Solution Manual For
Electricity and Magnetism Third Edition Edward M. Purcell and David J. Morin
Chapter 1-11 With Appendix
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically
have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams
include:Essay/Problem-Based Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific
legal principles and statutes to the facts presented in the scenario.Multiple Ch
Chapter 1
Electrostatics
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
(Version 1, January 2013)
1.34. Aircraft carriers and specks of gold
3 3
The volume of a cube 1 mm on a side is 10− cm . So the mass of this 1 mm cube is
2
1.93 · 10− g. The number of atoms in the cube is therefore
2
23 1.93 · 10− g 19
6.02 · 10 · = 5.9 · 10 . (1)
197 g
19
Each atom has a positive charge of 1 e = 1.6· 10− C, so the total charge in the cube
19 19
is (5.9 ·10 )(1.6 ·10− C) = 9.4 C. The repulsive force between two such cubes 1 m
apart is therefore
2 ( 3) 2
q 9 kg m (9.4 C) 11
9 · 10
8 2 9
The weight of an aircraft carrier is mg = (10 kg)(9.8 m/s ) 10 N. The above F is
therefore equal to the weight of 800 aircraft carriers. This is just another example of
the fact that the electrostatic force is enormously larger than the gravitational force.
1.35. Balancing the weight
2 2
Let the desired distance be d. We want the upward electric force e /4πϵ0d to equal
the downward gravitational force mg. Hence,
1 e
2 ( kg m )
3
(1.6 · 10− C)
19 2
2 9 2
d 9 · 10
4πϵ
which gives d = 5.1 m. The non-infinitesimal size of this answer is indicative of the
feebleness of the gravitational force compared with the electric force. It takes about
51
3.6· 10 nucleons (that’s roughly how many are in the earth) to produce a gravitational
6
force at an effective distance of 6.4· 10 m (the radius of the earth) that cancels the
electrical force from one proton at a distance of 5 m. The difference in these distances
12
accounts for a factor of only 1.6 ·10 between the forces (the square of the ratio of the
distances). So even if all the earth’s mass were somehow located the same distance
39
away from the electron as the single proton is, we would still need about 2· 10
nucleons to produce the necessary gravitational force.
1
, 2 CHAPTER 1. ELECTROSTATICS
1.36. Repelling volley balls
Consider one of the balls. The vertical component of the tension in the string must
equal the gravitational force on the ball. And the horizontal component must equal
the electric force. The angle that the string makes with the horizontal is given by
tan θ = 10, so we have
Ty Fg mg
= 10 =⇒ = 10 =⇒ = 10. (4)
Tx Fe q2/4πϵ 0r
2
Therefore,
( )
1 s 2 C2
8.85 · 10 −12
2 2 2 2
(4πϵ0)mgr
10
12 2 6
= 8.17 · 10− C =⇒ q = 2.9 · 10− C. (5)
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically have a more
structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based
Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific legal principles and statutes to the facts presented
in the scenario.Multiple Ch
1.37. Zero force at the corners
(a) Consider a charge q at √ a particular corner. If the square has side length ℓ, t h√e n
one of the other q’s is 2 ℓ away, two of them are ℓ away, and the −Q is ℓ/ 2
away. The net force on the given q, which is directed along the diagonal touching
it, is (ignoring the factors of 1/4πϵ0 since they will cancel)
2 2
q Qq ◦q
F= √ + 2 cos 45 2 − √ . (6)
( 2 ℓ)2 ℓ (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
(b) To find the potential energy of the system, we must sum over all pairs of charges.
Four pairs involve the charge Q, four involve the edges of the square, and two
involve the diagonals. The total potential energy is therefore
( 2 ) √ ( )
1 (−Q)q q q 4 2q q q
4· 2· √
4πϵ ℓ 2ℓ 4πϵ0ℓ
in view of Eq. (7). The result in Problem 1.6 was “The total potential energy
of any system of charges in equilibrium is zero.” With Q given by Eq. (7), the
system is in equilibrium (because along with all the q’s, the force on the Q
charge is also zero, by symmetry). And consistent with Problem 1.6, the total
potential energy is zero.
