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Chapter 1

Electrostatics

1.1 Static fields and forces

1.1.1 Applications of Coulomb’s law

1. Four point charges, each of charge q and mass m, are locatedat the four corners of a square of side L.

L L

q q

xq q

y

L

L

(a) Find the magnitude of the force on one of the charges.

(b) Use the force in part (a) to find the velocity of one of thecharges a long time after the four charges are releasedfrom rest in the original configuration.

Solution:

(a) The force on the upper right-hand charge due to theother three charges at the corners of a square with sides

1

2 1. ELECTROSTATICS

of length L is

F =q2

L2i+

q2

L2j+

q2

2L2

(i+ j√

2

)

= q2[1 + 2

√2

2√2L2

](i+ j). (1.1)

The magnitude of this force on any of the four charges is

F = q2(1 + 2√2)/2L2. (1.2)

(b) If the four charges are released from rest, we can writefor the acceleration of any of the charges

F

m= a =

dv

dt=

dv

dL′dL′

dt, (1.3)

where L′ is the side length of the square at any timeduring the motion. The center of the square remainsfixed, and the distance, r, of a charge from the center isrelated to L′ by L′ =

√2r, so

dL′

dt=

√2dr

dt=

√2v.

Then, a =√2v

dv

dL′ ,

(1.4)

and the squared velocity after a long time is given by

V 2 =∫ ∞

L

√2a dL′ =

[q2(1 + 2

√2)√

2m

] ∫ ∞

L

dL′

L′2

=q2(1 + 2

√2)√

2mL. (1.5)

The velocity is the square root of this:

V = q

[(1 + 2

√2)√

2mL

] 12

. (1.6)

2. Four point charges, each of charge q and mass m, are locatedat the four corners of a square of side L.

1.1. STATIC FIELDS AND FORCES 3

(a) Find the potential energy of this configuration.

(b) Use conservation of energy to find the velocity of oneof the charges a long time after the four charges arereleased from rest in the original configuration.

Solution:

(a) In the configuration of four point charges q at the fourcorners of a square of side L, each of the four charges is adistance L from two other charges, and a distance

√2L

from a third charge. Consequently, the potential energyis given by

U0 =1

2

∑i,j �=i

[qiqj

|ri − rj|]=

4

2

[2q2

L+

q2√2L

]

=

√2q2

L(1 + 2

√2). (1.7)

(b) The kinetic energy of the four charges (each of mass m)at a long time after their release (so that L → ∞) isgiven by

4(1

2mv2

)= U0 =

√2q2

L(1 + 2

√2), (1.8)

so v =

[q2(1 + 2

√2)√

2mL

] 12

, as in Prob. 1. (1.9)

3. Four point charges, each of charge q, are fixed at the fourcorners of a square of side L. Find the electric field a distancez above the plane of the square on the perpendicular axis ofthe square.

Solution:The four point charges q are located at the corners of a squarewith sides of length L. The distance from each charge to apoint z above the square, on the perpendicular axis of the

square, is√z2 + L2/2. The horizontal fields cancel, and the

4 1. ELECTROSTATICS

L q

q

(z 2 + L 2/2) 1/2

q

q

LL

L

θz

magnitude of the vertical field is given by

Ez = E cos θ =4qz

(z2 + L2/2)32

. (1.10)

4. Show that the electric field a distance r from a long straightwire with a uniform linear charge density λ is given by

E =2λr

r. (1.11)

Solution:By symmetry, the electric field of a long straight wire isperpendicular to the wire. The field a distance r from thewire is given by

E(r) =∫

(r− r′)dq′

|r− r′|3 = λr∫ ∞

−∞dz

(z2 + r2)32

=λr

r

∫ π2

−π/2cos θdθ =

2rλ

r. (1.12)

We made the substitution z = r tan θ in doing the integral.

5. Two long parallel wires, each with uniform linear charge den-sity λ, are a distance a apart. The origin of coordinates isthe midpoint between the two wires.

(a) Find the electric field in terms of the vectors r and a.

(b) Write down the x and y components of E, taking a inthe x direction.


