) E = − 2q − ẑ = ẑ (agrees with Ex. 2.1). 3/2 4π0 + 2 2 4π0 z 2 + d 2 3/2 , d 2 z + 2 1 √ 21 λ √1 2z − (−L+√1z2 +L2 ) 12 √z21+L2 2z ẑ (b) E = − 4π" 2 2 2 0 z 2 +L2 (L+ z +L ) λ 1+ 1 √ 1 1 1 , √ 1 1 2Lλ 2 2 2 2 √ −L− z +L √ (agrees (b) E = − 2z − ẑ = √ √ 2z with ẑ Ex. 2.1). λ √ z √ −L+ z +L =4π − 4π" 2 2 2 + 2L2 (−L +4π"z02z+ zL22 + ) 2L2 zẑ2 + L2 0 0 (L z 2+ +L2 z + L )(z22 +Lz2 )−L ( ) √ √ λ+ z −L + z 2 +-L2 − L − z 2 + 2Lλ 1 . L2 , √ = −σ 1 √ 1 ẑ = ẑ (agrees with Ex. 2.2). σ2 z2 2 2 2 2 (z +1L−)√ −L 4π0 zProb. z +2.6). L2 0 √ z 2 +2L ẑ (agrees with (c) E = − 2"4π 2 R +z 2z − 1 ẑ = 2" 0 R2 + z 2 0 σ 1 1 σ z the in − (a)1 is ẑ−q, 0 , which, naively, suggests E = −∇V √ charge 2z (c) EIf = −right-hand = then V 1 −= √ ẑ (agrees with Prob. 2.6). = 0, in contradiction 2 + z2 2 20 2 toRProb. 20 is that R z 2 know V on the z axis, and from this we cannot with the answer 2.2b. The point we + only ∂V hope to compute Ex = − ∂V ∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that If the right-hand charge in (a) is −q, then V = 0 ,sowhich, naively, E =is −∇V = 0, in Ex = Ey = 0. But now E points in the x direction, knowing V onsuggests the z axis insufficient tocontradiction determine E. 2 with the answer to Prob. 2.2. The point is that we only know V on the z axis, and from this we cannot hope ∂V to compute Ex = − ∂V ∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that Ex = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insufficient to determine E. Problem 2.26 " c #2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. b h Z √2h 2πσ 1 √ σh √ ( 2h) = dr = r 4π 2 2 0 0 0 √ (where r = r / 2) 1 V (a) = 4π0 c σ2πr h r̄ r a ! 34 1 V (b) = 4π0 Z √2h 0 2πσ 1 √ 4π0 2 Z √ σ2πr r̄ CHAPTER 2. ELECTROSTATICS dr (where r̄ = q h2 + r 2 − √ 2h r ) r 2h p dr √ 0 h2 + r 2 − 2h r √ q 2h q √ √ √ σ h = √ h2 + r 2 − 2h r + √ ln(2 h2 + r 2 − 2h r + 2 r − 2h) 2 20 2 = 0 √ √ √ σ h h = √ h + √ ln(2h + 2 2h − 2h) − h − √ ln(2h − 2h) 2 20 2 2 √ ! h i √ √ σ h σh 2+ 2 σh √ ln(2h + 2h) − ln(2h − 2h) = √ = √ ln = ln 40 40 2 20 2 2− 2 √ √ i σh σh h = ln(1 + 2). ∴ V (a) − V (b) = 1 − ln(1 + 2) . 20 20 (2 + √ 2)2 ! 2 3 Problem 2.27 Problem 2.27 Cut the Cut cylinder into slabs, shownasinshown the figure, the cylinder intoasslabs, in theand figure, and use result of result Prob. of 2.25c, z →with x and σ x→and ρ dx: use Prob.with 2.25c, z→ σ → ρ dx: z− L 2 √z+L/2 % &√ ' ρ R2 + x2 −2x dx2 R + x − x dx 2"0 z−L/2 z−L/2 V = 2ρ0 V = # √ √ √ ( √ 2 z+L/2 2 )*z+L/2 = 2ρ0 12 = x ρR21 +xx2 R +2R+2 ln(x R2 ++x2 )R−2 x x2 ++ R2 ln(x + x2z−L/2 )−x * 2"0 2 z−L/2 8 9 3 r 2 3 9 8 r 2 q q L + R2 + z+ L 2 < = z+ q q ( L +2 )R2 + z+ L 2 = < 2 2 ρ z+ 2 L L 2 L 2 L 2 2 ( 4 2 5−2zL2 ) . 