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Classical Electrodynamics: Lecture notes: Electric charge interaction

K K Likharev

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Classical ElectrodynamicsProblems with solutions

Konstantin K Likharev

Chapter 1

Electric charge interaction

Problem 1.1. Calculate the electric field created by a thin, long, straight filament,electrically charged with a constant linear density λ, using two approaches:

(i) directly from the Coulomb law, and(ii) using the Gauss law.

Solutions:

(i) From the translational and axial symmetries of the problem, it is clear that E(r) =nρE(ρ), where ρ is the shortest distance from the observation point to the filament1.Let us select the plane of drawing so that it contains both the filament and theobservation point, and take the line of the filament for axis z (see the figure below).

Then, according to the linear superposition principle, the field’s magnitude maybe calculated as

∫ ∫ ∫ρ θ ρρ

= = =+

ρ=−∞

=+∞

=−∞

=+∞

=−∞

=+∞E dE dE dE

z( ) cos

( ),

z

z

z

z

z

z

2 2 1/2

1 I regret using the same letter (ρ) as for the charge density per unit volume (used in the lecture notes, but not inthis problem), but both notations are traditional. The difference between these notions will be always very clearfrom the context.

doi:10.1088/978-0-7503-1407-7ch1 1-1 ª Konstantin K Likharev 2018

where dE is the magnitude of the elementary contribution to the field, created by asmall segment dz of the filament, with electric charge λdz. According to Eq. (1.7) ofthe lecture notes,

λπε ρ

=+

dE dzz

14

1,

02 2

so that the total field

∫ ∫ρ λρπε ρ

λπε ρ

ξξ

λπε ρ

=+

≡+

= *−∞

+∞ +∞E

dz

z

d( )

4 ( ) 2 (1 ) 2. ( )

02 2 3/2

0 0 2 3/20

For the last transition, I have used the well-known value (1) of this table integral—see, e.g. Eq. (A.32b)2.

(ii) Taking a round cylinder of radius ρ and length l with its axis on the filament forthe Gaussian volume, we ensure that on its round walls the electric field E isconstant and perpendicular to the volume boundary, while the field flux through thecylinder’s ‘lids’ is zero. As a result, Eq. (1.16) yields

πρ λε

=lEl

2 ,0

immediately giving the same result (*).

We see that for this highly symmetric problem both solution ways are straightfor-ward, but the Gauss law makes the calculations easier.

Problem 1.2. Two thin, straight parallel filaments, separated by distance ρ, carry equaland opposite uniformly distributed charges with a linear density λ—see the figure below.Calculate the electrostatic force (per unit length) of the Coulomb interaction between thewires. Compare the result with the Coulomb law for the force between the point charges.

Solution: Using the result (*) of problem 1.1, we obtain

ρ λ ρ λπε ρ

= = =Fl

qEl

E( )

( )2

.2

0

Note that the force drops with distance as 1/ρ, rather than as 1/r2 for point charges.Such different scaling of interaction in systems of different dimensionality is verytypical for physics at large.

2Actually, this integral may be easily worked out by substitution ξ ≡ tan φ, giving dξ = dφ/cos2φ = dφ (1 +tan2φ) = dφ (1 + ξ2), so that dξ/(1 + ξ2)3/2 = dφ/(1 + ξ2)1/2 = cos φ dφ = d(sin φ).

Classical Electrodynamics: Problems with solutions

1-2

Problem 1.3. A sphere of radius R, whose volume had been charged with a constantdensity ρ, is split with a very narrow, planar gap passing through its center.Calculate the force of mutual repulsion of the resulting two hemispheres.

Solution: Since the gap is very narrow, we may neglect its effect on the distribution ofthe electric field E inside the sphere, and use Eq. (1.22) of the lecture notes:

ρε

= =E r E rr

E r n( ) ( ), with ( )3

,r0

where r is the distance of the observation point from the sphere’s center O—see thefigure above. Acting on an elementary volume dV = d3r of the sphere’s material,with the elementary charge dQ = ρd3r, this electric field produces a radially directedforce of magnitude

ρε

ρ ρε

= = ≡dF E r dQr

d rrd r( )

3 3.

0

32

0

3

Due to the axial symmetry of the problem, the net (repulsive) Coulomb force Facting on each hemisphere has to be normal to the gap—in the figure above, directedalong the axis z. Hence only the z-component of the elementary force dF,

θ ρε

θ= =dF dFr

d rcos3

cos ,z

2

0

3

where θ is the polar angle of the radius-vector r of the elementary volume d3r (see thefigure above), can contribute to the net force F. As a result, we may use the standardspherical coordinates, with the origin in the sphere’s center, to calculate the net forcemagnitude as

∫ ∫ ∫ ∫ρε

φ θ θ θ ρε

π

π ρε

= = = =

=

π π

>F F dF d d r dr

R

R

3cos sin

32

12 4

12.

zz

z

R

0

2

0 0

2

0

/2

0

32

0

4

2 4

0

Another way to represent this result is to express it via the charge Q = (2π/3)R3ρ ofeach hemisphere:

πε πε= ≡F

QR

QR

14

34

14

,0

2

20

2

ef2

Classical Electrodynamics: Problems with solutions

1-3

showing that the effective distance of their interaction (as point charges) is Ref =(2/√3)R ≈ 1.16 R.

One may wonder whether the above calculation properly excludes the effect of theelectric field of a hemisphere on itself, since the field we have used is evidentlyproduced by the sphere as a whole. The proper response to this concern is that due toNewton’s second law, the internal forces between elementary charges of the samehemisphere compensate each other, so that these forces are automatically canceledat the integration over the hemisphere’s volume:

∫ ∫ ∫ ∫ρ ρ ρ ρ= + = + = => > > >

d r d r d r d rE E E E E F( ) 0 .z z z z0

full3

0other self

3

0other

3

0other

3

Problem 1.4. A thin spherical shell of radiusR, which had been charged with a constantareal density σ, is split into two equal halves by a very narrow, planar cut passingthrough the sphere’s center. Calculate the force of electrostatic repulsion between theresulting hemispheric shells, and compare the result with that of the previous problem.

