fall 2010 h7b section material - wordpress.com

Fall 2010 H7B Section Material

Kevin Grosvenor

Last Edited: June 6, 2011

Contents

1 Kinetic Theory 11.1 Stirling’s Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Boltzmann Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Maxwell Speed Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Equipartition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.7 Mean Free Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The Second Law 72.1 Efficiency and Coefficient of Performance . . . . . . . . . . . . . . . . . . . . . . 72.2 First and Second Laws in Practice . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Different Ways to Expand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Midterm 1 Review 133.1 Negative Temperature and the Second Law . . . . . . . . . . . . . . . . . . . . . 133.2 Stirling Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.3 Volume and Linear Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 The Exponential Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.5 Diesel Autoignition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.6 Entropy Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Vectors and Vector Calculus 174.1 Single Variable Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.3 Basic Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5 Work and Energy 275.1 Energy of Charge Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.2 Classical Electron Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6 Poisson’s Equation 336.1 Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2 Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

i

ii CONTENTS

7 Capacitors 377.1 Force on Cylindrical Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

8 Current 418.1 Resistance of Concentric Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . 428.2 Non-Ohmic Device: Vacuum Diode . . . . . . . . . . . . . . . . . . . . . . . . . . 42

9 Circuit Analysis 459.1 Simple Bridge Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

10 Midterm 2 Review 4910.1 Analysis of Tetrahedral Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4910.2 Thick Resistive Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5110.3 Potential of a Continuous Charge Distribution . . . . . . . . . . . . . . . . . . . 5210.4 Charge near Conducting Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5310.5 Non-evenly Charged Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Chapter 1Kinetic Theory

The material in this chapter deals with a more detailed and rigorous description of the informa-tion in Chapter 18 of Giancoli. However, this route will take us through some basic statisticalmechanics, which is technically outside the scope of this class, so hang on to your horses! Forexample, we will take for granted the statistical definition of temperature, Eqn. (1.4.3). First, Iwould like to get some preliminary mathematics out of the way.

1.1 Stirling’s Approximation

lnN ! = lnN+ln(N−1)+· · ·+ln 1 ≈∫ N

1lnx dx =

(x lnx−x

)∣∣N1

= N lnN−N+1 ≈ N lnN−N .The first approximation is in turning the sum of logs into an integral. Imagine drawing a bargraph where the height of the bar centered at the integer n is lnn (draw it). The sum is the areaof the bars. The integral ought to be a good approximation if N >> 1. A better approximationcan be found in An Introduction to Thermal Physics by Daniel Schroeder pp.389-391. Let usbox this important equation for future reference:

lnN ! ≈ N lnN −N . (1.1.1)

1.2 Lagrange Multipliers

Suppose we want to extremize a scalar function of one variable, f(x). You know the drill: find thepoints where the first derivative vanishes. For a function of multiple variables, f(xi), we requirethe partial derivatives of f with respect to each variable, xi, vanish. That is,∇f ≡

∑i xi

∂f∂xi

= 0.Note that this is a vector equation.

Now, suppose we want to extremize f(xi), but subject to a constraint that the variables arerequired to satisfy, g(xi) = C, where C is a constant. We could solve for one of the variables interms of the other ones using the constraint equation, plug this in to f(xi), which now becomesa function of one fewer variable, and extremize that as usual. However, this is often practicallyand theoretically inconvenient. A more elegant procedure is to introduce a Lagrange multiplier.

The idea goes as follows: imagine drawing the level sets of f , which are the lines (twovariables), surfaces (three variables), etc., of points where f takes a particular constant value.By definition, ∇f at a point is perpendicular to the level set at that same point since ∇f is thedirection of most rapid change of f . Now, imagine superposed on this diagram the particularlevel set of g taking the value C. In order for f to be extremized here at some value, C ′, thetwo level sets g = C and f = C ′ should be tangential at their point of intersection (i.e. locally

1

2 CHAPTER 1. KINETIC THEORY

parallel). In other words, ∇f ||∇g, or ∇f = λ∇g or ∇(f − λg) = 0 for some constant λ, whichis called the Lagrange multiplier. Take a look at the Wikipedia page on Lagrange multipliersfor a simple diagram if you have trouble visualizing this.

If we need multiple constraints to hold, gj(xi) = Cj , then we introduce a Lagrange multiplierfor each constraint and require

∇(f −

∑j λjgj

)= 0 . (1.2.1)

1.3 Entropy

Consider, for concreteness, a gas with N total particles and having total internal energy E.Suppose there are N1 particles with energy E1, N2 with energy E2, and so on: Ni with energyEi. What is the total number of ways of dividing the energy up in this particular way? Well,we must choose N1 particles out of N to have energy E1. There are

(NN1

)= N !

N1!(N−N1)! ways to

do this. Then, out of the remaining N − N1 particles, we must choose N2 to have energy E2.

There are(N−N1

N2

)= (N−N1)!

N2!(N−N1−N2)! ways to do this. The pattern continues. The total number

of configurations, called the multiplicity function, or Ω, is the product of all of these factors:

Ω(Ni) =N !

N1!(N −N1)!(N −N1)!

N2!(N −N1 −N2)!· · · = N !∏

iNi!. (1.3.1)

Assuming that the gas (or whatever system it may be) is sufficiently isolated from the rest ofthe universe (this is call the microcanonical ensemble approximation), any one of the gasparticles is equally likely to have each of the energies available. In this case, the entropy isdefined to be

σ ≡ ln Ω = lnN !−∑i lnNi! . (1.3.2)

Note that this entropy is unitless. Entropy in standard units often has units of Boltzmann’sconstant, k, and is defined as S = kσ.

If we use Stirling’s approximation, assuming that Ni >> 1, we get

σ ≈ N lnN −N −∑i(Ni lnNi −Ni) . (1.3.3)

1.4 Boltzmann Distribution

Which distribution of energies is most probable? It must be the one with the greatest number ofpossible configurations. Thus, it maximizes Ω, which is equivalent to saying that it maximizesthe entropy, σ = ln Ω. However, the system is subject to the two constraints on total number ofparticles and total energy:∑

iNi = N,∑iNiEi = E. (1.4.1)

Thus, introducing the Lagrange multipliers, α and β, one for each constraint, the extremizationcondition, Eqn. (1.2.1), reads

∇(σ − α

∑iNi − β

∑iNiEi

)= 0. (1.4.2)

If we write this as ∇(σ−αN − βE) = 0, assume that we have managed to write σ as a functionof E (and not of the separate Ni’s), then we can consider the effect of varying E (at fixed N) bywriting (∇E)N

[(∂σ∂E

)N− β

]= 0, where the subscript N means “at constant N”. Note that all

1.5. MAXWELL SPEED DISTRIBUTION 3

I have done here is a change of differentiating variable: schematically, ∂f∂x = ∂y∂x

∂f∂y . In statistical

mechanics, one learns that the statistical definition of temperature is

1

T≡(∂S

∂E

)N

. (1.4.3)

See Schroeder pp.85-88 for a discussion of this. Thus, we have

β =1

kT. (1.4.4)

Plug into Eqn. (1.4.2) the Stirling result for σ, Eqn. (1.3.3). The ith component of the equationreads

− lnNi + α+ βEi = 0 =⇒ Ni = e−αe−βEi . (1.4.5)

Define the partition function to be

Z ≡∑i e−βEi . (1.4.6)

Recall that N =∑iNi = e−α

∑i e−βEi = e−αZ. Thus, the fraction of particles with energy Ei,

which is the same as the probability for a particle to have energy Ei, is

P (Ei) ≡NiN

=e−βEi

Z. (1.4.7)

This defines the Boltzmann distribution.

1.5 Maxwell Speed Distribution

Consider an ideal gas, wherein the particles, of mass m, are non-interacting and the only internalenergy of the particles comes from their translational kinetic energy: E = 1

2mv2. From the

previous discussion, the probability for any one particle to have velocity v is proportional to theBoltzmann factor, e−βE = e−mv

2/2kT . Let δN be the number of particles having a velocitywithin a small three-dimensional volume, d3v, of the velocity v. Then, δN ∝ e−mv

2/2kT andδN ∝ d3v:

δN = Ce−mv2/2kT d3v, (1.5.1)

where C is some constant, yet to be determined.This defines the velocity distribution, whereas we are interested in the speed distribution.

Convert d3v ≡ dvx dvy dvz into spherical coordinates: d3v = v2 dv sin θ dθ dφ, where θ and φ arethe standard polar and azimuthal angles and v ≡ |v| is the speed, the magnitude of the velocity.Note that the rest of the terms in δN are independent of θ and φ (i.e. are spherically symmetric).

Therefore, when we integrate over angles, we just pick up a factor of∫ π

0sin θ dθ

∫ 2π

0dφ = 4π:

dN ≡∫ π

0

∫ 2π

0

δN = 4πCe−mv2/2kT v2 dv. (1.5.2)

If we integrate over all speeds, v, we had better get N , the total number of particles:

N = 4πC

∫ ∞0

e−mv2/2kT v2 dv. (1.5.3)

Change the integration variable to x ≡(m

2kT

)1/2v:

N = 4πC

(2kT

m

)3/2 ∫ ∞0

e−x2

x2 dx = 4πC

(2kT

m

)3/2√π

4= C

(2πkT

m

)3/2

, (1.5.4)


where the integral was computed in the document on Gaussian integrals.Therefore, finally, the Maxwell speed distribution is defined as

f(v) ≡ dN

dv= 4πN

(m

2πkT

)3/2

v2 e−mv2/2kT . (1.5.5)

This “derives” Giancoli (18-6) p.480. Note that f(v) = dNdv since, if we integrate f , we’re

supposed to get N .I would like to quickly rewrite this result. A constant important in Quantum Mechanics

is Planck’s constant, h, or ~ = h/2π. A quantity useful in determining when quantum effects

become important is the quantum density, nQ =(mkT2π~2

)3/2. We can write Eqn. (1.5.5) as

f(v) = 4πmV

h

n

nQv2 e−mv

2/2kT , (1.5.6)

where V stands for volume. Usually, quantum effects can be neglected as long as the gas issufficiently dilute: n << nQ.

1.6 Equipartition Theorem

The result Giancoli (18-4) p.478 says that each translational degree of freedom has 12kT worth

of internal energy associated with it. Thus, in three dimensions, the average kinetic energy isK = 3

2kT . An alternative to the “derivation” in Giancoli is to simply calculate v2 = v2rms directly

from the Maxwell speed distribution. This is done in what should be Giancoli (18-7c) p.482:vrms =

√3kT/m. Thus,

K = 12mv

2rms = 3

2kT . (1.6.1)

Suppose the gas particles are actually diatomic gas molecules (e.g. N2 or O2). Still neglectinginteractions between the molecules, there are additional energy degrees of freedom beyond thetranslational ones: the molecule may (1) rotate, or (2) vibrate.

Each vibrational mode is actually accompanied by two energy degrees of freedom: one kineticand one potential. It turns out that vibrational modes have quite high energies and are onlyexcited at quite high temperatures.

However, for most diatomic gases, even at room temperature, the rotational modes arepresent. There are three independent axes about which the molecule may rotate. Quantummechanics forbids rotation about the central axis of the diatomic molecule. Thus, there are onlytwo rotational modes and each is accompanied by an energy Erot,i = 1

2Iω2i . I have assumed here

that the moment of inertia, I, is the same for both rotations, only for simplicity, not necessity.The effect is to multiply δN in Eqn. (1.5.1) by C ′e−Iω

2/2kT d2ω (the possible vectors ωonly span a plane since the third rotation axis is forbidden by quantum mechanics.) Thus, theangular integral in d2ω yields a factor of 2π. Let us focus only on the piece introduced by therotational degrees of freedom since they are completely separate from the translational degreesof freedom. Then, we require

2πC ′∫ ∞

0

e−Iω2/2kT ω dω = 1, (1.6.2)

since the speed integration already gives N . This implies 2πC ′ = IkT . Then, we can compute

ω2rms = ω2:

ω2 =I

kT

∫ ∞0

ω3 e−Iω2/2kT dω =

I

kT

(2kT

I

)2 ∫ ∞0

x3 e−x2

dx =2kT

I. (1.6.3)

1.7. MEAN FREE PATH 5

Finally, therefore, the average rotational energy is

Erot = 12Iω

2rms = kT , (1.6.4)

which is 2× 12kT , one factor of 1

2kT per rotational degree of freedom.

In general, each quadratic degree of freedom (qdof) has associated with it 12kT worth of

internal energy. This is the equipartition theorem.

1.7 Mean Free Path

In principle, it is not difficult to write an expression for finding the extra factor of 1/√

2 in themean free path that one is supposed to get after the “careful calculation” of vrel mentioned inGiancoli. However, you are likely to find that it will test your integration abilities as well as yourpatience a fair amount. Let v be the velocity of the particle whose mean free path we mean tocalculate and let u be the velocity of the particle with which it collides. The relative velocity isvrel = v− u and the relative speed is

vrel =√v2 + u2 − 2vu cos θ, (1.7.1)

where θ is the angle between the two velocities.

