WiSe 2012 23.10.2012

Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl für Theoretische Physik IDepartment für PhysikFriedrich-Alexander-UniversitätErlangen-Nürnberg

Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

Tutorial 2

Tutorial 2.1 Differential operators in cylindrical coordinates

In cartesian coordinates the nabla operator is defined as follows:

~∇ = ~ex∂

∂x+ ~ey

∂

∂y+ ~ez

∂

∂z

Cylindrical coordinates (ρ, φ, z) are defined as follows:

x = ρ cosφ, y = ρ sinφ, z = z

a) Transform the nabla operator into cylindrical coordinates. Proceed with the following steps:

• Use the chain rule to write the partial derivatives ∂/∂x, ∂/∂y and ∂/∂z in terms of ∂/∂ρ,∂/∂φ, ∂/∂z

• Write ~ex, ~ey and ~ez in terms of ~eρ, ~eφ, ~ez

Result: ~∇ = ~eρ∂∂ρ + ~eφ

1ρ∂∂φ + ~ez

∂∂z

b) Calculate for a vector field ~A the expression ~∇ · ~AResult: ~∇ · ~A = 1

ρ∂∂ρ(ρAρ) + 1

ρ∂∂φ(Aφ) + ∂

∂z (Az)

Tutorial 2.2 Newton’s theorem

a) Prove the electrostatic analog of Newton’s theorem:

For a spherically symmetric charge (or mass, in the case of gravity) distribution ρ(r),the radial component of the electric field, Er = ~E · ~r/r, is given by

Er =Q(r)

r2with Q(r) = 4π

∫ r

0ρ(R)R2dR ,

i. e. the same as if the charge in the sphere of radius R is located at the center of thesphere.

Calculate also the associated electrostatic potential.

Note that the Poisson equation in spherical coordinates reads

−4πρ = ∇2ϕ =1

r2∂

∂r

(r2∂ϕ

∂r

)+

1

r2 sinϑ

∂

∂ϑ

(sinϑ

∂ϕ

∂ϑ

)+

1

r2 sin2 ϑ

∂2ϕ

∂φ2.

b) As an application of Newton’s theorem, consider a charge-free spherical cavity concentric withthe center of a spherically symmetric charge distribution. What is the electric force on a testcharge inside this hole?

Tutorial 2.3 Stable rest position in an electrostatic field

Consider a time-independent electric field ~E(~x) with the property that for a point particle with chargeq > 0 the position ~x = 0 is a stable rest position with a linear reset force. That is, for the force ~F = q ~E,the following properties hold:

• ~F is zero at ~x = 0.

• In the vicinity of ~x = 0 ~F forces the particle back to ~x = 0. ~F (~x) can be regarded as a linearfunction of ~x for small displacements.

The magnetic field is zero.

Show that the charge density ρ(~x), that creates the field ~E, cannot have a zero at ~x = 0. Whatsign does ρ(0) have?

Hint: The relation ~∇ × ~E = 0 follows from the induction law. Use that fact to show that Aij :=− ∂∂xj

Ei(~x = 0) is a symmetric matrix. Now choose the coordinate axes parallel to the direction of theprincipal axes of Aij and calculate in this coordinate system the charge density ρ(~x).

Tutorial 2.4 Penning trap

Consider the motion of a particle that has a charge q and mass m in a constant uniform magnetic field~B = Bez and an electric quadrupole potential (U0 > 0)

ϕ(~x) = − U0

2r20(x2 + y2 − 2z2) , ~x = (x, y, z) .

a) Show that the non-relativistic equation of motion for the particle in the x–y plane for the caseU0 = 0 leads to oscillatory motion. Determine the cyclotron frequency ωc characterizing theoscillation. It is favorable to introduce a complex variable ξ := x+ iy.

b) Determine the electric field ~E(~x) = −~∇ϕ(~x) and verify that ~E is solenoidal, i.e., ~∇ · ~E(~x) = 0.c) Show that the magnetic field does not couple to the motion along the z-direction, and determine

the characteristic frequency ωz for the corresponding harmonic oscillations in the quadrupolefield.

d) Solve the complete equations of motion in the x–y plane and show that the general solutionis a superposition of two oscillatory motions with a perturbed cyclotron frequency ω′c and themagnetron frequency ωM . Provide conditions such that the orbits are stable. Discuss the caseωz ωc in particular.