Solution: A thin cut does not alter substantially the electric field distribution, so ithas the same radial direction and spherically symmetric distribution, E(r) = nrE(r),as for the uncut shell. The function E(r) may be readily found using the Gauss lawapplied to spheres with r < R and r > R:

σ ε=

<<

⎧⎨⎩E rr R

R r R r( )

0, for ,/ , for .2

02

Hence the average field applied to the shell equals3

σε

= − + + =EE R E R( 0) ( 0)

2 2,

0

providing distributed normal forces F, with the following density (force per unit area):

σ σε

= = =dFdA

dQdA

E E2

.2

0

Due to the axial symmetry of the problem, only the ‘vertical’ component(meaning the direction perpendicular to the cut plane—see the figure above) of

3A strict proof of the correctness of such averaging will be given in the model solution of problem 2.1.

Classical Electrodynamics: Problems with solutions

1-4

these elementary forces contribute to the total force F acting on each hemisphere—for example the top one:

∫ ∫

∫ ∫

θ

φ σε

θ θ θ π σε

= =

= =*

π π

⎛⎝⎜

⎞⎠⎟F

ddA

d rdFdA

d r

d R dR

Fcos

2cos sin

2.

( )zhemisphere

2

hemisphere

2

0

2

0

/2 2

0

22 2

0

In order to compare this result with the solution of the previous problem, let usassume that the radius R and the total charge Q of each hemisphere are the same inboth cases:

π σ π ρ= =R R Q223

.2 3

In this notation, Eq. (*) may be rewritten as

πε πε= ≡ = ≈F

QR

QR

R R R1

412

14

, with 2 1.41 ,0

2

20

2

ef2 ef

while for the case of volume-distributed charge, the similarly defined effective distanceequals (2/√3)R ≈ 1.16 R. This difference is natural, because in the case of thinhemisphere shells, the elementary charges are, on the average, farther from each other.

Problem 1.5. Calculate the distribution of the electrostatic potential created by astraight, thin filament of finite length 2 l, charged with a constant linear density λ, andexplore the result in the limits of very small and very large distances from the filament.

Solution: Due to the limited (axial) symmetry of the problem, applying the Gausslaw to it is not very productive, so let us resort to direct summation (actually,integration) of component charge fields. Let us select the reference frame so that thefilament coincides with segment [−l, +l] of axis z, and the observation point r hasCartesian coordinates {ρ, 0, z}—see the figure above. Then the field source point rʹhas coordinates {0, 0, zʹ}, and Eq. (1.38) of the lecture notes, integrated across thefilament cross-section, reads

∫ ∫ϕ λπε

λπε ρ

= ′− ′

= ′− ′ +−

+

−

+dz dzz z

rr r

( )4 4 [( ) ]

.l

l

l

l

0 02 2 1/2

Classical Electrodynamics: Problems with solutions

1-5

Introducing the new, dimensionless variable ξ ≡ (zʹ − z)/ρ, we may reduce theintegral to the table one—see, e.g. Eq. (A.29a):

∫ϕ λπε

ξξ

λπε

ξ ξ

λπε

ρρ

=+

= + +

= − + + −+ + − +

ρ

ρ

ρ

ρ

− −

+ −

− −

+ −d

l z l zl z l z

r( )4 ( 1) 4

ln ( 1)

4ln

[( ) ] ( )[( ) ] ( )

.

l z

l z

l z

l z

0 ( )/

( )/

2 1/20

2 1/2

( )/

( )/

0

2 2 1/2

2 2 1/2

For the observation points very close to the filament, and not very close to any ofits ends, the denominator of the fraction becomes much smaller than its nominator.Expanding the latter in the small parameter ρ/(l + z), and keeping only the leadingterm, we obtain the result following from the solution of problem 1.1:

ϕ λπε ρ

λπε ρ

λπε

ρ

ρ

≈ −+

= − = − +

< ≪ −

l zl z

l z

z l l l z

r( )4

ln2( )/2( ) 4

ln4( )

2ln const,

for , and , .

02

0

2 2

20

On the other hand, at large distances from the filament the nominator anddenominator of the fraction become very close to each other. Expanding bothexpressions in the Taylor series in the small parameter l/r, where r ≡ (z2 + ρ2)1/2 is thedistance from filament’s middle point, in the first non-vanishing approximation weobtain

ϕ λπε

λπε

λπε

≈ +−

≈ + ≈ ≫⎛⎝⎜

⎞⎠⎟

l rl r

lr

lr

r lr( )4

ln1 /1 / 4

ln 12

42

, for ,0 0 0

i.e. just Eq. (1.35) for a point charge q = 2lλ.

Problem 1.6. A thin plane sheet, perhaps of an irregular shape, carries electriccharge with a constant areal density σ.

(i) Express the electric field’s component normal to the plane, at a certain distancefrom it, via the solid angle Ω at which the sheet is visible from the observation point.(ii) Use the result to calculate the field in the center of a cube, with one face chargedwith constant density σ.