Then, the average relative speed is

vrel =

∫d3v d3u e−m(v2+u2)/2kT vrel∫d3v d3u e−m(v2+u2)/2kT

=

(m

2πkT

)3

8π2

∫ ∞0

dv v2 e−mv2/2kT

∫ ∞0

duu2 e−mu2/2kT×

×∫ π

0

dθ sin θ√v2 + u2 − 2vu cos θ. (1.7.2)

The 8π2 factor is 4π × 2π, where 4π comes from the angular integral in v and the 2π from

the azimuthal angular integral in u. If we rescale variables as always, x =(m

2kT

)1/2v and

y =(m

2kT

)1/2u, and define z = cos θ, then, we get

vrel =

(m

2πkT

)3

8π2

(2kT

m

)7/2 ∫ ∞0

dxx2 e−x2

∫ ∞0

dy y2 e−y2

×

×∫ 1

−1

dz√x2 + y2 − 2xyz. (1.7.3)

The z integral is straightforward:

vrel =8

3π

(2kT

m

)1/2 ∫∫ ∞0

dx dy xy[(x+ y)3 − |x− y|3

]e−(x2+y2). (1.7.4)

The integrand is symmetric under the interchange x ↔ y. Thus, if we convert to polar coordi-nates, (x, y) = r(cosφ, sinφ), instead of integrating φ from 0 to π/2, we can go from 0 to π/4and just multiply by 2. In this range, x > y and so |x − y|3 = (x − y)3. Multiplying out the

integrand gives 2xy2(3x2 + y2)e−(x2+y2). Note that dx dy = r dr dφ and so an extra factor of r


is added. Denote the integral by I for convenience. Then,

I = 2

∫ π/4

0

2 cosφ sin2 φ (3 cos2 φ+ sin2 φ) dφ

∫ ∞0

r6e−r2

dr

= 4

∫ π/4

0

cosφ sin2 φ (3− 2 sin2 φ) dφ︸︷︷︸Iφ

∫ ∞0

r6e−r2

dr︸︷︷︸Ir

. (1.7.5)

Let us first calculate the angular integral. Define w = sinφ so that dw = cosφdφ and if φ goesfrom 0 to π/4, then w goes from 0 to 1/

√2. Then,

Iφ = 4

∫ 1/√

2

0

w2(3− 2w2) dw

= 4w3∣∣1/√2

0− 8

5w5∣∣1/√2

0

= 45

(1√2

)5[5 · 2− 2

]= 8

5√

2. (1.7.6)

Now, on to the radial integral. Unfortunately, the Gaussian integrals document only goes up tor4e−r

2

. But, it gives∫∞

0r4 e−ar

2

dr = 38

√πa5 . Taking − d

da of both sides of this equation andthen plugging in a = 1 yields

Ir = 15√π

16 . (1.7.7)

It follows thatI = IφIr = 3

√π

2√

2. (1.7.8)

Plugging this back into Eqn. (1.7.4) gives the average relative speed:

vrel =8

3π

3√π

2√

2

(2kT

m

)1/2

=√

2

(8

π

kT

m

)1/2

=√

2 v . (1.7.9)

Chapter 2The Second Law

The material in Chapter 19 of Giancoli is important, but most of it is straightforward, I hope,(e.g. Q = mc∆T or Q = mL or ∆Q

∆t = kQT1−T2

` , etc.) The stuff I find interesting, such asequipartition and heat and work seem to me to be somewhat out of place. Equipartition, I feel,most naturally fits in kinetic theory and heat and work along with the second law (in conjunctionwith the first law). For example, the material on isothermal and adiabatic expansion in Chapter19 is actually applied in Chapter 20, which is more important. So, on to Chapter 20.

2.1 Efficiency and Coefficient of Performance

Unfortunately, the three main measures of “efficiency” (generalized) introduced in this chaptermay seem like a jumble of random formulas. Let us try to organize them so that we fully under-stand why different measures pertain to different systems.

Heat Engine: (Fig. 20-2 p.530) The thing we want is the work, W . The thing we have toput in is the heat from the high temperature reservoir, QH . The exhaust heat, QL, into thelow temperature reservoir has to come along because of the second law (e.g. Kelvin-Planckstatement, Fig. 20-6 p.532); there is nothing we can do about that. Therefore, our measure of“efficiency” should be

e =want

input=

W

QH, (2.1.1)

from which the important formula, e = 1− QLQH

follows since W = QH−QL in order for ∆Eint = 0

since the gas is run through a cycle (i.e. it must end where it started).

Refrigerator: (Fig. 20-9 p.536) The thing we want is QL (i.e. draw heat out of the coldreservoir, the inside of the refrigerator, that is already cooler than the hot reservoir, the outside).The thing we have to put in is the work, which is why we must plug the refrigerator into anelectrical outlet. Therefore, our measure of “efficiency”, which in this case is called coefficientof performance, should be

COP =want

input=QLW

. (2.1.2)

Heat Pump: (Fig. 20-11 p.538) The thing we want is QH (i.e. the heat dumped into the hotreservoir, which is the already-warm inside of the house). The thing we must put in is the work,W . Hence,

COP =want

input=QHW

. (2.1.3)

7

8 CHAPTER 2. THE SECOND LAW

The latter are not called efficiencies because they are not numbers between 0 and 1.

2.2 First and Second Laws in Practice

Question: Consider two identical bodies (heat reservoirs) with constant (temperature-independent)total heat capacities, C. The initial temperature of bodies A and B are Ta and Tb, respectively,with Ta > Tb.

(a) An ideal (reversible and adiabatic, i.e. isentropic) heat engine operates between the tworeservoirs until they reach thermal equilibrium. What are the final temperatures of the tworeservoirs? What is the maximum amount of work obtainable from the engine?

(b) Suppose that there is no heat engine and that the bodies come into thermal equilibrium byheat conduction. What is the final temperature of the two reservoirs in this case? What isthe change in entropy of the two-body system?

(c) Suppose the bodies have now come to thermal equilibrium via heat conduction, as in (b).We would like to restore the initial conditions. We have a very large heat reservoir at afixed temperature, T0, which is sufficiently large that its temperature will not be affectedby the proccesses we consider. We break the thermal contact between the two bodies andoperate two heat engines, one for each body, between the bodies and the larger outside heatreservoir. What is the minimum amount of work which has to be expended to restore theinitial conditions?

2.2. FIRST AND SECOND LAWS IN PRACTICE 9

Solution:

(a) Since the two bodies reach thermal equilibrium, their final temperatures will be the same,Tf . Since the process is reversible, we can use Eqn. 20-8 p.539:

∆S =

(∫ Tf

Ta

+

∫ Tf

Tb

)C dT

T= C ln

T 2f

TaTb. (2.2.1)

Note that we used d Q = C dT here.

The second law implies ∆S ≥ 0. Thus, Tf ≥√TaTb. Since the process is also adiabatic,

and hence isentropic, ∆S = 0. Thus,

Tf =√TaTb . (2.2.2)

A heat engine must work through cycles, which implies that ∆Eint = 0. Hence, W =QH −QL = Qa −Qb, since A is the hotter of the two bodies. Hence,

W = C(Ta − Tf )− C(Tf − Tb) = C(Ta + Tb − 2Tf ). (2.2.3)

As above, the second law implies Tf ≥√TaTb and thus,

W ≤ C(Ta + Tb − 2

√TaTb

)= C

(√Ta −

√Tb)2. (2.2.4)

(b) When d Q worth of heat flows from A to B, the temperature of A drops by dT = d Q/C.Since the bodies are identical, the temperature of B rises by the same amount. Therefore,they will reach the same final temperature, which is just halfway between the two:

Tf = 12 (Ta + Tb) . (2.2.5)

Neglecting possible changes in volume,

∆S = C lnT 2f

TaTb= C ln

(Ta + Tb)2

4TaTb= C ln

(1 +

(Ta − Tb)2

4TaTb

)≥ 0 . (2.2.6)

(c) To restore the initial conditions, the entropy of the bodies must be decreased by the amount∆S above (i.e. since entropy is a state variable). For the process to be reversible, the entropyof the outer reservoir must be increased by ∆S so that ∆Stot = 0. Thus, the minimum workrequired is

Wmin = T0∆S = CT0 ln

(1 +

(Ta − Tb)2

4TaTb

). (2.2.7)


2.3 Different Ways to Expand

Question: Consider a rigid, thermally insulated container of volume 2V divided in half by athermally insulating partition. Fill the left half with N particles of an ideal, monoatomic gas inequilibrium at temperature T0. The particles have mass m. Now compare three ways of allowingthe gas to expand and fill the entire container:

(a) If the gas expands adiabatically (the partition is moved slowly to the right, so that the gasis always in equilibrium), what is the final temperature?

(b) If the gas expands suddenly (the partition is instantaneously removed), what is the finaltemperature when equilibrium is reached?

(c) Let the gas leak slowly via a small hole of area A poked through the partition.

(i) Estimate the rate at which the particles leak through the hole.

(ii) After a short time, plug up the hole and let the dilute gas on the right side achieveequilibrium at Tf . Is Tf > T0, Tf < T0 or Tf = T0? If either of the first two cases iscorrect, calculate Tf .

2.3. DIFFERENT WAYS TO EXPAND 11

Solution:

(a) Along an adiabat, PV γ is constant. Since PV = nRT and n is constant, TV γ−1 is constant.

Hence,TfT0

=(V0

Vf

)2/3, since γ = 5/3 for an ideal, monoatomic gas. Thus,

Tf =

(V0

Vf

)γ−1

Ti =

(1

2

)2/3

T0 ≈ 0.63T0 . (2.3.1)

(b) The gas does no work because there is no piston to be moved! Since the system is thermallyinsulated, no heat flows in or out. Hence, 3

2Nk∆T = ∆Eint = Q −W = 0 and so ∆T = 0.Thus,

Tf = T0 . (2.3.2)

(c) (i) Gas particles will be passing through the hole in all sorts of directions and with all sortsof different speeds. For a quick estimate, let’s just consider particles travelling straighttowards the hole (i.e. perpendicular to the wall) and with speed v ∼

√kT/m (neglect

factors of order 1, e.g.√

8/π). In time dt, all the particles in a cylinder of cross-sectional area A and length v dt will pass through the hole. The number of particles,dN , in this volume is the density N/V multiplied by the volume, Av dt. Hence, if N isthe number of particles on the left, it decreases at the rate

dN

dt∼ −N

VAv = −NA

V

√kT

m. (2.3.3)

(ii) In (i), just for the estimate, we let every particle move with the average speed, v.However, we know that there is a Maxwell distribution of speeds. Naturally, the fasterparticles pass through the hole at a higher rate. But, these also happen to be themore energetic particles. Therefore, the relative proportion of high energy particlesto low energy particles will be greater on the right than on the left and so the finaltemperature on the right will be higher than the initial temperature, T0. Similarly,the final temperature on the left will be lower than T0. This is the main mechanismbehind Joule-Thomson cooling.

To calculate this, we must be more careful than in (i). Let θ be the angle to thenormal at the hole (so straight through corresponds to θ = 0). Consider the particlesin a little bit of volume, dV = R2(v dt) sin θ dθ dφ at the angle θ. The number density

of particles with speed between v and v + dv is f(v) dvV ≡ N

V P (v) dv, where f(v) is

the Maxwell speed distribution and P (v) dv ≡ f(v) dvN measures the probability that a

particle have speed between v and v + dv. Most of these particles will not be movingtowards the hole. From their point of view the apparent area of the hole is A cos θ(e.g. at θ = π/2, which is when the particle is just grazing the surface of the partition,this apparent area should be zero, as it is since cos π2 = 0). The probability that the

particle’s velocity points in the correct direction (towards the hole), is A cos θ4πR2 , where R

is the radial distance to the hole. Hence, the number of particles escaping is

d4Nesc =

(prob. for

right dir.

)(particle density

w/ right speed

)(volume)

=A cos θ

4πR2

NP (v) dv

VR

2(v dt) sin θ dθ dφ. (2.3.4)


Note that the 4 in d4Nesc is there because there are four differentials on the right:dt, dv, dθ and dφ. Integrating over φ ∈ [0, 2π) just gives a factor of 2π. Integratingsin θ cos θ over θ ∈

[0, π2 ) gives 1

2 . Note: θ cannot exceed π/2. Thus,

d2Nesc =AN

4VvP (v) dv dt. (2.3.5)

We need to compare the rate at which energy flows to the right versus the rate at whichparticles flow to the right. The ratio of the two will be the average energy per particleon the right side. The rate at which energy is flowing to the right is given by

d2Eesc = d2Nesc × 12mv

2 =ANm

8Vv3P (v) dv dt. (2.3.6)

Integrating over v yields

dNesc

dt=AN

4V

∫ ∞0

vP (v) dv =AN

4Vv =

AN

4V

√8

π

kT0

m. (2.3.7)

To calculate dEesc

dt , we need v3. The integral∫∞

0x5 e−x

2

dx = 1 will be useful. Theresult is

dEesc

dt=ANm

8V

√8

π

kT0

m

4kT0

m= 2kT0

dNesc

dt. (2.3.8)

This says that, on average, each particle that enters to the right takes with it 2kT0

worth of energy. Once the particles on the right reach thermal equilibrium at Tf , theenergy of each particle is 3

2kTf . Setting this equal to 2kT0 yields a temperature higherthan T0 as expected:

Tf = 43T0 . (2.3.9)

Chapter 3Midterm 1 Review

3.1 Negative Temperature and the Second Law

Let us take the following as the fundamental statement of the second law:

∆S = ∆Ssyst + ∆Senv > 0. (3.1.1)

Suppose it were possible to have a heat reservoir at negative absolute temperature (that’s neg-ative on the Kelvin scale!), is the Kelvin-Planck statement of the second law still equivalent tothe one above? Explain.

SOLUTION:

No! Usually the hotter reservoir loses entropy as it gives up heat, QH , to an engine since∆SH = −QH/TH < 0, assuming that it’s temperature remains constant. The point is that theentropy gained by the cooler reservoir exceeds this loss as long as the cooler reservoir actuallyabsorbs a sufficient amount of heat QL. That is, QH cannot be completely converted to usefulwork.

However, if the hotter reservoir is the one with negative temperature, then, since TH < 0,we have ∆SH = −QH/TH > 0 if and only if QH < 0. In other words, the negative-temperature(hotter) reservoir can spontaneously lose heat while at the same time increase entropy! Thus, thestatement, Eqn. (3.1.1), does not prevent all of the heat given off by the negative-temperaturereservoir from being turned into work.

3.2 Stirling Cycle

The Stirling cycle proceeds as (1) isothermal expansion at T2 from V1 to V2; (2) isochoric coolingat V2 from T2 to T1 < T2; (3) isothermal compression at T1 from V2 to V1; (4) isochoric heatingat V1 from T1 to T2. Compare the efficiency of the Stirling cycle with the Carnot cycle workingbetween the same two extreme temperatures, T1 and T2.

SOLUTION:

Heat enters the cycle during steps (4) and (1) and exits during steps (2) and (3):

−Q2 = Q4 = nCV (T2 − T1), Q1 = nRT2 ln V2

V1, Q3 = −nRT1 ln V2

V1. (3.2.1)

13

14 CHAPTER 3. MIDTERM 1 REVIEW

Thus, we have

Qin = nCV (T2 − T1) + nRT2 ln V2

V1, Qout = nCV (T2 − T1) + nRT1 ln V2

V1. (3.2.2)

The efficiency is

eStirling =Qin −Qout

Qin=

nR(T2 − T1) ln(V2/V1)

nCV (T2 − T1) + nRT2 ln(V2/V1). (3.2.3)

Divide top and bottom by nRT2 ln(V2/V1) and write eCarnot = T2−T1

T1:

eStirling =eCarnot

1 + CV eCarnot

R ln(V2/V1)

. (3.2.4)

3.3 Volume and Linear Expansion

Suppose that some 3D substance expands non-isotropically with different coefficients of linearexpansion in the three different directions, αx, αy and αz. Let β be the volume coefficient ofexpansion. Determine and prove the relation between β and the α’s.

SOLUTION:

β = 1VdVdT = d lnV

dT = d ln(xyz)dT = d ln x

dT + d ln ydT + d ln z

dT = 1xdxdT + 1

ydydT + 1

zdzdT = αx + αy + αz.

3.4 The Exponential Atmosphere

[Schroeder 1.16 ]

(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest,the pressure holding it up from below must balance both the pressure from above and theweight of the slab. Use this fact to find an expression for dP/dz, the variation of pressurewith altitude, in terms of the density of air.