Solution:

(a) For the configuration of two wires a distance a apart,the electric field is

E =2λ(r− a/2)

|r− a/2)|2 +2λ(r+ a/2)

|r+ a/2)|2 . (1.13)

(b) The electric field in Cartesian coordinates is

Ex =2λ(x− a/2)

(x− a/2)2 + y2+

2λ(x+ a/2)

(x+ a/2)2 + y2(1.14)

Ey =2λy

(x− a/2)2 + y2+

2λy

(x+ a/2)2 + y2. (1.15)

6. A straight wire of length L has a uniformly distributedcharge Q. Find the electric field on the axis of the wire adistance z from the center of the wire for the cases

(a) z > L/2.

(b) −L/2 < z < L/2. (Hint: Use symmetry for part (b).)

–L/2 L/20 z´ z

z – z´

Solution:

(a) The field on the axis of the uniformly charged wire,a distance z from the center of the wire, is given forz > L/2 by

Ez =Q

L

∫ +L2

L2

dz′

(z − z′)2

=Q

L

[1

z − L/2− 1

z + L/2

]

=Q

z2 − L2/4. (1.16)

6 1. ELECTROSTATICS

(b) For −L/2 < z < L/2, the point z is a distance (L/2−z)from the end of the wire. The wire can be thought of astwo parts.

–L/2

2z L/2 – z L/2 – z

L/20 z

The part of the wire from z′ = z − (L/2 − z) = 2z −L/2 to z′ = L/2 is symmetric about the point z. Thismeans that the field due to that portion of the wire willcancel. The remaining part of the wire has a length L′ =L − 2(L/2 − z) = 2z and a charge Q′ = 2zQ/L. Themidpoint of this part of the wire is at z0 = (2z − L/2−L/2)/2 = z−L/2. Thus the electric field from this partof the wire is

Ez =Q′

[(z − z0)2 − L′2/4]

=2Qz

L[(z − z + L/2)2 − z2]

=2Qz

L[L2/4− z2]. (1.17)

7. (a) Find the electric field a distance z along the axis of auniformly charged ring of charge Q and radius R.

(b) Integrate the field for a ring to find the electric field adistance z on the axis of a uniformly charged disk ofcharge Q and radius R.

(c) Find the electric field of the charged disk for

(1) z = 0+ (just above the disk).

(2) z � R. (Expand the square root.)

Solution:

(a) Every point on the uniformly charged ring is the samedistance from a point a distance z along the axis of thering, and a line from any point on the ring makes the


same angle θ with the z-axis. Thus the electric field atz is

Ez =Q cos θ

r2=

Qz

(z2 +R2)32

. (1.18)

(b) The disk has a surface charge density σ = Q/πR2. It canbe considered as a collection of rings, each of radius r′

with a charge

dq = 2πr′σdr′ =2Qr′dr′

R2. (1.19)

The electric field a distance z along the axis of the diskis given as (using part (a))

Ez =∫ R

0

2Qr′

R2

zdr′

(z2 + r′2)32

=2Q

R2

[1− z√

z2 +R2

]. (1.20)

(c) (i) For z = 0+,

Ez =2Q

R2= 2πσ. (1.21)

(ii) For z � R, we write E as

Ez =2Q

R2

⎡⎣1−

(1 +

R2

z2

)− 12

⎤⎦ . (1.22)

Using the binomial theorem, we get

Ez =2Q

R2

[1−

(1− R2

2z2+ · · ·

)].

� Q

z2. (1.23)

This limit, equal to the field of a point charge Q, can beused as a check on the original result.

8. A straight wire of length L has a uniformly distributedcharge Q. Find Ex and Ey a distance d from the wire in theconfiguration shown in the figure on the next page. Expressyour answers in terms of the angles θ1 and θ2.

8 1. ELECTROSTATICS

θ2θ1

d

Ex

Ey

L

Solution:The parallel component of the electric field a distance d froma uniformly charged straight wire of length L is given by theintegral

Ex =Q

L

∫ x2

x1

dx′ cos θr2

, (1.24)

where x1 and x2 are the two endpoints of the wire. Let

x′ = − d cot θ

dx′ = d csc2 θdθ

r = d csc θ. (1.25)

Then

Ex =Q

Ld

∫ π−θ2

θ1cos θ dθ =

Q

Ld(sin θ2 − sin θ1), (1.26)

where θ1 and θ2 are the angles shown.For the perpendicular component of E, the same substi-

tution for z, leads to

Ey =Q

Ld

∫ π−θ2

θ1sin θ dθ =

Q

Ld(cos θ2 + cos θ1). (1.27)

9. Four point charges, each of charge q, are fixed at the fourcorners of a square of side L.

(a) Find the potential a distance z above the plane of thesquare on the perpendicular axis of the square.

