Introduction to electrodynamics solution manual 3rd

More sophisticateddefinitions of convergence permit one to work with series of this form, but it is better to sum the series first and then differentiate the second method.]” • Page 51, Pro

David J Griffiths

Introduction to Electrodynamics, 3rd ed

Author: David Griffiths

Date: September 1, 2004

• Page 4, Prob 1.15 (b): last expression should read y + 2z + 3x.

• Page 4, Prob.1.16: at the beginning, insert the following figure

• Page 8, Prob 1.26: last line should read

From Prob 1.18: ∇ × v a =−6xz ˆx + 2z ˆy + 3z2ˆ z⇒

• Page 9, Prob 1.31, line 2: change 2x3 to 2z3; first line of part (c): insert

comma between dx and dz.

• Page 12, Probl 1.39, line 5: remove comma after cos θ.

• Page 13, Prob 1.42(c), last line: insert ˆz after ).

• Page 14, Prob 1.46(b): change r
to a.

• Page 14, Prob 1.48, second line of J: change the upper limit on the r

integral from∞ to R Fix the last line to read:

• Page 15, Prob 1.49(b), last integration “constant” should be l(x, z), not

l(x, y).

• Page 17, Prob 1.53, first expression in (4): insert θ, so da = r sin θ dr dφ ˆθˆθˆθ.

• Page 17, Prob 1.55: Solution should read as follows:

z dz = z

22



02

3(2− 2y) − 2y1

sin3θ +

cos θ sin θ



dθ = − cos θ sin θ

cos2θ +sin2θ

• Page 21, Probl 1.61(e), line 2: change = z ˆz to +z ˆz.

• Page 25, Prob 2.12: last line should read

r V(r)

In the figure, r is in units of R, and V (r) is in units of 4π q

0R

• Page 30, Prob 2.28: remove right angle sign in the figure.

• Page 42, Prob 3.5: subscript on V in last integral should be 3, not 2.

• Page 45, Prob 3.10: after the first box, add:

a2+ b2 −1

a −1b



.

• Page 45, Prob 3.10: in the second box, change “and” to “an”.

• Page 46, Probl 3.13, at the end, insert the following: “[Comment: nically, the series solution for σ is defective, since term-by-term differen-

Tech-tiation has produced a (naively) non-convergent sum More sophisticateddefinitions of convergence permit one to work with series of this form,

but it is better to sum the series first and then differentiate (the second

method).]”

• Page 51, Prob 3.18, midpage: the reference to Eq 3.71 should be 3.72.

• Page 53, Prob 3.21(b), line 5: A2should be σ

40R ; next line, insert r2after1

2R

• Page 55, Prob 3.23, third displayed equation: remove the first Φ.

• Page 58, Prob 3.28(a), second line, first integral: R3 should read R2

• Page 59, Prob 3.31(c): change first V to W

• Page 64, Prob 3.41(a), lines 2 and 3: remove 0 in the first factor in the

expressions for Eave; in the second expression change “ρ” to “q”.

• Page 69, Prob 3.47, at the end add the following:

Alternatively, start with the separable solution

V (x, y) = (C sin kx + D cos kx)

Ae ky + Be −ky

Note that the configuration is symmetric in x, so C = 0, and V (x, 0) =

0⇒ B = −A, so (combining the constants)

V (x, y) = A cos kx sinh ky.

But V (b, y) = 0, so cos kb = 0, which means that kb = ±π/2, ±3π/2, · · · ,

or k = (2n − 1)π/2b ≡ α n , with n = 1, 2, 3, (negative k does not yield

a different solution—the sign can be absorbed into A) The general linear

• Page 74, Prob 4.4: exponent on r in boxed equation should be 5, not 3.