(b) Assuming the air to be an ideal gas, derive the barometric equation dPdz = −mgkT P , where m

is the mass of an air molecule, T the temperature and P the pressure.

(c) Assuming that the atmosphere is isothermal, solve the barometric equation to obtain pressureas a function of height.

SOLUTION:

(a) Let A be the area of the top and bottom faces of the slab. Let ρ(z) be the density of air atheight z. We assume that dz is sufficiently small that we may treat ρ as constant throughoutthe slab. Balancing pressures yields

dP ≡ P (z + dz)− P (z) = −ρ(z) gAdz

A= −ρ(z) g dz. (3.4.1)

This gives us the differential equation

dP

dz= −ρ(z) g . (3.4.2)

3.5. DIESEL AUTOIGNITION 15

(b) ρ ≡(NV

)m = mP

kT . Plugging this into Eqn. (3.4.2) yields the barometric equation

dP

dz= −mg

kTP (z) . (3.4.3)

(c) If T is constant then, we may write Eqn. (3.4.4) as

d lnP

dz=

1

P

dP

dz= −mg

kT= const. (3.4.4)

This implies that

lnP (z) = −mgzkT

+ const. (3.4.5)

Plugging in z = 0 implies const. = lnP (0) and thus,

P (z) = P (0) e−mgz/kT . (3.4.6)

3.5 Diesel Autoignition

[Schroeder 1.37 ] In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of itsoriginal volume. Estimate the temperature of the air after compression, and explain why a Dieselengine usually does not require spark plugs.

SOLUTION:

Assume the air is ideal and diatomic so that γ = 7/5. Suppose that the process is sufficientlyfast that no heat transfer occurs (adiabatic). Then, PV γ is constant, or TV γ−1 is constant.Thus,

Tf =(ViVf

)γ−1Ti = 202/5 Ti ≈ 3.314Ti . (3.5.1)

Plugging in an initial temperature of Ti ≈ 300K gives

Tf ≈ 994 K . (3.5.2)

This is high enough for the fuel to autoignite.

3.6 Entropy Changes

(a) Two identical cars, each of mass m, are travelling at speed v directly towards each otherwhen they collide and are brought to rest. Calculate the resulting changein entropy of theenvironment.

(b) An Aluminum cup, of mass mAl, is at temperature TAl and is filled with water of massmw and temperature Tw > TAl. Calculate the total change in entropy of this system afterreaching thermal equilibrium.

SOLUTION:


(a) Assume that the cars are brought instantaneously to rest and that all of their initial energyis “lost” in the form of heat (no sound, etc.) and that the transfer of energy itself occursinstantaneously. Furthermore, assume that the temperature of the surroundings does not

change (i.e. the environment is a sufficiently large heat bath). Then, ∆S = Q/T = mv2/T ,

where T is the temperature of the environment.

(b) The cup and water is a closed system and thus loses and gains no heat. That is, the heatlost by the water is precisely the heat gained by the cup. Hence,

QAl +Qw = mAlcAl(Tf − TAl) +mwcw(Tf − Tw) = 0. (3.6.1)

Solving for Tf yields

Tf =mwcwTw +mAlcAlTAl

mwcw +mAlcAl. (3.6.2)

The total change in entropy is

∆S = mAlcAl

∫ Tf

TAl

dT

T+mwcw

∫ Tf

Tw

dT

T

= mAlcAl lnTfTAl

+mwcw lnTfTw

,

(3.6.3)

where Tf is given in (3.6.2).

Chapter 4Vectors and Vector Calculus

There are a number of good sources for a relatively quick introduction to vector calculus (e.g. theMIT notes: web.mit.edu/6.013_book/www/navigate2.html). However, you should understandthat it will take a fair amount of investment of time and effort on your part to really get facilewith this material. It will be worth it. With a solid understanding of vector calculus, you willfind in Electricity and Magnetism (E & M) a kind, trusted friend. On the other hand, withoutthis foundation, you will find her to be a formidable foe.

Okay, let’s get to it. This chapter is organized as follows. Section 1 is a brief review of singlevariable calculus. This should be very familiar. If it is not, then you best get familiar with itquickly and work through some problems. Some are provided at the end of certain sections.Section 2 is about vectors and operations among them as well as a discussion about line, areaand volume elements in various coordinate systems. Section 3 extends calculus to arbitrarydimensionality, with a special focus on the familiar three dimensions in which we appear to live.Pay special attention to the integral theorems. These are general mathematical theorems withfar-reaching consequences in E & M.

4.1 Single Variable Calculus

Let’s start with the one of many possible equivalent definitions of a derivative. Denote the realnumbers by R (that’s the usual numbers you’re familiar with: integers, fractions, irrationals likeπ, etc.) Let f(x) be a real-valued function of one real variable x, i.e. x ∈ R and f(x) ∈ R forall x. This is often denoted by f : R → R since it takes a number x in R and spits out anothernumber f(x) also in R. We define the derivative of f at x by taking a point slightly displacedfrom x, namely x+ ε, and taking the slope of the line connecting those two points on the graphof f(x):

df

dx≡ limε→0

f(x+ ε)− f(x)

ε. (4.1.1)

Note that ε could very well be negative here. It should not matter how one lets ε approach 0.However, it can happen that the limit does depend on how one approaches 0. In those cases,the derivative simply does not exits at that value of x. A simple example is f(x) = |x| whosederivative is undefined at x = 0.

Denote the sets of all possible maps f : R → R by C(R). Not all of these have well-definedderivatives everywhere. Denote those which cannot be differentiated everywhere by C0 (wewill drop the (R) part for brevity.) Denote those with everywhere-well-defined derivatives andcan be differentiated only once by C1. The subscript “1” is meant to indicate that you candifferentiate these functions only once. Likewise define Cn(R) as those functions which can be

17

web.mit.edu/6.013_book/www/navigate2.html

18 CHAPTER 4. VECTORS AND VECTOR CALCULUS

differentiated only n times. One neat interpretation of the derivative is that it is a linear operatorddx : Cn → Cn−1.1 All linearity means is that, if a, b ∈ R, then

d

dx

[af(x) + bg(x)

]= a

df

dx+ b

dg

dx. (4.1.2)

In fact, the derivative needs to satisfy more than that; it must satisfy the Liebniz rule in orderto be, in general, what is called a derivation. The Liebniz rule says

d

dx

[f(x) g(x)

]= f(x)

dg

dx+df

dxg(x). (4.1.3)

This is just the usual “product rule”.

Finally, there is the chain rule:

d

dxf(g(x)) =

dg

dx· dfdg, (4.1.4)

where df/dg means, for example, define g = g(x), write f(g(x)) as a function of g and takethe derivative with respect to g. As an example, let f(x) = x2 and g(x) = 2x + 1. Then,f(g(x)) = (2x+1)2 = 4x2 +4x+1, whose derivative is 8x+4. We can also set g = g(x) = 2x+1and write f(g(x)) = f(g) = g2. Thus, df/dg = 2g, which we then write back in terms of x asdf/dg = 2(2x+ 1). Finally, dg/dx = 2 and so

d

dxf(g(x)) =

dg

dx· dfdg

= 2 · 2(2x+ 1) = 8x+ 4. (4.1.5)

Okay, so that is all we have to say about derivatives except that we will often denote df/dxby f ′ for short. We may denote higher derivatives, dnf/dxn, by f (n)(x), or just a bunch ofapostraphes if n is not so large.

Also, before we move on, I should mention the exponential function f(x) = ex since it is soubiquitous. We will take Euler’s number, e, to be defined by the limit:

e ≡ limε→0

(1 + ε)1/ε. (4.1.6)

This is in fact one fundamental definition of this number, which takes the explicit value e ≈ 2.718.See Exercise 4 for how to use this to calculate the derivative of exponentials and trigonometricfunctions.

Derivatives are useful in practice in defining Taylor expansions. As long as f(x) is sufficientlyhighly differentiable (i.e. f(x) ∈ Cn for sufficiently large n), then we can define its Taylorexpansion around a point x = x0 as

f(x) = f(x0) + f ′(x0) (x− x0) +1

2!f ′′(x0) (x− x0)2 +

1

3!f ′′′(x0) (x− x0)3 + · · · (4.1.7)

The exclamation marks mean, for example, 3! = 1 · 2 · 3 = 6. By convention, 0! ≡ 1. We canwrite (4.1.4) more succinctly using summation notation:

f(x) =

∞∑n=0

1

n!f (n)(x0) (x− x0)n. (4.1.8)

1There are obviously functions that are infinitely differentiable, in C∞. For example, the higher derivativesof a constant function are all identically zero, which is in C∞. For such functions d/dx may still be interpretedas a map from C∞ to C∞−1, except that C∞−1 is the same as C∞.

4.1. SINGLE VARIABLE CALCULUS 19

Of course, in practice, we usually truncate the sum at a fairly low order since the higher orderterms usually die off very quickly if x is sufficiently close to x0. If x0 = 0, then the Taylorexpansion is often called the Maclaurin series. Below is a small set of useful Maclaurin series.

1

1− x= 1 + x+ x2 + x3 + · · · (for |x| < 1), (4.1.9)

(1 + x)n = 1 + nx+n(n− 1)

2x2 + · · · , (4.1.10)

ex = 1 + x+1

2!x2 +

1

3!x3 + · · · , (4.1.11)

ln(1 + x) = x− 1

2x2 + · · · (for − 1 < x ≤ 1). (4.1.12)

And here are the three basic trigonometric functions

sinx = x− 1

3!x3 +

1

5!x5 − · · · , (4.1.13)

cosx = 1− 1

2x2 +

1

4!x4 − · · · , (4.1.14)

tanx = x+1

3x3 + · · · (for |x| < π/2). (4.1.15)

Now, let us move on to integration. First, the Fundamental Theorem of Calculus: if you needto integrate f(x) and you already know a function F (x) such that dF/dx = f(x), called theantiderivative of f , then ∫ b

a

f(x) dx = F (b)− F (a). (4.1.16)

That is actually how we do most integrals. Sometimes, F (x) is obvious (e.g. the integral of 2xis x2). Sometimes, it takes a fair amount of ingenuity and guess-and-check work.

Therefore, I will not delve into the intricacies of the definitions of integrals. However, I willmention the basic idea since it generalizes to higher dimensions: to integrate f(x) over a region,say x ∈ [a, b] (this means a ≤ x ≤ b), you:

topsep=0mm,leftmirgin=* Chop up the interval [a, b] into a large number, N , of bits of equallength, (b− a)/N ;

topsep=0mm,leftmiirgiin=* Draw all the N rectangles with height from the x-axis to thevalue of f(x) at the left hand endpoint of each interval;

topsep=0mm,leftmiiirgiiin=* Add up the areas of all of these rectangles. Then, you must(conceptually, not physically) take the limit as N →∞.

4.1.1 Exercises

1. Show that (4.1.2) (linearity) and (4.1.3) (Liebniz) follow directly from the derivative definition(4.1.1).

2. Consider the logarithm function f(x) = lnx. It should be a familiar result that f ′(x) = 1/x.Derive this from the basic definition (4.1.1).

[Hint: use ln(xy) = lnx+ ln y, lnxy = y log x, ln e = 1 and the fact that you can interchangethe limits and logarithms.]

3. Using Exercise 2, prove that ex is its own derivative.


4. Define i to be some symbol that squares to −1. Obviously, i /∈ R. The complex numbers, C,are all expressions of the form z = x + iy where x, y ∈ R. We can extend our definitions ofderivatives with hardly any change to maps f : C → C. Euler’s formula is useful in dealingwith trigonometric functions: eix = cosx+ i sinx. Combine this with Exercise 3 to calculatethe derivatives of cosx and sinx.

5. If you are of the opinion that geometry is more fundamental than algebra, then Exercise 4 maybe unsatisfactory: why should we trust that Euler got his formula right? The alternative is togo through a lot of trigonometry. One must first prove that cos(x+y) = cosx cos y−sinx sin yand sin(x + y) = sinx cos y + cosx sin y (see Wikipedia article on trigonometric additionformulas, particularly around formula 41). Knowing this, show that

sin y − sinx = 2 cosy + x

2sin

y − x2

. (4.1.17)

Plug in y = x+ ε, divide by ε and take the ε→ 0 limit, which you should see is the derivativeof sinx. Finally, you need to show that 1

θ sin θ → 1 as θ → 0. There is a way to show thisby proving the inequality 0 < 1− 1

θ sin θ < 1− cos θ. It is actually easy to show, but requiresdrawing circles and triangles! Anyway, obviously, 1 − cos θ → 0 as θ → 0. Use this to finishoff the proof that d

dx sinx = cosx.

[Note: the limit of a product is the product of the limits. By the way, if we agree to takeEuler’s formula as fundamental, then proving all these trigonometric identities becomes almosttrivial. Try the addition formulas.]

4.2 Vectors

For our purposes, a vector will be something that has several components (usually three; orfour in relativity). We must be able to add two vectors (component by component) and wemust be able to multiply a vector by a real number. There are a slew of other requirements,but they are usually trivially satisfied, at least for the main vector space we will care about:R3, or Rn for general dimension. The symbol Rn means the set of all n-component expressions,A = (A1, A2, . . . , An), such that each component is a real number (i.e. Ai ∈ R for i = 1, . . . , n.)

In three-dimensional space, let us replace x, y, z with x1, x2, x3 in order to make it easier togeneralize to any dimension. We often denote the unit vector in the direction of xi by xi, whosecomponents are all zero except for the ith one, which is a one. Then, A may be written, in ageneral dimension, n,

A =

n∑i=1

Aixi. (4.2.1)

So that we don’t have to keep writing summation signs everywhere, we will usually follow Ein-stein’s convention that repeated indices are summed over, unless stated otherwise. Then, (4.2.1)becomes neater:

A = Aixi. (4.2.2)

The dimensionality is nowhere to be found now, so you must make sure you know what it isfrom context.

Next, we introduce the Kronecker delta symbol:

δij =

1 if i = j,

0 if i 6= j.(4.2.3)

Rn has what’s called an inner product structure. This is a map (·, ·) : Rn×Rn → R. That is, youtake two vectors, put one in the first slot and the other in the second slot of (·, ·), and you will

4.2. VECTORS 21

get a real number, called their inner product. It is also often called their dot product, especiallyin three dimensions, and we will often denote it by A ·B. It is defined by

A ·B = δijAiBj = AiBi, (4.2.4)

where you need to keep in mind that repeated indices are summed over.In addition, R3 has a special structure called a cross product. This is given by a map × :

Rn × Rn → Rn, so it takes two vectors and spits out another vector.2 In order to talk aboutcross products, we must introduce the Levi-Civita symbol, and to do that, we must understandcyclic indices. My housemate has a wonderful mnemonic for this. Think of a clock that goesfrom 1 through 3 instead of 1 through 12. Starting at any of the numbers, if you traverse theclock in a clockwise fashion, then the order is declared to be cyclic. If you traverse the clockin the counter-clockwise direction, then the order is anti-cyclic. So, (123), (231) and (312) arecyclic, whereas (132), (213), (321) are anti-cyclic. By the way, and for example, (231) is thepermutation that sends 1 (the first slot) to 2 (the first number appearing), 2 to 3 and 3 to 1.

