solution. Our method has the advantage, among others, that it answers this question in a satisfactory and unambiguous way. It shows explicitly that al...

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arXiv:physics/0105082v2 [physics.ed-ph] 4 Nov 2003

C.J. Efthimiou1 and R.A. Llewellyn2 Department of Physics University of Central Florida Orlando, FL 32816

Abstract We propose a unified approach to addition of some physical quantities (among which resistors and capacitors are the most well-known) that are usually encountered in introductory physics such that the formulæ required to solve problems are always simply additive. This approach has the advantage of being consistent with the intuition of students. To demonstrate the effectiveness of our approach, we propose and solve several problems. We hope that this article can serve as a resource paper for problems on the subject.

1

Introduction

All introductory physics textbooks, with or without calculus, cover the addition of both resistances and capacitances in series and in parallel. The formulæ for adding resistances R = R1 + R2 + . . . , 1 1 1 + + ... , = R R1 R2

in series ,

(1)

in parallel ,

(2)

in series ,

(3)

in parallel ,

(4)

and capacitances 1 1 1 = + + ... , C C1 C2 C = C1 + C2 + . . . ,

are well-known and well-studied in all the books. In books with calculus there are often end-of-chapter problems in which students must find R and C using continuous versions of equations (1) and (4) [3, 7, 8, 9, 10]. However, we have found none which includes problems that make use of continuous versions of equations (2) and (3) [3, 4, 5, 7, 8, 9, 10]. Students who can understand and solve the first class of problems should be able to handle the second class of problems, as well. We feel that continuous problems that make use of all four equations should be shown to the students in order to give them a global picture of how calculus is applied to physical problems. Physics contains much more than 1 2

[email protected] [email protected]

1

mathematics. When integrating quantities in physics, the way we integrate them is motivated by the underlying physics. Students often forget the physical reasoning and they tend to add (integrate) quantities only in one way. In this paper, we introduce an approach to solving continuous versions of equations (2) and (3) that is as straightforward and logical for the students as solving continuous versions of equations (1) and (4). We then extend the logic to the addition of other quantities encountered in undergraduate introductory physics. This demonstrates that the method is not specific to resistors and capasitors but general and includes all quantities obeying similar addition laws. The organization of this article is as follows: section 2 discusses many physical quantities taken from introductory physics that obey similar addition laws. Among these quantities, resistance and capacitance are the most well-known to students. Inductance is also known but probably not mastered at the level it should be. Elasticity is somewhat known, but thermal resistance, diffusion resistance, and viscous resistance are almost unkown to students. As a result, for resistance and capacitance we only remind readers of the basic formulæ, while for the rest quantities we expand the discussion to some length so the students will become familiar with the physical background. In section 3, we present basic applications of the addition formulæ. This section is meant to demonstrate in a simple way how one chooses the correct addition formula (in series or in parallel), given a problem. In section 4 we solve several additional problems that make use of the main idea. In each problem, we have chosen one representative quantity. However, the reader must realise that the same problem can be stated for any of the quantities given in section 2, not just the chosen quantity. We hope that this article will motivate teachers to explain to students the subtle points between ‘straight integration’ as taught in calculus and ‘physical integration’ to find a physical quantity.

2

Basic Formulæ

2.1

Resistance

The basic formula to compute resistance is the formula of a uniform cylindrical resistor: R = ρ

L , A

where ρ is the resistivity of the material, L is the length of the conductor and A the crosssectional area. Written as conductance3 , this is G =

1 , R

is G = σ

A , L

where σ = 1/ρ is the conductivity of the material. 3

Often the term conductivity is used for G. However, the term conductance is in uniform lingustic aggreement with the rest of the terminology.

2

2.2

Capacitance

The basic formula to compute capacitance is that of a parallel-plate capacitor filled with a uniform dielectric material: A C = ε0 κ , d where κ is the dielectric constant of the dielectric, A the area of the plates, and d the distance between the plates. We are going to call the inverse of the capacitance D =

1 , C

the incapacitance of the capacitor. For a parallel-plate capacitor D =

2.3

1 d . ε0 κ A

Inductance

Calculation of inductance is usually based on the definition ΦB , I

L =

(5)

and therefore some discussion is necessary regarding our point of view. It is well-known that the inductance for a solenoid is given by L = µ0 N 2

A , ℓ

(6)

where N is the number of turns of the solenoid, A is the cross-section and ℓ the length of the solenoid. This expression can be used as the basic formula when we compute the inductance of another geometry which involves an inductor made from a single wire by twisting it in a particular geometry and creating a number N of turns. When we have some geometry in which no obvious way to count ‘turns’ exists, we must use a different formula. Clearly, the corresponding basic formula must come from a simple geometry giving rise to a uniform magnetic field. In figure 1 we see an inductor with such properties. The inductor consists of two finite-plane wires that carry opposite currents with uniform linear current density. Due to the complete analogy with a parallel-plate capacitor, we call it the parallel-plate inductor. An infinite sheet with uniform linear current density Js creates a uniform magnetic field in space with value B = µ0 Js /2. The direction of the field is found using the right-hand rule. When two infinite sheets are given, the total field is the sum of two fields. Given the directions of the currents of the parallel-plate inductor, the magnetic field adds to zero outside the plates and to B = µ0 Js between the plates. Of course, in the case of finite plates this result is only approximate, being a good approximation when d ≪ w. To compute the inductance of the