1.38. Oscillating on a line
If the charge q is at position (x, 0), then the force from the right charge Q equals
Qq/4πϵ0(ℓ x) , where the minus sign indicates leftward. And the force from the
2
left charge Q equals Qq/4πϵ0(ℓ + x) . The net force is therefore (dropping terms of
, 3
2
order x )
( )
Qq 1 1
F (x) = − −
(ℓ − x)2
4πϵ0 (ℓ + x)2
( )
Qq 1 1
≈ − −
4πϵ0ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq
(1 + 2x/ℓ) (1 2x/ℓ)
4πϵ0ℓ2
Qqx
= − . (9)
πϵ ℓ3
This is a Hooke’s-law type force, being proportional to (negative) x. The F = ma
equation for the charge q is
Qqx Qq
= ẍ = x. (10)
− = m ẍ
πϵ ℓ3 πϵ0mℓ3
The frequency of small oscillations is the square root of the (negative of the) coefficient
of x, as you can see by plugging in x(t) = A cos ωt. Therefore ω = Qq/πϵ0mℓ3. This
frequency increases with Q and q, and it decreases with m and ℓ; these make sense. As
2
far as the units go, Qq/ϵ0ℓ has the dimensions of force F (from looking at Coulomb’s
law), so ω has units of F/mℓ. This correctly has units of inverse seconds.
ALTERNATIVELY: We can find the potential energy of the charge q at position (x, 0),
and then take the (negative) derivative to find the force. The energy is a scalar, so we
don’t have to worry about directions. We have
( )
Qq 1 1
U (x) = + . (11)
4πϵ0 ℓ − x ℓ+x
2
We’ll need to expand things to order x because the order x terms will cancel:
( )
Qq 1 1
U (x) = +
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq x x2 x x2
≈ 1+ + 2 + 1− + 2
4πϵ0ℓ ℓ ℓ ℓ ℓ
( 2)
Qq 2x
4πϵ0ℓ ℓ2
The constant term isn’t important here, because only changes in the potential energy
matter. Equivalently, the force is the negative derivative of the potential energy, and
the derivative of a constant is zero. The force on the charge q is therefore
dU Qqx
F (x) = − =− , (13)
dx πϵ ℓ3
in agreement with the force in Eq. (9).
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams
typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types
of law exams include:Essay/Problem-Based Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require
students to apply specific legal principles and statutes to the facts presented in the scenario.Multiple Ch
1.39. Rhombus of charges
We’ll do the balancing-the-forces solution first. Let the common length of the strings
be ℓ. By symmetry, the tension T is the same in all of the strings. Each of the two
charges q is in equilibrium if the sum of the vertical components of the electrostatic
, 4 CHAPTER 1. ELECTROSTATICS
forces is equal and opposite to the sum of the vertical components of the tensions.
This gives
( ) 2
qQ q q2 2 qQ
2
4πϵ0ℓ 2 16πϵ
Similarly, each charge Q is in equilibrium if
( ) 2
qQ Q Q2 2 qQ
2
4πϵ0ℓ2 16πϵ 2πϵ
(15)
The righthand sides of the two preceding equations are equal, so the same must be
2 3 2 3 2 2 3
true of the lefthand sides. This yields q / sin θ = Q / cos θ, or q /Q = tan θ, as
desired.
Some limits: If Q q, then θ 0. And if q Q, then θ π/2. Also, if q = Q, then
θ = 45◦. These all make intuitive sense.