Tutorial 2.5 Cylindrical capacitor

a) Using Gauß’ law, calculate the capacitance of two concentric conducting cylinders of length L,when L is large compared to their radii a, b (a < b). Apply the result to calculate the innerdiameter of the outer conductor in an air-filled coaxial cable whose center is a cylindrical wireof diameter 1 mm and whose capacitance is 0.5× 10−6µF cm−1.

b) For the cylindrical capacitor from part a) calculate the total electrostatic energy and expressit alternatively in terms of the equal and opposite charges Q and −Q placed on the capacitorplates and the potential difference between the plates.Sketch the energy density of the capacitor’s electrostatic field as a function of the appropriatelinear coordinate.

c) Two long, parallel, cylindrical conductors of radii a1 and a2 are separated by a distance d, whichis large compared with either radius. Show that the capacitance per unit of length is givenapproximately by

C

L'[4 ln

(d

a

)]−1,

where a is the geometrical mean of the two radii.(i) What gauge wire (state the radius in millimetres) would be necessary to make a two-wire

transmission line with a capacitance of 0.1 pF cm−1, if the separation of the wires is 0.5cm, 1.5 cm, and 5.0 cm?

d) Calculate the attractive force per unit length between the two conductors in a parallel cylindricalcapacitor for:(i) fixed charges on each conductor,(ii) a fixed potential difference between the conductors.

Due date: Thursday, 25.10.12, at 13:00

WiSe 2012 23.10.2012

Prof. Dr. A.-S. SmithDipl.-Phys. Ellen FischermeierDipl.-Phys. Matthias Sabaam Lehrstuhl für Theoretische Physik IDepartment für PhysikFriedrich-Alexander-UniversitätErlangen-Nürnberg

Theoretische Physik 2: Elektrodynamik(Prof. A.-S. Smith)

Solutions to Tutorial 2

Solution of Tutorial 2.1 Differential operators in cylindrical coordinates

a)x = ρ cosφ, y = ρ sinφ, z = z

Use the chain rule:

∂f

∂ρ=∂x

∂ρ

∂f

∂x+∂y

∂ρ

∂f

∂y+∂z

∂ρ

∂f

∂z= cosφ

∂f

∂x+ sinφ

∂f

∂y

Analogously:

∂f

∂φ= −ρ sinφ

∂f

∂x+ ρ cosφ

∂f

∂y

∂f

∂z=

∂f

∂z

So with

∂

∂ρ= ∂ρ = cosφ ∂x + sinφ ∂y

1

ρ

∂

∂φ=

1

ρ∂φ = − sinφ ∂x + cosφ ∂y

we get

∂x = cosφ ∂ρ −1

ρsinφ ∂φ

∂y = sinφ ∂ρ +1

ρcosφ ∂φ

∂z = ∂z

Now the unit vectors:

~eρ =

∣∣∣∣∂~x∂ρ∣∣∣∣−1 ∂~x∂ρ , . . .

1

With~x = x~ex + y~ey + z~ez = ρ cosφ ~ex + ρ sinφ ~ey + z~ez

follows

~eρ = cosφ ~ex + sinφ ~ey

~eφ = − sinφ ~ex + cosφ ~ey

~ez = ~ez

=⇒ ~ex = cosφ ~eρ − sinφ ~eφ, ~ey = sinφ ~eρ + cosφ ~eφ

All together:~∇ = ~ex∂x + ~ey∂y + ~ez∂z = . . . = ~eρ∂ρ + ~eφ

1ρ∂φ + ~ez∂z

b) It holds:

• ~ex ⊥ ~ey ⊥ ~ez and ~eρ ⊥ ~eφ ⊥ ~ez due to ~ei · ~ej = δij

• ∂φ~eρ =

− sinφcosφ

0

= ~eφ ∂φ~eφ =

− cosφ− sinφ

0

= −~eρ ∂φ~ez = 0

• ∂ρ~eφ = ∂ρ~eρ = ∂ρ~ez = ∂z~eφ = ∂z~eρ = ∂z~ez = 0

and therefore ~∇ · ~A in cylindrical coordinates is

~∇ · ~A = (~eρ∂ρ + ~eφ1ρ∂φ + ~ez∂z) · (Aρ~eρ +Aφ~eφ +Az~ez) =

= ∂ρAρ + 1ρAρ + 1

ρ∂φAφ + ∂zAz = 1ρ∂∂ρ(ρAρ) + 1

ρ∂∂φ(Aφ) + ∂

∂z (Az)

Solution of Tutorial 2.2 Newton’s theorem

a) Poisson’s equation for the charge density simplifies since the electrostatic potential dependssolely on the radial distance,

4πρ(r) = −∇2ϕ(r) = − 1

r2d

dr

(r2

dϕ

dr

)=

1

r2d

dr(r2Er).

where Er = −~r/r · ~∇φ denotes the radial component of the electric field. Integrating once

4π

∫ r

0ρ(R)R2dR = r2Er(r)

orEr(r) =

Q(r)

r2with Q(r) = 4π

∫ r

0ρ(R)R2dR.