3

I

B

d

I

w

ℓ Figure 1: A parallel-plate inductor. parallel-plate inductor, we must compute the flux ΦB in the definition (5). We notice that the magnetic field is perpendicular to the area A = ℓd. In this case ΦB = B A = µ0 Js ℓd and I = Js w. Therefore

ℓd ΦB = µ0 . I w Notice that the only difference is that the ‘solenoid’ is now made from only one turn, i.e. N = 1. Therefore, in problems where the current is distributed continuously only ‘one turn’ should be counted. This issue can cause to curious students considerable frustration since they can raise the following objection: “In the calculation of the inductance, we concluded that N = 1 since one ‘turn’ was used (incorrectly) in the computation of flux, namely only the flux through one cross-section was used. However, there is an infinite number of cross-sections and an integration has to be done before inserting the flux into equation (5). Therefore, the result is not correct.” In this calculation, we believe that it is not easy to persuade the student that she is wrong and no flux has been lost. And unfortunately, similar situations arise in all continuous problems. Books fall short in providing any answers to this question—they just present the solution. Our method has the advantage, among others, that it answers this question in a satisfactory and unambiguous way. It shows explicitly that all flux has been counted. This is easily understood if the reader studies the problems of sections 4.12 and 4.13. Below, we present an independent calculation to confirm that the result found above for the inductance of the parallel-plate inductor is correct. The alternative calculation will be based on the energy stored in the inductor. The magnetic energy density for the parallel-plate inductor is L =

uB =

B2 1 = µ0 Js2 . 2µ0 2

The total energy stored in the magnetic field is then U = uB w d ℓ =

4

1 ℓd 2 µ0 I . 2 w

However, for any inductor, the energy stored in the magnetic field is also given by U = LI 2 /2. When this is compared to the previous result, we conclude that L = µ0

ℓd , w

(7)

i.e. the same result as found above. We are going to call the inverse of the inductance K =

1 , L

the dedutance of the inductor. Therefore, the deductance of a parallel-plate inductor is 1 w . µ0 ℓ d

K =

2.4

Thermal Resistance

Imagine a cylinder made from a uniform conducting material whose bases are kept at different temperatures. Then, due to the temperature difference ∆T between the bases, heat will flow from one base to the other. The rate according to which heat is flowing, i.e. ∆Q , ∆t

Ith =

is called the thermal current. It is known, see for example [2, 8], that Ith = σth A

∆T , L

(8)

where L is the length of the cylinder, A is its cross-section, and σth is a constant characteristic of the material, called the thermal conductivity. We define the inverse of the thermal conductivity ρth = 1/σth to be the thermal resistivity of the material. Equation (8) is sometimes referred to as Fourier’s law for the flow of energy.

A

L The thermal resistance of the cylinder is then defined by Rth =

5

∆T . Ith

(9)

Notice the analogy with the standard resistance: R = ∆V /I. Potential difference is the reason electric current flows. Here, temperature difference is the reason behind the thermal current. Comparing the two formulæ (8) and (9) we have written above, we arrive at Rth = ρth

L , A

(10)

an expression almost identical to that of the electric resistance for the cylinder. We define the thermal conductance as Gth =

1 . Rth

This implies that for the uniform cylinder Gth = σth

2.5

A . L

(11)

Diffusion Resistance and Viscous Resistance

The thermal conductivity discussed in the last section is only a particular example from a more general category of processes known as transport phenomena [2]. Transport phenomena are irreversible processes that occur in systems that are not in statistical equilibrium. In these systems, there is a net transfer of energy, matter, or momentum. Fourier’s law stated in the previous section was a law for the flow of energy. We will write similar laws for the flow of matter and the flow of momentum. Imagine a cylinder filled with gas such that the particle densities n1 and n2 of the gas at the bases are kept constant at different values. Then, due to the density difference ∆n between the bases, particles will flow from one base to the other. The rate according to which particles are flowing, i.e. ∆n Idif f = , ∆t is called the particle current. It is known, see for example [2], that Idif f = σdif f A

∆n , L

(12)

where L is the length of the cylinder, A is its cross-section, and σdif f is a constant characteristic of the material called the diffusion coefficient. Another name, in the spirit of what we have been discussing, would be diffusion conductivity. The inverse of σdif f , ρdif f = 1/σdif f , is named the diffusion resistivity of the material. Equation (12) is sometimes referred to as Fick’s law. The diffusive resistance of the cylinder is then defined by Rdif f =

∆n . Idif f

Its inverse gives the diffusive conductance: Gdif f = 6

1 Rdif f

.

(13)

Now imagine that the thermal agitation (speed) of the molecules at the two bases of the cylinder is different. Then, due to the speed difference ∆v between the bases, momentum will flow from one base to the other. The rate according to which speed is flowing, i.e. Ivis =

∆v , ∆t

is called the momentum current. It is known, see for example [2], that I = σvis A

∆v , L

(14)

where L is the length of the cylinder, A is its cross-section, and σvis is a constant characteristic of the material called the coefficient of viscosity. Another name, again in the spirit of what we have been discussing, would be viscous conductivity. The inverse of σvis , ρvis = 1/σvis , is the viscous resistivity of the material. The viscous resistance of the cylinder is then defined by ∆v . (15) Rvis = I Its inverse gives the viscous conductivity: Gvis =

1 . Rvis

Formulæ (10) and (11) we derived in the previous section are also applicable in the present cases with the appropriate index changes. Although it is not the topic of our article, we mention that ultimately all transport phenomena are related via the microscopic description. In particular we can find the following expressions for the coefficients: 1 n kB vave lf ree , 2 1 vave lf ree , = 3 1 = n m vave lf ree , 3

σth = σdif f σvis

where lf ree is the mean free path of the molecules, vave the average velocity of the molecules, n the number of molecules per unit volume, m the mass of one molecule, and kB Boltzmann’s constant. These expressions should be compared with that of the electric conductivity which is more familiar: 1 q 2 lf ree , σ = n 2 m vave where q is the charge of a carrier.