• Page 75, Prob 4.7: replace the (defective) solution with the following:

If the potential is zero at infinity, the energy of a point charge Q is

(Eq 2.39) W = QV (r) For a physical dipole, with −q at r and +q

If you do not (or cannot) use infinity as the reference point, the result still

holds, as long as you bring the two charges in from the same point, r0(or

two points at the same potential) In that case W = Q [V (r) − V (r0)],and

• Page 91, Problem 5.10(b): in the first line µ0 I2/2π should read µ0I2a/2πs;

in the final boxed equation the first “1” should be a s

• Page 92, Prob 5.15: the signs are all wrong The end of line 1 should

read “right (ˆ z),” the middle of the next line should read “left (−ˆz).” In

the first box it should be “(n2− n1)”, and in the second box the minussign does not belong

5

• Page 114, Prob 6.4: last term in second expression for F should be +ˆz ∂B x

∂z

(plus, not minus)

• Page 119, Prob 6.21(a): replace with the following:

The magnetic force on the dipole is given by Eq 6.3; to move the dipole

in from infinity we must exert an opposite force, so the work done is

(I used the gradient theorem, Eq 1.55) As long as the magnetic field goes

to zero at infinity, then, U = −m · B If the magnetic field does not go

to zero at infinity, one must stipulate that the dipole starts out orientedperpendicular to the field

• Page 125, Prob 7.2(b): in the box, c should be C.

• Page 129, Prob 7.18: change first two lines to read:



s + a s

• Page 131, Prob 7.27: in the second integral, r should be s.

• Page 132, Prob 7.32(c), last line: in the final two equations, insert an I immediately after µ0

• Page 140, Prob 7.47: in the box, the top equation should have a minus

sign in front, and in the bottom equation the plus sign should be minus

• Page 141, Prob 7.50, final answer: R2should read R2

• Page 143, Prob 7.55, penultimate displayed equation: tp should be ·.

• Page 147, Prob 8.2, top line, penultimate expression: change a2to a4; in(c), in the first box, change 16 to 8

• Page 149, Prob 8.5(c): there should be a minus sign in front of σ2in thebox

• Page 149, Prob 8.7: almost all the r’s here should be s’s In line 1 change

“a < r < R” to “s < R”; in the same line change dr to ds; in the next

line change dr to ds (twice), and change ˆr to ˆs; in the last line change r

to s, dr to ds, and ˆr to ˆ s (but leave r as is).

6

pression for R change 2.01 to 2.10.

• Page 175, Prob 9.34, penultimate line: α = n3 /n2 (not n3/n3)

• Page 177, Prob 9.38: half-way down, remove minus sign in k2

x + k2+ k2=

−(ω/c)2

• Page 181, Prob 10.8: first line: remove ¿.

• Page 184, Prob 10.14: in the first line, change (9.98) to (10.42).

• Page 203, Prob 11.14: at beginning of second paragraph, remove ¿.

• Page 222, Prob 12.15, end of first sentence: change comma to period.

• Page 225, Prob 12.23 The figure contains two errors: the slopes are for v/c = 1/2 (not 3/2), and the intervals are incorrect The correct solution

is as follows:

• Page 227, Prob 12.33: first expression in third line, change c2to c.

7

219

Vector Analysis

(a) From the diagram, IB + CI COSO3= IBI COSO1+ ICI COSO2'Multiply by IAI.

Similarly: IB + CI sin 03=IBI sin 01+ICI sin O2, Mulitply by IAI n.

If n is the unit vector pointing out of the page, it follows that

Ax(B + e) =(AxB) + (Axe) (Cross product is distributive.)

(b) For the general case, see G E Hay's Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and

Section 8 (cross product)

suppose A = ~ and C is perpendicular to A, as in the diagram

Ax(BxC)

ICI sin 82

IBlsin81A

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane For example,

we might pick the base (A) and the left side (B):

2 CHAPTER 1 VECTOR ANALYSIS

To make a unit vector out of it, simply divide by its

ft- - IAX BI AXB = '7X+ '7y + '7z 16 A 3 2 I

= x[Ay(BxCy - ByCx) - Az(BzCx - BxCz)] + yO + zO

(I'll just check the x-component; the others go the same way.)

= x(AyBxCy - AyByCx - AzBzCx + AzBxCz) + yO + zOo

B(A.C) - C(A.B) =[Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)] x + 0y + 0 z

= x(AyBxCy + AzBxCz - AyByCx - AzBzCx) + yO + zOo They agree.

Ax(BxC) =

Ax(BXC)+Bx(CxA)+Cx(A-xB) =B(A.C)-C(A.B)+C(A.B)-A(C.B)+A(B.C)-B(C.A) = o.