The Levi-Civita symbol is defined to be

εijk =

1 if (ijk) is cyclic,

−1 if (ijk) is anti-cyclic,

0 if any index is repeated.

(4.2.5)

Roughly speaking, the Levi-Civita symbol is to the cross product what the Kronecker delta isto the dot product:

A×B = εijkxiAjBk. (4.2.6)

The following identity is quite important:

εijkεi`m = δj`δkm − δjmδk`. (4.2.7)

This will allow you to deal with situations involving multiple cross products. In fact, this canbe extended to any dimension. In n+ 1 dimensions, we write

εii1···inεij1···jn =

∣∣∣∣∣∣∣∣∣δi1j1 δi1j2 · · · δi1jnδi2j1 δi2j2 · · · δi2jn

......

. . ....

δinj1 δinj2 · · · δinjn

∣∣∣∣∣∣∣∣∣ , (4.2.8)

where the vertical lines surrounding the matrix means “take the determinant”.A number of very important vector identities can be proven using these formulas. You are

asked to prove one in Exercise 2 and Exercise 1.3.1.Now, we need to discuss how to change coordinate systems. We have expressed everything in

fixed Cartesian coordinates because it is simplest to do so in general. However, if you’re workingon a problem that has spherical symmetry, you should probably use spherical coordinates; oruse cylindrical coordinates for cylindrical symmetry. There are many more coordinate systemsthat are useful in particular situations, but these are the only ones we will study.

Let me do a simple two dimensional case to illustrate the point. Suppose you want to computethe area of a circle of radius R on the xy-plane. You chop up the circle into a huge number oflittle rectangles each width dx and length dy, and so each with area da = dx dy. You must nowintegrate this over the circle. The trouble is in the limits of integration. The integration region iny depends on the value of x: if x = 0, then y ∈ [−1, 1]R, but if x = 3R/5, then y ∈ [−1, 1]4R/5.The integral is actually

A =

∫ R

−Rdx

∫ √R2−x2

−√R2−x2

dy = 2

∫ R

−Rdx√R2 − x2 = πR2, (4.2.9)

2Actually, the result of a cross product is what’s called a pseudovector, but no matter.


where this integral is rather difficult to compute.It would be much easier if we used circular coordinates, where r is the radial distance from the

center of the circle and φ is the angle from the positive x-axis in the counter-clockwise direction.In these coordinates, the natural bit of area is the “end of a pie slice” from radius r to r + drand subtending the angle between φ and φ + dφ. Working in radians, the area of the pie sliceup to radius r is a(r) = dφ

2ππr2 = 1

2r2 dφ. Up to r + dr, the area is a(r + dr) = dφ

2ππ(r + dr)2 =12 (r2 +2r dr)dφ, where we drop terms higher than linear order in each differential. The differenceis the bit of area: da = a(r + dr) − a(r) = r dr dφ. Now, the limits are easy: r ∈ [0, R] andφ ∈ [0, 2π):

A =

∫ R

0

r dr

∫ 2π

0

dφ = πR2. (4.2.10)

So, in two dimensions, we can have a line element (the length of a curve) or an area element.Each is given by

Element Cartesian Circular

Line (ds) dx x + dy y dr r + r dφ φArea (da) dx dy r dr dφ

In three dimensions, we have spherical and cylindrical coordinates instead of circular ones. Inaddition to line and area elements, we can have volume elements. I will not bother to write theCartesian case. The other two are given by

Element Spherical Cylindrical

Line (ds) dr r + r dθ θ + r sin θ dφ φ drr + r dφ φ+ dz zArea (da) r2 sin θ dθ dφ r r dz dφ r

Volume (dv) r2 sin θ dr dθ dφ r dr dφ dz

Note that in three dimensions, both the line and area elements are vectors. Also, the areaelement in spherical coordinates is specifically for a bit of area lying on the surface of a sphere(constant r), and similarly for cylindrical coordinates.

4.2.1 Exercises

1. Derive (4.2.7) from (4.2.8) in the case n = 2.

2. Use (4.2.7) to prove the “BAC-CAB” rule:

A× (B×C) = B(A ·C)−C(A ·B). (4.2.11)

3. Calculate the surface area and volume of a sphere and a cylinder using (4.2).

4.3 Basic Vector Calculus

The multivariable version of a single derivative is the “del” operator. It is a vector operator, ora vector of operators:

∇ ≡ xi∂i ≡ xi∂

∂xi, (4.3.1)

where the symbol ∂/∂xi means take the derivative only with respect to xi while keeping all othervariables fixed. Note that I have taken care to write the vector xi to the left of ∂i to rememberthat it is not being differentiated. It makes no difference in Cartesian coordinates, since thoseare fixed, but it does make a difference in, for example, spherical coordinates which change frompoint to point.

4.3. BASIC VECTOR CALCULUS 23

If you act ∇ on a pure function (scalar function), then it is called the gradient, which is avector function. If you dot product ∇ with a vector function, f(x) = fi(x) xi, where each fi(x)is a scalar function, then you get the divergence, which is a scalar function. Finally, if you crossproduct ∇ with a vector function, then you get the curl, which is a (pseudo) vector function.

Next, we discuss how to write ∇ in different coordinates. Suppose we change from theCartesian coordinates (x, y, z) into a general set of coordinates (u, v, w) (for example, u = r,v = θ and w = φ for spherical coordinates). Then, we define the coordinate multipliers as

ζu =

√(∂x

∂u

)2

+

(∂y

∂u

)2

+

(∂z

∂u

)2

, (4.3.2)

and similarly define ζv and ζw. Then, the general del is

∇ =u

ζu

∂

∂u+

v

ζv

∂

∂v+

w

ζw

∂

∂w. (4.3.3)

Note that for Cartesian coordinates, ζx = ζy = ζz = 1 and (4.3.3) reproduces the usual Cartesiandel.

As practice, consider spherical coordinates x = r sin θ cosφ, y = r sin θ sinφ and z = r cos θ.Then,

ζr =√

(sin θ cosφ)2 + (sin θ sinφ)2 + (cos θ)2 = 1, (4.3.4)

ζθ =√

(r cos θ cosφ)2 + (r cos θ sinφ)2 + (−r sin θ)2 = r, (4.3.5)

ζφ =√

(−r sin θ sinφ)2 + (r sin θ cosφ)2 = r sin θ. (4.3.6)

Therefore, the spherical del is

∇ = r∂

∂r+θ

r

∂

∂θ+

φ

r sin θ

∂

∂φ. (4.3.7)

There are a few exercises at the end to help you become facile with the del operator. The firstchapter of Griffiths also has a wealth of relevant problems.

The formula for ∇ alone is not sufficient to give us the divergence and curl in arbitrarycurvilinear coordinates (except for Cartesian coordinates). Thus, I will just summarize below thegradient, divergence and curl in arbitrary coordinates, (u, v, w) with corresponding coordinatemultipliers, ζu, ζv and ζw:

∇f =u

ζu

∂f

∂u+

v

ζv

∂f

∂v+

w

ζw

∂f

∂w,

∇ ·A =1

ζuζvζw

[∂

∂u(Auζvζw) +

∂

∂v(ζuAvζw) +

∂

∂w(ζuζvAw)

],

∇×A =u

ζvζw

[∂

∂v(ζwAw)− ∂

∂w(ζvAv)

]+

v

ζuζw

[∂

∂w(ζuAu)− ∂

∂u(ζwAw)

]+

w

ζuζv

[∂

∂u(ζvAv)−

∂

∂v(ζuAu)

],

∇2f =1

ζuζvζw

[∂

∂u

(ζvζwζu

∂f

∂u

)+

∂

∂v

(ζuζwζv

∂f

∂v

)+

∂

∂w

(ζuζvζw

∂f

∂w

)],

(4.3.8)

where ∇2f is called the Laplacian of f and is defined as the divergence of the gradient:

∇2f ≡ ∇ · (∇f). (4.3.9)


Now, we discuss some of the important integral theorems. The Fundamental Theorem forGradients is the generalization of the Fundamental Theorem of Calculus. Recall that the lattersays that ∫ b

a

df

dxdx = f(b)− f(a). (4.3.10)

The generalization of the dfdx is ∇f , which is a vector. The generalization of the 1D integral is

what is called a path integral. Choose some (sufficiently smooth) path, P, that takes you frontpoint a to point b (each now has several coordinates). Chop up the path into infinitely manyinfinetisimal bits of length d` and define the differential vector, d`, to have length d` and bedirected locally tangentially to your chosen path in the direction going from a to b. The thingthat generalizes df

dx is (∇f) · d` (a dot product between two vectors). Then, the FundamentalTheorem for Gradients reads ∫ b

a

(∇f) · d` = f(b)− f(a) . (4.3.11)

We have deliberately not written out the path, P, in the above equation. It turns out that forsufficiently tame spaces, the result of the path integral is independent of the path (c.f. Exercise3). Thus, in a practical situation, one would choose a path that makes the integral as simple aspossible.

Next, is the Fundamental Theorem for Curls. This can be viewed as a generalization ofthe Fundamental Theorem for Gradients as follows: consider the points b and a as the boundaryof the path P. The boundary is often denoted ∂P. Then, the path integral of ∇f is equal tothe sum of the values of f at the boundary. The minus sign for f(a) simply indicates that ais the starting point of the path. Now, the curl is a vector function, which we may want tointegrate over some two-dimensional surface, S, as a measure of how much that vector functionpierces that surface (i.e. the flux). To do so, we must defined a bit of area on the surface: da,and the vector version, da, whose magnitude is da and whose direction is locally perpendicularto the surface and in the direction in which we are calculating the flux. Let P = ∂S be theboundary of that surface traversed in a right-handed sense relative to the direction in which weare calculating the flux. Then, the Fundamental Theorem for Curls reads∫

S(∇×A) · da =

∮P

A · d` . (4.3.12)

Finally, there is the Fundamental Theorem for Divergences. The divergence is a scalarfunction, which we may want to integrate over some three-dimensional volume, V. Let S = ∂Vbe the boundary surface. Then, ∫

V(∇ ·A) dτ =

∮S

A · da . (4.3.13)

Actually, all of these are particular cases of a more general mathematical statement called Stoke’sTheorem. The proof of this theorem is actually fairly simple when stated in the language ofdifferential geometry, but too tedious and requires too much legerdemain for my liking whenproved in “more basic” terms (see Purcell’s proof of Gauss’ Theorem, which follows from thedivergence theorem and the fact that ∇ ·E = 4πρ, one of Maxwell’s equations; c.f. Exercise 4).

4.3.1 Exercises

1. Use (4.2.7) to prove the follow, which gives us the wave equation from Maxwell’s equationsin vacuum:

∇× (∇×A) = ∇(∇ ·A)−∇ · ∇A. (4.3.14)

4.3. BASIC VECTOR CALCULUS 25

2. Determine the del operator in cylindrical coordinates.

3. Prove the following statement: The path integral∫ b

a(∇f) · d` is path-independent if and

only if the path integral of ∇f around any closed loop vanishes. The latter is often denoted∮(∇f) · d` = 0.

[Note: you will have to assume that f is single-valued ; that is, that the value that f takesat a point is the same as the value it takes after I go around any arbitrary loop and comeback to the same initial point. This is not as self-evident an assumption as it may seem. Forexample, the angle function on a circle, which assigns an angle in radians to a point on thecircle is infinitely multi-valued : 0, ±2π, ±4π, etc. are all assigned to the same point!]

4. Starting from Maxwell’s equation, ∇ ·E = 4πρ, and the Divergence Theorem, Eqn. (4.3.13),derive Gauss’ Law:

∫E · da = 4πQenc.

5. (Griffiths 1.12 ) The height of a certain hill (in feet) is given by

h(x, y) = 10(2xy − 3x2 − 4y2 − 18x+ 28y + 12), (4.3.15)

where y is the distance (in miles) north, x the distance east of LeConte Hall.

(a) Where is the top of the hill located?

(b) How high is the hill?

(c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile east ofLeConte Hall? In what direction is the slope steepest at that point?

6. (Griffiths 1.16 ) Sketch the vector function v = rr2 and compute its divergence. The answer

may surprise you... can you explain it?

7. Calculate the curl of the vector functions below. Which ones are possible static electric fieldsand what what would be the corresponding charge density producing that field?

(a) A = x2 x + 2xz2 y− 2xz z,

(b) B = xy x + 2yz y + 2zx z,

(c) C = y2 x + (2xy + z2) y + 2yz z.

8. Calculate the line integral of the function v = y2 x + 2x(y + 1) y from the point a = (1, 1, 0)to the point b = (2, 2, 0) along two paths (a) first go horizontally from a to c = (2, 1, 0), thengo vertically to b; (b) go diagonally along the shortest path from a to b. What is

∮v · d` for

the loop that goes from a to b along path (a) first, and then back along path (b)? Does vrepresent a static electric field?

SOLUTION:

Along the horizontal part of path (a), d` = dx x and so v · d` = y2 dx. Along this path, y isalways 1, and so v · d` = dx. Integrating this from x = 1 to x = 2 gives 1.

Along the vertical part of path (a), d` = dy y and so v ·d` = 2x(y+ 1) dy. Along this path,x is always 2, and so v ·d` = 4(y+ 1) dy. Integrating this from y = 1 to y = 2 gives 10. Thus,the total line integral along path (a) is

∫(a)

v · d` = 11.

Along path (2), d` = dx x + dy y = dx(x + y) since dx = dy. We also have v = x2 x +2x(x + 1) y. Thus, v · d` = x2 dx + 2x(x + 1) dx = (3x2 + 2x) dx whose integral from x = 1to x = 2 is

∫(b)

v · d` = 10.


∮(a)−(b)

v ·d` =∫

(a)v ·d`−

∫(b)

v ·d` = 1. Since this is not zero, v cannot represent a static

electric field.