5 r 2 ρ z+ R + z+ − z− R + z− +R ln 2 L L L L ( ) ( ) ( ) ( ) 2 2 4 r2 40 : = 2 2 R +(z+ 2 ) −(z− ) 2 R +(z− ) +R ln −2zL . 4"0 :(z+ 2 ) 2 2 2 z− L + R2 + z− L 2 ; ( L +2 )R2 +(z− L; 2 z− ) 2 (Note: L "# $! x !" " ! "# $ z+L/2 R = ! $ dx 2 2 2 L2 L2 2' 2 2 2 −(Note: z + L2− &z++zL−'2L2+ &z=−−z L 2− zL −2 4 + z − L zL +2 4 = −2zL.) = −z − zL − + z − zL + L4 = −2zL.) 2 2 4 s (s 2 2 2 , , L 2& + ' '2 z + . .z2 − L2 & ∂V ẑρ L L2 L 2 2 2 2 z + −2 R + z −2 z − L2 E = −∇V = −ẑ = −∂V Rẑρ+ z +2 + qL − qL 2 2 / / ∂z= −ẑ 40 = − 2 z+ 2 + 2 z− 2 − R + E = −∇V L L '2 − R + '2 ∂z 4$0 2R + z +22 & 2R + z −2 2 & R + z + L2 R + z − L2 L L z+ 2 z− 2 " ) 1 +0 q 1+ q z− L # z+ L 2 L 2 1 2 2+ 1 (+z−qL2 )2 2 L 2 1 + R2 + z+q R ( ) 2 2 L 2 2 2 R +(z+ 2−) R +(z− 2 )− 2L q +R 2 q 2 2 − +R 2+ / 2+ / ' − 2L z + L2 + RL z +2L2 & z L−'2L2 + RL z −2L2 & L 2 z + 2 + R + z{z+ 2 z − 2 + R + }z − 2 | $! 1 " # 1 q q − 1 1 2 / − / L 2& '22 + '2 R2 + z +2L2 & R z −22 R + z + L2 R + z − L2  s  s 2 2     , L , L ẑρ .2  E=− 2 ẑρ R2  + z + – − 2L .R22 + z − – − 2L 4E0  2 z+ 2 z − L  − 2L =− − 2 R2 + 2 R2 +  4$0  2 2   , c .2 , .2 ρ  L L  ẑ. = L − R2 + z + + R2 + z − 2$0 2 2 Problem 2.28 z $ Orient axes so P is on z axis. P V = 1 %ρ dτ. = 2 Here√ρ is constant, dτ = r sin θ dr dθ dφ, z θ r ! 1 Hello CHAPTER 2. ELECTROSTATICS 35   s 2 s 2 ρ  L L  ẑ. = L − R2 + z + + R2 + z − 20 2 2 z # Problem 2.28 Orient axes so P is on z axis. R ρ Here ρ√is constant, dτ = r2 sin θ dr dθ dφ, 1 V = 4π0 r dτ. r = z 2 + r2 − 2rz cos θ. ρ V = 4π 0 Rπ 0 √ ∴V R 2 √r sin θ dr dθ dφ ; z 2 +r 2 −2rz cos θ r P z θ r "y φ R 2π 0 dφ = 2π. x! √ √ π 1 r2 + z 2 − 2rz cos θ 0 = rz r2 + z 2 + 2rz − r2 + z 2 − 2rz 1 2/z , if r z. sin θ 1 dθ = rz z 2 +r 2 −2rz cos θ ρ = 4π · 2π · 2 0 √ ( ) n 3 o Rz 1 2 RR 1 2 ρ ρ 1z R2 −z 2 z2 2 r dr + r dr = + = R − . z r 0 z 3 2 20 3 z 0 q 3q q 1 z2 z2 2 But ρ = 4 πR , so V (z) = R − = 3 − 3 20 4πR3 3 8π0 R R2 ; V (r) = Contents 3 q 8π0 R 3− r2 R2 . X Problem 2.29 R R 1 1 ∇2 V = 4π ∇2 rρ dτ = 4π ρ(r0 ) ∇2 r1 dτ (since ρ is a function of r0 , not r) 0 0 R 1 = 4π ρ(r0 )[−4πδ 3 (r − r0 )] dτ = − 10 ρ(r). X 0 Problem 2.30. Problem 2.30. (a) (a)Ex. Ex.2.5: 2.4: EEabove = 2σ2″σ00n̂; n̂;EEbelow =−−2σ2″σ00n̂n̂(n̂ (n̂always alwayspointing pointingup); up);EEabove −EEbelow =σ0″σ0n̂. n̂.X! above= below= above− below= Ex. Ex.2.6: 2.5: At Ateach eachsurface, surface,EE==00one oneside sideand andEE==σ0″σ0 other otherside, side,so so∆E ∆E==σ0″σ0. .X! 22 σR Prob. r̂ = σσr̂ ; Einin==00; ;so Prob.2.11: 2.11: EEout = σR so∆E ∆E== σ0″σ0r̂.r̂.X! out= “0rr2 2 r̂ = 0″0 r̂ ; E 0 ! s (b) (b) ! R H! Outside: E·da E·da==E(2πs)l E(2πs)l== 1″10QQenc = σ”10(2πR)l (2πR)l⇒ ⇒EE== σ”σ0RRsŝŝ== σ”σ0ŝŝ(at (atsurface). surface). enc= Outside: 0 0 0 s 0 c σ !2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is Inside: soEE== ∆E enc Inside: QQenc ==0,0, so 0.0. ∴∴ ∆E ==σ0″of0ŝ.ŝ. X!material may be protected all copyright laws as they currently exist. No portion this % ” #$ under reproduced, in any form or by any means, without permission in writing from the publisher. l 2 (c) Vout = RR0 r2σσ = Rσ (at surface); Vin = Rσ ; so Vout = Vin . X Rσ 0 0 (c) Vout = “0 r = Rσ “0 (at surface); Vin = “0 ; so Vout = Vin . ! ∂Vout ∂Vin ∂Vin R2 σ σ out = 0 ; so ∂V = − σ0σ. X ∂rout = − 0Rr22σ = − 0σ (at surface); ∂r ∂rout − ∂r ∂Vin ∂V ∂Vin ∂V = − = − (at surface); = 0 ; so − ∂r “0 r 2 “0 ∂r ∂r ∂r = − “0 . ! c Problem 2.31 * ‘ ( ) −q q 1 √q + −q √1 . = −2 + = + rij 4π”0 a a 4π”0 a 2a 2 , + 2 q 1 . ∴ W4 = qV = −2 + √ 4π”0 a 2 1 (a) V = 4π” 0 & qi 1 ) −q2 * 1 ) q2 q2 * (1) (4) − + + − 36 CHAPTER 2. ELECTROSTATICS Problem 2.31 P qi n q2 ∴ W4 = qV = 4π0 a 1 (a) V = 4π 0 1 rij = 4π0 −q −q √q a + 2a + a 1 −2 + √ 2 o = 4πq0 a −2 + √12 . (1) r . 2 2 −q q2 1 1 √q (b) W1 = 0, W2 = 4π ; W = − 3 a 4π0 a ; W4 = (see (a)). 0 2a n o 2 1 2q 2 1 1 q √1 − 1 − 2 + √1 √ Wtot = 4π −1 + = −2 + . 0 a 2 2 4π0 a 2 (2) r(4) − + r+ −r (3) Problem 2.32 Conservation of energy (kinetic plus potential): 1 1 1 qA qB 2 2 mA vA + mB vB + = E. 2 2 4π0 r At release vA = vB = 0, r = a, so E= 1 qA qB . 4π0 a When they are very far apart (r → ∞) the potential energy is zero, so 1 1 1 qA qB 2 2 mA vA + mB vB = . 2 2 4π0 a Meanwhile, conservation of momentum says mA vA = mB vB , or vB = (mA /mB )vA . So 1 1 2 mA vA + mB 2 2 s vA = mA mB 2 1 qA qB 2π0 (mA + mB )a 2 vA = 1 2 mA ; mB mA mB 2 (mA + mB )vA = s vB = 1 qA qB . 4π0 a 1 qA qB 2π0 (mA + mB )a mB . mA Problem 2.33 From Eq. 2.42, the energy of one charge is W = ∞ ∞ X 1 1 1 (−1)n q 2 q 2 X (−1)n qV = (2) = . 