Classical Electrodynamics: Problems with solutions

1-6

Solutions:

(i) Let us place the coordinate origin 0 at the charged plane’s point that is closest tothe field observation point (at distance z from the plane)—see the figure above.Then, according to Eq. (1.7) of the lecture notes, the normal (z-) component of theelectric field component induced by charge element dQ = σd2ρ (where ρ is the 2Dradius-vector within the plane) equals

θπε

θ σπε

ρ θ= = = *dE dEdQr

dr

cos1

4cos

4cos , ( )z

02

0

2

2

where r = (ρ2 + z2)1/2 is the distance from the elementary area d2ρ and theobservation point, and cos θ = z/r. But as the figure above shows, the productd2ρ cos θ equals d2ρ′—the projection of the elementary area d2ρ on the plane normalto the vector r. In turn, the ratio d2ρ′/r2 is just dΩ—the solid angle at which theelementary area d2ρ is visible from the field measurement point4. As a result,Eq. (*) may be rewritten simply as

σπε

= ΩdE d4

,z0

and its integration over the charged sheet yields the requested result:

σπε

= Ω **E4

. ( )z0

(ii) Since all six faces of a cube are visible from its center at equal angles Ω, and theirsum constitutes the full solid angle 4π, each Ω is equal to 4π/6, and Eq. (**) yields

σπε

π σε

= ≡E4

46 6

.z0 0

Due to the evident symmetry of the system, the field cannot have other Cartesiancomponents, so that we may rewrite the result as

σε

=E6

.0

Problem 1.7. Can one create, in a non-vanishing region of space, electrostatic fieldswith the Cartesian components proportional to the following products of Cartesiancoordinates {x, y, z},

yz xz xyxy xy yz

(i) { , , },(ii) { , , }?

4All this calculation is absolutely similar to the one made at the proof of the Gauss law in section 1.2 of thelecture notes—see the transition from Eq. (1.13) to Eq. (1.14).

Classical Electrodynamics: Problems with solutions

1-7

Solution: Let us calculate the curl of both supposed fields, using the definition of thatoperator—see, e.g. Eq. (A.54):

∇ × = ∂∂

−∂∂

∂∂

− ∂∂

∂∂

− ∂∂

⎧⎨⎩⎫⎬⎭

Ey

E

zEz

Ex

E

xEy

E , , .z y x z y x

For the field (i) we obtain ∇ × E ∝ {x − x, y − y, z − z} ≡ 0, while for the field(ii) ∇ × E ∝ {z − 0, 0 − 0, y − x} vanishes only at one point (x = y = z = 0). However,according to the homogeneous Maxwell equation (Eq. (1.28) of the lecture notes),the curl of the electrostatic field has to equal zero at any point where it exists; hencethe field (i) can exist in a region of finite size, while the field (ii) cannot.

The fact that the field (i) has zero divergence as well, i.e. requires ρ(r) ≡ 0 withinthe region of its existence, does not prevent it from being realistic, because the fieldmay be created by electric charges outside that particular region. Note also that theelectric field (ii) may be induced by a magnetic field changing (linearly) in time—seesection 6.1 of the lecture notes, but such E cannot be called an electrostatic field.

Problem 1.8. Distant sources have been used to create different electric fields on twosides of a wide and thin metallic membrane, with a round hole of radius R in it—seethe figure below. Besides the local perturbation created by the hole, the fields areuniform:

= × <>≫

⎧⎨⎩E zE z

E r n( ), at 0,, at 0.r R z

1

2

Prove that the system may serve as an electrostatic lens for charged particlesflying along axis z, at distances ρ ≪ R from it, and calculate the focal distance f ofthe lens. Spell out the conditions of validity of your result.

Solution: Let us select the orientation of the mutually perpendicular axes x = ρcos φand y = ρsin φ so that for the particular flying particle y = 0. At the axis of this axiallysymmetric system (i.e. at ρ = 0), the partial derivatives ∂Ex/∂x and ∂Ey/∂y have to beequal due to the symmetry, so that the Maxwell equation (1.27), ∇ · E = 0, yields

∂∂

= − ∂∂ρ ρ= =

Ex

Ez

2 .x z

0 0

Classical Electrodynamics: Problems with solutions

1-8

Since at the axis, the value of Ex has to vanish, its Taylor expansion at x = 0 startswith the linear term, so that at small x

≈ ∂∂

∣ = − ∂∂

∣ρ ρ<<=

= =E xEx

x Ez2

.x x Ry

x z

00 0

According to this expression, at sufficiently small x, the field Ex is so small thatparticle’s deviation from its initial trajectory during its flight through the perturbedfield region (of the length Δz ~ R) is negligible. However, during this flight theparticle picks up a transverse momentum created by the field:

∫ ∫ ∫

∫

= = =

= − ∂∂

= − −−∞

+∞

ρ=

p F dt q E dt q Edtdz

dz

qxv

Ez

dz qxv

E E2 2

( ).

x x x x

z

z

z02 1

(Here the change of the particle’s longitudinal velocity vz during its flight through thehole was neglected—which is legitimate if the length of the uniform field region ismuch larger than R, as implied by the problem’s conditions.) If this momentum, andhence the acquired radial velocity vx = px/m, are directed toward the axis, which istrue at the proper sign of the combination q(E2 − E1),

5 it would lead to the particlehitting the system’s axis at the distance

≡ = =−

=−

=−

=−( )

z f v t vxv

vx

p m

mvq E E

Tq E E( ) /

2( )

4( )

,z f zx

zx

z2

2 1 2 1

where T = mvz2/2 is particle’s initial kinetic energy. The most important feature of

this result is that the initial coordinate x of the particle has dropped out of it. Thismeans that particles with any small x (and, by axial symmetry, all particles flyingsufficiently close to axis z, and parallel to it) will be directed to the same focal point f.This is the key property of any lens—in this case the electrostatic one.

It is remarkable that this result does not depend on the exact field distribution inthe hole region. (This distribution may be calculated analytically—for example,using the degenerate ellipsoidal coordinates, to be discussed in section 2.4 of thelecture notes.) It is valid if two strong conditions,

≪ ≪R fT

qE,

2

are satisfied. The first of them has allowed us to treat x as a constant during theintegration of force Fx over time, while the second one, to neglect the change of vz

5 If the sign is opposite, the system will disperse the parallel particle beam, but in a highly ordered way—withthe distance from the axis proportional to the initial distance x. So the system will still work as a lens, though adiverging (‘negative’) one, with f < 0.