If this is zero, then either A is parallel to C (including the case in which they point in oppositedirections, or

one is zero), or else B.C = B.A = 0, in which case B is perpendicular to A and C (including the case B =0)

Conclusion:Ax(BxC) = (Ax B) xC <=:=}either A is parallel to C, or B is perpendicular to A and C.Problem 1.7

~= (4x+6y+8z) - (2x+8y+7z) = !2x-2y+ zl

~ =yl4 + 4 + 1=@J

. ~ 12A 2A lAI

= COS2 cpAyBy + sincpcoscp(AyBz + AzBy) + sin2 cpAzBz + sin2 cpAyBy - sin cpcoscp(AyBz + AzBy) + COS2cpAzBz

Moreover, if R is to preserve lengths for all vectors A., then this condition is not only sufficient but also

necessary For suppose A = (1,0,0) Then ~j,k (~i RijRik) AjAk =~i RilRil, and this must equal 1 (since we

Problem 1.9 y

x z

(a)INo change.! (Ax = Ax, Ay = Ay, Az = Az)

(c) (AxB) -t (-A) X(-B) =(AxB) That is, if C = AxB, Ie -t C I No minus sign, in contrast to

Angular momentum (L=rxp) and torque (N =rxF) are pseudovectors

changes sign under inversion of coordinates.

(a)Vf =2xx + 3y2y + 4z3z

4 CHAPTER1 VECTOR ANALYSIS

~ =(x - x') x+ (y - y') y + (z - Zl)Zj ~ = Vex -X')2 + (y - y')2 + (z - ZI)2.

(a) V(~2)= tx[(X_X')2+(y_y')2+(Z_Z')2Jx+tyOy+ tzOz = 2(X_X')X+2(y_y')y+2(z_Z')Z = 2~.

(b) V(k) = tx [(x - X')2 + (y - y')2 + (z - ZI)2]-!x + tyo-! y + tz O-! z

1

()-J!.2( ,

)~ 1()-J!.2( ,

)~ 1()-J!.2( ,

)~

=-2 2 x-x X-2 2 y-y Y-2 2 z-Z Z

=-O-~[(x - x') x+ (y - y') y + (z - Zl) Ii]=_(1/1-3)~ =-(I/1-2)i.

(c) /x(~n) = n1-n-lfi = n1-n-l(H21-x)= n1-n-lix, soIV(1-n)=n1-n-l i.1

y =+y cosif> + z sinif>; multiply by sinif>: ysinif> =+y sin if>cos if>+ Z sin2 if>.

Z=-y sin if> +Z cos if>;multiply by cos if>:z cos if>=-y sin if>cosif> +Z COS2 if>.

So ~ =cosif>; ~ = - sinif>; ~v=sinif>; ~~ =cosif> Therefore

(VJ)y = U = ~~ + M~v =+cosif>(VJ)y +sinif>(VJ)z

This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the

origin How, then, can V.v = O? The answer is that V.v =0 everywhere except at the origin, but at the origin our calculation is no good, since r = 0, and the expression for v blows up In fact, V.v is infinite at

that one point, and zero elsewhere, as we shall see in Sect 1.5

Vy =cosif>Vy+sinif>vz; Vz = -sinif>vy +cosif>vz.

~ - ~ ,!,+

!lJL "!'-(~!bl. + ~oz ) ,!, + (!lJL !bl. + !lJL 8Z) . ,!, U I . P b 114

8y - 8y cos'l' o'g sm'l' - oy 8y 8z 8y cos'l' 8y 8y 8z oy sm'l' se resu t m ra

~

v = yx + xy; or v = yzx + xzy + xy Zjor v = (3x2z - Z3) X+ 3y + (x3- 3xz2)z;

= J (~x+ ~y + ~z) +g (Ux+ Uy+ ¥Zz)= J(Vg)+ g(VI). qed

(iv)V.(AxB) =!Ix (AyBz - AzBy) + !ly (AzBx - AxBz) + !lz (AxBy - AyBx)

(v) Vx (fA) =(O(~:%)- O(~~v») x+ (8(~~,.) - 8(~:%»)y + (8(~:v) - 8(~:,.»)Z

- [(Ay¥Z - AzU) x + (AzU - Az¥Z)y + (AxU - AyU) z]

= J(VxA) - Ax (VI) qed

6 CHAPTER 1 VECTOR ANALYSIS

=~ {x [J- +X(-~)(}-)32X] +yx [-~(}-)32Y] +ZX [-~(}-)32Z]}

= ~{;- ~ (X3+ xy2 + XZ2)}=~ {; - ~ (X2+ y2 + Z2)} = ~(; - ;) = O.