2 2 4π0 na 4π0 a 1 n n=1 (The factor of 2 out front counts the charges to the left as well as to the right of q.) The sum is − ln 2 (you can get it from the Taylor expansion of ln(1 + x): 1 1 1 ln(1 + x) = x − x2 + x3 − x4 + · · · 2 3 4 with x = 1. Evidently α = ln 2 . c 2 CHAPTER 2. ELECTROSTATICS 37 Problem 2.32 ! Problem 2.34 (a) W = 21 ρV dτ . From Prob. 2.21 (or P R 2 2 q 1 (a) W = 12 ρV dτ . From Prob. 2.21 (or Prob. 2.28): V = 2ρ0 R2 − r3 = 4π 3 − Rr 2 & $ R% 0 2R 1 1 q r2 W = ρ 3 − 2 4πr2 dr 3 Z R 2 R4π$ R 5 R 3 0 2R 0 1 1 q r2 qρ r 1 r qρ W = ρ 3 − 2 4πr2 dr = 3 − 2 = R3 − % 2 2 4π0 2R 0 R 40 R 3 R 5 0 40 R qρ5 2 qR2 q 1 3q = = R = 4 2 3 5$ 5$ 4π$ 5 R 2 πR 0 0 3 0 qρ 2 qR q 1 3q = R = = . 50 50 43 πR3 4π0 5 R ! (b) W = “20 E 2 dτ . Outside (r > R) E = R q 1 q 1 *$ (b) W = 20 E 2 dτ . Outside (r > R) E = 4π 2 r̂ ; Inside (r R. (c) W = 20 V E·da + V E 2 dτ , where V is large enough to enclose all the charge, but otherwise S 1 q *$ % arbitrary. Let’s use a sphere of radius a > R. Here V = 4π0 r . &% & 1 q 1 q $0 r2 sin W (Z ) 2 2= 2 Z R Z a 4π$ r 4π$ r 0 0 0 1 q 1 q 1 q 2r=a W = r2 sin θ dθ dφ + E 2 dτ + (4πr/ dr) 2 2 4π0 r 4π0 r2 4π r 0 0 R $0 q2 1 q2 4π r=a = 4π + + 2 2 5R a 2 (4π$ ) a (4π$ ) ( 2 2 0 0 0 q 1 q 4π 1 1 2 / 0 = 4π + + 4πq − 2 1 q 1 1 1 1 2 (4π0 )2 a (4π0 )2 5R (4π0 )2 r R = = + − + 4π$ 2 a 5R a R 4π 0 1 q2 1 1 1 1 1 3 q2 = + − + = .X 4π0 2 a 5R a R 4π0 5 R As a → ∞, the contribution from the surf ” # 2 1 q 1 q2 while As a → ∞, the contribution from the surface integral 4π0 2a goes to 4π” zero, the picks volume ( 6a − 1) up integral the slack. 0 2a 5R 2 1 q 6a Problem 2.33 4π0 2a ( 5R − 1) picks up the slack. Problem 2.35 dW = dq̄ V = dq̄ q̄ = 4 3 r3 πr ρ = q 3 3 R 1 4π0 q̄ , r (q̄ = charge on sphere of radius r). (q = total charge on sphere). 4πr2 3q dq̄ = 4πr dr ρ = 4 3 q dr = 3 r2 dr. R πR 3 3 1 qr 1 3q 2 1 3q 2 4 dW = r dr = r dr 3 3 4π0 R r R 4π0 R6 2 Z 1 3q 2 R 4 1 3q 2 R5 1 3q W = r dr = = .X 6 6 4π0 R 0 4π0 R 5 4π0 5 R 2 c ! r ” dq̄ q̄ c “2005 Pearson Ed protected under al reproduced, in any 38 CHAPTER 2. ELECTROSTATICS Problem 2.36 R 1 q (a) W = 20 E 2 dτ. E = 4π 2 (a < r a), E2 = 4π r 2 r̂ (r > b). 0 b 0 r 0 02 a 2 2 R R −q q2 1 1 2 ∞ 1 2 E1 · E2 = 4π0 E1 · E2 dτ = − 4π0 q b r4 4πr dr = − 4π . r 4 , (r > b), and hence 0b R 2 q 1 1 1 Wtot = W1 + W2 + 0 E1 · E2 dτ = 8π q 2 a1 + 1b − 2b = 8π a − b .X 0 0 Problem 2.37 z r q2 a q 1 q1 r̂; 4π0 r2 y q1 x E1 = b r E2 = 1 q2 r̂ ; 4π0 r 2 r = p Wi = 0 q1 q2 (4π0 )2 Z 1 r2 2 r 2 cos β r sin θ dr dθ dφ, where (from the figure) r2 + a2 − 2ra cos θ, Therefore Wi = q1 q2 2π (4π)2 0 Z cos β = (r − a cos θ) (r − a cos θ) r . r 3 sin θ dr dθ. It’s simplest to do the r integral first, changing variables to r : 2 r d r = (2r − 2a cos θ) dr ⇒ (r − a cos θ) dr = r d r . As r : 0 → ∞, r : a → ∞, so Z Z ∞ q1 q2 π 1 r Wi = d sin θ dθ. r2 8π 0 0 a q1 q2 8π0 a Z π The r integral is 1/a, so Wi = sin θ dθ = 0 q1 q2 . 