This result allows for a neat interpretation. The electric field corresponding to a sphericallysymmetric charge distribution at radial distance r is identical to one where the charge containedin the sphere of that radius is concentrated in the origin.The potential is obtained by integrating once more,

ϕ(r) =

∫ ∞r

Q(R)

R2dR.

[The point of the reference potential is at infinity, ϕ(∞) = 0.]b) The force on the test charge is proportional to the electric field. Newton’s theorem implies a

vanishing force at the inside, since the charge distribution equals zero there.

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Solution of Tutorial 2.3 Stable rest position in an electrostatic field

~E(~x) is a linear function of ~x in the vicinity of ~x = 0.

⇒ Ej(~x) = E0j −Ajlxl +O(r2) (1)

Because of ~E(0) = 0 ⇒ E0j = 0.

⇒ (curl ~E)l = εlmn∂mEn = −εlmn∂mAnjxj +O(r) =

= −εlmnAnjδmj +O(r) = −εlmnAnm +O(r) = 0

⇒ εlmnAnm = 0 ⇒ Anm = Amn (symmetric matrix)

For example for l = 1 : ε1mnAnm = ε123A32 + ε132A23 = A32 − A23 = 0 ⇒ A32 = A23 (Analogouslyfor the other components).Because Ajl is symmetric and real (because ~E is real) a principal axis transformation can be doneso that Ajl is diagonal in this new coordinate system, that means Ajl = 0 for j 6= l. The diagonalelements must be > 0 because otherwise the field in (1) would not show in the direction of the originfor a positive test charge and the resulting force would not be a linear reset force.

⇒ ρ(~x) = ε0 div ~E = ε0∂lEl = −ε0Alj ∂lxj︸︷︷︸=δlj

+O(r) = −ε0All +O(r)

⇒ ρ(0) = −ε0All < 0

At the origin there must be a nonzero, negative charge.⇒ It is not possible to place charges out of a charge-free cavity, so that there is a potential within thecavity that produces a stable rest position with a linear reset force.

Solution of Tutorial 2.4 Penning trap

For a particle in an electromagnetic field, Newton’s equation balances inertial forces and the Lorentzforce with contributions from the electric and the magnetic field,

m~x︸︷︷︸inertial

= −q~∇ϕ︸ ︷︷ ︸electric

+q

c~x× ~B︸ ︷︷ ︸

magnetic

.

In our case, ~B = B~ez and ϕ(~x) = −(U0/2R20)(x2 + y2 − 2z2).

a) Without quadrupole field, U0 = 0. The equations of motion in components are

mx =q

cyB , my = −q

cxB .

Complexify, ξ = x+ iy. Then with the cyclotron frequency ωc = qB/mc ,

ξ = −iωcξ , with the solution ξ(t) = ξ0 +Aeiαe−iωct

with a real amplitude A and phase α. Decomposing into cartesian components again, yields

x(t) = x0 +A cos(ωct− α) , y(t) = y0 −A sin(ωct− α) ,

i.e., circular orbits spinning counter-clockwise. The centers (x0, y0) are arbitrary.

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b) Taking the gradient of the potential yields the electric field,

~E(~x) = −~∇ϕ(~x) =U0

r20

xy−2z

,

the divergence of which vanishes,

−∇2ϕ = ~∇ · ~E =U0

r20(1 + 1− 2) = 0 .

Thus, the Laplace equation tells us that it is solenoidal, i.e., there are no charges.c) The motion in the z-direction is governed by

z =qEzm

= −2qU0

mr20z ≡ −ω2

zz ;

the magnetic field does not couple to the parallel motion. The solution is again a harmonicoscillation around the center of the trap,

z(t) = az cos(ωzt− ψ) .

d) Now, we consider both the magnetic and the quadrupol field,

mx =q

cyB +

qU0

r20x ,

my = −qcxB +

qU0

r20y .

In terms of the complex variable,

ξ = −iωcξ +ω2z

2ξ .

This second-order, linear differential equation with constant coefficients is solved using theAnsatz ξ(t) ∝ e−iωt. The characteristic frequency is then determined by

ω2 − ωcω +ω2z

2= 0 .

The two solutions are a reduced cyclotron frequency,

ω′c =ωc2

+

√ω2c

4− ω2

z

2≈ ωc −

ω2z

2ωc+O(ω4

z) ,

and the so-called magnetron frequency,

ωM =ωc2−√ω2c

4− ω2

z

2≈ ω2

z

2ωc+O(ω4

z) .