(b) Use the potential to find the electric field at the samepoint.


Solution:

(a) Each charge q, at the corner of a square with sides of

length L, is a distance√z2 + L2/2 from a point z along

the perpendicular axis of the square. (See the figure inProb. 3.)

Thus the potential at the point z is

φ =4q√

z2 + L2/2. (1.28)

(b) The electric field is

Ez = −∂zφ =4qz

(z2 + L2/2)32

. (1.29)

10. A straight wire of length L has a uniformly distributedcharge Q.

(a) Find the potential on the axis of the wire a distancez > L/2 from the center of the wire.

(b) Use the potential to find the electric field at the samepoint.

Solution:

(a) The potential on the axis of the uniformly charged wire,a distance z from the center of the wire, is given forz > L/2 by

φ =Q

L

∫ +L2

−L/2

dz′

z − z′=

Q

Lln

(z + L/2

z − L/2

). (1.30)

(b) The electric field is

Ez = − ∂zφ =Q

L

[1

z − L/2− 1

z + L/2

]

=Q

z2 − L2/4, as in Prob. 6(a). (1.31)

10 1. ELECTROSTATICS

11. (a) Find the potential a distance z along the axis of auniformly charged ring of charge Q and radius R.

(b) Integrate the potential for a ring to find the potential adistance z along the axis of a uniformly charged disk ofcharge Q and radius R.

(c) Use the potentials in parts (a) and (b) to find thecorresponding electric fields.

Solution:

(a) Every point on the uniformly charged ring is the samedistance

√z2 +R2 from a point a distance z along the

axis of the ring, so the potential at z is

φ =Q√

z2 +R2. (1.32)

(b) The uniformly charged disk of radius R has a surfacecharge density σ = Q/πR2. It can be considered as acollection of rings, each of radius r′ with a charge

dq = 2πr′σdr′ =2Qr′dr′

R2. (1.33)

The potential a distance z along the axis of the disk is givenas (using part (a))

φ =2Q

R2

∫ R

0

r′dr′√z2 + r′2

=2Q

R2

[√z2 +R2 − z

]. (1.34)

(c) The electric field of the ring is

Ez = −∂z

[Q√

z2 +R2

]=

Qz

(z2 +R2)32

. (1.35)

The electric field of the disk is

Ez = −2Q

R2∂z

[√z2 +R2 − z

]=

2Q

R2

[1− z√

z2 +R2

].

(1.36)


12. (a) Integrate the potential for a ring to find the potentiala distance d from the center of a uniformly chargedspherical shell of charge Q and radius R.

(b) Integrate the potential for a spherical shell to find thepotential a distance d > R from the center of a uniformlycharged sphere of charge Q and radius R.

(c) Use the potentials in parts (a) and (b) to find thecorresponding electric fields.

Solution:

(a) A uniformly charged spherical shell has a surface chargedensity

σ =Q

4πR2. (1.37)

The spherical surface can be thought of as composed ofrings of radius r = R sin θ, where θ is the angle from thez-axis. Each ring has a charge

dq = 2πrσR dθ =1

2Q sin θ dθ. (1.38)

The distance from the plane of a ring to the point d isz = d − R cos θ. Thus the potential at a point d ≥ Rfrom the center of the sphere is

φ =∫

dq√z2 + r2

=Q

2

∫ π

0

sin θ dθ√(d−R cos θ)2 +R2 sin2 θ

=Q

2

∫ π

0

sin θ dθ√d2 +R2 − 2Rd cos θ

=Q

2

[√d2 +R2 − 2Rd cos θ

−Rd

]π0

=Q

2Rd[(d− R)− (d+R)] =

Q

d, d ≥ R. (1.39)

The potential for d ≤ R is

φ =Q

2Rd[(R− d)− (d+R)] =

Q

R, d ≤ R. (1.40)


For d outside the sphere, the potential is the same as thatof a point charge. For d inside the sphere, the potentialis constant and hence the electric field is zero.