2.6

Elasticity

The concept of elasticity is more than a mere definition. The behavior of a rubber band or the behavior of a rod or a cable under stress is basically analogous to that of many springs connected together. 7

dx

x

A

L

Let’s imagine a uniform rod of length L and cross-sectional area A. We focus on an infinitesimal piece of length dx at distance x from one base. If dξ is the infinitesimal extension of this piece under the force F (x), then Hooke’s law states that dξ = − F (x) dℓ , where dℓ is the elasticity constant for the piece dx. We can write the above relation as dℓ dξ = − F (x) , dx dx where λ = dℓ/dx is the elasticity per unit length and ǫ = dξ/dx is the extension of the system per unit length, known as linear strain. It is known that approximately [2] 1 dξ = − F (x) , dx YA where A is the cross-section and Y is a constant characteristic of the material known as Young’s modulus. Combining the last expressions we conclude that λ =

dℓ 1 = . dx YA

If λ is constant then 1 L , Y A where L is the length of the system. The stiffness would be respectively: ℓ =

k = Y

A . L

We have thus obtained basic formulæ similar to those of resistors and capacitors that allow the computation of k and ℓ in any geometry. Most probably these formulæ are well-known to engineers, but they are not well-known among physicists. However, once written down, they look familiar and natural.

3

Addition Formulæ

Let R1 , R2 , ... be some resistors4 , C1 , C2 , ... be some capacitors, L1 , L2 , ... be some inductors, k1 , k2 , ... be some springs, R1 , R2 , ... be some transport conductors (either thermal, or diffusion, 4

By abuse of language, we identify the objects with their property.

8

or viscous). We will also indicate by R, C, L, k, R the equivalent resistance, capacitance, inductance, stiffness, and transport resistance, respectively, either when the objects are connected in series or in parallel. Either from introductory physics, or as a straightforward exercise on the definitions, the reader can persuade himself that when the objects are connected in series R 1 C L 1 k R

= R1 + R2 + . . . , 1 1 = + + ... , C1 C2 = L1 + L2 + . . . , 1 1 + + ... , = k1 k2 = R1 + R2 + . . . ,

and when they are in parallel 1 R C 1 L k 1 R

= = = = =

1 1 + + ... , R1 R2 C1 + C2 + . . . , 1 1 + + ... , L1 L2 k1 + k2 + . . . , 1 1 + + ... . R1 R2

Introducing the conductance, incapacitance, deductance, elasticity, and thermal conductance, we can rewrite them in a form that is always additive: P = P1 + P2 + · · · , where, if the elements are connected in series, P stands for any of R, D, L, ℓ, R , if the elements are connected in parallel, P stands for any of G, C, K, k, G . In the remaining section, we will demonstrate the application of the addition formulæ by examining specific examples from resistors and capacitors. Along with the solution, several comments are made to help the reader understand some of the implicit assumptions and other details in the solution. In the section that follows, more problems are discussed for the readers who wish to master the technique. 1. Problem [Cylindrical Resistor] The cylindrical resistor shown in figure 2 is made such that the resistivity ρ is a function of the distance r from the axis. What is the total resistance R of the resistor? Solution We divide the cylindrical resistor into infinitesimal resistors in the form of cylindrical shells of thickness dr. One of these shells is seen in red in figure 2. 9

A L

Gtr = σtr A L

basic formula

G=σ

A L

C = ε0 κ Ad

K=

k=Y L 1 µ0 N 2 A

thermal conductance stiffness deductance capacitance

I ΦB

K= Q ∆V

C= I ∆V

series

parallel

definition

G=

conductance

Itr ∆Ttr

Gtr = F ξ

k=

L A

Rtr = ρtr 1 L Y A

ℓ= A L

L = µ0 N 2 1 d ε0 κ A

D= basic formula

L R=ρA

thermal resistance elasticity inductance incapacitance resistance

∆Ttr Itr

Rtr = ξ F

ℓ= ΦB I

L= ∆V Q

D= ∆V I

definition

R=

transport conductors springs inductors capacitors resistors connection

Table 1: This table summarizes the additive physical quantities in the most common cases encountered in introductory physics. The quantities that are not usually defined in the introductory books are the conductivity G = 1/R, the incapacitance D = 1/C, the deductance K = 1/L, the elasticity constant ℓ = 1/k, and the the thermal conductivity Gtr = 1/Rtr . The index tr stands for transport and it should be interpreted as a generic name for any of the three cases of thermal conductivity, diffusion, or viscosity. 10

dr r a l Figure 2: The figure shows a cylindrical wire of radius a. A potential difference is applied between the bases of the cylinder and therefore electric current is running parallel to the axis of the cylinder. When the current is flowing along the axis of the cylinder, the infinitesimal resistors are not connected in series. Instead, all of the infinitesimal cylindrical shells of thickness dr are connected at the same end points and, therefore, have the same applied potential. In other words, the shells are connected in parallel and it is the conductance that is important. Specifically Z G = dG . cylinder

For the infinitesimal shell dG = σ(r)

2πrdr . ℓ

Therefore G =

2π dG = ℓ cylinder

Z

Z

0

a

σ(r) rdr .