4π0 a Of course, this is precisely the interaction energy of two point charges. Problem 2.38 (a) σR = q −q q ; σa = ; σb = . 2 2 4πR 4πa 4πb2 c Problem 2.34 ! 1 q (a) W = !20 E 2 dτ. E = 4π! 2 (a < r a), E2 = 4π! 2 r̂ (r > b). 0 b 0 r 0 r ” ” #02 a 2 # CHAPTER 2. ELECTROSTATICS 39 2 ! ! ∞ q2 −q 1 1 E1 · E2 = 4π! E1 · E2 dτ = − 4π! . q 2 b r14 4πr2 dr = − 4π! r 4 , (r > b), and hence 0 0 0b $ $ % % ! 2 q 1 1 2 1 1 WRtot = W1 + W − 1b .! = R8π! R 2b + #10 qE1 · E2 Rdτa = 8π!0 qR2 R a +1 b − 1 q q q 0 0 0 a q b (b) V (0) = − ∞ E·dl = − ∞ 4π dr − (0)dr − dr − (0)dr = + − . 2 2 b a 4π0 r R 0 r 4π0 b R a Problem 2.35 q −q q R0 (a) σR = ; σa = ; σ R=a . RR 1 q (c) σb → 0 (the charge “drains (0)2 = −b ∞ (0)dr 4πR2 off”); V4πa 4πb2 − a 4π 2 dr − R (0)dr = 0 r 1 q q − . 4π0 R a !a !0 ! R$ 1 q % !0 !b $ 1 q % q q# 1 “q Problem 2.39 + − dr − dr − (0)dr − (0)dr = (b) V (0) = − ∞ E·dl = − ∞ 4π! 2 2 b R a 4π!0 r 0 r 4π#0 b R a qa qb qa + qb (a) σa = − ; σb = − ; σR = . !0 !a ! R$ 1 q % 1 “q q 4πa2 4πb2 4πR2 (c) σb → 0 (the charge “drains off”); V (0) = − ∞ (0)dr − a 4π! − 2 dr − R (0)dr = 0 r 4π#0 R a 1 qa + qb (b) Eout = Problem2 2.36 r̂, where r = vector from center of large sphere. 4π0 r qa qb qa + qb σa = − ; qσb = − ; σR = . 2 2 2 1 q(a) 1 a b 4πa 4πb r (r ) is 4πR (c) Ea = r̂ , E = r̂ , where the vector from center of cavity a (b). a b b a b 4π0 ra2 4π0 rb2 1 qa + qb r̂, where r = vector from center of large sphere. (b) Eout = 4π# r2 0 (d) Zero. 1 qb(but not Ea or Eb ); force on qa and qb still zero. (e) σR changes (but not σa 1or σqba); Eoutside changes r̂a , Eb = r̂b , where ra (rb ) is the vector from center of cavity a (b). (c) Ea = 2 4π#0 ra 4π#0 rb2 Problem 2.40 (a) No. For example, if it is very close to the wall, it will induce charge of the opposite sign on the wall, (d) Zero. and it will be attracted. (b) No. Typically itRwill be attractive, seeσbfootnote 12changes for a extraordinary (e) σ changes (but not but σa or ); Eoutside (but not Ea counterexample. or Eb ); force on qa and qb still zero. Problem 2.41 Problem 2.37 Between plates, the E =plates 0; outside the 0plates = So σ/#0 = Q/#0 A. So Between the plates, E = the 0; outside E = σ/ = Q/E0 A. 2 2 0 2 0 PQ=2 #0 E 2 = Q2#0 Q = Q . P = E = = .2 2 2#0 A2 2 2 20 A2 2 20 A22 #0 A Problem 2.42 Problem 2.38 z! 1 Q 1 Q Inside, E = 0; outside, E= = 0;4π Inside, E outside, 2 r̂; soE = 4π! r 2 r̂; so 0 r “E 0 θ 1 Q Q )z ; σ = Q 2 . 