(b) The potential at a distance d ≥ R from the center of auniformly charged spherical shell is Δq/d, where Δq isthe charge on the shell. The potential does not dependon the radius R of the shell, as long as d ≥ R. A uni-formly charged sphere can be considered a collection ofthe uniformly charged shells, so that the potential dueto the uniformly charged sphere will be φ = Q/d, whereQ is the net charge on the sphere. (Q is the sum of allthe Δq on each spherical shell.)Note that the charged sphere need not be uniformlycharged as long as its charge distribution is sphericallysymmetric.

(c) Since the potentials in a and b are the same as thepotential of a point charge, the electric fields are thesame as the electric field of a point charge:

E =Qr

r2, r ≥ R. (1.41)

for r ≥ R.

1.1.2 Gauss’s law

13. (a) Use Gauss’s law to find the electric field inside and out-side a uniformly charged hollow sphere of charge Q andradius R.

(b) Integrate E to find the potential inside and outside thehollow sphere.

Solution:

(a) By symmetry, the electric field of the uniformly chargedhollow sphere is in the radial direction. We consider aGaussian sphere of radius r, concentric with the hollow


sphere. Then, for r ≤ R, Gauss’s law becomes

∮E ·dA = 4πQ

πr2E = 0

E = 0. (1.42)

For r ≥ R, Gauss’s law is

∮E ·dA = 4πQ

πr2E = 4πQ

E =Q

r2. (1.43)

(b) The electric field for r ≥ R is the same as that for a pointcharge, so the potential is the same as the potential ofa point charge:

φ(r) =Q

r. (1.44)

For r ≤ R, E = 0, so the interior of a uniformly chargedshell is an equipotential with

φ =Q

R. (1.45)

14. Find the potential energy of a uniformly charged shell ofcharge Q and radius R by integrating

(a) ρφ.

(b) E2.

Solution:

(a) The potential energy of a uniformly charged sphericalshell of charge Q and radius R is given by

U =1

2

∫φdq =

1

2

(Q

R

)Q =

Q2

2R. (1.46)


(b) The potential energy is also given by integrating E2:

U =1

8π

∫E2d3r =

1

8π

∫r≥R

Q2d3r

r4

=Q2

8π

∫ ∞

R

4πr2dr

r4

=Q2

2R, the same as in part (a). (1.47)

15. (a) Use Gauss’s law to find the electric field inside and out-side a uniformly charged solid sphere of charge Q andradius R.

(b) Integrate E to find the potential inside and outside thesolid sphere.

Solution:

(a) The charge density inside the uniformly chargedsphere is

ρ =3Q

4πR3. (1.48)

By symmetry, the electric field is in the radial direction.We consider a Gaussian sphere of radius r, concentricwith the uniformly charged sphere. Then, for r ≤ R,Gauss’s law becomes∮

E ·dA = 4πQ

πr2E =[

3Q

4πR3

] [4πr3

3

]

E =Qr

R3. (1.49)

For r ≥ R, Gauss’s law is∮E ·dA = 4πQ

πr2E = 4πQ

E =Q

r2. (1.50)


(b) The electric field for r ≥ R is the same as that for a pointcharge, so the potential is the same as the potential ofa point charge:

φ(r) =Q

r, r ≥ R. (1.51)

For r ≤ R,

φ(r)− φ(R) = −∫ r

RE ·dr

φ(r) =Q

R− Q

R3

∫ R

rrdr

=Q

R

[1− 1

2R2(r2 − R2)

]

=Q

2R3(3R2 − r2). (1.52)

16. Find the potential energy of a uniformly charged sphere ofcharge Q and radius R by integrating

(a) ρφ.

(b) E2.

Solution:

(a) The potential energy of a uniformly charged sphere ofcharge Q and radius R is given by (using the answer toproblem 15(b))

U =1

2

∫ρφd3r =

1

2

∫r≤R

(3Q

4πR3

) [Q(3R2 − r2)

2R3

]d3r

=3Q2

16πR6

∫ R

0(3R2 − r2)4πr2dr

=3Q2

4R6

(3R5

3− R5

5

)=

3Q2

5R. (1.53)


(b) The potential energy is also given by integrating E2:

U =1

8π

∫E2d3r

=1

8π

∫r≤R

(Qr

R3

)2

d3r +1

8π

∫r≥R

(Q

r2

)2

d3r

=Q2

2

[∫ R

0

r4dr

R6+∫ ∞

R

dr

r2

]=

Q2

2R

[1

5+ 1

]

=3Q2

5R, (1.54)

the same as in part (a).