Alternatively, we could just calculate the curl:

∇× v =

(∂vz∂y− ∂vy

∂z

)x +

(∂vx∂z− ∂vz

∂x

)y +

(∂vy∂x− ∂vy

∂x

)z

= 0 x + 0 y +[2(y + 1)− 2y

]z

= 2 z. (4.3.16)

The fact that this is nonzero already tells us that this cannot be a static electric field. Inaddition, we can use Stoke’s theorem to calculate the loop integral. If you draw the path,it should be clear that (a) − (b) goes around in the counter-clockwise direction. Thus, theright hand rule says that da points in the +z direction: da = da z = dx dy z. Let A denoteboth the triangular region whose boundary is the path (a) − (b). The area of this region is∫Ada = 1

2 . Then, by Stoke’s theorem,∮(a)−(b)

v · d` =

∫A

(∇× v) · da =

∫A

(2 z) · da z = 2

∫A

da = 1. (4.3.17)

9. Calculate the surface integral of v = 2xz x + (x+ 2) y + y(z2 − 3) z over the six faces of thecube that extend equally in the x, y and z directions from 0 to 2. Do this in two ways: (a) di-rect calculation of the integral over each face separately; and (b) using the divergence theorem.

SOLUTION:

(i) x = 0 face: da = dy dz (−x) and v = 2y+y(z2−3) z, so v ·da = 0. Thus,∫x=0

v ·da = 0.

(ii) x = 2 face: da = dy dz x and v = 4z x + 4 y + y(z2 − 3) z, so v · da = 4z dy dz. Thus,∫x=2

v · da = 4∫ 2

0dy∫ 2

0z dz = 16.

(iii) y = 0 face: da = dx dz (−y) and v = 2xz x + (x+ 2)y, so v · da = −(x+ 2) dx dz. Thus,∫y=0

v · da = −∫ 2

0(x+ 2) dx

∫ 2

0dz = 12.

(iv) y = 2 face: da = dx dz y and v = 2xz x + (x+ 2)y + 2(z2− 3)z, so v ·da = (x+ 2) dx dz.This is precisely the negative of (iii), so

∫y=2

v · da = −12.

(v) z = 0 face: da = dx dy (−z) and v = (x + 2) y − 3y z, so v · da = 3y dx dy. Thus,∫z=0

v · da = 3∫ 2

0dx∫ 2

0y dy = 12.

(vi) z = 2 face: da = dx dy z and v = 4x x + (x + 2) y + y z, so v · da = y dx dy. Thus,∫z=2

v · da =∫ 2

0dx∫ 2

0y dy = 4.

Altogether,∮

v · da = 0 + 16 + 12− 12 + 12 + 4 = 32.

∇ · v = 2z(y + 1) and so the volume integral over the interior of the cube is 2∫ 2

0dx∫ 2

0(y +

1)dy∫ 2

0z dz = 32, as it should be by the divergence theorem.

Chapter 5Work and Energy

A charge, q, located at a point, r, where there is an electric potential of ϕ(r), has an associatedelectric potential energy given by U(r) = qϕ(r). This is exactly analogous to the Earth havinggravitational potential energy by virtue of its orbiting around the Sun, in which case the potentialenergy is the Newton’s gravitational potential due to the Sun multiplied by the mass of the Earth.

Recall that the potential, ϕ, is only well-defined up to a constant. The more physical entityis the electric field, which, when multiplied with the charge gives the physical electric force.Adding an arbitrary constant to the potential, ϕ, does not change the associated electric field,E = −∇ϕ, since the gradient of a constant vanishes. Usually, we set this arbitrary constant tobe whatever it has to be to ensure that (a) ϕ→ 0 as r →∞, and (b) ϕ is everywhere continuous.Condition (a) is the usual case, but is not always possible (e.g. the potential for an infinite linecharge). Condition (b) is always the case, even when E is discontinuous.

5.1 Energy of Charge Distributions

We can build up charge distributions one bit at a time and calculate how much work it wouldtake to do this. The result would be the potential energy of the charge system that it gains(or loses) by virtue of the fact that they are distributed in a particular manner. We start withdiscrete charge distributions: a finite number, N , of point charges, q1, . . . , qN , sprinkled aroundin space at positions r1, . . . , rN . Let us start with the charges at infinity and infinitely far apart.The potential energy of each in this initial state is 0. Let us bring in the first charge, q1, to theposition r1. Since there are no charges around yet, this does not cost us any work. That is, thepotential energy of the first charge after this first move is still 0.

Next, we bring in q2 from ∞ to r2. It interacts with q1. The potential energy that q2 hasafter this move is q2 multiplied by the potential produced by q1 at the position r2. Denote thisby U2 ≡ U12 ≡ q2ϕ12 = q1q2

|r2−r1| . Since the potential energy of q2 prior to moving it was 0, energy

conservation implies that this quantity, U12, is precisely the work I would have had to do tomove q2 from ∞ to r2.

Next, we bring in q3 from ∞ to r3. Now, it interacts with both q1 and q2. Its potentialenergy at the end is U3 ≡ U13 + U23 = q3(ϕ13 + ϕ23) = q1q3

|r3−r1| + q2q3|r3−r2| .

This pattern continues until we have moved all N charges. The total potential energy of thesystem is just the sum of all of the Ui’s for i = 1, . . . , N :

U =

N∑i=1

Ui =

N∑i=1

∑j<i

qiqj|ri − rj |

. (5.1.1)

27

28 CHAPTER 5. WORK AND ENERGY

We could include the terms j > i in the sum as well, but then we have to divide by 2 becausethe summand is symmetric under the interchange i↔ j:

U =1

2

N∑i,j=1i 6=j

qiqj|ri − rj |

. (5.1.2)

It’s important that i 6= j because a charge does not feel its own field (in the nonrelativisticlimit). In any case, a term with i = j would technically be ∞! We can write

U =1

2

N∑i=1

qi

N∑j=1j 6=i

qj|ri − rj |

=1

2

N∑i=1

qiϕi(ri), (5.1.3)

where ϕi(ri) is the potential at the position, ri, of charge qi due to all of the charges except qi.When the charge distribution becomes continuous, we can no longer say how much charge

is at a particular point (technically that’s zero). So, usually, it no longer becomes important toensure that we calculate the potential due to every charge except the charge at a particular pointsince the latter is negligible. Furthermore, the sum becomes an integral over the charge density:

U =1

2

∫ρϕ dv. (5.1.4)

It is important to note that this integral is over all space. It just so happens that ρ vanisheseverywhere outside the charge distribution and so, if you want, you can restrict the integral tojust include the region where there is charge.

Let’s quickly interpret Eqn. (5.1.4) as a set of instructions: (1) take a point r in space andconsider a little bit of volume, dv, located at this point; (2) multiply dv by ρ(r) to get dq; (3)multiply dq by ϕ(r) and integrate over all space and then stick in a factor of 1

2 . Usually, its step(3) that is most formidable. First of all, you must know ϕ(r), which is not always very easy tocalculate (c.f. Problem 2.12 in Purcell). Secondly, you have to calculate the integral, which isgenerically difficult.

Recall that Eqn. (5.1.4) can be put in the form

U =1

8π

∫E2dv, (5.1.5)

where E2 ≡ |E|2, by writing ρ = 14π∇ ·E and using the identity

∇ · (ϕE) = ϕ∇ ·E + E · ∇ϕ. (5.1.6)

By the way, this relation has nothing to do with E&M; it’s purely a mathematical relation: youcan replace ϕ with any scalar function and E with any vector function. However, in the specificcase of E&M, we may replace ∇ϕ on the right hand side with −E, which turns the last terminto −E2. The first term on the right is the one we want to integrate. The integral of the leftside over all space vanishes by the divergence theorem∫

∇ · (ϕE)dv =

∮ϕE · da = 0, (5.1.7)

since the surface integral is over the boundary of all space, which is at r →∞, which is preciselywhere ϕ and E vanish.

You might be confused about the two equations for U : one involving ρ and ϕ and the otherjust E. Both integrals are integrals over all space, but, as we mentioned earlier, we can restrictthe first formula to an integral just over the region containing charge since ρ = 0 everywhere

5.2. CLASSICAL ELECTRON RADIUS 29

else. Generically, E 6= 0, even in regions where there is no charge. So, the first equation seems toindicate that the energy is stored in the charges themselves whereas the second formula indicatesthat the energy is spread all throughout the regions of space where there is an electric field! Atthis level of sophistication, there is no real answer to this puzzle: both are simply methods forcalculating the energy and we need not concern ourselves with exactly where the energy is stored.This is in the same vein as David Mermin’s “shut up and calculate!” perspective. However, onceyou consider radiation, the more useful interpretation is that the energy is stored in the fields!

One final comment: since U is quadratic in E, it does not satisfy superposition. If one chargedistribution, ρ1, produces a field E1, and another, ρ2 produces a field E2, then the combinedelectric potential energy of the system is not simply the sum of the potential energies of eachseparate entity:

U =1

8π

∫|E1 + E2|2dv =

1

8π

∫ (E2

1 + E22 + 2E1 ·E2

)dv

= U1 + U2 +1

4π

∫E1 ·E2 dv. (5.1.8)

The cross terms screw up superposition! This relates to the section in Purcell that talks aboutthe energy of crystals. In this case, it is reasonable from everyday experience to say that theenergy of two cubes of salt sitting next to each other is twice the energy of one cube of salt alone.That is superposition and it would not be the case unless the cubes of salt had zero net charge(i.e. they were net neutral). Only in this case can we neglect the cross terms, which is to saythat the two cubes of salt do not interact electrically with each other.

5.2 Classical Electron Radius

As far as we can tell, electrons are point particles and thus have no spatial extent. But, what ifwe model the electron as a ball of some spherically symmetric charge distribution, with a totalcharge −e and radius re? What would be the work, U , required to “assemble an electron”? Aclassical calculation is done by assuming that that work must have been provided by all of therest mass energy of the electron, which is mec

2, by Einstein’s famous equation. Setting U equalto mec

2 allows us to solve for re in terms of known stuff. This called the classical electron radius.Calculate U and re in the following scenarios:

(a) The charge is distributed evenly only over the surface of the sphere.

(b) The charge is distributed evenly over the volume of the sphere.

(c) ρ(r) ∼ Cr for some constant C, for r ≤ re.

(d) ρ(r) ∼ Cr2 for some constant C, for r ≤ re.

SOLUTION:

We will solve this problem for the general form ρ(r) ∼ Crn for n ≥ 0 and an integer. Note that(b) is the case n = 0, (c) is n = 1 and (d) is n = 2. What might be harder to realize is that (a)is equivalent to the limit n→∞.

Now, Gauss’ law gives the electric field outside the sphere:

E(r) = − e

r2r for r > re. (5.2.1)


In spherical coordinates, the gradient of a function that depends only on r is given by −∇ϕ =−dϕdr r. So, E = −∇ϕ implies dϕ

dr = er2 and thus ϕ(r) = − er + C, where C is a constant, which we

set to 0 in order for ϕ(r)→ 0 as r →∞. Thus,

ϕ(r) = −er

for r > re. (5.2.2)

In order to determine the electric field inside the sphere, we will need to determine C, whichwe do by requiring that the total charge in the sphere of radius re is −e. We will have to do aspherical volume integral. The differential bit of volume in spherical coordinates is dv = r2dr dΩ,where dΩ = sin θ dθ dφ and

∫dΩ = 4π. In our case, the integrand only depends on r, not θ or

φ, and so we can just pull out the factor of 4π from the beginning:

− e = 4π

∫ re

0

ρ(r) r2 dr = 4πC

∫ re

0

rn+2 dr =4πCrn+3

e

n+ 3. (5.2.3)

Solving for C gives us the charge density

ρ(r) = − (n+ 3)ern

4πrn+3e

= − (n+ 3)e

4πr3e

(r

re

)n. (5.2.4)

Now, taking a sphere of radius r < re as our Gaussian surface, Gauss’ law reads

4πr2E(r) =

∮E · da = 4πqenc

= (4π)2

∫ r

0

ρ(r′) r′2 dr′

= −4π(n+ 3)e

rn+3e

∫ r

0

r′n+2 dr′

= −4π(n+ 3)e

∫ r/re

0

xn+2 dx

= −4πe

(r

re

)n+3

. (5.2.5)

Note that we changed variables to x ≡ r′/re in the penultimate line. All n + 3 powers of re inthe denominator are used up: n + 2 of them by r′n+2 and one by dr′. Solving for the electricfield yields

E(r) = E(r) r = − e

r2

(r

re

)n+3

r = − e

r2e

(r

re

)n+1

r for r < re. (5.2.6)

Again, E = −∇ϕ implies dϕdr = e

rn+3e

rn+1 and thus,

ϕ(r) =e

(n+ 2)re

(r

re

)n+2

+ C for r < re. (5.2.7)

Now, we will determine C not by requiring that ϕ(r) → 0 as r → ∞ since we are not workingin that region anymore, and, in any case, we have already ensured that. Now, we must ensurethat ϕ is continuous. Therefore, it must join correctly with ϕ(r) for r > re, which, at re, is justϕ(re) = −e/re. Setting Eqn. (5.2.7), evaluated at r = re, equal to −e/re and solving for C yields

C = − e

re

(1 +

1

n+ 2

)= −

(n+ 3

n+ 2

)e

re. (5.2.8)

5.2. CLASSICAL ELECTRON RADIUS 31

Therefore, the potential is

ϕ(r) = − e

(n+ 2)re

[n+ 3−

(r

re

)n+2]. (5.2.9)

Now, we can calculate the potential energy:

U =1

2

∫ρϕ dv =

1

2· 4π

∫ re

0

−(n+ 3)e

4πr3e

(r

re

)n −e(n+ 2)re

[n+ 3−

(r

re

)n+2]r2 dr

=(n+ 3)e2

2(n+ 2)re

∫ 1

0

xn+2(n+ 3− xn+2

)dx

=(n+ 3)e2

(2n+ 4)re

(xn+3 − x2n+5

2n+ 5

)∣∣∣∣10

(5.2.10)

The term in the big parentheses evaluated at the appropriate points is just 2n+42n+5 . The factors

of 2n+ 4 cancel from top and bottom leaving us with

U =

(n+ 3

2n+ 5

)e2

re. (5.2.11)

Now, let us calculate U using the electric field:

U =1

8π

∫E2 dv =

1

2

[∫ re

0

e2

r4e

(r

re

)2n+2

r2 dr +

∫ ∞re

e2

r4r2 dr

]=

e2

2re

[∫ 1

0

x2n+4 dx+

∫ ∞1

dx

x2

]=

e2

2re

(x2n+5

2n+ 5

∣∣∣∣10

− 1

x

∣∣∣∣∞1

)=

e2

2re

2n+ 6

2n+ 5

=

(n+ 3

2n+ 5

)e2

re. (5.2.12)

At this point we breathe a sigh of relief for this is identical to Eqn. (5.2.11).Setting this equal to mec

2 and solving for re yields

re =

(n+ 3

2n+ 5

)e2

mec2. (5.2.13)

The quantity e2/mec2 is actually the thing that is usually called the classical electron radius.