(c) (Va.V')Vb= (X2tx + 3XZ2ty - 2xz tz) (xyx + 2yzy + 3xzz)

= X2(yx + Oy + 3zz) + 3XZ2(xx+ 2zy + Oz) - 2xz (Ox + 2yy + 3xz)

=IX2 (y + 3Z2) x + 2xz (3z2 - 2y) Y - 3X2ZZI

[Bx(VxA )]x =B (~-~y ax oy )-B z(~-~oz ax )

[(A.V)B]x = (Ax tx + Ay ty + Az tJBx = Ax °:Xz+ Ay °.:vz+ Az °fzz

[(B.V )A]x = B oAz + B oAz + B oAz x ax y oy z oz

- A oB" - AoBz - A oBz + A !ll! + B oA"- B oAz - B oAz + B ~

+A x& oBz +A !ilk +A y~ ZhoBz +B ~ x& +B y~ oAz +BZhM

= B oAz +A oBz +B (oA"- ~ +~ ) +A (OB"- ~ +~ )

+Bz(-~+~+~Tz ax Tz ) +Az( -~+!ll!Tz ax +~Tz )

= [V(A.B)]x (same for y and z)

(vi) [Vx(AxB)]x = ty(AXB)z - tz(AxB)y = ty(AxBy - AyBx) - tz(AzBx - AxBz)

- oAz B + A ~ - ~ B - A oBz - ML B - A !ll! + oAz B + A !ll!

VxA = x (ty(3Z)- tz(2Y))+y (tz(x) - t.,(3z))+z (t.,(2Y)- ty(x)) = 0; B.(VxA) = 0

VxB = x (ty(O) - tz(-2X)) + y (tz(3y) - t.,(0)) + z (t.,(-2X) - ty(3Y)) = -5 z; A.(VxB) = -15z

?

(b)A.B =3xy- 4xy =-xy ; V(A.B) =V(-xy) = xt.,( -xy) + y ty(-xy) = -yx - xy

Ax(VxB) = I x 2y 3z 1= x(-lOy) + y(5x); Bx(VxA) =0

0 0 -5

(B.V)A = (3y to: - 2xty) (xx + 2yy + 3zz) =x(3y) + y( -4x)

Ax(VxB) + Bx(VxA) + (A.V)B + (B.V)A

=-10yx + 5xy + 6yx - 2xy + 3yx - 4xy =-yx - xy =V.(A.B) ./

(c) VX (AxB) = x (ty (-2X2 - 6y2)- tz (9ZY))+ y (tz (6xz) - t., (-2x2 - 6y2)) + z (t., (9zy) - ty (6xz))

V.A = t.,(x) + ty(2y) + tz(3z) = 1 +2 +3= 6; V.B = t.,(3y) + ty(-2x) = 0

(B.V)A - (A.V)B + A(V.B) - B(V.A) =3yx - 4xy - 6yx + 2xy -18yx + 12xy =-21yx + lOxy

8 CHAPTER 1 VECTOR ANALYSIS

- ax 8y - 8y 8x + ayoz - oz oy oz ax - ax 8z - , y equ 1 y 0 cross- enva Ives.

2xy3z4 3x2y3z4 4x2y3z3

= x(3 4x2y2z3 - 4 3x2y2z3) + y(4 2xy3z3 - 2 4xy3z3) + z(2 3xy2z4 - 3 2xy2Z4)=O ./

(1/2) - z + (z2/2) Finally, the z integral is J; Z2(~- Z+ Z;)dz = Jo1(; - Z3 + ~4) dz = (~ - ~4+ f~)lb= i-t+fo=~

(a) Segment 1: x: 0 -t 1, y = z = dy = dz = O.JVT.dl = J;(2x)dx = x21~= 1.