17. A conducting shell with charge q and radius A is locatedinside, and at the center, of a conducting shell with charge−q and radius B. Find the potential energy of this systemby integrating

(a) ρφ.

(b) E2.

Solution:

(a) Since the combined system has no charge, the potentialon the outer sphere is zero. The potential on the innersphere is given by

φ(A) =∫ B

A

qdr

r2=

q

A− q

B. (1.55)

Thus the potential energy is

U =1

2qφ(A) =

q2

2

[1

A− 1

B

]. (1.56)

(b) The only electric field is between the two spherical shells,where E = q/r2. Then

U =1

8π

∫d3rE2 =

1

2

∫ B

Ar2dr

q2

r4=

q2

2

[1

A− 1

B

]. (1.57)

18. (a) Use Gauss’s law to find the electric field inside and out-side a long straight wire of radius R with uniform chargedensity ρ.


(b) Integrate E to find the potential inside and outside thewire with the boundary condition φ(R) = 0.

Solution:

(a) The long straight wire has radius R and uniform chargedensity ρ. By symmetry, the electric field is perpendic-ular to the axis of the wire. We consider a Gaussiancylinder of length L and radius r, concentric with thewire. Then, for r ≤ R, Gauss’s law becomes∮

E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)

(b) We have to pick some radius for which φ = 0. We chooseφ(R) = 0, so that the surface of the wire is at zeropotential. Then, for r ≤ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R2πρrdr = πρ(R2 − r2).

(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)

19. Consider a large sheet of aluminum foil with surface chargedensity σ, assumed to be uniformly distributed on thesurfaces of the aluminum sheet.

(a) Calculate the electric field close to the foil and far fromany edge.


(b) Show that using the formula for the electric field fora sheet of charge gives the same answer as using theformula for the electric field just outside a conductor,when each formula is applied to this problem.

Solution:

(a) If we stay close to the sheet and far from an edge, wecan consider the large sheet to be an infinite plane. Thesymmetry of the infinite plane sheet requires that E beperpendicular to the plane of the sheet, and not dependon position parallel to the sheet. The appropriate Gaus-sian surface is a “Gaussian pillbox,” a flat box withidentical parallel ends (I and II) of arbitrary area A andany shape. (Gauss meant a flat box used to carry healingpills, and not the deadly military fort of the same gen-eral shape.) The pillbox is oriented parallel to the sheetof charge and is bisected by the sheet. E is parallel to theside (III) of the pillbox, so the only contribution to theGauss’s law integral is from the two flat surfaces.

E┴

E┴

σI

IIIII

A

From the symmetry of the sheet, E has the same magni-tude, but opposite direction, on either side of the sheet.Gauss’s law then leads to∮

E ·dA = 2E⊥∫IdA = 2E⊥A = 4πσA, (1.62)

soE⊥ = 2πσ, (1.63)

for an infinite sheet of charge with constant surfacecharge density σ. Note that E is independent of the dis-tance from the infinite plane sheet, as long as the sheetstill looks infinite.

1.1.3. VECTOR DIFFERENTIAL OPERATORS 19

(b) At the surface of a conductor, the field at surface II ofthe pillbox will be zero, so the field at the surface of aconductor with surface charge σ� is

E⊥ = 4πσ�. (1.64)

Aluminum is a conductor, so the electric field at its sur-face will be given by Eq. (1.64), but half of the surfacecharge on the aluminum sheet will be on one face of thesheet and half will be on the other face. Thus the sur-face charge σ� on either face will be half the total surfacecharge σ, and each equation will give the same electricfield.

1.1.3 Vector differential operators

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .The basic vector differential operator is ∇. It is written as an in-verted Greek Delta, and generally pronounced as “Del,” although“Nabla” is sometimes used. It is given in Cartesian coordinates(r = xi + yj+ zk) as

∇ = i∂x + j∂y + k∂z. (1.65)

However, it is usually simpler to let ∇ act on functions of rwithout using any specific coordinate system.

When ∇ acts on a scalar as ∇φ, it is called the gradient ofthe scalar. When ∇ is dotted into a vector as ∇·V, it is calledthe divergence of the vector. When ∇ is crossed into a vectoras ∇×V, it is called the curl of the vector. Another importantvector differential operation is the directional derivative a ·∇,which acts as the derivative in the direction defined by the unitvector a. Since a ·∇ is a scalar, it can act on a vector as (a ·∇)V,as well as on a scalar.