For example, if σ(r) = σ0 ar , then G = 2σ0

πa2 , ℓ

where σ0 = 1/ρ0 . The resistance is therefore R =

ρ0 ℓ . 2 πa2

2. Problem [Truncated-Cone Resistor] A resistor is made from a truncated cone of material with uniform resistivity ρ. What is the total resistance R of the resistor when the potential difference is applied between the two bases of the cone? Solution This is a well-known problem found in many of the introductory physics textbooks [3, 7, 8, 9, 10]. We can partition the cone into infitesimal cylindrical resistors of length dz. One 11

representative resistor at distance z from the top base is seen in figure 3. The area of the resistor is A = πr 2 and therefore its infinitesimal resistance is given by dR = ρ

dz . πr 2

From the figure we can see that z r−b h = ⇒ dz = dr . h c−b c−b

b z r

dz h

c

Figure 3: A truncated cone which has been sliced in infitesimal cylinders of height dz. The infinitesimal resistors are connected in series and therefore R =

Z

cone

dR = ρ

h π(c − b)

Z

b

c

h dr = ρ . 2 r πbc

(16)

Comment: However, this solution, which is common in textbooks [3, 7, 8, 9, 10], tacitly assumes that the disks used in the partition of the truncated cone are equipotential surfaces. This is of course not true, as can be seen quite easily. If they were equipotential surfaces, then the electric field lines would be straight lines, parallel to the axis of the cone. However, this cannot be the case as, close to the lateral surface of the cone, it would mean that the current goes through the lateral surface and does not remain inside the resistor. Therefore, the disks are not equipotential surfaces. One way out of this subtlety is to assume that the disks are approximate equipotential surfaces as suggested in [9]. This is the attitude we adopt in this article as our intention is not to discuss the validity of the partitions used in each problem, but to emphasize the unified description of resistances and capacitances as additive quantities. Similar questions can be raised and studied in the majority of the problems mentioned in the present manuscript. A reader with serious interests in electricity is referred to the article of of Romano and Price [6] where the conical resistor is studied. Once that article is understood, the reader can attempt to generalize it to the rest of the problems of our article.

3. Problem [Cylindrical Capacitor] In introductory physics, the capacitance of a cylindrical capacitor is found using the definition C = Q/∆V , where Q is the charge on the positive plate of the capacitor and 12

∆V the absolute value of the potential difference between the two plates. However, this problem asks to compute the capacitance using only the formulæ giving the capacitance of a parallel plate, plane capacitor. Solution As shown in the left side of figure 4, the capacitor is partitioned into small cylindrical capacitors for which the distance between the plates is dr. For such small capacitors, the formula of a parallel-plate capacitor is valid. We notice though that all infintesimal capacitors are connected in series. Therefore dD = and D =

Z

cylinder

dD =

1 dr . ε0 2πrh

1 a 1 Z b dr = ln . 2πε0 h a r 2πε0 h b

The total capacitance is then C =

2πε0 h 1 = . D ln ab

dr b

b r

a

a

h

h

dz

z

Figure 4: A cylindrical capacitor with radii a and b and height h. In the left picture, we have sliced it in infinitesimal cylindrical shells, while in the right picture we have sliced it in infinitesimal annuli. 13

Comment: One might be tempted to partition the cylindrical capacitor into infinitesimal capacitors as seen in the figure to the right (blue section). Such capacitors look simpler than the infitesimal cylindrical shell we used above. Furthermore, R they are connected in parallel (notice that each capacitor is carrying an infinitesimal charge dQ and cylinder dQ = Q) and therefore it is enough to deal with capacitance, R C = cylinder dC, and not incapacitance D.

However, with a minute’s reflection the reader will see that in order to use the parallel-plate capacitor formula in the infinitesimal case, the distance between the plates must be infinitesimal which indicates that the infinitesimal capacitors must be connected in series. In the proposed (blue) slicing, the distance between the plates of the infinitesimal capacitor is finite, namely b − a. The infinitesimal capacitor is still a cylindrical capacitor of infitesimal height and therefore its capacitance should be expressed in a form that is not known before the problem is solved. In other words, 2πε0 dz ln(b/a)

dC = from which 2πε0 C = ln(b/a)

Z

0

h

dz =

2πε0 h. ln(b/a)

(17)

(18)

However, the expression (17) is unkown until the result (18) is found.

4. Problem [Truncated-cone Capacitor] A capacitor is made of two circular disks of radii b and c respectively placed at a distance h such that the line that joins their centers is perpendicular to the disks. Find the capacitance of this arrangement (Seen in figure 3). Solution We partition the capacitor into infinitesimal parallel-plate capacitors of distance dz and plate area A = πr 2 exactly as seen in figure 3. These infitesimal capacitors are connected in series and therefore the incapacitance is the relevant additive quantity: dD =

1 dz . ε0 πr 2

Notice that the computation is identical to that of R with final result: D =

πbc 1 h ⇒ C = ε0 . ε0 πbc h

(19)

When b = c, we recover the result of the parallel-plate capacitor.