Numbering the surfaces as in Fig 1.29:

(ii) da = -dydzx,x = O.v.da = O.Jv.da = O.

(iii)da = dxdzy,y = 2 v.da = 4zdxdz.Jv.da = JJ4zdxdz = 16.

(iv)da = -dxdzy,y = O.v.da = O.Jv.da = O. .

(vi) da = -dxdyz,z = O.v.da = O.Jv.da = O

=>J v.da = 8 + 16 + 24 = 48 /

Problem 1.33

Vxv = x(O- 2y) + y(O- 3z) + z(O- x) = -2yx - 3zy - xz.

10 CHAPTER 1 VECTOR ANALYSIS

(2) x =0; z=2- y; dx = 0, dz = ~dy, y: 2 -t O.v.dl = 2yzdy.

Jv.dl = J202y(2- y)dy = - J:(4y - 2y2)dy= - (2y2- ~y3)I~ = - (S - ~ S) = -i.

(3) x =y =0; dx =dy= 0; z: 2 -t O.v.dl = O.Jv.dl = O So §v~dl = -i ./

Problem1.34

(i) da = dy dz x, x = 1; y,z: 0 -t 1 (Vxv).da = (4z2- 2)dydz; J(Vxv).da = J;(4z2 - 2)dz

= (~z3 - 2z)l~ =~ - 2= -~.

(ii) da = -dxdyz, z = OJx,y: 0 -t 1 (Vxv).da = 0; J(Vxv).da = O.

(iii) da = dxdzy, y = 1; x,z: 0 -t 1 (Vxv).da = 0; f(Vxv).da = O.

(iv) da = -dxdzy, y = 0; x, z : 0 -t 1 (Vxv).da = 0; J(Vxv).da = O.

(v) da = dxdyz, z = 1; x,y: 0 -t 1 (Vxv).da = 2dxdy; J(Vxv).da = 2.

=>f(Vxv).da = -~ + 2 = ~ ./

Problem 1.35

(I used Stokes' theorem in the last step.)

IvB,(VXA)dr= IvV.(AxB)dr+ IvA,(VXB)dr= t(AXB).da-+ IvA,(VXB)dr. qed.

Problem 1.361 r = ,jx2 + y2 + Z2; -I ( ' );

0=cos y'x'+y'+" q, = tan-I m.

from Fig 1.36 The most systematic approach is to study the expression:

r =xx+ y Y + zz =r sin 0 cos q, x + r sin 0 sin q,y +r cos 0 z.

If I only vary r slightly, then dr = fj-,.(r)dris a short vector pointing in the direction of increase in r To make

it a unit vector, I must divide by its length Thus:

f= I~I; {h 1;1; J>= 1;1'

~ =-rsinOsinq,x + rsinOcosq,y; 1~12=r2 sin2 Osin2 q, +r2 sin2 OCOS2 q, =r2 sin2 0

=}IIi = cosOcos q,x+ cosOsinq,y - sinOz.

J>=- sinq,x + cosq,y

sinOf = sin2 0 cosq,x + sin2 Osin q,y + sin OcosOz.

Add these:

Multiply (1) by cosq" (2) by sinq" and subtract:

Multiply (1) by sinq" (2) by cosq" and add:

Subtract these:

12 CHAPTER 1 VECTOR ANALYSIS

f(V 'VI )dr = J( 4r )(r2 sin (Jdr d(J d4» = (4) JoRr3dr Jo" sin (Jd() n" d4>= (4) ( ~4)(2)(211") =1411" R41

Jvl.da =J(r2f).(r2 sin (Jd(J d4>f) =r4 Jo" sin(J d(JJ:" d4>= 411"R4./ (Note: at surface of sphere r = R.)

(b) V'V2 = ~ tr (r2~) =0 => IJ(V'v2)dr =0I

Jv2.da = J (~f) (r2 sin (Jd(J d4>f) = Jsin (J d(J d4>=1411".1

They don't agree! The point is that this divergence is zero except at the origin, where it blows up, so our

V.v = ~ tr(r2 rcos(J) + rs~nOt(J(sin(Jrsin(J) + rs~nO:4>(r sin (Jcos4»

= :13r2 cos (Jr + -J:-rSIn (J r 2 sin (Jcos (J+ -r Ins~(J r sin (J (- sin'I' A.)