Classical Electrodynamics: Problems with solutions

1-9

during the focusing process. Note, however, that a violation of the second conditiondoes not ruin the focusing effect; it only makes the expression for f somewhat bulkier.

Problem 1.9. Eight equal point charges q are located at the corners of a cube ofside a. Calculate all Cartesian components Ej of the electric field, and their spatialderivatives ∂Ej/∂rj′, at cube’s center, where rj are the Cartesian coordinates orientedalong the cube’s sides—see the figure below. Are all your results valid for the centerof a planar square, with four equal charges in its corners?

Solution: According to Eq. (1.33) of the lecture notes, the Cartesian components ofthe field and their derivatives may be expressed via the derivatives of the electrostaticpotential:

ϕ ϕ= − ∂∂

∂∂

= − ∂∂ ∂ *

′ ′E

r

E

r r r, . ( )

jj

j

j

j j

2

Due to the cube’s symmetry, at its center the whole vector E should vanish. (Indeed,if it did not, the vector would be directed toward either some face or some corner ofthe cube, but that would violate the equivalence of all faces and corners.) Hence,according to the first of Eqs. (*), all field components, i.e. the partial derivatives ofthe scalar potential, have to vanish at the center. So, if we take this point for theorigin (r = 0, i.e. r1 = r2 = r3 = 0), the Taylor expansion6 of function ϕ(r1, r2, r3),apart from an inconsequential constant ϕ(0, 0, 0), should start with quadratic terms.Due to the cube’s symmetry with respect to coordinate swaps, the coefficients atthese terms have to be independent of the coordinates:

∑ ∑ϕ ϕ− = + + **= ′=

′≠

′ ( )r r rA

rB

r r O r( , , ) (0, 0, 0)2 2

, ( )j j j

j j1

3

, 1

3

j j j j1 2 32 3

where A and B are constants:

ϕ ϕ= ∂∂

∣ ≡ −∂∂

∣ = ∂∂ ∂

∣ ≡ −∂∂

∣ ≠ ′ ***= =′

=′

=Ar

E

rB

r r

E

rj j, while , for . ( )

j

j

j j

j

j jr r r r

2

2 0 0

2

0 0

6 See, e.g. Eq. (A.14b).

Classical Electrodynamics: Problems with solutions

1-10

But because of the system’s symmetry, the potential also has to be invariant withrespect to the change of sign of any single coordinate, rj → −rj, which geometricallycorresponds to its mirror reflection in the plane of two other coordinate axes7. AsEq. (**) shows, this is only possible if B = 0.

Moreover, since there is no charge in the vicinity of the cube’s center,the potential has to satisfy the Laplace equation (1.42), in Cartesian coordinatesreading

∑ ϕ∂∂

= ****= r

0. ( )j 1

3

j

2

2

Comparing this requirement with the first of Eqs. (***), we obtain A = 0 as well, sothat all derivatives ∂Ej/∂rj′ at r = 0 have to equal zero.

For the field in the center of a planar square, with four similar charges in itscorners, the situation is somewhat different, due to the reduced symmetry of thesystem. The electric field E at the center (and hence all its Cartesian components) stillhas to equal zero, because its in-plane component (say, E12) cannot be directedtoward any side of the square, and its normal component E3 cannot be directed toeither side of the plane, due to the r3 ↔ −r3 symmetry. However, the potential maynot remain the same if the coordinate r3 directed normally to square’s planeis swapped with one of in-plane coordinates (r1 or r2). As a result, Eq. (**), stillwith B = 0 due to the mirror symmetry, has to be generalized as

ϕ ϕ− = + + ′ +( ) ( )r r rA

r rA

r O r( , , ) (0, 0, 0)2 2

,j1 2 3 12

22

32 4

and the Laplace equation (****) imposes only the following requirement: 2A +Aʹ = 0, i.e.

∂∂

= ∂∂

= − ∂∂

Er

Er

Er

12

.1

1

2

2

3

3

This relation is similar to the one used for the solution of problem 1.8 (where it istrue due to the axial symmetry of the system).

Problem 1.10. By a direct calculation, find the average electric potential of thespherical surface of radius R, created by a point charge q located at a distance r > Rfrom the sphere’s center. Use the result to prove the following general mean valuetheorem: the electric potential at any point is always equal to its average value onany spherical surface with the center at that point, and containing no electric chargesinside it.

7Actually, because of this symmetry, the higher terms of the expansion (**) should start with O(rj4), rather

than O(rj3).

Classical Electrodynamics: Problems with solutions

1-11

Solution: Using the evident axial symmetry of the problem (see the figure above),and Eq. (1.35) of the lecture notes, we obtain:

∮ ∫ ∫ϕπ

ϕ θ ππ

ϕ θ θ θπε

θ θ≡ Ω = =′

π πd d

qr

d1

4( )

24

( )sin12 4

sin ,ave0 0 0

where r′ is the distance between the point charge and the observation point:

θ′ = + −r R r Rr( ) 2 cos .2 2 2

The integral may be readily worked out by the introduction of a new variableξ ≡ cos θ (so that sin θ dθ = dξ):

∫ϕπε

ξξ πε

ξ

πε πε

=+ −

=′ −

+ −

=−

+ − − + + =

ξξ

− = += −

{ }

q d

R r Rr

qr Rr

R r Rr

qRr

R r Rr R r Rrq

r

12 4 [ 2 ]

12 4

2( 2 )

[ 2 ]

12 4

2( 2 )

[ 2 ] [ 2 ]4

.

ave0 1

1

2 2 1 20

2 2 1 2 1

1

0

2 2 1 2 2 2 1 2

0

We see that the average indeed coincides with the potential’s value in the middleof the sphere. (Notice that this result is only valid for the case r > R.) Now the proofof the mean value theorem is elementary using the linear superposition principle:since the relation in question,

πεϕ=q

r4,

0ave

holds for each point charge located outside the sphere, it is also true for any systemof such charges.