4

Problems with Solutions

In this section we pose and solve a number of problems that will help the reader become fluid in the application of the quantities in table 1. A few of the problems are well-known, standard ones found in all textbooks; these problems are re-examined and solved in this section using our technique.

14

4.1

Spherical Capacitor

Re-derive the well-known expression for the capacitance of a spherical capacitor C = 4πε0

ab , b−a

(where a, b are the radii of the spheres with b > a) by partitioning it into infinitesimal capacitors. Solution We partition the capacitor into spherical shells of thickness dr. The infinitesimal shells are connected in series and therefore their incapacitance is 1 dr dD = . ε0 4πr2 The total incapacitance is thus Z Z 1 b−a 1 dr = , D = dD = 2 4πε0 sphere r 4πε0 ab sphere from which we find the well-known formula for the capacitance: C = 4πε0

dr

ab . b−a

r b a

Figure 5: The partition for a spherical capacitor into infinitesimal shells.

4.2

Spherical Capacitor with Dielectric

Show that the capacitance of a spherical capacitor which is filled with a dielectric having dielectric constant κ(r) = cr n , where r is the distance from the center and c, n are constants, is given by c C = 4πε0 ln(b/a) for n = −1 and an+1 bn+1 C = 4πε0 c(n + 1) n+1 b − an+1 15

for n 6= −1. Solution As in the previous problem, we partition the capacitor into spherical shells of thickness dr. The infinitesimal shells are connected in series and therefore their incapacitance is dr 1 . ε0 κ(r) 4πr2

dD = The total incapacitance is thus D =

Z

sphere

dD =

1 4πε0

dr = 2 sphere κ(r)r

Z

For n = −1, D =

1 4πε0 c

For n 6= −1, 1 D = 4πε0 c

4.3

Z

a

b

dr rn+2

Z

b

a

=

1 4πε0 c

Z

a

b

dr . rn+2

1 b dr = ln . r 4πε0 c a

b r−n−1 bn+1 − an+1 1 1 = . = 4πε0 c −(n + 1) 4πε0 c(n + 1) bn+1 an+1 a

Cylindrical Capacitor with Dielectric I

Show that the capacitance of a cylindrical capacitor which is filled with a dielectric having dielectric constant κ(r) = cr n , where r is the distance from the axis and c, n 6= 0 are constants, is given by an bn . C = 2πε0 hcn n b − an Solution We partition the capacitor into coaxial infinitesimal cylindrical shells of radius r and thickness dr which are connected in series (see figure 4). The infinitesimal incapacitance of such a shell is dD =

dr dr 1 1 . = ε0 κ(r) 2πrh 2πε0 hc rn+1

Therefore D =

1 2πε0 hc

since n 6= 0. Therefore C =

Z

a

b

dr rn+1

=

b n − an 1 , 2πε0 hc n an bn

an b n 1 , = 2πε0 hcn n D b − an

i.e. exactly the advertised formula.

4.4

Cylindrical Capacitor with Dielectric II

Show that the capacitance of a cylindrical capacitor which is filled with a dielectric having dielectric constant κ(z) = cz n , where z is the distance from the base and c, n ≥ 0 are constants, is given by chn+1 . C = 2πε0 (n + 1) ln(b/a) 16

Solution We partition the capacitor into coaxial infinitesimal annuli of thickness dz which are connected in parallel (see figure 4). Each infinitesimal capacacitor has the geometry of a cylindrical capacitor and therefore its infinitesimal capacitance is given by dC = 2πε0 κ(z)

2πε0 c n dz = z dz . ln(b/a) ln(b/a)

Therefore 2πε0 c C = ln(b/a)

Z

h

z n dz =

0

2πε0 c hn+1 , ln(b/a) n + 1

since n ≥ 0.

4.5

Truncated-Cone Capacitor I

(a) Two metallic flat annuli are placed such that they form a capacitor with the shape of a hollow truncated cone as seen in figure 6. Partition the capacitor into infinitesimal capacitors and show that the capacitance is given by "

h c−a b−a C = 2πε ln − ln a(c − b) c+a b+a

#

.

Show that this result reduces to that of a cylindrical capacitor for c = b. Also, show that it agrees with the result for a parallel-plate capacitor with a = 0. (b) Now, fill the two bases with disks of radius a and argue that the capacitance of the hollow truncated cone equals that of the truncated cone minus the capacitance of the parallelplate capacitor that we have removed. This means that the capacitance of the hollow truncated cone should equal bc − a2 C = πε0 . h How is it possible that this result does not agree with that of part (a)? Solution (a) We divide the truncated cone into annuli of height dz. These are parallel-plate capacitors connected in series. Therefore 1 dz dD = . ε0 π(r2 − a2 ) From the similar triangles see on the left side of figure 6, we see that r−b h z = ⇒ dz = dr . h c−b c−b

(20)

Therefore h dr h 1 = dD = ε 0 π c − b r 2 − a2 2πε0 a(c − b)

1 1 − r−a r+a

dr ,

and D

=

h 2πε0 a(c − b)

Z

b

c

1 1 − r−a r+a

h dr = 2πε0 a(c − b)

17

c−a b−a ln . − ln c+a b+a

When c = b we see that D = 0/0 and therefore the result cannot be found by simple substitution. However we can use L’ Hospital’s rule5 : D =

b−a ln c−a − ln b+a 2a h h c−a h 1 d h . = . lim c+a = lim ln = 2 2 2 2πε0 a c→b c−b 2πε0 a c→b dc c + a 2πε0 a b − a πε0 b − a2