= 3cos(J + 2cos(J.- sin 4>=5cos(J- sin 4>

J(V.v)dr =J(5 cos (J- sin 4» r2 sin (Jdr d(J d4>=JoRr2 dr Jo£ [J:" (5 cos (J- sin 4» d4>] d(J sin (J

Jv.da =J(rcos(J)R2 sin(Jd(Jd4>=R3 Jo;; sin(Jcos(Jd() J:" d4>= R3 U)(211")=1I"R3.

Jv.da =f(r sin (J)(r dr d4»=JoRr2 dr J:" d4>= 211"~3.

Total: Jv.da =11"R3 + ~11"R3 =~11"R3 ./

Problem 1.40I Vt =(cos(J+ sin (Jcos4»f + (- sin(J+ cos(Jcos4»8 + Si~(J(- si~(Jsin4»cb

\72t = v.(Vt)

= ~ tr (r2(cos(J+sin (Jcos4») + rs~n (Jto (sin(J( - sin (J+ cos(Jcos 4») + rs~nO:4>(- sin 4»

= r s~n0 [2 sin (JJos (J+ 2 sin2 (Jcos 4>- 2 sin (JJos (J + COS2(JCOS4>- sin2 (Jcos 4>- cos 4>]

=>I\72t =0I

Check: r cos (J= z, r sin (Jcos 4>= x => in Cartesian coordinates t =x + z Obviously,Laplacian is zero.

JVt.dl = J:dr =2

Segment 2: (J= ~,r = 2,4>: 0-t~. dl =rsin(Jd4>cb=2d4>cb.

Vt.dl =(-sin4»(2d4» = -2 sin 4>d4> JVt.dl = - Jo;;2sin4>d4>= 2cos4>l! = -2

Segment 3: r = 2, 4J= ~;0: ~-t O.

Total: J:Vt.dl =2- 2 + 2= m Meanwhile,t(b)- tea) =[2(1+0)]- [O( )] =2 ./

S COg4J- ~sin 4J=COS2 4Jx +COg4Jsin 4Jy + sin2 4Jx - sin 4Jcos4Jy=x(sin2 4J+COS2 4J) = x.

So Ix = cos4JS- sin 4J~.I

Multiply first by sin4J,second by COg4J,and add:

s sin4J+ ~cas 4J=sin 4JCOg4Jx + sin2 4Jy- sin 4JCOg4Jx+ cos24Jy =y(sin2 4J + COS24J)= y.

SoIy =sin 4Js + cas 4J~.I 1z = z.1

top: z=5, da=sdsdifJzj v.da=3zsdsdifJ=15sdsdifJ Jv.da=15Jo sdsJ02d4J=1511".

bottom: z = 0, da = -sdsd4Jzj v.da = -3zsds.difJ= O Jv.da = O.

back:4J= ~, da = dsdz~j v.da = ssin4Jcos4Jdsdz= O Jv.da = O.

left: 4J= 0, da = -dsdz~j v.da = -ssin4Jcos4Jdsdz= O Jv.da = O.

front: s = 2, da = s d4Jdz s; v.da = s(2 + sin2 4J)sd4Jdz = 4(2 + sin2 4J)d4Jdz.

+~ (ts(S2 sin ifJcosifJ)- :4>(s(2 + sin2 ifJ))) Z

14 CHAPTER 1 VECTOR ANALYSIS

(d) 11 (if a> b), 0 (if a < b).I

So, xd~15(x)= -15(x). Qed

(b) J~oolex) ~:dx = f(x)O(x)l~oo - J~oofxO(x)dx = f(oo) - Jooo1xdx = f(oo) - (f(oo) - feD))

Problem 1.48

Second method: integrating by parts (use Eq 1.59)

IF2 is a gradient; FI is a curl I IU2= ~ (X2 + y2 + Z2)I would do (F2 =VU2).

}

Ax = tz2x+h(x,y)j Az = -tzx2 +j(y,z)

Ay = tx2y + k(y, z); Ax = -txy2 + l(x, y)

(b) ~ (c): same as (c) ~ (b), only in reverse; (c) => (a): same as (a)=> (c)

{ F da = { F da.