Problem 1.11. Two similar thin, circular, coaxial disks of radius R, separated bydistance 2d, are uniformly charged with equal and opposite areal densities ±σ—seethe figure below. Calculate and sketch the distribution of the electrostatic potentialand the electric field of the disks along their common axis.

Classical Electrodynamics: Problems with solutions

1-12

Solution: Let us start from calculating the electrostatic potential of one disk, with aconstant areal charge density σ, at the axis point separated by distance z from disk’splane—see the figure below.

In the polar coordinates {ρ, φ} within the plane of the disk (with angle φ referred toan arbitrary horizontal axis), the elementary disk area is ρdφdρ, and its electriccharge is σρdρdφ, so that Eq. (1.38) of the lecture notes takes the form

∫ ∫ϕ σπε

φ ρ ρ=π

z ddr

( )4

,R

0 0

2

0

where r is the distance between the elementary charge’s location and the observationpoint on disks’ common axis. As the figure above shows, r = (ρ2 + z2)1/2, so that thefunction under the integral is independent of φ, and the integral may be easilyworked out:

∫ ∫ϕ σπε

π ρ ρρ

σε

ρρ

σε

=+

= ++

= + −ρ

ρ

=

=z

d

z

d z

zR z z( )

42

( ) 4( )

( ) 2[( ) ].

R R

0 0 2 2 1/20 0

2 2

2 2 1/20

2 2 1/2

Now using this formula and the linear superposition principle, we may readily writedown an expression describing the potential created by both disks, at distance z fromthe center of the system (in this new reference frame, the disk center positions are ±d):

ϕ σε

=+ − − −

− + + + +⎪ ⎪

⎪ ⎪⎧⎨⎩

⎫⎬⎭

R z d z d

R z d z d2

[ ( ) ]

[ ( ) ].

0

2 2 1/2

2 2 1/2

This function is plotted in the figure below for three values of the ratio R/d. Forsmall values of this ratio, the potential is clearly separated into two peaks, ofopposite polarity, created by each disk. On the other hand, at R ≫ d the result tendsto the one for two infinite planes, with ϕ between the disks being a linear function ofz, with the slope corresponding to the electric field—see the model solution ofproblem 1.14 below.

Classical Electrodynamics: Problems with solutions

1-13

Now the electric field at the axis (which has only one, vertical component due tothe axial symmetry of the problem) may be calculated by differentiation of theelectrostatic potential—see Eq. (1.33) of the lecture notes:

ϕ

σε

= − ∂∂

= − −+ −

+ − + ++ +

− −⎧⎨⎩

⎫⎬⎭

Ez

z d

R z dz d

z d

R z dz d

2 [ ( ) ]sgn( )

[ ( ) ]sgn( ) .

z

02 2 1/2 2 2 1/2

In the figure below, this function is plotted for the same three values of the R/dratio as the potential in the figure above. The plots show that at R/d = 5, the fieldbetween the disks is already pretty uniform, although its magnitude is still noticeablysmaller than that (σ/ε0) between two infinite planes. Note also that the ratio does notaffect the electric field’s jump by ±σ/ε0 as the observation point crosses a disk—justas it should be, according to Eq. (1.24) of the lecture notes.

On the technical side, this solution illustrates again the advantage of calculating theelectrostatic potential (a scalar function) first, and only then the electric field from it.

Classical Electrodynamics: Problems with solutions

1-14

Problem 1.12. In a certain reference frame, the electrostatic potential created by anelectric charge distribution is

ϕ = + −⎛⎝⎜

⎞⎠⎟

⎧⎨⎩⎫⎬⎭C

r rrr

r( )1 1

2exp ,

0 0

where C and r0 are constants, and r ≡ ∣r∣ is the distance from the origin. Calculate thecharge distribution in space.

Solution: According to the Poisson equation (see Eq. (1.41) of the lecture notes), thecharge density may be calculated as

ρ ε ϕ= − ∇r r( ) ( ).02

Since the potential distribution is spherically symmetric, ϕ(r) = ϕ(r), the generalexpression for the Laplace operator of such scalar function in spherical coordinates8

is reduced to

ϕ ϕ∇ = *⎛⎝⎜

⎞⎠⎟r

rddr

rddr

( )1

. ( )22

2

so that by performing straightforward differentiation we obtain a very simple,exponential charge density distribution:

ρ ε= − − !⎧⎨⎩

⎫⎬⎭Cr

rr

r( )2

exp . (INCOMPLETE )0

03

0

Note, however, that since Eq. (*) is invalid at r = 0, this point has to be exploredseparately. At r → 0 the potential distribution tends to C/r and, as Eq. (1.35) of thelecture notes shows, such a potential is created by a point charge q = 4πε0C, locatedat the origin. As a result, the complete solution of our problem is

ρ ε πδ δπ

= − − = − −⎡⎣⎢

⎧⎨⎩⎫⎬⎭

⎤⎦⎥

⎡⎣⎢

⎧⎨⎩⎫⎬⎭

⎤⎦⎥C

rrr

qr

rr

r r r( ) 4 ( )1

2exp ( )

18

exp ,003

0 03

0

with the total charge9

∫ ∫ ∫

∫

ρπ π

π

ξ ξ ξ

= = − − = − −

= − − =

∞

∞

⎛⎝⎜

⎧⎨⎩⎫⎬⎭

⎞⎠⎟

⎛⎝⎜

⎧⎨⎩⎫⎬⎭

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Q d r qr

rr

d r qr

rr

r dr

q d

r( ) 11

8exp 1

18

4 exp

112

exp{ } 0.