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically have a more
structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based
Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific legal principles and statutes to the facts presented
in the scenario.Multiple Ch
ALTERNATIVELY: To solve the exercise by minimizing the electrostatic energy, note
that the only variable terms in the sum-over-all-pairs expression for the energy are the
ones involving the diagonals of the rhombus. The other four pairs involve the sides of
2
the rhombus which are of fixed length. The variable terms are q /4πϵ0(2ℓ sin θ) and
2
Q /4πϵ0(2ℓ cos θ). Minimizing this as a function of θ yields
( 2 )
d q Q2 2 cos θ 2 sin θ q2 3
y sin θ cos θ
e
e
1.40. Zero potential energy
Let’s first consider the general case where the three charges don’t necessarily lie on
2
the same line. Without loss of generality, we can put the two electrons on the x axis a
unit distance apart (that is, at the values x = 1/2), as shown in Fig. 1. And we may
-2 -1 -e -e 1 2 x assume the proton lies in the xy plane. For an arbitrary location of the proton in this
plane, let the distances from the electrons be r1 and r2. Then setting the potential
Figure 1 energy of the system equal to zero gives
( 2 2 )
1 e e e2 1 1
4πϵ
One obvious location satisfying√this requirement has the proton on the y axis with
r1 = r2 = 2, that is, with y = 15/2 ≈ 1.94. In general, Eq. (17) defines a curve in
the xy plane, and a surface of revolution around the x axis in space. This surface is
the set of all points where the proton can be placed to give U = 0. The surface looks
something like a prolate ellipsoid, but it isn’t.
Let’s now consider the case where all three charges lie on the x axis. Assume that
the proton lies to the right of the right electron. We then have r1 = x − 1/2 and
r2 = x + 1/2, so Eq. (17) becomes
√
1 1 2 2± 5
+ = 1 =⇒ x − 2x − 1/4 = 0 =⇒ x = . (18)
x − 1/2 x + 1/2 2
The negative root must be thrown out because it violates our assumption that x > 1/2.
(With x < 1/2, the distance r1 isn’t represented by x − 1/2)√. So we find x = 2.118.
The distance from the right electron at x = 1/2 equals (1 + 5)/2. The ratio of this
Electricity and Magnetism Third Edition Edward M. Purcell and David J. Morin
Chapter 1-11 With Appendix
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically
have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams
include:Essay/Problem-Based Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific
legal principles and statutes to the facts presented in the scenario.Multiple Ch
Chapter 1
Electrostatics
Solutions manual for Electricity and Magnetism, 3rd edition, E. Purcell, D. Morin.
(Version 1, January 2013)
1.34. Aircraft carriers and specks of gold
3 3
The volume of a cube 1 mm on a side is 10− cm . So the mass of this 1 mm cube is
2
1.93 · 10− g. The number of atoms in the cube is therefore
2
23 1.93 · 10− g 19
6.02 · 10 · = 5.9 · 10 . (1)
197 g
19
Each atom has a positive charge of 1 e = 1.6· 10− C, so the total charge in the cube
19 19
is (5.9 ·10 )(1.6 ·10− C) = 9.4 C. The repulsive force between two such cubes 1 m
apart is therefore
2 ( 3) 2
q 9 kg m (9.4 C) 11
9 · 10
8 2 9
The weight of an aircraft carrier is mg = (10 kg)(9.8 m/s ) 10 N. The above F is
therefore equal to the weight of 800 aircraft carriers. This is just another example of
the fact that the electrostatic force is enormously larger than the gravitational force.
1.35. Balancing the weight
2 2
Let the desired distance be d. We want the upward electric force e /4πϵ0d to equal
the downward gravitational force mg. Hence,
1 e
2 ( kg m )
3
(1.6 · 10− C)
19 2
2 9 2
d 9 · 10
4πϵ
which gives d = 5.1 m. The non-infinitesimal size of this answer is indicative of the
feebleness of the gravitational force compared with the electric force. It takes about
51
3.6· 10 nucleons (that’s roughly how many are in the earth) to produce a gravitational
6
force at an effective distance of 6.4· 10 m (the radius of the earth) that cancels the
electrical force from one proton at a distance of 5 m. The difference in these distances
12
accounts for a factor of only 1.6 ·10 between the forces (the square of the ratio of the
distances). So even if all the earth’s mass were somehow located the same distance
39
away from the electron as the single proton is, we would still need about 2· 10
nucleons to produce the necessary gravitational force.