(Note: sign change because for §F .da, da is outward, whereas for surface II it is inward.)

(b) ~ (c): same as (c) ~ (b), in reverse; (c)~ (a): same as (a)~ (c)

Problem 1.52

(a) To find t:

(2) g~ = (2xy + Z2)

16 CHAPTER 1 VECTOR ANALYSIS

From (1) & (3) we get M =2yz =>J =yz2 + g(y) =>t =y2x + yz2 + g(y), so :~ = 2xy + z2 + ~ =

2xy + Z2 (from (2» => ~ = 0.We may as well pick 9= OJthenIt =xy2 + YZ2.1

8W - 8Wy = !!.l.+ X2 - !!.i.8 = x2 => !!.l. 8 - !!.i

Surface consists of four parts:

(2) Left: da =-r dr d();Pj 4>=O v da=(r2cos() sin 4» (r dr d())=O f v da =O.

v dl= (ayx + bxy)' (dxx + dyy + dzz) =aydx + bxdYi X2 + y2=R2 =>2xdx + 2ydy = 0,

18 CHAPTER 1 VECTOR ANALYSIS

(Vxv)x = /y(6) - -/z(y + 3x) =O Therefore J(Vxv) da=O .f

Start at the origin

Stokes' theorem says this should equal J(Vxv) da

Vxv = rs~nO [:0 (sin0 3r) - :<jJ (-r sin 0 COSO)]i+ ~ [Si~O:<jJ(rcos20) - :r(r3r)] 9

+ ~ [:r (-rr cos0 sin 0) - :0 (rcos20)] 4>

= ~[3rcosO]i+rsmu -[-6r]O+ -[-2rcosOsinO + 2rcosOsinO]q,r r

(1) Backface:da=-rdrdO4>j (Vxv).da=O J(Vxv).da=O

1 'If /2

Surface consists of two parts:

R4 sin2 () d() d</J.

20 CHAPTER 1 VECTOR ANALYSIS

Problem 1.60

Ie (VT) dT = ITe da Since e is constant, take it outside the integrals: e I VT dT=e IT da But e

equals corresponding component on the right, and hence

!VTdT = !Tda qed

identity (1) says da (v x c) =e (da x v) = -c (v x da) Thus Ie (V xv) dT = - Ie (v x da) Takee

outside, and again let e be X, y, z then:

!(VXV)dT = - !v x da Qed

(c) Let v =TVU in divergencetheorem: IV.(TVU)dT = ITVU.da Productrule #(5)::} V.(TVU) =

TV.(VU) + (VU) (VT) = T\72U + (VU) (VT) Therefore

!(T\72U + (VU) (VT») dT= !(TVU). da Qed

!(T\72U - U\72T) dT=!(TVU - UVT) da qed

(e) Stoke's theorem: I(Vxv), da = §v ill Let v = cT By Product Rule #(7): Vx(eT) =

and let c ~ x, y, and z to prove:

!VT x da = - f T ill Qed

to zero, and the z component of f is cos0, so

1

1r/2 I I

§da=al - a2 =j:.O

(d) For one such triangle, da = ~(r x ill) (since r x dl is the area of the parallelogram, and the direction is

(e) Let T =c .r, and use product rule #4: VT =V(c .r) =c x (Vxr) + (c V)r But Vxr = 0, and

(c V)r = (cxtx + Cyty + c:t:)(xi + yy = zz) = Cxi + Cyy+ c: Z= c So Prob 1.60(e)says

f Tdl = f(c r)dl = - !(VT) x da = - ! c x da = -c x ! da = -c x a = a x c. Qed

Problem 1.62

(1)

V-v =~~r2 ar (r2. !r) = ~~(r)r2 ar =I ~.I r

For a sphere of radius R:

f v da = f (-k i) (R2 sin 0 dOd4>i) =R f sin 0 dO d4>=47rR.

}

Evidently there is no delta function at the origin

Vx (rnr) = r2 ar ( r2rn) = r2 ar (rn+2)= -(n + 2)rnH = (n+ 2)rn-I r2

(except for n = -2, for which we already know (Eq 1.99) that the divergence is 47ro3(r».