3

03

0

3

03

0 0

2

0

2

For the reader’s reference, such a ρ(r) is a reasonable approximation of the chargedistribution in an atom, describing the screening of the positive nuclear charge

8 See, e.g. Eq. (A.67).9 The last step of this calculation uses the table integral (A.34d) with n = 2.

Classical Electrodynamics: Problems with solutions

1-15

q = Ze by Z negatively charged electrons. Such an approximation, as well as a moreaccurate Thomas–Fermi model of the screening10, essentially ignore the quantummechanics of the electrons, and as the result work reasonably well only for veryheavy atoms (Z ≫ 1).

Problem 1.13. A thin flat sheet, cut in the form of a rectangle of size a × b, iselectrically charged with a constant areal density σ. Without an explicit calculationof the spatial distribution ϕ(r) of the electrostatic potential induced by this charge,find the ratio of its values at the center and at the corners of the rectangle.

Hint: Consider partitioning the rectangle into several similar parts and using thelinear superposition principle.

Solution: Selecting the Cartesian coordinates as shown in the figure above, we mayuse Eq. (1.38) of the lecture notes to calculate the potential at the origin (i.e. in one ofthe rectangle’s corners) as

∫ ∫ϕ σπε

σπε

=+

≡dx dyx y

aI

41

( ) 4,

a b

00 0 0 2 2 1/2

0

where I is the following dimensionless integral:

∫ ∫ξ ζξ ζ

≡+

I d d1

( ),

b a

0

1

0

/

2 2 1/2

with ξ ≡ x/a, ζ ≡ y/a.Now let us calculate what potential would be induced, at the same point, only by

the adjacent quarter of the rectangle, of size (a/2) × (b/2), marked in the figure abovewith darker shading:

∫ ∫ϕ σπε

′ =+

dx dyx y4

1

( ).

a b

00 0

/2

0

/2

2 2 1/2

10 See, e.g. Part QM, problems 8.22 and 8.23.

Classical Electrodynamics: Problems with solutions

1-16

Introducing dimensionless variables in a similar but modified way: ξ ≡ x/(a/2),ζ ≡ y/(a/2), we may reduce this result to the same dimensionless integral I:

ϕ σπε

′ = aI

( /2)4

,00

so that even without the calculation of the integral11 we see that ϕ0′ = ϕ0/2.Next, we may note that due to the linear superposition principle, the potential ϕA

at the center of the rectangle may be represented as a sum of four potentials inducedby such quarters in their corners. Due to the symmetry, all these partial potentialsare equal to ϕ0′, so that finally

ϕ ϕ ϕ= ′ =4 2 .A 0 0

This solution illustrates the power of scaling arguments, broadly used, forsymmetric geometries, in all fields of physics.

Problem 1.14. Calculate the electrostatic energy per unit area of a system of twothin, parallel planes with equal and opposite charges of a constant areal density σ,separated by distance d.

Solution: From the similar problem solved in the lecture notes (see figure 1.4 and itsdiscussion), it is clear that the electric field everywhere should be normal to theplanes and constant within each of three ranges:

=

< −

− < < +

+ <

−

+

⎧⎨⎪⎪

⎩⎪⎪

E z d

E d z d

E d z

E n

, at /2,

, at /2 /2,

, at /2 ,

z 0

where the axis z is selected as shown in the figure below, and E−, E0, and E+ are somescalar constants. These constants may be calculated by applying the Gauss law to threepillboxes with all three possible combinations of lid positions, giving three equations:

σ ε σ ε− = + − = − − =+ − + −E E E E E E/ , / , 0,0 0 0 0

whose solution yields

σε

= = = − *+ −E E E0, . ( )00

11 Just for reference, this integral may be worked out analytically: = + + + −Iba

a b ab

ba

ln( )

sinh2 2 1/2

1 .

Classical Electrodynamics: Problems with solutions

1-17

(An even simpler way to obtain Eq. (*) is to employ the linear superpositionprinciple by summing up the fields of the same magnitude (1.23), ∣E∣ = ∣σ∣/2ε0,induced by each of the planes. Due the difference of plane charge signs, the fields addup between the plates, but cancel each other outside the system.)

Thus, the field exits only between the planes, producing the electrostatic potential

∫ϕ ϕ ϕ σε

= − ′ ′ = + ×<>= =

⎧⎨⎩E z dzz z dd z z d

( ), for /2,

( /2)sgn( ), for /2,z

z

z00

00

so that the potential difference (called voltage) between the planes is

ϕ ϕ σε

σ≡ + − − = ≡⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟V

d dd

C2 2,

0 0

where C0 = ε0/d is the specific capacitance of this simple system (which is equivalentto a plane capacitor—for more, see chapter 2).

Now using Eq. (*) in Eq. (1.65) for the potential energy of the field, we obtain

∫ε ε σε

= = = =−

+UA

E dzE

d dC V

2 2 2 2.

d

d0

2

22 0 0

2 2

0

02

(We will repeatedly run into this expression in chapter 2.) Note that at fixed σ (of anysign) the energy grows with d. This is natural, because the oppositely charged planesattract each other, so that following the force (reducing d) corresponds to the waytoward the potential energy’s minimum.

Problem 1.15. The system analyzed in the previous problem (two thin, parallel,oppositely charged planes) is now placed into an external, uniform, normal electricfield Eext = σ/ε0—see the figure below. Find the force (per unit area) acting on eachplane, by two methods:

(i) directly from the electric field distribution, and(ii) from the potential energy of the system.

Solutions:

(i) In order to calculate the force F acting on the top plane, we have to neglect thefield of this plane itself12 and add the external field directed up and equal to σ/ε0, and

12Different elementary charges of the same plane do Coulomb-interact, but the elementary forces betweenthem are directed along the plane, and cancel at summation, due to the 2nd Newton’s law.