It has a value of ∼ 2.82× 10−15 m. The numerical prefactor, which depends on n, never reallydeviates too much from 1. It’s a maximum at n = 0 (constant volume charge density), whenit equals 3/5, and it’s a minimum at n = ∞ (constant surface charge density), when it equals1/2. We have probed electrons to smaller distance scales than that and have not observed anystructure. This clearly experimentally excludes these models!

Chapter 6Poisson’s Equation

A standard electrostatic problem is: given some charge distribution and appropriate boundaryconditions for the potential, calculate the potential throughout all of space. An example is:calculate the potential through all space due to a point charge (placed at the origin, say) withthe boundary condition that ϕ → 0 as r → 0. A less trivial example is to find ϕ everywherein space produced by a system of conductors each of which sits at some fixed value of ϕ. Theimportant thing is that the solution to this problem is unique. Thus, it doesn’t matter how youget the solution; you could just guess it somehow. As long as it is a solution, it is the solution.

6.1 Conductors

We will only consider perfect conductors, in which the conduction electrons are completely freeto move within the confines of the object. For the time being, we will assume that the conductorand whatever charges are around it or embedded in it, are in steady state. This means thatthere is no net motion of charge locally anywhere (e.g. no currents in or on the conductor). Inthis case, there are three important things to remember about conductors:

1. Within the material of the conductor, the electric field vanishes.

- This is true no matter what charges lie around or are even embedded in the conductor.

- If, on the contrary, there were a nonzero electric field within the conducting material, thenthe free charges would feel forces and would move about, thus violating the steady stateassumption.

2. Any net charge on a conductor must reside on the surfaces of the conductor (this possiblyincludes internal surfaces if there are cavities in the material).

topsep=0mm, libel=-, leftmirgin=* If there were a charge in the bulk (not the sur-face) of the conductor, then we could consider asufficiently small Gaussian sphere (lying entirelywithin the material of the conductor) surround-ing this charge. Gauss’ law implies that there is anonzero flux through this surface. However, sinceE = 0 everywhere in the material of the conductor,∫

E · da across this Gaussian sphere must vanish.CONTRADICTION!

33

34 CHAPTER 6. POISSON’S EQUATION

3. The electric field at the surfaces of a conductor is always perpendicular to the conductingsurface.

topsep=0mm, libel=-, leftmirgin=* If it were not, then the charges on the surface wouldfeel a force along the surface and would start tomove along the surface, thus violating the steadystate assumption.

Now, consider the following problem: below is an irregularly shaped conductor with irregularlyshaped cavities each containing a point charge. The conducting material itself has a net charge of−5C. In equilibrium, find the total charge residing on each of the four surfaces of the conductor.

Figure 6.1: Conductor with cavities containing point charges.

SOLUTION:

Surround cavity i = 1, 2, 3 with a Gaussian surface, Si, that encloses both the point chargecontained in the cavity, Qi, as well as all of the surface of the cavity, and hence the charge onthat surface, Qsi . But, choose Si to always stay within the conducting material itself (i.e. in thegrey region). Since Si is in the grey region and E = 0 there, 0 = 1

4π

∮Si

E · da = qenc = Qi +Qsi .Thus, Qsi = −Qi and so the charge on the surface surround the −1C charge is +1C, the chargeon the surface surrounding the −3C charge is +3C, and the charge on the surface surroundingthe +2C charge is −2C. Therefore, the total charge on the inner surfaces is 1+3−2 = 2C. Thereis a total of −5C in the grey region and thus −7C must reside on the outermost surface. Sincethe surfaces are all irregular and the charges are not placed in the center of the cavities (well,they don’t really have centers since they’re irregular), these charges are not evenly distributed,but we at least now the totals.

6.2 Image Charges

As stated previously, it doesn’t matter how you find a solution to an appropriately constructedelectrostatic problem, as long as you find it, you’re done, since the answer is unique. The methodof image charges is just one way of taking andantage of this.

The prime example is the problem of finding the electric field (and/or potential) everywherein space produced by a fixed point charge, Q, sitting still at some fixed distance, h, above aninfinite perfectly conducting plane. Let the plane lie at z = 0.

Now, let me introduce the guidelines for playing with image charges:

1. You must never put an image charge into the region where you are computing the potential;and

6.2. IMAGE CHARGES 35

2. The image charges must add up to the correct total in each region.

You can get quite a lot of mileage from just these two rules. For example, let us calculate thepotential in the z > 0 region. Then, we are not allowed to place any image charges in this region;we can only place image charges in the z ≤ 0 region. The latter region includes the plane. Youknow that some charge will build up on the plane. In fact, you know that the total chargeinduced on the plane is −Q (why?) Thus, the total charge of the image charges that we place inthe z ≤ 0 region must be −Q. The simplest idea would be to just have one image charge of −Q.If we place this directly below the real Q charge and at the same distance, h, from the plane,then we get the wonderful result that every point on the plane sits at ϕ = 0. This is because thepoints on the plane are equidistant to the real and image charges and thus ϕ = Q

r + −Qr = 0.

This is exactly what we want because ϕ must vanish on the plane since it extends to infinity,where ϕ is assumed to vanish. Hence, the potential produced in the z > 0 region by Q at z = hand −Q at z = −h satisfies Poisson’s equation, ∇2ϕ = −4πρ, in the z > 0 region and has thecorrect value on the boundary of this region, namely it vanishes on the plane and at ∞. Byuniqueness, this is the solution!

What is the potential in the region z < 0? Well, we can only place image charges now in thez ≥ 0 region. The total charge there is 0 since there is Q at z = h and −Q on the plane itself.It follows that the potential vanishes in the z < 0 region (actually, it could just be constant,but it must paste properly with ϕ in the z > 0 region, which vanishes on the plane, so it mustbe zero). Actually, ϕ = 0 in the z < 0 region can be seen by the fact that there are no chargesthere, and so ∇2ϕ = 0, and ϕ must vanish on the plane and at ∞. Clearly, ϕ = 0 satisfies theseconditions. Uniqueness takes care of the rest.

Another neat example of a problem that can be solved using image charges is that of a pointcharge, q, that sits some distance, a, from the center of a grounded (ϕ = 0) conducting sphereof radius R < a. To calculate the potential everywhere outside the sphere, place an unknownimage charge, q′, a distance b < R from the center of the sphere and between the center of thesphere and q. Determine what q′ and b have to be in order to satisfy ϕ = 0 on the surface of thesphere. If you find it too difficult to ensure that the ϕ = 0 everywhere on the sphere, just ensurethat ϕ = 0 at the two points on the sphere where the sphere intersects the line connecting thecenter, q′ and q. Anyway, this is sufficient to determine q′ and b. Calculate the potential energyof this configuration.

SOLUTION:

The potential at the point on the sphere between q and q′ is ϕ = qa−R + q′

R−b = 0. The potential

on the point on the other point on the sphere mentioned in the problem is ϕ = qa+R + q′

R+b = 0.These are two equations and we have two unknowns: q′ and b. Solving for them yields

q′ = −Raq, b =

R2

a. (6.2.1)

Note that, unlike the case of the plane, we did not know beforehand what the image chargeought to be. But, we can run the argument backwards now and conclude that the total inducedcharge on the sphere must be q′ = −Ra q.

Let q be at a variable distance, r > R, from the center of the sphere. The electric force on qis just the one due to the image charge, which is

F = qE′ = q

( −Rr q(r − R2

r

)2 r

)= − q2Rr

(r2 −R2)2r. (6.2.2)

Let us calculate the work it would take to move the charge from r = a to r =∞ along the directradial path at a constant speed. The differential bit of path is d` = dr r. Since the speed is

36 CHAPTER 6. POISSON’S EQUATION

constant, the force on the charge must vanish. Hence, the force which I impart on it must beequal and opposite to the electric force exerted on it by the image charge. Thus, the differentialbit of work that I do is

dW = Fme · d` =q2Rr dr

(r2 −R2)2. (6.2.3)

Hence, the total work I do is

W = q2R

∫ ∞a

r dr

(r2 −R2)2= −q

2R

2

1

r2 −R2

∣∣∣∣∞a

=q2R

2(a2 −R2). (6.2.4)

The work I would have done to do this in reverse (i.e. bring the charge in from ∞) is justnegative of this. That result is the potential energy of this configuration. Hence,

U = − q2R

2(a2 −R2). (6.2.5)

Chapter 7Capacitors

Capacitors can be very confusing if you don’t have keep your wits about you. I think the bestway to deal with them is to work on problems. I have based the problem below on Purcell3.23 p.117. But, before we look at the problem, let us isolate some important concepts aboutcapacitors here.

We usually deal with the mutual capacitance between two pieces of conducting material (e.g.two parallel plates, two concentric spheres, etc.) Of course, an object has a self capacitance (c.f.your homework problem about the charged prolate spheroid: Purcell 3.20 p.116). But, moreoften, it is the mutual capacitance that we care about and we usually drop “mutual”.

The capacitance of two pieces of conductor can be calculated by placing a charge +Q onone and −Q on the other, calculating the resulting potential difference, V ≡ ∆ϕ, and definingC ≡ Q/V . Note that ∆ϕ is defined as ϕ+ − ϕ−, where ϕ+ is the potential at the conductorwhereupon we have placed the positive charge +Q and ϕ− is for the negative conductor. Often,one uses Gauss’ law to first calculate E, and then derives ϕ from that. The final result forC should never depend on Q. This means that Q is fiducial : we placed it there just for thepurpose of calculation, but the final answer is independent of Q. In fact, C is a purely geometricquantity: it changes only if you change the shape and/or dimensions of the conductors. Indeed,as you know, the capacitance of a parallel plate capacitor with plates of area A and separationdistance s is just C = A/4πs, which is purely geometric and only changes if we change the areaor the separation between the planes.

Hopefully, the discussion thus far has not been too confusing. You will put the above proce-dure to use in the problem below for concentric cylinders. I think that the real confusion ariseswhen we calculating forces on a conductor in the intermediate region of a capacitor, the classicalexample being a plane sliding between the planes of a parallel plate capacitor. Note that thecapacitor could just be a self capacitor, namely just one piece of conductor (e.g. in the problembelow, the inner cylinder of a coaxial cylindrical capacitor is slid out). The procedure is: (1)calculate the electrostatic potential energy, U , as a function of some penetration distance, then(2) F = −∇U . The latter relation is actually the general relation between a conservative forceand the associated potential energy. So, it holds for gravity as well, but not for friction, whichis non-conservative.

One complication is that we can either charge up the capacitor via a battery, then disconnectthe battery, then insert the other conductor; or, we can keep the battery connected the wholetime. The first case keeps Q constant whereas the second case keeps V constant. For the firstscenario, we should probably use U = Q2/2C, whereas we should use U = CV 2/2 for the second.The first scenario tends to be the simpler one because in the second case, the battery does thework and part of that work is spent on changing Q. The difference between these two methodsis explored in the problem.

37

38 CHAPTER 7. CAPACITORS

One other source of confusion, which will undoubtedly remain, is the fact that we usuallyignore complicated edge effects to calculate U , and thus F. Nevertheless, we recognize that thephysical source of the force is the very edge effects that we neglected! Believe it or not, there isa similarity between this issue and one we faced earlier: the two different methods of calculatingthe work done in forming a charge distribution. On the one hand, 1

2

∫ρϕ dv indicates that the

energy is stored locally in the charge. On the other hand, 18π

∫E2 dv indicates that the energy

is sotred globally in the field! Likewise, force is a local concept. This is especially clear for aconservative force, which is written as F = −∇U . Derivatives have to do with how functionsbehave locally. Yet, we are calculating U globally, in the electric field!

Strictly speaking, when we calculate 12

∫ρϕ dv and 1

8π

∫E2 dv, we are not calculating the

exact same physical thing since the former lies in the charge whereas the latter lies in thefield. Nevertheless, they are numerically identical. A similar concept pertains to the force in acapacitor.

7.1 Force on Cylindrical Capacitor

A cylinder of 1.00-inch outer diameter hangs, with its axis vertical, from one arm of a beambalance. It is completely surrounded by a stationary cylinder, coaxial, with inner diameter 2.72inches. A battery is connected between the inner and outer cylinders with a voltage differenceof 6 kilovolts.

(a) Find the capacitance of coaxial cylinders of radii a and b (with a < b), and length L. Assumethat L >> b − a so that end effects may be neglected. Check your result by showing thatyou get the correct limit when the gap between the cylinders is very small compared withthe radii.

(b) Calculate the magnitude and direction of the electric force experienced by the inner cylinderif you try to lift the inner cylinder out. Do this in the following two scenarios:

(i) The battery is never disconnected; and

(ii) The battery is disconnected before you pull up.

SOLUTION:

(a) Place +Q on the inner cylinder and −Q on the outer cylinder. We will compute the electricfield, E, between the cylinders. Symmetry tells us that E just points radially outwardfrom the central axis (i.e. cylindrically radially outward, not spherically radially outward).Consider a Gaussian cylinder of length L and radius r such that a < r < b. It encloses allof the charge, Q, on the inner cylinder. Then,

2πrLE(r) =

∮E · da = 4πQenc = 4πQ. (7.1.1)

Therefore, the electric field is

E =2Q

L

r

r. (7.1.2)

In cylindrical coordinates, the gradient of a function, f , that depends only on r and not onz or the azimuthal angle, φ, is given by ∇f = df

dr r. Therefore,

− dϕ

drr = −∇ϕ = E =

2Q

L

r

r=⇒ ϕ(r) = −2Q

Lln r + C, (7.1.3)

7.1. FORCE ON CYLINDRICAL CAPACITOR 39

where, as usual, C is just some constant.

Note that there is no way of making ϕ → 0 as r → ∞, unless we allow C to be infinite!But the value of C is irrelevant here since we only care about the potential difference:

V = ∆ϕ = ϕ+ − ϕ− = ϕ(a)− ϕ(b) =2Q

Llnb

a. (7.1.4)

The capacitance is

C =Q

V=

L

2 ln(b/a). (7.1.5)

(b) Let z be the unit vector pointing up. Hopefully, you remember that the electric force tendsto suck the inner cylinder in, and thus points in the −z direction. If you pull out the innercylinder a distance ∆L, you have effectively decreased the length of the coaxial cylindercapacitor to L−∆L. The capacitance changes to C + ∆C = L−∆L

2 ln(b/a) and so

∆C = − ∆L

2 ln(b/a). (7.1.6)

(i) Here is the common mistake people make. The electrostatic energy stored in thecapacitor is U = 1

2CV2. Since the battery is never disconnected, V remains constant.