It is useful to know (commit to memory) the following actionsof ∇ on the position vector r, and its magnitude r:

1. dφ = dr ·∇φ

2. ∇φ(r) = rdφdr



Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)





Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)



81372-1 corrected headers.indd 1 6/5/18 3:46 PM


3. ∇(p · r) = p, for a constant vector p

4. (p ·∇)r = p

5. ∇· r = 3

6. ∇× r = 0

7. ∇× (∇φ) = 0

These actions of ∇ will be proven in subsequent problems.Two important integral theorems are the divergence

theorem ∫∇·Vd3r =

∮dS ·V, (1.66)

and Stokes’ theorem∮V ·dr =

∫dS · (∇×V). (1.67)

In the surface integrals, the vector differential area dS is aninfinitesimal area that can be approximated by a plane, withits vector direction being perpendicular to the plane. By con-vention for the integral over a closed surface (indicated by thenotation

∮), the positive direction of the vector dA is out of the

closed surface. The surface integral in Stokes’ theorem is overan open surface bounded by the closed contour for the line inte-gral (also indicated by

∮, but with a different meaning than for

the surface integral). The sign convention for Stokes’ theorem isthe right-hand rule that, if your fingers of your right hand cir-cle around the closed loop, your thumb will point in the positivedirection for the vector differential area dS.

There are also two algebraic identities that will be of greatuse:

a · (b× c) = b · (c×a) = c · (a×b), (1.68)

that is, the triple scalar product is invariant to cyclic permutation(a → b → c → a) or to an interchange of the dot and the cross.

a× (b× c) = b(a · c)− c(a ·b). (1.69)


This expansion of the triple vector product is called the bacminus cab rule. These two identities should also be committedto memory. Their use will be essential in solving the problems.

Operations on combinations of functions of position can besimplified by using the two distinct properties of ∇:

I. ∇ is a differential operator.

II. ∇ is a vector.

Because ∇ is a differential operator, it acts on functions one at atime, just as in d(uv) = u dv+v du. We also follow the conventionthat the differential operator acts only on functions to its right, sothe order in which ∇ appears must be to the left of the functionsit acts on and to the right of the other functions. As a vector,∇ must behave, in any expansion, like any other vector. In everycase, strict adherence to these two properties of ∇ will lead tothe correct evaluation of vector derivatives. We will use the twoproperties of ∇ in solving the problems.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20. The relation

dφ(r) = dr ·∇φ(r) (1.70)

can be considered a definition of the gradient of a scalarfunction. Use this definition to find ∇φ(r) in Cartesian co-ordinates.

Solution:In Cartesian coordinates, dr = idx + jdy + kdz, and incalculus, dφ(x, y, z) = (∂xdx+ ∂ydy + ∂zdz)φ(x, y, z). Then

dφ(r) = dr ·∇φ(r) becomes

(∂xdx+ ∂ydy + ∂zdz)φ(x, y, z) = (idx+ jdy + kdz)

·∇φ(x, y, z)

= (∇xdx+∇ydy +∇zdz)

× φ(x, y, z), (1.71)



Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)





and we see that

(∇φ)x = ∂xφ, (∇φ)y = ∂yφ, (∇φ)z = ∂zφ, (1.72)

or ∇φ = (i∂x + j∂y + i∂z)φ. (1.73)

21. Use Eq. (1.70) to derive the relations

(a) ∇φ(r) = rdφdr.

(b) ∇(r ·p) = p, for a constant vector p.

Solution:

(a) ∇φ(r) is a vector whose only possible direction is the rdirection. This means it can be given as ∇φ(r) = rψ(r),where ψ(r) is a scalar function of r. Then we can write

dφ(r) = dr ·∇φ(r) = dr · rψ(r)= drψ(r). (1.74)

This last equality means that

ψ(r) =dφ(r)

dr,

and rψ(r) = ∇φ(r) = rdφ(r)

dr. (1.75)

(b) For a constant vector p,

d(r ·p) = dr ·p = dr ·∇(r ·p). (1.76)

We see from the second equality that

∇(r ·p) = p. (1.77)

22. Use the divergence theorem for an infinitesimal sphere ofradius R to show that ∇· r = 3.


Solution:Applying the divergence theorem to the vector r gives, foran infinitesimal sphere of radius R, gives∫

∇· rd3r =∮

dS · r =∮

RdS = 4πR3. (1.78)

For an infinitesimal sphere we can take the divergence of rout of the integral to get∫

∇· rd3r = ∇· r∫d3r = (4πR3/3)∇· r. (1.79)

Comparing Eqs. (1.78) and (1.79) shows that ∇· r = 3.