Classical Electrodynamics: Problems with solutions

1-18

the field of the bottom plane, directed down and equal to σ/2ε0—see figure 1.4 and(1.24) of the lecture notes. The net field equals σ/2ε0 and is directed up, so that theforce per unit area is

σε

σε

= = *FA

qA 2 2

, ( )0

2

0

and is directed up. A similar calculation for the bottom plane yields force of thesame magnitude, but directed down.

(ii) Using the linear superposition principle, we may calculate the total field in eachspatial region by adding the external field to that of the planes, which was calculatedin the previous problem:

σε

σε

σε

= + ×< −

− − < < ++ <

= ×< −

− < < ++ <

⎧⎨⎪⎩⎪

⎫⎬⎪⎭⎪

⎧⎨⎪⎩⎪

z dd z dd z

z dd z dd z

E n n

n

0, at /2,1, at /2 /2,

0, at /2 ,

1, at /2,0, at /2 /2,1, at /2 .

z z

z

0 0

0

Using these values, let us calculate the potential energy U from Eq. (1.72) of thelecture notes, artificially limiting the integration volume to a pillbox of an arbitraryarea A, with some fixed thickness d0 > d. The second integral does not depend on d,and may be dropped (included into an arbitrary constant that may be always addedto the energy), and the first integral yields

ε σε

σε

= − ≡ −⎛⎝⎜

⎞⎠⎟

UA

d d d2

( ) const2

.0

0

2

0

2

0

Now the vertical force (per unit area) exerted at the top plane may be calculated asF/A = −∂(U/A)/∂d, giving the same result (*) as the first method. The same is true forthe lower plane whose vertical coordinate is (const −d), so that the ratio U/A shouldbe differentiated over (−d) rather than d.

Problem 1.16. Explore the relation between the Laplace equation (1.42) and thecondition of minimum of the electrostatic field energy (1.65).

Solution: Let us consider a small variation δϕ of the electrostatic potential inside acharge-free volume V limited by a closed surface S, such that the variation vanishesat the surface, but is otherwise arbitrary. Let us calculate the correspondingvariation of the electric field energy (1.65):

∫ ∫∫ ∫

δ δ ε ϕ ε δ ϕ ϕ

ε ϕ δ ϕ ε ϕ δϕ

∇ ∇ ∇

∇ ∇ ∇ ∇

= = ⋅

= ⋅ = ⋅*

⎜ ⎟⎛⎝

⎞⎠U d r d r

d r d r

2( )

2( )

( ) ( )( )V V

V V

0 2 3 0 3

03

03

Classical Electrodynamics: Problems with solutions

1-19

The last step above has exploited the main rule of the variational calculus that theoperations of small variation and the usual differentiation are interchangeable13.Now let us work out the resulting 3D integration by parts—just as was done at thederivation of Eq. (1.65) of the lecture notes, i.e. by applying the divergence theorem(A.79) to the vector function f = δϕ∇ϕ:

∮ ∫ ∫ ∫δϕ ϕ δϕ ϕ δϕ ϕ δϕ ϕ∇ ∇ ∇ ∇ ∇= ⋅ ≡ ⋅ + ∇d r d r d r d r( ) ( ) ( ) .S

nV V V

2 3 3 2 3

The surface integral on the left-had side of this relation has to equal zero, due to ourcondition δϕ∣S = 0, and the first volume integral on the right-hand side equals that inthe last form of Eq. (*). Hence, that equality may be rewritten as

∫δ ε δϕ ϕ= − ∇U d r.V

02 3

Since the variation δϕ is arbitrary, this expression shows that the only way for thepotential energy of the field to have the lowest possible value (just as it does in anystable equilibrium in classical mechanics) and hence for the variation δU to equalzero, is to have the Laplace operator of the field equal zero, i.e. to satisfy the Laplaceequation (1.42) in all charge-free spatial regions.

Problem 1.17. Prove the following reciprocity theorem of electrostatics14: if twospatially confined charge distributions ρ1(r) and ρ2(r) create respective distributionsϕ1(r) and ϕ2(r) of the electrostatic potential, then

∫ ∫ρ ϕ ρ ϕ=d r d rr r r r( ) ( ) ( ) ( ) .1 23

2 13

Hint: Consider integral ∫ ⋅ d rE E1 23 .

Solution: Applying Eq. (1.33) of the lecture notes to E1(r), let us transform theintegral mentioned in the hint as

∫ ∫ ϕ∇⋅ = − ⋅d r d rE E E .1 23

1 23

Now we may use the rule of spatial differentiation of a vector-by-scalar functionproduct15 to continue as follows:

∫ ∫ ∫ϕ ϕ ϕ∇ ∇ ∇− ⋅ = ⋅ − ( )d r d r d rE E E( ) .1 23

1 23

1 23

Next, we may use the inhomogeneous Maxwell equation (1.27) in the first integral,and the well-known divergence theorem16 to transform the second integral to that of

13 This rule has a simple geometric meaning—see, e.g. Part CM section 2.1, in particular figure 2.2.14 This is only the simplest one of the whole family of reciprocity theorems in electromagnetism. (Sometimes itis called the ‘Green’s reciprocity theorem’, but historically it is fairer to reserve the last name for thegeneralization to surface charges, discussed in section 2.10 of the lecture notes.)15 See, e.g. Eq. (A.74a), with f = ϕ1 and g = E2.16 See, e.g. Eq. (A.79).

Classical Electrodynamics: Problems with solutions

1-20

(ϕ1E2)n over some distant, closed surface S that limits the volume of our spatialintegration. As a result, our expression becomes

∫ ∮εϕ ρ ϕ− ( )d r d rE

1.

S n0

1 23

1 22

Since the charge (and hence the field) distributions are space-confined, we mayalways select the surface S so distant that the surface integral is negligible17, and ourchain of transformations may be summarized as

∫ ∫εϕ ρ⋅ =d r d rE E

1.1 2

3

01 2

3

Now repeating the same calculation with swapped indices, we arrive at thereciprocity theorem.