1 EQ = 12z4π! f = σ(Eave ave Eave = 12 4π =0σ(E R2 r̂; 4πR 2 r̂; f 2. ave )zz; σ = 4πR 0 R ! !$ Q % 1 $ 1 Q % R RF Q = f1z da1= Q 4πR cos θ R2 sin θ dθ dφ 2 Fz = fz da = z4πR cos2 θ 2R24π! sin0 θRdθ dφ 2 2 4π R2 0 $ $ Q %2 $ 1 %( $ % % ! Q2 2 (π/2 1 2 Q 2 1 π/2 2θ dθ R 1 Q 2 2π π/2 sin Q 2 sin θ = θ cos = = . Q 2 = π/2 Q Q 2 1 1 1 1 π!0 4R = 2 2π!0 4R. 0= = 20 4πR 2π 0 2!0sin4πR θ cos θ dθ0= π0 4R 32πR2 #0 2 sin θ 0 2π0 4R 32πR2 0 Problem 2.43 #2005 c Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright as they No portion this material may be law: Say the charge on the inner cylinderlaws is Q, for acurrently length exist. L. The field isofgiven by Gauss’s R reproduced, in1any form or1 by any means, without Q 1 permission in writing from the publisher. E·da = E · 2πs · L = 0 Qenc = 0 Q ⇒ E = 2π0 L s ŝ. Potential difference between the cylinders is Z b Q V (b) − V (a) = − E·dl = − 2π 0L a Z b 1 Q ds = − ln s 2π 0L a As set up here, a is at the higher potential, so V = V (a) − V (b) = 2πQ0 L ln c b a . b . a cylinder is Q, for a length L. The field is given by Gauss’s law: !Say the charge on the inner E·da = E · 2πs · L = !10 Qenc = !10 Q ⇒ E = 2π!Q0 L 1s ŝ. Potential difference between the cylinders is ” b Q V (b) − V (a) = − E·dl = − 2π”0 L a ” b Q 1 ds = − ln 2π”0 L a s As set up here, a is at the higher potential, so V = V (a) − V (b) = 2π!Q0 L ln 2π”0 0L 40 % &. C = VQ = 2π! b , so capacitance per unit length is ln( a ) ln ab # $ b . a %b& a . CHAPTER 2. ELECTROSTATICS Problem 2.400 L 2π0 . C = VQ = 2π b , so capacitance per unit length is ln( a ) ln ab “0 2 E A”. (a) W = (force)×(distance) = (pressure)×(area)×(distance) = 2 Problem 2.44 ( ‘ 2 (b) W = (energy per unit volume)×(decrease in volume) = “00 E22 (A”). Same as (a), confirming that the (a) W = (force)×(distance) = (pressure)×(area)×(distance) = E A. 2 energy lost is equal to the work done. 2 Problem 2.41 (b) W = (energy per unit volume)×(decrease in volume) = 0 E2 (A). Same as (a), confirming that the energyFrom lost Prob. is equal to the the field workatdone. 