23. (a) Use Stokes’ theorem for an infinitesimal surface to showthat the curl of a gradient equals zero. (∇×∇φ = 0)

(b) Use this result to show that ∇× r = 0.

Solution:

(a) Applying Stokes’ theorem for an infinitesimal surface toa gradient gives∫

dS · (∇×∇φ) =∮dr ·∇φ =

∮dφ = 0. (1.80)

For an infinitesimal surface, the factor ∇×∇φ can betaken out of the surface integral to give ∇×∇φ = 0.

(b) The position vector r is a gradient that can be seen by

∇r2 = rd(r2)dr

= 2r. As a gradient its curl is zero(∇× r = 0).

24. Show that (p ·∇)r = p.

Solution:We can consider (p ·∇)r to be the bac of bac minus cab.We will derive the results here step by step to emphasize themethod.

Step 1: We write the three vectors in the order p,∇, rbecause ∇ acts only on r, and not on p:

(p ·∇)r = p ∇ r p ∇ r. (1.81)



Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)





We have written the three vectors down twice to correspondto the identity bac = abc + cab, which we will use.

Step 2: We write the appropriate algebraic identity abovethe equation. (Saving time by keeping this in your head leadsto errors that are hard to detect.)

b(a · c) = a× (b× c) + c(a ·b)(p ·∇)r = p ∇ r p ∇ r. (1.82)

We use this step to make the identifications on the left-handside: b ∼ r, a ∼ p, c ∼ ∇. This tells us where to put thedots and the crosses. It also tells us that the sign betweenthe terms is positive.

Step 3: All that is left to do is to put in the dots and crossesaccording to the picture in step 2:

b(a · c) = a× (b× c) + c(a ·b)(p ·∇)r = −p× (∇× r) + ∇(p · r) = p. (1.83)

We have interchanged the r and ∇ in the first term(introducing a minus sign) to give ∇× r, which equals zero.

It is important to write down these three steps in turn. Leav-ing out step 2 wastes a lot of time in looking for invisibleerrors.

25. Expand the curl of a cross product, ∇× (A×B).

Solution:We use the bac minus cab rule in expanding ∇× (A×B).Since ∇ is a differential operator, there are two terms, onewith A constant and one with B constant:

∇× (A×B) = ∇× (A×B)|A constant

+∇× (A×B)|B constant. (1.84)

We treat these two terms separately. ForA constant, the firststep is to write the three vectors in the order A∇B, since


∇ only acts on B. This is written down twice, anticipatingthe use of the bac minus cab rule, giving

∇× (A×B) = A ∇ B A ∇ B. (1.85)

The second step is to write the bac minus cab equation abovethe line, so this intermediate step looks like:

a× ( b× c) = b(a · c)− c(a ·b)∇× (A×B)|A constant = A ∇ B A ∇ B. (1.86)

We use this step to relate the a, b, and c of the bac minus cabrule to the vectors ∇, A, and B, so a ∼ ∇, b ∼ A, c ∼ B.This association tells us where to put the dots and crossesto preserve the vector algebra. This third step leaves

a× (b× c) = b (a · c) − c(a ·b)∇× (A×B)|A constant = A(∇·B) − (A ·∇)B, (1.87)

with the sign determined by bac minus cab.There will be two more terms for B constant. These could

be found by repeating the same process as when A washeld constant, but there is a simpler way. We note thatthe original expression ∇× (A×B) changes sign under theinterchange A �� B. Then, the two terms with B constantjust follow by interchanging A and B with a sign change inEq. (1.87), which gives the four terms in the full expansion:

∇× (A×B) = A(∇·B)− (A ·∇)B−B(∇·A)

+ (B ·∇)A. (1.88)

26. Expand the gradient of a dot product, ∇(A ·B).

Solution:

We consider ∇(A ·B) to be the bac of bac minus cab, so theintermediate step for A constant is

b(a · c) = a× (b× c) + c(a ·b)∇(A ·B)|A constant = A ∇ B A ∇ B. (1.89)



Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)





The left-hand side of this equation shows that the appropri-ate identification is a ∼ A, b ∼ ∇, c ∼ B, which tells ushow to put in the dots and crosses on the right-hand side, so

∇(A ·B)|A constant = A× (∇×B) + (A ·∇)B. (1.90)

The other two terms with B constant follow by just inter-changing A and B since the original expression ∇(A ·B) issymmetric in A and B. This gives the final result

∇(A ·B) = A× (∇×B) + (A ·∇)B+B× (∇×A)

+ (B ·∇)A. (1.91)

27. (a) Find the electric field (for r = 0) of an electric dipole p,given by Ep = −∇[(p · r)/r3].

(b) Find the magnetic field (for r = 0) of a magneticdipole μ, given by Bμ = ∇× [(μ× r)/r3].

Solution:

(a)

Ep = −∇[p · rr3

]= −∇(p · r)

r3− (p · r)∇

(1

r3

)

=3(p · r)r− p

r3. (1.92)

(b)

Bμ = ∇×(μ× r

r3

)=

∇× (μ× r)

r3− (μ× r)×∇

(1

r3

)

=μ(∇· r)− (μ ·∇)r

r3+

3(μ× r)× r

r5

=3μ− μ

r3+

3(μ · r)r− 3μ

r3

=3(μ · r)r− μ

r3(1.93)


28. Show that the operator L = −ir×∇ satisfies L×Lφ =iLφ.[Hint: LetE = −∇φ. Expand (r×∇)× (r×E) in bac-cab.]

Solution:We want to show that L×Lφ = iLφ, where L =− ir×∇.This corresponds to −(r×∇)× (r×∇φ) = r×∇φ. Weintroduce E = −∇φ, and then will show that

(r×∇)× (r×E) = −r×E. (1.94)

We usea× (b× c) = b(a · c)− c(a ·b), (1.95)

withr×∇ = a, r = b, E = c. (1.96)

We take the vector derivative twice, first holding E constantand then holding r constant. For E constant,

(r×∇)× (r×E) = [E · (r×∇)]r− E[(r×∇) · r]= [(E× r) ·∇)]r− E[r · (∇× r)]

= E× r. (1.97)

For r constant,

(r×∇)× (r×E) = r[(r×∇) ·E]− [r · (r×∇)]E

= r[r · (∇×E)]− [(r× r) ·∇]E

= 0. (1.98)

Adding the results of Eqs. (1.97) and (1.98) gives

(r×∇)× (r×E) = −r×E, (1.99)

which confirms that L×Lφ = iLφ.

29. The potential of a point charge is φ = q/r.

(a) Find the electric field, given by E = −∇φ.

(b) Find the curl, and (c) the divergence of E for r �= 0.



Solution:


E ·dA = 4πQ

2πrLE = 4π2r2ρL

E = 2πρr, r ≤ R. (1.58)


2πrLE = 4π2R2ρL

E =2πρR2

r, r ≥ R. (1.59)


φ(r) = −∫ r

RE ·dr = −

∫ r


(1.60)For r ≥ R,

φ(r) = −∫ r

RE ·dr = −

∫ r

R

2πR2ρdr

r= 2πρR2 ln(R/r).

(1.61)




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Solved Problems in Classical Electromagnetism is a valuable tool to help students learn to do physics while using concepts they learn in their

courses. Students who are taking or have already taken an advanced EM course will find the book to be a useful adjunct to their textbook, giving added practice in applying what they are learning. For students who are taking an undergraduate EM course and want to get more depth, this book can help them achieve that aim and also help them prepare for graduate work. Beginning students, or those not even taking a course at the moment, can benefit from these problems and learn just from working on them with the help of the solutions. In each chapter, the problems start out relatively easy and then get progressively more advanced, helping students to go just as far as they can at their present level.

The book includes a number of review sections to assist students without previous advanced training in working out the problems. The first review section is a comprehensive development of vector calculus that will prepare students to solve the problems and provide a strong foundation for their future development as physicists. The problems are drawn from the topics of electrostatics, magnetostatics, Maxwell’s equations, electromagnetic radiation, and relativistic electromagnetism.

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