This is just C = ε0 A/h for a plane capacitor. The case a = 0 is obtained in the same way: D

= =

c−a − ln b−a ln c+a h h d b+a lim = lim 2πε0 (c − b) a→0 a 2πε0 (c − b) a→0 da −2c h −2b h = − lim , 2 2 2 2 2πε0 (c − b) a→0 c − a b −a πε0 cb

c−a b−a ln − ln c+a b+a

as found previously. b

a z r−b

r

c−b

dz

h

c

Figure 6: A partition of the truncated cone in infinitesimal slices. (b) We use a parallel-plate capacitor with circular plates of radius a at a distance h to fill the plates of our capacitor. This capacitor has capacitance πa2 . Cadd = ε0 h The conical capacitor we have thus created has capacitance Ctotal = ε0

πcb . h

The original capacitor and the one we added are connected in parallel since the same voltage is applied at their plates. Therefore, according to the superposition principle Ctotal = C + Cadd ⇒ C = ε0

π(cb − a2 ) . h

Apparently this result does not agree with that of part (a). The reason is subtle but easy to find. The superposition principle states that if a problem in electricity can be split in two other problems, then the solution to the original problem is the sum of the solutions of the partial problems. But is our problem the exact sum of the two partial ones? 5

L’ Hospital’s rule states that: if limx→x0 f (x) = limx→x0 g(x) = 0 and the limit limx→x0 f ′ (x)/g ′ (x) exists, then limx→x0 f (x)/g(x) = limx→x0 f ′ (x)/g ′ (x).

18

Let’s assume that each plate of the truncated cone has a charge of absolute value Q and constant charge density equal to σ = Q/πb2 on the top plate and equal to σ ′ = −Q/πc2 on the bottom plate. Q splits into Q1 and Q2 on the plates of the hollow truncated cone and the cylinder, respectively, in proportion to the areas of the plates. On the top plate of the hollow truncated cone we have Q1 = σ π(b2 − a2 ) and on the top plate of the cylinder Q2 = σ πa2 . On the bottom plate of the hollow truncated cone we have Q′1 = σ ′ π(c2 − a2 ) and on the top plate of the cylinder Q′2 = σ ′ πa2 . However, Q′1 and Q′2 are not −Q1 and −Q2 (except when b = c). The only way to ensure this is to change the charge densities on the plates. But then the problem is not a simple addition of two other problems.

a

b

a

b +

=

c

c

Figure 7: The truncated cone is the sum of the hollow truncated cone plus a cylinder.

4.6

Truncated-Cone Capacitor II

A capacitor with the shape of a hollow truncated cone is now formed from two ‘cylindrical’ shells. Find its capacitance C. Solution As in the previous problem, we partition the capacitor into infinitesimal annuli. However, in this case they behave as infinitesimal cylindrical capacitors connected in series. The plates have radii a and r, and the height is dz. Therefore dC = 2πε0

dz . ln ar

Using equation (20), we can find h C = 2πε0 c−b

Z

c b

ha dr . = 2πε0 ln ar c−b

Z

c/a

b/a

dt , ln t

where in the last equation we have made a change of variables t = r/a. The integral Z x dt li(x) ≡ ln t 0

is known as the logarithmic integral [1]. Using this definition, we can write the previous result in the form: C = 2πε0

ha [li(c/a) − li(b/a)] . c−b

When b = c, we find C = 0/0 and therefore we should use L’ Hospital’s rule to compute the result: Z c/a d dt C = 2πε0 ha lim . c→b dc 0 ln t

19

Recall now that d dX

Z

X

f (t)dt = f (X) . 0

Therefore, C = 2πε0 ha

1 h = 2πε0 , a ln(b/a) ln(b/a)

i.e. the result of a cylindrical capacitor with radii a and b.

4.7

Hollow Cylindrical Conductor

A conductor has the shape of a hollow cylinder as seen in figure 4. Show that the resistance when the voltage is applied between the inner and outer surfaces is given by R =

b ρ ln . 2πh a

Solution We split the conductor into infinitesimal cylindrical shells (left picture of figure 4) which are connected in series and have infinitesimal resistance: dr dR = ρ . 2πrh The total resistance of the conductor is then: Z c Z ρ c dr ρ = ln . R = dR = 2πh b r 2πh b

4.8

Hollow Truncated-Cone Conductor I

(a) A conductor has the shape seen in figure 6. Show that the resistance when the voltage is applied between the upper and lower bases is given by "

h c−a b−a R = ρ ln − ln 2πa(c − b) c+a b+a

#

.

Show that this result reduces to that of a solid truncated-cone for a = 0. (b) Argue now that the resistance of the hollow truncated-conical wire is the difference between the the resistance of the truncated-conical wire and a cylindrical wire of radius a. This implies that bc − a2 . R = ρ h Explain why this does not agree with part (a). Solution (a) We partition the conductor into infinitesimal cylindrical conductors of length dz and cross-sectional area π(r2 − a2 ) as seen in figure 6. These infinitesimal conductors are connected in series. They have resistance dR = ρ

dz dr h , = ρ π(r2 − a2 ) π(c − b) r2 − a2

20

where we used equation (20). Therefore Z c c−a h b−a h dr ln . = ρ − ln R = ρ π(c − b) b π(r2 − a2 ) 2π(c − b)a c+a b+a For a = 0, R = 0/0. We thus must use L’ Hospital’s rule: R

=

ρ

c−a − ln b−a ln c+a h c−a h d h b−a b+a ln = ρ lim = ρ lim − ln . 2π(c − b) a→0 a 2π(c − b) a→0 da c+a b+a πcb

(b) Using the superposition principle for figure 7, we would have written down D = Dcone − Dcylinder =

1 cb − a2 , ρ h

since they are connected in parallel. Then R = ρ

h . π(cb − a2 )

This does not agree with the result of part (a) since the sum of two partial problems is not the problem we are studying. One can verify this by checking the current densities on the top and bottom faces.