2.4, height z above the center of a square loop (side a) is Problem 2.45 1 4λaz ) E= ẑ. % 2& 4π” a2 (side a) is a 0 2 From Prob. 2.4, the field at height z above the center loop z of + a4square z2 + 2 ! ” da 2 ! ” da 2 1 4λaz da over a from 0 to ā: ẑ. Here λ → σ 2 (see figure), and we integrate q E= 2 2 4π0 z 2 + a ” ! z2 + a 4 2 ” ā 1 a2 a da Here λ → (see figure), and we&integrate over a from ) Eσ =da . Let u = 0 to 2σz , soā:a da = 2 du. 2 % 2 2 4π”0 4 0 z 2 + a4 z 2 + a2 Z ” +√ ,-ā2 /4 * 2 ā 2 1 a da du a2 ā /4 σz 2u + z2 q E = = 12σz . Let u = , so a da = 2 du. −1 √ a2 4σz0 = tan 4π a2 0 2 2 z2 + 2 4π” π”0 z4 z 2u + 0 0 z +(u4+ z ) 2 z 0 ) ” !#ā2 /4 + ā2 , . Z ā2 /4 / √ 2 12σ du σz 2 2u + z 2 2 +z √ − tan−1 == 4σztan−1 = (1) ; tan−1 4π π0 z z (u +zz 2 ) 2u + z 2 π”0 0 0 0 q ( ! * ) ā2 0 2 2σ 2 + z 2σ a2 π −1(1) ; = tan−1 − tan−1 ẑ. 1+ 2 − π0 zE = π”0 tan 2z 4 a ” a+da ! 1 % 2 & 2σ ! r tan−1 (∞) −# π = 2σ π − π = σ . ! a → ∞ (infinite plane):”E = π! 4 π!0 2 4 2!0 0 2 2σ a π σ a2 −1 −1 p E= tan 1 + −12 √− ẑ = tan ẑ. 2 + (a2 series: and as azTaylor z ‘ a (point charge): 1 4+ x − π4 , π π0 Let f (x) = tan2z /2) 0 expand4z −1= f (0) + ” 2σ +π1 x2πf “” (0) + · π xf (0) 2σ f (x) a → ∞ (infinite plane): E = π tan (∞) − = π0 22− 4 = 2σ0 ·. ·X 4 0 √ −1 z a (point charge): Let f (x) + x − π4 Inc., , andUpper expand asRiver, a Taylor series: c = tan #2005 Pearson 1Education, Saddle NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any without permission in writing from the publisher. 1 means, 0 2 00 f (x) = f (0) + xf (0) + x f (0) + · · · 2 1 1√1 1√ 1 0 Here f (0) = tan−1 (1) − π4 = π4 − π4 = 0; f 0 (x) = 1+(1+x) 2 1+x = 2(2+x) 1+x , so f (0) = 4 , so f (x) = 2 a 2σ Thus (since 2z 2 = x 1), E ≈ π 0 c 1 a2 4 2z 2 1 x + ( )x2 + ( )x3 + · · · 4 2 q 1 σa 1 = 4π 2 = 4π z 2 . X 0 z 0 5 1√ 1√1 1 Here f (0) tan−1 (1) − π4 = π4 − π4 = 0; f ” (x) = 1+(1+x) CHAPTER 2. = ELECTROSTATICS 411+x , so f ” (0) = 14 , so 2 1+x = 2(2+x) 1 x + ( )x2 + ( )x3 + · · · 4 ! “∂ 1 ∂ 1 ∂ k sin θ cos φ 2 sin θ cos θ sin φ 2 3k 2σ q 1 1 1 σa2k a2 a2 1 ρ = (since ∇·E = r + sin θ + = Thus = x ” 1), E ≈ = . ! 0 2z 2 0 2z 2θ ∂θ 4π#0 z 2 4π#0r z 2 r2 ∂r r π#0 r4sin r sin θ ∂φ r 3 2 Problem12.42 1 2k sin φ(2 sin θ cos θ − sin θ) 1 (−k sin θ sin φ) = 0 2 3k + $ + r sin θ %& r r r sin θ# 1 ∂ $ A %r 1 ∂ B sin θ cos φ ρ =”0 ∇·E = “0 r2 + θ ∂φk0 k 22∂r 0 2 r r r sin = 2 3+ 2 sin φ(2 cos θ − sin θ) − sin(φ = 2 3 + rsin φ(4 cos2 θ − 2 + 2 cos2 θ − 1) ‘ r “0r 1 1 B sin θ = “0 2 A + (− sin φ) = 2 (A − B sin φ). 