Since Eqn. (7.1.5) says that C ∝ L, if we pull the inner cylinder out a bit, therebydecreasing L, we also decrease C and hence, U . But, systems tend to want to decreaseU , so without further help from me, the inner cylinder should keep sliding outward.This is like nudging a ball off the top of a hill since it has a lower gravitational potentialenergy at the bottom than at the top. This indicates that the electric force is repulsive,rather than attractive, which is what the answer should be!

More mathematically, we have

F = −∇U = −∆U

∆Lz = −V

2

2

∆C

∆Lz =

V 2

4 ln(b/a)z, (7.1.7)

which points upwards, when we know it should point downwards!

The mistake is that we have neglected the work done by the battery. That is, inpulling the inner cylinder out, I have to work against the battery. Since V remainsthe same, but C changes, Q must change. It changes by ∆Q = (∆C)V . The workdone by the battery in moving that charge is Ubat = (∆Q)V = (∆C)V 2. Half of this,12 (∆C)V 2 is spent decreasing the electric field between the cylinders, countering theeffect of Eqn. (7.1.7). The remaining half provides the force against which you needto pull up. It should be clear that this amounts to simply changing the sign of Eqn.(7.1.7)!

First, converting V = 6000 volts into statvolts gives V ≈ 20 statvolts. Second,b/a = 2.72/1.00 = 2.72 ≈ e, where e is Euler’s number, whose logarithm is ln e = 1.Thus,

F = − V 2

4 ln(b/a)z = − (20 statvolt)2

4 ln(2.72/1.00)z ≈ 100 dynes down . (7.1.8)

(ii) Now, the charge remains constant, but V changes. So, we should use U = Q2/2C. Iwill rewrite Eqn. (7.1.6) as

∆C

∆L= −C

L. (7.1.9)

40 CHAPTER 7. CAPACITORS

Then, the force is

F = −∆U

∆Lz = −Q

2

2

(− 1

C2

∆C

∆L

)z = − Q2

2CLz. (7.1.10)

The charge is whatever it was when the battery was still connected:

Q = CV =LV

2 ln(b/a). (7.1.11)

Plugging this into Eqn. (7.1.10) yields

F = − V 2

4 ln(b/a)z , (7.1.12)

as in Eqn. (7.1.8).

Chapter 8Current

In electrostatic equilibrium, we argued that the electric field must vanish within the material ofa conductor. This no longer holds out of equilibrium or when the equilibrium is not static. Anexample of the latter is when there are steady currents in the conductor (non-steady currentswould be out of equilibrium altogether). Since E may no longer vanish in the material, thematerial will no longer be an equipotential. This may seem alarming at first since we said thata perfect conductor ought to be an equipotential so that it cost absolutely no energy to movea free charge between any two arbitrary points, there being no resistance to charge movement.However, one should keep in mind that the potential difference between points in the conductorin the dynamic (non-static) equilibrium case is externally applied (e.g. by attaching a batteryacross the object). So, for example, if we think of the particles in the air as if they neverinteract at all, then they just move about freely according to their thermal velocities and wouldbe evenly distributed everywhere. However, in the presence of an external potential, which inthis case is gravity, the particles become denser near the earth’s surface. Of course, in a circuit,usually charges don’t bunch up in some region, but this is only because they have usually havean open pathway with constant conductivity. The air particles, on the other hand, are stuckabove ground. In any case, we will consider a situation in the problem below where the chargedensity is not constant.

The main starting point, or culminating point (depending on your perspective), of this areaof study is Ohm’s law. This is not properly a law; in particular, in the grand scheme of things,a generic piece of material will not follow Ohm’s law. However, plenty of important materialsdo, so the effort in learning it is not wasted. Ohm’s law relates current density to electric field:J = σE, where σ is the conductivity. Remember that I is related to J via a surface integral overa cross-sectional area perpendicular to the current flow: I =

∫J · da. In this treatment, σ is just

a positive number (with units of 1/sec), and thus J is parallel to E. In general, many materialsare anisotropic σ actually becomes a 3× 3 matrix.

If you consider a cylindrical wire with conductivity σ with metallic planar endcaps that sitat different voltages, say V and 0, you can use Ohm’s law, J = σE, to derive V = IR, whereyou discover that R = `/σA = ρ`/A, where ρ = 1/σ is the resistivity, ` is the length of the wireand A is the cross-sectional area of the wire. But, the relation V = IR doesn’t always hold;in particular, it will generically not hold in a substance whose charge density is not constant intime, or whose current density is not divergenceless, since there is no constant current of whichto speak, so what on earth would V = IR even mean?

But, why does it even work so well in the case of the wire? Have you stopped a second tothink about this? We essentially have a plate on one end and another on the other end and weare saying that E is constant and runs parallel to the length of the wire. Essentially it looks likethe electric field between infinite parallel planes. But, the planes are very far apart compared

41

42 CHAPTER 8. CURRENT

to their size: ` >>√A!!! Usually, for parallel planes, we have to assume ` <<

√A in order to

assume infinite planes.The ultimate resolution, has to do (gasp!) with uniqueness. In the region of the wire, the

potential satisfies Laplace’s equation, ∇2ϕ = 0. This is assuming that the wire is neutral, ofcourse. Two boundary conditions are that ϕ = V on one end and ϕ = 0 on the other. But whatabout the surface along the length of the wire? There, we require that no current leak out. Inother words, J cannot have any component perpendicular to the surface of the wire along itslength: J · n = 0, where n is the normal vector perpendicular to the surface of the wire alongits length. Since J = σE, this implies E · n = 0 or ∂ϕ

∂n ≡ n · ∇ϕ = 0. The former boundaryconditions, indicating the value of ϕ at the endpoints is what is generally called a Dirichletboundary condition. This latter type, indicating the value of the derivative of ϕ is called aNeumann boundary condition. Clearly, the following function works: ϕ(x) = V0

(1 − x

`

),

with x being the coordinate along the length of the wire starting from x = 0 where ϕ = V0 tox = ` where ϕ = 0. Since the solution is unique, this is the solution. But this is precisely thesame potential as in the case of infinite parallel planes!

8.1 Resistance of Concentric Spheres

Consider two concentric metal spheres, the inner one of radius a and the outer one of radius b.By some method, that you need not worry about, the inner sphere is kept at potential V0 andthe outer one at 0. The region between the two spheres is filled with homogeneous isotropicmaterial of conductivity σ. Calculate the resistance of this system.

SOLUTION:

By Gauss’ law, we know that E = 0 for r > b and r < a. This means that ϕ is constant in thoseregions. Since ϕ = 0 at r = b is constant for r > b and is everywhere continuous, it follows thatϕ = 0 for r ≥ b. Similarly, ϕ = V0 for r ≤ a.

Gauss’ law also tells us that the field between the spheres goes like 1/r2. Therefore, ϕ goeslike 1/r. So, we need a function, ϕ(r), whose r dependence is 1/r, and which takes the value 0at r = b and V0 at r = a. The unique solution is

ϕ(r) = V0

1r −

1b

1a −

1b

for a ≤ r ≤ b. (8.1.1)

The corresponding electric field is

E(r) = −∂ϕ∂r

r =abV0

b− ar

r2. (8.1.2)

Fortuitously, the total current going through a sphere of radius r between a and b is constantand given by

I =

∫J · da = σE(r) 4πr2 =

4πσabV0

b− a. (8.1.3)

Therefore, the resistance is

R =V0

I=

b− a4πσab

. (8.1.4)

8.2 Non-Ohmic Device: Vacuum Diode

Please review Purcell section 4.2 (p.126-127), especially the last paragraph regarding the vacuumdiode pictured in Figure 4.2. In this problem, we will work out how the electron charge density

8.2. NON-OHMIC DEVICE: VACUUM DIODE 43

and their speed vary between the diode plates. We will also work out the relationship betweenthe steady current and the voltage across the diode and we will discover that this device doesnot follow Ohm’s law!

[Griffiths 2.48 p.107 ] In a vacuum diode, electrons are “boiled” off a hot cathode, at potentialzero, and accelerated across a gap to the anode, which is held at positive potential V0. Thecloud of moving electrons within the gap (called space charge) quickly builds up to the pointwhere it reduces the field at the surface of the cathode to zero. From then on a steady currentI flows between the plates.

Suppose the plates are large relative to the separation (A >> s2), so that edge effects canbe neglected. Then V , ρ and v (the speed of the electrons) are all functions of x alone.

(a) Write Poisson’s equation for the region between the plates.

(b) Assuming the electrons start from rest at the cathode, what is their speed at point x, wherethe potential is V (x)?

(c) In the steady state, I is independent of x. What, then, is the relation between ρ and v?

(d) Use these three results to obtain a differential equation for V , by eliminating ρ and v.

(e) Solve this equation for V as a function of x, V0 and s. Plot V (x) and compare it to thepotential without space charge. Also, find ρ and v as functions of x.

(f) Show that I = KV3/20 and find the constant K. (This is called the Child-Langmuir Law.

It holds for other geometries as well, whenever space charge limits the current. Notice thatthe space charge limited diode is nonlinear - it does not obey Ohm’s law.)

44 CHAPTER 8. CURRENT

Chapter 9Circuit Analysis

There is one big concept in circuit analysis and that is the concept of Thevenin equivalents(resistance and voltage). For the purposes of this class, we have only two main tools in thiscontext: (1) simplifying parallel and series resistors; and (2) Kirchoff’s node and loop laws.

Two resistors are in series if all the current that comes out of one has to go into the other.Basically, there is nothing else in between except, obviously, a piece of wire connecting the two.Two resistors are in parallel if their tops are connected and their bottoms are connected withnothing else in between. The potential is the same on the top of one as on the top of theother (and similarly for their bottoms). Purcell Figure 4.14 and 4.15 (p.151) are quintessentialexamples of series and parallel, respectively. Remember that two resistors in series, R1 and R2,can be replaced by one equivalent resistor with resistance Req = R1 + R2. On the other hand,

the equivalent resistance of two in parallel is Req =[R−1

1 +R−12

]−1. Using these two things may

simplify a circuit considerably for you (c.f. Purcell Figure 4.16 p.152).What does this mean? Why is this useful? Take a look at Figure 4.16 again. In the original

circuit on the left, imagine attaching a battery across the two white circles denoting the start andend of the circuit of resistors. How much current will this entire circuit draw from the battery?The answer is the same as the current that would be drawn by the single equivalent reistor onthe right of the figure if the battery were attached across it!

Often, this procedure will not simplify an entire circuit of resistors down to one equivalentresistor. This is because it is possible for a pair of resistors to be neither in series nor in parallel(c.f. Purcell Figure 4.17 p.152). Nevertheless, it is probably best to simplify the circuit as muchas possible using this first before moving on to using a combination of Ohm’s and Kirchoff’slaws. Recall that Kirchoff’s node law says that whatever current goes into a node must comeout, otherwise charge would be building up at that node! Kirchoff’s loop law states that thesum of the voltage drops across a loop in the circuit must be 0 so if voltage is lost somewherealong the loop, then it must be gained somewhere else.

The usual procedure that I use to solve a circuit problem goes as follows:

(1) Set the potential to be V at the beginning of the circuit and 0 at the end and introduce anunknown potential for each and every node of the circuit;

(2) Decide on a direction for the current on each branch (if you happen to choose the wrongdirection then you’ll just get a negative current, so don’t worry about getting the directionscorrect);

(3) Express the currents in terms of the voltages using Ohm’s law;

(4) Use Kirchoff’s node law at each of the nodes whose potential is unknown;

45

46 CHAPTER 9. CIRCUIT ANALYSIS

(5) This should give you one equation per node with unkown potential and so determining eachpotential should be a matter of solving a bunch of simultaneous equations. If any of theequations are actually degenerate for some reason, then use some of the loop laws.

(6) Determine the total current, I, entering or leaving the circuit (using Ohm’s law, now thatyou know the potentials). Then, Req = V/I.

9.1 Simple Bridge Circuit

Calculate the equivalent resistances, RAB (between points A and B) and RCD (between pointsC and D), for the circuit below. Check your result in the case when all the resistors have thesame resistance and explain intuitively why your result is correct (or incorrect as the case maybe).

Figure 9.1: Simple bridge circuit.

SOLUTION:

If we consider RCD, then the entrance point is C and the exit point is D. In this case, R1 is inseries with R2 and can be replaced by R1 + R2, and similarly for R3 and R4. Then, these twoequivalents are in parallel with R5. Hence,

RCD =1

1R1+R2

+ 1R3+R4

+ 1R5

. (9.1.1)

When all the resistors are the same, R0, then RCD = R0/2 . This makes sense since if we inject

some fixed current, I, into C, then I/2 will go through CD, I/4 throug CAD and I/4 throughCBD, by Ohm’s law.

If we consider RAB , then no two of the resistors are in series or in parallel! Define the potentials

VA = V, VB = 0, VC =?, VD =?. (9.1.2)

Denote the current IAC to be the current from A to C and similarly across the other fourbranches. Using Ohm’s law, write

IAC =VA − VCR1

=V − VCR1

,

IAD =VA − VDR2

=V − VDR2

,

ICB =VC − VBR3

=VCR3

, (9.1.3)

IDB =VD − VBR4

=VDR4

,

ICD =VC − VDR5

.

9.1. SIMPLE BRIDGE CIRCUIT 47

Kirchoff’s node law applied to node C reads

IAC = ICB + ICD =⇒ V − VCR1

=VCR3

+VC − VDR5

. (9.1.4)

Applied to node B, it gives

IAD + ICD = IDB =⇒ V − VDR2

+VC − VDR5

=VDR4

. (9.1.5)

Thus, we have two equations for the two unknowns, VC and VD. The solution is quite a mess,but the procedure is rather simple: use the first equation to solve for VD in terms of VC (andV and the R’s, of course), then plug this into the second equation and solve for VC . Or youcould write the equation in matrix form and the solution involves inverting a two by two matrix.Anyway, the solution is

VC =

( 1R1

(1R2

+ 1R4

+ 1R5

)+ 1

R2R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)V,

VD =

( 1R2

(1R1

+ 1R3

+ 1R5

)+ 1

R1R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)V.

(9.1.6)

Let’s pause for a moment to check that this result makes sense if all of the resistors are identicaland equal to R0, say. Symmetry under reflection across a line between A and B maps C onto Dand thus VC must equal VD. Furthermore, reflecting across a line connecting C and D indicatesthat they should actually both be exactly midway between VB = 0 and VA = V . Hence, weshould find that VC = VD = V/2. Indeed, if you plug in R1 = · · · = R5 = R0, you will findVC = VD = V/2.