4.9

Hollow Truncated-Cone Conductor II

A conductor has the shape seen in figure 6. Show that the resistance when the voltage is applied between the inner and outer surfaces is given by R =

ρ c−b . 2πha li(c/a) − li(b/a)

Show that, for c = b, this result agrees with that of problem 4.7. Solution We partition the conductor into infinitesimal annuli connected in parallel. The corresponding infinitesimal conductivity is h dr dz . dS = 2πσ r = 2πσ ln a c − b ln ar From this, the total conductivity is found to be Z c dz ha S = 2πσ [li(c/a) − li(b/a)] , r = 2πσ ln c −b b a which can be inverted to give to total resistance R =

4.10

ρ c−b . 2πha li(c/a) − li(b/a)

Wedge Conductor

A conductor has the shape of a truncated wedge as seen in figure 8. Show that the resistance of the conductor when the voltage is applied between the left and right faces is R =

ρ ℓ ln(c/b) , a c−b 21

z c b

z−b

c−b

y ℓ ℓ

a

y

dy

x Figure 8: A wedge partitioned in infinitesimal layers. while the resistance when the voltage is applied between the top and bottom faces is ρ c−b . a ℓ ln(c/b)

R =

Solution (a) We partition the conductor into infinitesimal layers of thickness dy along the y-direction. These layers are infinitesimal resistors with the shape of square cylinders connected in series. Their resistance is dR = ρ

dy . za

From the figure we see that z−b ℓ ℓ y = ⇒ y = (z − b) ⇒ dy = dz. ℓ c−b c−b c−b Then dR = ρ

ℓ dz , a(c − b) z

and ℓ R = ρ a(c − b)

c

Z

b

ℓ dz = ρ ln c/b . z a(c − b)

(b) When the voltage is applied between the top and bottom faces, the infinitesimal resistors are connected in parallel. Now the current flows through area a dy and the length it travels is z: ady aℓ dz = σ . z c−b z

dS = σ Therefore S = σ from which

aℓ c−b

Z

b

R = ρ

c

aℓ c dz = σ ln , z c−b b

c−b 1 . aℓ ln(c/b)

22

4.11

Toroidal Inductor

Find the inductance for a section of angular span φ of a toroidal inductor of radii a and b, height h, and N number of turns. Solution We partition the toroid into infinitesimal solenoids all run by the same current and thus connected in series. This is seen in figure 9.

s

b a

I

φ

I h

r

infinitesimal solenoids

dr

Figure 9: A toroidal inductor. All infinitesimal inductors have the same number of turns N and they have the simple geometry of a solenoid. Therefore, each has an infinitesimal inductance dL = µ0 N 2

h dr , s

where s = φr is the length of the inductor at distance r from the center. Then Z Z h b dr h b L = dL = µ0 N 2 = µ0 N 2 ln . φ a r φ a For a full circle, φ = 2π and

h b ln , 2π a a well-known result [3, 8, 9], usually found by computing flux. Notice that, using our method, it is not necessary to know the value of the magnetic field in order to find the inductance. L = µ0 N 2

4.12

Parallel-Plate Inductor

Split the parallel-plate inductor into convenient infinitesimal inductors. Then make use of equation (5) to again derive equation (7). Solution The inductor is split in parallel infinitesimal slices as seen in figure 10. Each slice is similar to a turn of a solenoid; it is carrying an infinitesimal current dI = Js dx. The infinitesimal slices have a deductance of dK = where ΦB = BLd = µ0 Js Ld. Therefore dK =

dI , ΦB

1 dx . µ0 Ld

23

The total deductance is then

Z

K = and

w 0

L =

w 1 dx = , µ0 Ld µ0 Ld 1 Ld = µ0 . K w

I

d

dx I w L

Figure 10: A parallel-plate inductor can be split into infinitesimal inductors connected in parallel.

4.13

Coaxial Inductor I

Think of a coaxial cable made of two cylindrical shells of radii a and b. If the cable has length h, compute its inductance. (The currents are uniformly distributed along the cross-sections of the wires.) Solution Standard Calculation: The standard computation found in introductory physics books (e.g. [9]) uses the definition (5). The magnetic field between the two cylinders can be easily found using Amp`ere’s law: B =

µ0 I , 2πr

where r is the distance from the center. Then we take a cross-section between the two cylinders (seen in blue in the left picture of figure 11). We split this cross-section into infinitesimal strips of width dr that are penetrated by infinitesimal flux µ0 Ih dr . dΦB = B h dr = 2π r The total flux through the whole cross-section is ΦB =