3k 3k 0 0 sin 2θ θ −r1) = r φ cos 2θ). = 2 1 +r sin φ(2r cos (1 + sin r r2 Problem 2.46 f (x) = Problem 2.43 1 Q From Prob. 2.12, the field inside a uniformly charged sphere is: E = 4π# Problem 2.47 3 r. So the force per unit volume 0 R ) * ) Q *) Q * Q 2 1 Q 3 From the field inside E =z 4π force per unit volume 3 r. So r =a #uniformly r,charged and thesphere force is: in the direction on the dτ is: is f =Prob. ρE =2.12, 4 0 R 4πR3 0 πR3 Q 4π# 0 R3 3 Q 2 Q 3 is f = ρE = 4 πR3 4π0 R3 r = 0 4πR3 r, and the force in the z direction on dτ is: %2 $ 3 3 Q dFz = fz dτ = r cos θ(r2 sin θ dr dθ dφ). 2 3 3 “0Q 4πR dFz = fz dτ = r cos θ(r2 sin θ dr dθ dφ). 3 0 4πR The total force on the “northern” hemisphere is: %2 + R is: + π/2 $ + 2π + the “northern” The total force on hemisphere Q 3 3 r dr cos θ sin θ dθ dφ Fz = fz dτ = “0 4πR 2 Z3 R 0 Z π/2 0 Z Z 2π 0 3 Q$ -π/2 . %2 $3 4 % , cos2 θ sin Fz = fz dτ = 3 3 Q 0 r dr R 0 sin θ — θ dθ 0 dφ 3Q2 0 = 4πR . (2π) = 3 “0 4πR 2 4 2 !-0 64π”0 R2 π/2 3 Q R4 sin2 θ 3Q2 = (2π) = . 0 4πR3 4 2 0 64π0 R2 Problem 2.44 + + 1 σ 1 σ σR 1 σ Problem 2.48 2 Vcenter = Z da = Z R da = 4π” R (2πR ) = 2″ 4π” η 4π” 1 σ0 1 σ 0 1 σ 0 σR 0 Vcenter = da = da = (2πR2 ) = r 4π0 4π0 R 20 (4π/0 R Z + 2 da = sin2 θsin dθ,θ dθ, da2πR = 2πR 1 1σ σ Vpole V=pole = da ,da with , with 2 2 2 2 2 2 2 2 2 2 4π0 4π”r η r η= =R R+ +R R− −2R2R coscos θ= 2R2R (1(1 −− coscos θ).θ). θ= 0 Z + 2 π/2 -π/2 1 σ(2πR ) 2π/2 sin θ sin dθ θ dθ σR σR√ √ ) π/2 1 σ(2πR √ √ √ √ = = = √= √(2 1(2 − cos 1 −θ)cos θ)4π0 4π” 0 0 1 −θ cos θ2 220 2″0 R0 2R 20 0 1 − cos √ √ σR σR σR σR σR σR √= √ = √= √ (1 − (1 0) − = 0) . Vcenter = =( 2 − . ∴ Vpole ∴ V−pole − Vcenter ( 1). 2 − 1). 20 2″0 20 2″0 20 2″0 r R R θ Problem Problem 2.492.45 determine the electric and outside the sphere, Gauss’s FirstFirst let’s let’s determine the electric field field insideinside and outside the sphere, usingusing Gauss’s law: law: / 0 + Z + 4 I Z Z r + r ( 4 πkr 2 2 3 πkr (r < (r R), R).> R). 0 c $2005 Pearson Education, Inc., Upper 4 Saddle River, NJ. All rights reserved. This material is So Eprotected = 4k0 r2under r̂ (r all R). exist. No portion of this material may be copyright asr 2they currently 0 reproduced, in any form or by any means, without permission in writing from the publisher. c 42 CHAPTER 2. ELECTROSTATICS Method I : 2 2 Z Z 0 R kr2 0 ∞ kR4 E 2 dτ (Eq. 2.45) = 4πr2 dr + 4πr2 dr 2 0 40 2 R 40 r2 ) 2 (Z R ∞ Z ∞ 0 k 1 πk 2 R7 1 πk 2 R7 6 8 8 7 = 4π r dr + R dr = +R − = +R 2 2 40 80 7 r R 80 7 0 R r W = = 0 2 Z πk 2 R7 . 70 Method II : W = 1 2 Z ρV dτ (Eq. 2.43). ( ) R Z r 2 r kR4 kr k 1 r3 4 For r Ω11 n = 12 : Ω11 + 11 = 48.5757 + 11 = 59.5757 < Ω12 Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at the center. c