Note that if some parts of these two charge distributions reside on some surface(s)S, and may be well described by surface charge densities σ1(r) and σ2(r) (as is veryinstrumental, for example, in systems with good conductors, to be discussed inchapter 2 of the lecture notes), the reciprocity theorem may be rewritten as

∫ ∫ ∫ ∫ρ ϕ σ ϕ ρ ϕ σ ϕ+ = +d r d r d r d rr r r r r r r r( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ,V S V S

1 23

1 22

2 13

2 12

where ρ1(r) and ρ2(r) are the remaining, ‘genuinely volume’ parts of the distributions.

Problem 1.18. Calculate the energy of electrostatic interaction of two spheres, ofradii R1 and R2, each with a spherically symmetric charge distribution, separated bydistance d > R1 + R2.

Solution: According to Eq. (1.55) of the lecture notes, applied sequentially to each of thespheres, the energy Uint of their interaction may be calculated in either of two ways18:

∫ ρ ϕ= *U d rr r( ) ( ) , ( )intsphere 1

1 23

∫ ρ ϕ= **U d rr r( ) ( ) . ( )intsphere 2

2 13

But as was discussed in section 1.2 of the lecture notes (see Eqs. (1.19) and (1.20)),the electric field, and hence the electrostatic potential ϕ, created by a sphericallysymmetric charge distribution ρ(r) outside of its boundary coincides with that of the

17 Indeed, we may take S in the shape of a spherical shell, with radius R much larger than the spatial scales ofdistributions ρ1(r) and ρ2(r). At such large distances, their fields are close to those of point charges, so thataccording to Eqs. (1.6) and (1.35), ϕ1 ∝ Q1/R and E2 ∝ Q2/R

2, so that the integral over surface S, of areaA ∝ R2, is decreasing with R as 1/R (or faster if any of the net charges Q1,2 vanishes).18 Their equivalence is strictly confirmed by the reciprocity theorem (see the previous problem) even if thecharge distributions ρ1,2(r) partly or fully overlap—although in our current case, they do not.

Classical Electrodynamics: Problems with solutions

1-21

full charge Q of the sphere concentrated in its center. Applying this fact to, forexample, the first sphere, we may rewrite Eq. (**) as

∫∫

ρ ϕ ϕπε

ρ

= =−

≡

U d rQ

Q d r

r r rr r

r

( ) ( ) , where ( )4

,

with ( ) ,

intsphere 2

2 1(point) 3

1(point) 1

0 1

1sphere 1

13

where r1 is the position of the center of sphere 1. This means that the interactionenergy, however calculated, cannot depend on whether the charge of sphere 1 isdistributed as given, or compressed to point r1, i.e. should be invariant with respectto the replacement

ρ δ→ −Qr r r( ) ( ).1 1 1

Making this replacement in Eq. (*), we obtain

ϕ=U Q r( ).int 1 2 1

Now applying the same argument to the potential created by the sphericallysymmetric distribution ρ2(r),

∫ϕ ϕπε

ρ= =−

≡Q

Q d rr rr r

r( ) ( )4

, with ( ) ,2 2(point) 2

0 22

sphere 22

3

where r2 is the position of the center of sphere 2, we finally obtain:

πε= ≡ −U

Q Q

dd r r

4, with .int

1 2

01 2

In plain English, the spheres interact as if their electric charges were concentrated attheir centers. Remember, however, that this result is only applicable at d > R1 + R2, sothat the charge distributions ρ1(r) and ρ2(r) do not overlap—even partly.

Problem 1.19. Calculate the electrostatic energy U of a (generally, thick) sphericalshell, with a charge Q uniformly distributed through its volume—see the figurebelow. Analyze and interpret the dependence of U on the inner cavity’s radius R1, atfixed Q and R2.

Classical Electrodynamics: Problems with solutions

1-22

Solution: Calculating the only (radial) component E of the electric field E = nrE (say,using the Gauss law), we readily obtain

πε= ×

⩽ <− − < <

<

⎧⎨⎪⎩⎪

EQ

r

r R

r R R R R r RR r

4

0, for 0 ,

( )/( ), for ,1, for ,0

2

1

313

23

13

1 2

2

so that Eq. (1.65) for the electrostatic energy yields

∫ ∫ ∫ ∫

∫ ∫

ε ε π ε ππε

πεξ α

αξ

ξξ

ξ πεα

α α α αα

= = = −−

+

= −−

+ =

≡ − + −−

+

α

∞ ∞

∞

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢⎢

⎛⎝⎜

⎞⎠⎟

⎤⎦⎥⎥

⎡⎣⎢⎢

⎛⎝⎜

⎞⎠⎟

⎤⎦⎥⎥

U E dr E r drQ r R

R Rdrr

drr

QR

d d QR

f

f

2 24

24

4

8 1 8( ),

with ( )1/5 (9/5)

(1 )1,

R

R

R

0 2 3 0

0

2 2 0

0

2 313

23

13

2

2 2

2

0 2

1 3 3

3

2

2 1

2

0 2

3 5 6

3 2

1

2

2

where α ≡ R1/R2 ⩽ 1.A plot of the function f(α), i.e. of the normalized electrostatic energy as a function

of R1 at fixed Q and R2, is shown in the figure below. The function reachesits maximum f(0) = 6/5 (so that U is given by Eq. (1.66) of the lecture notes, withR = R2) at α = 0, i.e. for a solid sphere, and tends to f(1) = 1, giving

πε=U

QR8

min

2

0 2

at α → 1, i.e. for the ultimately thin spherical shell. This is very natural, because theelementary charges of the sphere repulse each other and try to go apart as far aspossible, in particular increasing its inner radius.

Classical Electrodynamics: Problems with solutions

1-23