Z

dΦB

µ0 Ih = 2π

Then, L = ΦB /I and therefore

Z

a

b

µ0 Ih b dr = ln . r 2π a

b µ0 h ln . 2π a Obviously the standard computation falls short of explaining why only one cross-section has been used and no integration over the angular coordinate has been performed. New Calculation I: Let’s partition the cable into infinitesimal inductors with the shape of the wedge as seen in the middle picture of figure 11. The angular span of each wedge is dφ and each plate carries current dI = I dφ 2π . The wedges are connected in parallel. L =

24

dr b

b

dφ

a r

b r

a

a

dr

h

h

h

Figure 11: A coaxial cable made from two cylindrical shells with radii a and b and height h. In the right picture, we have sliced it into infinitesimal cylindrical shells, while in the middle picture we have sliced it into infinitesimal wedges. Each wedge has an infinitesimal deductance of dK = The total detuctance is K =

I dI = dφ . ΦB 2πΦB

I 2πΦB

Z

2π

dφ =

0

I . ΦB

and therefore we recover the result of the previous solution. We now see why only the flux of a single cross-section should be included. New Calculation II: The computation we presented in the previous paragraph was no shorter than the standard one, as it also must compute the flux for one cross-section. It only had the advantage of explaining why the flux of a single cross-section must be included. However, now we give a short calculation based on our technique, by-passing the use of flux and relying solely on the basic formula (7). We partition the cable into infinitesimal cylindrical shells seen in the right picture of figure 11 which behave as parallel-plate inductors connected in series. The distance between the plates is dr, the length of each plate is h, and the width of each plate is 2πr. Therefore dL = µ0 The total inductance is then h L = µ0 2π

Z

a

b

h dr . 2πr

h b dr = µ0 ln . r 2π a

25

4.14

Coaxial Inductor II

In the coaxial inductor of the previous problem, imagine that the current is flowing from the inner cylindrical shell towards the outer shell. Compute the inductance of this configuration of this configuration. Solution This is another problem that presents the power of our approach. Using the standard computations, students would have great difficulty in solving this problem, as they must first compute the magnetic field. However, our method reduces the problem to a an almost trivial calculation! We again partition the cable into cylindrical shells of width dr. These are now infinitesimal inductors that are connected in parallel, each having a deductance dK =

1 dr . µ0 2πrh

Therefore the total deductance is K =

Z

1 1 dK = µ0 2πh

Z

b

a

1 1 b dr = ln , r µ0 2πh a

which of course implies that L = µ0 2π

5

h . ln ab

Conclusions

There is probably no need for additional problems. The reader has certainly uncovered the pattern. All the quantities we have used —let the generic symbol P stand for any of them— follow a simple additive law P =

X

Pi ,

discrete case ,

i

P =

Z

dP ,

continuous case .

For elements in the shape of a uniform cylinder of length L and cross-section A whose material is characterised by the constant p (corresponding to quantity P ), P given by L , if P stands for R, D, L, ℓ, R , A P = 1 A , if P stands for G, C, K, k, G . pL p

For infinitesimal elements

dL , if P stands for R, D, L, ℓ, R , A dP = 1 dA , if P stands for G, C, K, k, G . p L p

In all cases of identical geometry, the results will be identical. In fact, in many instances above we could have saved some computations but we avoided doing so in order to present the 26

big picture first. Once the reader is aware of the global picture, she can easily use it to transfer a result for a quantity P in some geometry to a quantity P ′ in a similar geometry. An example of this is as follows. A sauna room has the shape of the wedge seen in figure 8. The insulation of the wall at y = ℓ has deteriorated and, as a result, the wall is at a lower temperature with respect to the wall at y = 0. An engineer would like to know the thermal resistance of the room. Assume that all other walls are at the same temprature with the wall at y = 0.

It is obvious that heat will flow from the left wall to the right wall. One might want to divide the room in infinitesimal layers parrallel to the side walls and continue as usual but this is not necessary. This problem has already been solved in section 4.10. Based on the results there, we can write dowm immediately that Rth =

ρth ℓ ln(c/b) . a c−b

Here is another problem A cylindrical cable of radius R and lenght h is made of a huge number of small filaments such that the cable may be considered continuous. The filaments have been arranged in such a way that Young’s modulus for the cable is Y (r) = c r, where r is the distance from the center of the cable and c some constant. What is the stiffness of the cable?

The stiffeness of a cylindrical shell of radius r is given by dk = Y (r)

2πrdr . h

The stiffness of the cable would then be k =

2πc R3 . 3h

The reader is invited to construct similar problems for quantities in table 1.

References [1] M. Abramowitz, I.A. Stegun, Handbook of Mathematical Functions with Formulas, Dover. [2] M. Alonso, E.J. Finn, Physics, Addison-Wesley. [3] D. Haliday, R. Resnick, J. Walker, Fundamentals of Physics, 6th ed., John-Wiley & Sons. [4] E. Hecht, Physics, Brooks/Cole 1996. [5] P. Nolan, Fundamentals of College Physics, Wm. C. Brown Communications 1993. [6] J.D. Romano, R.H. Price, The Conical Resistor Conundrum: A Potential Solution, Am. J. Phys. 64 (1996) 1150.

27

[7] R.A. Serway, Physics for Scientists and Engineers, 4th ed., Saunders College Publishing. [8] P.A. Tipler, Physics for Scientists and Engineers, 3rd ed., Worth Publishers. [9] R. Wolfson, J.M. Pasachoff, Physics for Scientists and Engineers, 3rd ed., Addison-Wesley. [10] H.D. Young, R.A. Freedman, University Physics, 9th ed., Addison-Wesley 1996.

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