Stability requires both frequencies to be real, i.e., 2ω2z < ω2

c , or in terms of the magnetic field

B2 >2m2c2ω2

z

q2=

4mc2

qr20U0 .

The general solution then reads

ξ(t) = Aeiαe−iω′ct +Beiβe−iωM t ,

or in cartesian components,

x(t) = A cos(ω′ct− α) +B cos(ωM t− β) ,

y(t) = −A sin(ω′ct− α)−B sin(ωM t− β) .

4

Ωc =20ΩM

Fig. 1: Orbits in a Penning trap for com-mensurate frequencies

In general, Penning trap parameters are chosen such thatthe radius of magnetron motion exceeds that of the cy-clotron motion by a factor of the order 100 and ωM ωz ωc. The orbits consist then of a rapid cyclotron motion ofsmall radius whereas the centers slowly follow another cir-cular trajectory. The Penning trap was co-invented by theGerman-born American physicist Hans Georg Dehmelt, whoreceived the Nobel prize in 1989 together with WolfgangPaul (Paul trap).Linguistic note: pen means ’Laufstall’, ’Einzäunung’.

Solution of Tutorial 2.5 Cylindrical capacitor

a) To find the capacitance of the cylindrical capacitor that is formed by the two cylinders, we givethem equal and opposite charge and find the ratio of this charge to the potential differencebetween them. So let the inner and the outer cylinders have a uniformly-spread total chargeof +Q and −Q respectively (Fig. 1). Since the length L of the cylinders is much larger thantheir radii, the electric field between the cylinders can be approximated by that for the caseof infinitely long cylinders, and we can ignore the end effects. Now imagine a cylinder (heredenoted by S(ρ)) between the two given cylinders, i.e., coaxial with the two given cylinders, andwith a radius ρ satisfying a < ρ < b. Applying Gauß’ law for this cylinder:∮

S(ρ)

~E · d~S = 4πQ∮S(ρ)

~E · d~S =

∫sheet

E(ρ)~eρ · ~eρ dS = E(ρ)

∫sheet

dS = E(ρ) 2πρL

E(ρ) 2πρL = 4πQ

E =2λ

ρ; ~E =

2λ

ρ~eρ .

where "sheet" denotes the curved part of S(ρ), and λ is the line charge density. The potentialdifference between the two cylinders is then

Fig. 2: Cylindrical capacitor.

Φ(a)− Φ(b) = −∫ a

b

~E · d~l = 2λ

∫ b

a

~eρρ· ~eρ dr = 2λ ln

(b

a

),

Electric capacitance of the cylindrical capacitor is therefore

C =Q

U=

L

2 ln(b/a).

From this, it follows that the radius of the outer cylinder is

b = a exp

(L

2C

).

Therefore, for the given values,2a = 1 mm

C

L= 0.5× 10−6µF cm−1.

In CGS system of units

5

1 F = 9× 1011 cm.

ThereforeC

L= (0.5× 10−6) (9× 105) = 4.5× 10−1, which gives 2b = 2× 0.5 exp

(109

)= 3.04mm

b) The potential energy of a system of n charged conductors is

W =1

2

n∑i=1

QiVi .

In our case, the number of conductors is n = 2. Hence,

W =1

2QV1 +

1

2(−Q)V2 =

1

2Q(V1 − V2) =

1

2QU ,

where U is the potential difference. By definition

C =Q

U.

Therefore, the electrostatic energy is

W (Q) =1

2

Q2

C

W (U) =1

2CU2 .

We have already shown that the capacitance of the cylindrical capacitor equals

C =L

2 ln(b/a),

so its energy in terms of the charge Q is simply

W (Q) =ln(b/a)

LQ2 ,

and in terms of the potential difference U is

W (U) =L

4 ln(b/a)U2 .

Fig. 3: Energy density in the cylinder capacitor as afunction of the distance from the axis of thecapacitor. We set b→∞.

The relation for the energy density reads

ω =1

8π

∣∣∣ ~E∣∣∣2 .The electric field in the cylindrical capacitorequals

~E =2λ

ρ~eρ , a < ρ < b ,

where ~eρ is the unit vector in the directionof the cylindrical coordinate ρ. The energydensity is then

ω =1

2π

λ2

ρ2, a < ρ < b .