R2 sin 0 dOd4>i are both in the i directions, so v x da = O .;

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.

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.)

(b) This time the "vertical" components cancel, leaving

term-t 0, and the z term -+ _4 1 ~kz.1I"fOz z

E = ~ )"(27rr)z A

47r£0 (r2 +Z2)3/2 z.

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 0".27rr dr =) 27rr, so ) =O"dr is the "line charge" of each ring.

According to Frob 2.7, all shells interior to the point (Le at smaller r) contribute as though their charge

E( )r -r,- 1

Qint-47T€Or2 where Qint is the total charge interior to the point Outside the sphere, all the charge is interior, so

(b) By Gauss'slaw: Qenc= €ofE da = €o(kR3)(41rR2)= I41r€okR5. I

By direct integration: Qenc = JpdT =JoR(5€okr2)(41rr2dr) =201r€ok JoR r4dr = 41r€okR5.,(

Problem 2.10

Think of this cube as one of 8 surrounding the charge Each of the 24 squares which make up the surface

of this larger cube gets the same flux as every other one, so:

Qenc= 1 Adp=} IE = P dy<0 100 I (for y >d).

=0 (since Vx (~) = 0,from Prob 1.62)

(since p depends on r', not r)

so E2 is a possible electrostatic field.

Let's go by the indicated path:

z

E dl =(y2 dx + (2xy + z2)dy + 2yz dz)k

Step II: x =Xo, Y : 0 -+ Yo, z = O dx = dz = O.

E.dl=k(2xy + z2)dy =2kxoY dy.

fII E dl =2kxo frioy dy =kxoY5.

(xo, Yo, ZO)

III

y

III

and for r <R: V(r) = - J:: (~~) df- J~ (4;fO -J/:sf)df = -dto [:k-:h (r22R2)]

( r2 )

Wh . R VV - -L- 8 (1)A - -L- 1 A E - VV - -L- 1 A ,(

en r >, - 411"fO 8r ;:: r - - 411"fO f=2"r,so - - - 411"fO f=2"r.

(In this form it is clear why a = 00 would be no good-likewise the other "natural" point, a = 0.)

VV = L2'xE (In (l.))§= L2,Xi§ = -E.,(

Using Eq 2.22 and the fields from Prob 2.16:

V(b) - V(O) = - Jofb E.dl = - Jofa E dl- J. a b E dl = _L2fO Jofa S ds - e£ 2fO a s J b ids

= - (~) 8;I: + ~ Insl: = 1- ~:: (1 + 2ln (~) ) 1

(a)Iv = ~ 2q .

Ifthe right-hand charge in (a) is -q, thenI V=0I, which,naively,suggestsE = - VV = 0, in contradiction

with the answer to Prob 2.2b The point is that we only know V on the z axis, and from this we cannot

hope to compute Ex = - ~~ or Ey = - ~~ That was OK in part (a), because we knew from symmetry that

Problem 2.26

V(a) = ~ rv'2h (a21rr)ch= 21ra~(V2h) =ah.

(where r =1-/V2)1

30 CHAPTER 2 ELECTROSTATICS

Cut the cylinder into slabs, as shown in the figure, and

use result of Prob 2.25c, with z -t x and C1-t P dx:

Prob 2.11: Eout = ~f<or = £.f; Ein = 0; so 6.E = £.f ,f<0 <0

'IF": : 5R Outside: §E da=E(211's ) l = 1 Qenc = 1 (211'R)l =} E = £.B.§ = £.§ (at surface<0 <0 <0 s <0 ).'..~ / ' Inside: Qenc = 0,so E =O : 6.E =.£.§ ,f<0

l

(c)Vout =R2u = Ru (at surface ) <or <0 ; \In = Ru ; so Vout= \In- ,f<0

“a < r < R” to “s < R”; in the same line change dr to ds; in the next

line change dr to ds (twice), and change ˆr to ˆs; in the last...

to zero at infinity, then, U = −m · B If the magnetic field does not go

to zero at infinity, one must stipulate that the dipole starts out orientedperpendicular to the field... ·.

• Page 147, Prob 8.2, top line, penultimate expression: change a2to a4; in(c), in the first box, change 16 to

• Page 149, Prob 8.5(c):