The total current leaving the circuit is I = ICB + IDB . Using the third and fouth equationsin Eqn. (9.1.3) and plugging in our solutions for VC and VD gives

I =

[1

R3

( 1R1

(1R2

+ 1R4

+ 1R5

)+ 1

R2R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)

+1

R4

( 1R2

(1R1

+ 1R3

+ 1R5

)+ 1

R1R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)]V. (9.1.7)

Finally, the equivalent resistance is RAB = V/I, which is

RAB =

[1

R3

( 1R1

(1R2

+ 1R4

+ 1R5

)+ 1

R2R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)

+1

R4

( 1R2

(1R1

+ 1R3

+ 1R5

)+ 1

R1R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)]−1

. (9.1.8)

When the resistors are all identical, we already showed that we get VC = VD = V/2. Sincethere is no potential difference across R5, there is no current there and thus we could just aswell drop it altogether and the circuit reduces to the two branches, ACB and ADB in parallelwith each other, each with 2R0 resistance. Hence, the equivalent resistance should be RAB =[

12R0

+ 12R0

]−1= R0. This is precisely the result one finds if one plugs in R1 = · · · = R5 = R0

into the above formula: RAB = R0 .

48 CHAPTER 9. CIRCUIT ANALYSIS

Chapter 10Midterm 2 Review

10.1 Analysis of Tetrahedral Circuit

Calculate the equivalent resistance, RAB , between nodes A and B of the tetrahedral circuit ofresistors given below. Check to make sure that your answer gives the correct limit in the casewhen all of the resistors are identical and discuss why your answer is correct in this limit (orincorrect, as the case may be).

If you are unable to answer the question with generic resistors, answer it with identicalresistors, in which case the question is significantly easier.

Figure 10.1: Tetrahedral circuit of resistors.

SOLUTION:

Let us first answer the question in the simpler case when all of the resistors are identical, sayR0. In this case, we can make use of the symmetry of the system. Consider reflecting thesystem across the plane that intersects the tetrahedron at the dotted line and the line AB asshown in the diagram below. Vertices A and B are unchanged, but C and D are switched. Hadthere been a current flowing in the original circuit between C and D, for example from C to D,that current would switch direction under this reflection. However, since all of the resistors areactually identical, the system itself will not have changed under this reflection and therefore, nocurrents could possibly change either! It follows that there cannot be any current between Cand D and thus we may drop the resistor between C and D altogether, leaving us with exactlythe same circuit as the simple bridge circuit in Sec. 9.1 when all of the resistors are identical(with slight vertex and resistor relabeling). In that case, we found the equivalent resistance to

be RAB = R0/2 .

49


Figure 10.2: Reflection symmetry of the tetrahedral system.

In the general case, we set the voltages

VA = V, VB = 0, VC =?, VD =?. (10.1.1)

Using Ohm’s law, we write the currents in terms of the voltages:

IAB =V

R6, ICB =

VCR3

,

IAC =V − VCR1

, IDB =VDR4

, (10.1.2)

IAD =V − VDR2

, ICD =VC − VDR5

.

Kirchoff’s node laws at C and D read

IAC = ICB + ICD =⇒ V − VCR1

=VCR3

+VC − VDR5

, (10.1.3a)

IAD + ICD = IDB =⇒ V − VDR2

+VC − VDR5

=VDR4

. (10.1.3b)

We group like terms in these two equations to yield(1

R1+

1

R3+

1

R5

)VC −

1

R5VD =

1

R1V, (10.1.4a)

− 1

R5VC +

(1

R2+

1

R4+

1

R5

)VD =

1

R2V. (10.1.4b)

At this point, we have two equations for two unknowns, VC and VD. We just have to solve it.The solution is

VC =

( 1R1

(1R2

+ 1R4

+ 1R5

)+ 1

R2R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)V,

VD =

( 1R2

(1R1

+ 1R3

+ 1R5

)+ 1

R1R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)V.

(10.1.5)

Notice that this is exactly the same as Eqn. (9.1.6), for the simple bridge circuit. This is NOTa coincidence, as we’ll mention momentarily.

When the resistors are all R0, we already know that there should be no current between Cand D. This implies that we ought to have VC = VD in this case. Indeed, if we plug in R0 forall the resistors in the above formulas for VC and VD, we get VC = VD = V/2. So, we can besure that we will get the correct answer in this limit!

10.2. THICK RESISTIVE CYLINDER 51

The total current flowing out of the circuit is I = IAB + ICB + IDB . Plugging in our resultsfor VC and VD and dividing V by Iout yields the equivalent resistance:

RAB =

[1

R6+

1

R3

( 1R1

(1R2

+ 1R4

+ 1R5

)+ 1

R2R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)

+1

R4

( 1R2

(1R1

+ 1R3

+ 1R5

)+ 1

R1R5(1R1

+ 1R3

+ 1R5

)(1R2

+ 1R4

+ 1R5

)− 1

R25

)]−1

(10.1.6)

First, let us check that we get RAB = R0/2 in the case when R1 = · · · = R6 = R0, which is theresult we found first. Well, indeed, that’s what you get if you plug R0 in for all the resistors inthe above formula! Finally, note that this result is just the result you would get if you put R6

in parallel across AB of the bridge circuit in Section 9. Hopefully, you can see just by lookingat the tetrahedron, that that is precisely this circuit.

10.2 Thick Resistive Cylinder

Consider two concentric metal cylinders with radii a and b (with a < b) and length ` >> a, b.The region between the cylinders is filled with a material with resistivity ρ. Calculate the resis-tance of this object.

SOLUTION:

We must determine the electric field that is produced when one places a voltage differencebetween the two cylinders. Let ϕ(a) = V0 and ϕ(b) = 0; we must determine ϕ(r). Well,hopefully you remember that the electric field of a cylindrical object falls off like E ∼ 1/r, whichimplies ϕ ∼ ln r. So, we require a function, ϕ(r), whose r-dependence is ln r and which takesthe value V0 at r = a and 0 at r = b. The unique solution is

ϕ(r) =ln(r/b)

ln(a/b)V0. (10.2.1)

Alternatively, you might recall that the potential between the cylinders in the case when thereis +Q charge on the inner cylinder and −Q on the outer clinder is ϕ(r) = − 2Q

` ln r + C, where

C is an arbitrary constant. If you want ϕ(b) = 0, then you must have C = 2Q` ln b and so

ϕ(r) = − 2Q` ln(r/b). Then, replace Q by CV0, where C is the capacitance of concentric cylinders,

which is C = `2 ln(b/a) . This gives the same result.

Therefore, the electric field is

E = −dϕdr

r =V0

ln(b/a)

r

r. (10.2.2)

The current density is simply J = E/ρ. The current flowing out of a a cylinder with radius rsuch that a < r < b is

I =

∫J · da = 2πr`

V0

ρr ln(b/a)=

2π`V0

ρ ln(b/a). (10.2.3)

Therefore, the resistance is

R =V0

I=ρ ln(b/a)

2π`. (10.2.4)

Note that, in this case, the longer the cylinder, the lower the resistance, as opposed to a cylindricalwire. This makes sense because the current is actually flowing radially outwards in this case, asopposed to the wire where it flows along the length of the tube.


10.3 Potential of a Continuous Charge Distribution

Calculate the potentials that follow. You may use the fact that the antiderivative of 1√1+x2

is

ln(x+√

1 + x2).

(a) At a point P some distance r perpendicularly away from the center of a finite line of length2s and constant charge density λ.

(b) At a point P some distance h away from the center of a square loop of sidelength 2s andconstant linear charge density λ.

SOLUTION:

(a) Let x be the direction along the line going from x = −s to x = s. Take a generic differentialpiece of the line between x and x+dx. The differential bit of charge in this piece is dq = λdx.The distance between this piece and the point at which we are calculating the potential isR =

√x2 + r2. This piece contributes a bit of potential, dϕ, at the point P :

dϕ =dq

R=

λ dx√x2 + r2

. (10.3.1)

The total potential is thus

ϕ =

∫dϕ = λ

∫ s

−s

dx√x2 + r2

= λ

∫ s

−s

dx

r√

1 + (x/r)2. (10.3.2)

Change variables to ξ ≡ x/r so that dξ = dx/r and if x ranges between −s and s, then ξgoes from −s/r to s/r. Thus, we have

ϕ = λ

∫ s/r

−s/r

dξ√1 + ξ2

= λ ln

sr +

√1 +

(sr

)2− sr +

√1 +

(sr

)2 . (10.3.3)

We may write this as

ϕ = λ ln

√s2 + r2 + s√s2 + r2 − s

. (10.3.4)

In the limit that r << s, the line looks essentially infinitely long since the point P is soclose to it. Thus, up to a constant, we had better get ϕ ≈ −2λ ln r, which is the result thatfollows from E = 2λ r

r for an infinite line (via Gauss’ law). Indeed, we can write

ϕ = λ lns√

1 + (r/s)2 + s

s√

1 + (r/s)2 − s

≈ λ lns[1 + 1

2 (r/s)2]

+ s

s[1 + 1

2 (r/s)2]− s

= λ ln2s+ r2

2sr2

2s

= λ ln[1 +

(2sr

)2]≈ λ ln(2s/r)2

= −2λ ln(r/2s). (10.3.5)

10.4. CHARGE NEAR CONDUCTING PLANES 53

The first ≈ follows by expanding the square root, and the second from the the fact thats/r >> 1 and so we can just drop the 1 in the ln.

In the limit r >> s, the line looks essentially like a point since the point P is so far awayfrom it. The total charge of this point is q = 2λs. We had better get ϕ ≈ Q

r = 2λsr in this

limit. Indeed,

ϕ = λ lnr√

1 + (s/r)2 + s

r√

1 + (s/r)2 − s

≈ λ lnr[1 + s

r + 12

(sr

)2]r[1− s

r + 12

(sr

)2] (10.3.6)

Note that all we have done here is expand the square root and factor r out of the numeratorand denominator. Since s/r << 1, we may drop the quadratic terms proportional to (s/r)2.Then, we approximate 1

1−(s/r) ≈ 1 + sr . We get

ϕ ≈ λ ln1 + s

r

1− sr

≈ λ ln(1 + s

r

)2≈ λ ln

(1 + 2s

r

)≈ 2λs

r. (10.3.7)

We made use of the expansion ln(1 + ε) ≈ ε for ε << 1.

(b) We can think of the square loop as four lines. The distance between P and each of the fourlines is r =

√h2 + s2. Plugging this r into Eqn. (10.3.3) and multiplying by 4 gives

ϕ = 4λ ln

√2s2 + h2 + s√2s2 + h2 − s

. (10.3.8)

10.4 Charge near Conducting Planes

Place a point charge +q at the coordinates a2 (1, 1, 0) (a > 0).

(a) Suppose there is an infinite conducting plane at x = 0 (i.e. in the yz-plane). Calculate thenet force (magnitude and direction) felt by the point charge.

(b) Suppose that we fold the y < 0 half of the infinite plane by 90 until it lies in the x > 0 partof the xz-plane. What is the net force on the point charge now?

SOLUTION:

(a) Place an image charge −q at the image point a2 (−1, 1, 0). The force felt by +q due to the

conducting plane is the same as the force it would feel in the presence of the image charge,which is just given by Coulomb’s law:

F = − q2

a2x . (10.4.1)


(b) Let us think of the folded infinite plane as two infinite planes intersection orthogonally (oneis the yz-plane and the other is the xz-plane). Part (a) indicates that we should place a−q image charge at a

2 (−1, 1, 0), which ensures that the potential on the yz-plane vanishes.We also would like to place a −q image charge at a

2 (1,−1, 0), which, together with justthe origin +q charge ensures that the potential vanishes on the xz-plane. However, the +qcharge taken together with both the image charges so far do not make the potential vanish onneither plane! We require one more image charge +q at a

2 (−1,−1, 0). This is drawn below,with black being +q and white −q. Also drawn below is a Mathematica plot of the electricfield with the red lines indicating the position of the planes. Note that the field vectors areperpendicular to the planes, as they should be.

Figure 10.3: Image charge setup and corresponding electric field.

Again, the force on the original +q is the force it would feel in the presence of the three image

charges. The forces due to the two negative image charges are F1 = − q2

a2 x and F2 = − q2

a2 y,

while the force due to the positive image charge is F3 = q2

2a2

(x+y√

2

). Note that the 2a2 in the

denominator of F3 comes from the fact that the distance between the two positive chargesis a√

2. Also, note that the vector in parenthesis in F3 is the unit vector in the northeastdirection in the diagrams above. Adding these three forces yields

F = −2√

2− 1

2· q

2

a2

(x + y√

2

)≈ −0.914

q2

a2

(x + y√

2

). (10.4.2)

10.5 Non-evenly Charged Cylinder

Calculate the electric field and potential everywhere produced by an infinitely long solid cylinderof radius R and whose volume charge density is ρ(r) = σ/4πr, where σ is some constant withunits of surface charge density, and r is the radial distance from the central axis of the cylinder.

SOLUTION:

Firstly, I want to write ρ in a way that makes sense for both r ≤ R and r > R. Define theHeaviside theta function, θ(x), to take on the value 0 with x ≤ 0 and 1 when x > 1. Then,ρ(r) = σ

4πr θ(R− r), which vanishes for r > R, by virtue of the theta function.

Since ρ depends only on r, the system possesses cylindrical symmetry, which implies thatE and ϕ depend only on r and E points in the radial direction: E(r) = E(r) r. Consider aGaussian cylinder concentric to the solid cylinder and having radius r and length `. Recall thata differential bit of volume in cylindrical coordinates is rdr dφ dz. Since ρ does not depend on φ

10.5. NON-EVENLY CHARGED CYLINDER 55

or z, their integrals just give a factor of 2π`. Thus, Gauss’ law reads

2πr`E(r) =

∫E · da = 4πQenc = 4π · 2π`

∫ r

0

ρ(r′) r′ dr′. (10.5.1)

Cancel 2π` from both sides, divide by r, and plug in for ρ:

E(r) =4π

r

σ

4π

∫ r

0

θ(R− r′)r′

r′ dr′ =σ

r

∫ r

0

θ(R− r′) dr′ =σr<r, (10.5.2)

where r< is the lesser of the two, r and R. In other words,

E(r) =

σ r, r ≤ R,σR rr , r > R.

(10.5.3)

We must have E = −dϕdr r. In the inner region, this implies ϕ(r) = −σr + C and in the outerregion, ϕ(r) = −σR ln r + C ′, where C and C ′ are two constants, which are determined by achoice of where ϕ goes to zero. Since the potential goes like ln r in the outer region, there is noway to make it vanish as r →∞, unless we allow C ′ to be infinite. Instead, let us choose to setthe zero of potential to be at the surface of the cylinder: r = R. This forces us to set C = σRand C ′ = σR lnR. Therefore,

ϕ(r) =

σ(R− r), r ≤ R,−σR ln(r/R), r > R.

(10.5.4)

Below is a plot of ϕ, normalized so that we plot the unitless quantity, ϕ/σR, versus the unitlessquantity, r/R.

Figure 10.4: Potential of the solid cylinder.

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