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c) As the separation of the cylinders d is much larger than their radii a1 and a2 (Fig. 3), if theyare charged, then we can assume that the electric field of each distribution will not affect theother one much, and so the charge distributions can be assumed to be uniform. (Note: Whenthe separation is comparable to their radii, this assumption is not valid.) Hence, let this uniformcharge per unit length be +λ on the cylinder with radius a1 and −λ on the one with radius a2.From part a), the potential of a cylinder with a uniform line charge distribution λ is

Φ(ρ) = 2λ ln

(A

ρ

),

(with the axis of the cylinder aligned with the z-axis of coordinate system and A an arbitraryradial distance for which Φ(A) = 0).

Φ = Φ1 + Φ2 = 2λ ln

(A√

x2 + (y − d/2)2

)− 2λ ln

(A√

x2 + (y + d/2)2

)

= λ ln

(x2 + (y + d/2)2

x2 + (y − d/2)2

).

Fig. 4: Separated cylindrical conductors.

The surfaces of the cylinders are equipotentialbecause the cylinders are conductors. For thecylinder 1 we have

x = a1 cosϕ1

y =d

2+ a1 sinϕ1 .

Therefore the potential Φ(1) (of cylinder 1) is

Φ(1) = λ ln

(a21 cos2 ϕ1 + (a1 sinϕ1 + d)2

a21 cosϕ21 + a21 sinϕ2

1

)= λ ln

(a21 + 2da1 sinϕ1 + d2

a21

)= λ ln

(1 +

2a1d

sinϕ1 +(a1d

)2)+ 2λ ln

(d

a1

).

Because d a1 we can approximate the first logarithmic term by

ln

(1 +

2a1d

sinϕ1 +(a1d

)2)' ln 1 = 0 .

With this approximation the potential Φ(1) is

Φ(1) ' 2λ ln

(d

a1

).

Similarly, we can determine the potential over the surface of cylinder 2

x = a2 cosϕ2

y =− d

2+ a2 sinϕ2

Φ(2) = λ ln

(a22

a22 − 2da2 sinϕ2 + d2

)' −2λ ln

(d

a2

).

7

Hence the potential difference is

U = Φ(1)− Φ(2) = 2λ ln

(d2

a1a2

)= 4λ ln

(d

a

),

where a is the geometric mean of the radii

a =√a1a2 .

Therefore the capacitance per unit length is

C

L=λ

U=

[4 ln

(d

a

)]−1.

(i) We are givenC

L= 0.1 pF cm−1 = 9× 10−2

and we can easily express a in terms of the capacitance per unit length and the wireseparation

a = d exp

(− L

4C

).

Therefore, for separations d of 0.5 cm, 1.5 cm, and 5.0 cm, we get a wire radius of 0.31 mm,0.93 mm, and 3.11 mm, respectively.Note that for the given capacitance, the condition d a1, a2 is satisfied, and hence we areallowed the use of the approximate formula that we have derived.

d) (i) From part b), the total electrostatic energy is W (Q) = 12Q2

C . Therefore, the electrostaticenergy per unit of length (for two parallel, very long conductors of radii a1 and a2 whichare separated by a distance d (d a1, a2)) is

WQλ(d) =λ2

2(C/L)= 2λ2 ln

(d

a

),

where λ is the charge per unit of length. The force per unit of length between conductorsreads

F = −∂WQλ(d)

∂d= −2λ2

d.

(ii) The electrostatic energy per unit of length expressed in terms of the potential differencebetween the two cylinders is

WUλ(d) =1

8

U2

ln (d/a).

When we fix the potential difference between the cylinders and change their separation, thesource of voltage (battery) does some work, namely in increasing or decreasing the chargeon the cylinders of the capacitor. In this case, therefore, the change in the electrostaticenergy for the capacitor is not only due to the mechanical work involved (i.e., moving ofthe capacitor’s plates) as it was case in (i), but also due to the source work which changesthe charge on the plates. That is,

∆WU (d) = ∆Wmech + ∆Wsource ,

where ∆Wmech is the work done due to the moving of the plates, and ∆Wsource is the workdue to the charging or the discharging of the plates. The source work is

∆Wsource = ∆Q U

∆Q = U ∆C

∆Wsource = ∆C U2 .

8

So, for the force per unit length, we can write

F = −∂Wmechλ∂d

=− ∂WUλ(d)

∂d+∂Wsourceλ

∂d

=1

8

U2

d ln2(d/a)− 1

4

U2

d ln2(d/a)= −1

8

U2

d ln2(d/a),

where for the capacitance per unit length of the cylindrical conductors we have used theexpression derived earlier

C

L=λ

U=

[4 ln

(d

a

)]−1.

On comparing this expression with the one obtained in part (i)

−1

8

U2

d ln2(d/a)= − 1

8d ln2(d/a)× (4λ ln(d/a))2 = −2λ2

d,

we see that the expressions are the same.

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