1
, 2 CHAPTER 1. ELECTROSTATICS
1.36. Repelling volley balls
Consider one of the balls. The vertical component of the tension in the string must
equal the gravitational force on the ball. And the horizontal component must equal
the electric force. The angle that the string makes with the horizontal is given by
tan θ = 10, so we have
Ty Fg mg
= 10 =⇒ = 10 =⇒ = 10. (4)
Tx Fe q2/4πϵ 0r
2
Therefore,
( )
1 s 2 C2
8.85 · 10 −12
2 2 2 2
(4πϵ0)mgr
10
12 2 6
= 8.17 · 10− C =⇒ q = 2.9 · 10− C. (5)
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically have a more
structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based
Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific legal principles and statutes to the facts presented
in the scenario.Multiple Ch
1.37. Zero force at the corners
(a) Consider a charge q at √ a particular corner. If the square has side length ℓ, t h√e n
one of the other q’s is 2 ℓ away, two of them are ℓ away, and the −Q is ℓ/ 2
away. The net force on the given q, which is directed along the diagonal touching
it, is (ignoring the factors of 1/4πϵ0 since they will cancel)
2 2
q Qq ◦q
F= √ + 2 cos 45 2 − √ . (6)
( 2 ℓ)2 ℓ (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
(b) To find the potential energy of the system, we must sum over all pairs of charges.
Four pairs involve the charge Q, four involve the edges of the square, and two
involve the diagonals. The total potential energy is therefore
( 2 ) √ ( )
1 (−Q)q q q 4 2q q q
4· 2· √
4πϵ ℓ 2ℓ 4πϵ0ℓ
in view of Eq. (7). The result in Problem 1.6 was “The total potential energy
of any system of charges in equilibrium is zero.” With Q given by Eq. (7), the
system is in equilibrium (because along with all the q’s, the force on the Q
charge is also zero, by symmetry). And consistent with Problem 1.6, the total
potential energy is zero.
1.38. Oscillating on a line
If the charge q is at position (x, 0), then the force from the right charge Q equals
Qq/4πϵ0(ℓ x) , where the minus sign indicates leftward. And the force from the
2
left charge Q equals Qq/4πϵ0(ℓ + x) . The net force is therefore (dropping terms of
, 3
2
order x )
( )
Qq 1 1
F (x) = − −
(ℓ − x)2
4πϵ0 (ℓ + x)2
( )
Qq 1 1
≈ − −
4πϵ0ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
(1 + 2x/ℓ) (1 2x/ℓ)
4πϵ0ℓ2
Qqx
= − . (9)
πϵ ℓ3
This is a Hooke’s-law type force, being proportional to (negative) x. The F = ma
equation for the charge q is
Qqx Qq
= ẍ = x. (10)
− = m ẍ
πϵ ℓ3 πϵ0mℓ3
The frequency of small oscillations is the square root of the (negative of the) coefficient
of x, as you can see by plugging in x(t) = A cos ωt. Therefore ω = Qq/πϵ0mℓ3. This
frequency increases with Q and q, and it decreases with m and ℓ; these make sense. As
2
far as the units go, Qq/ϵ0ℓ has the dimensions of force F (from looking at Coulomb’s
law), so ω has units of F/mℓ. This correctly has units of inverse seconds.
ALTERNATIVELY: We can find the potential energy of the charge q at position (x, 0),
and then take the (negative) derivative to find the force. The energy is a scalar, so we
don’t have to worry about directions. We have
( )
Qq 1 1
U (x) = + . (11)
4πϵ0 ℓ − x ℓ+x
2
We’ll need to expand things to order x because the order x terms will cancel:
( )
Qq 1 1
U (x) = +
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq x x2 x x2
≈ 1+ + 2 + 1− + 2
4πϵ0ℓ ℓ ℓ ℓ ℓ
( 2)
Qq 2x
4πϵ0ℓ ℓ2
The constant term isn’t important here, because only changes in the potential energy
matter. Equivalently, the force is the negative derivative of the potential energy, and
the derivative of a constant is zero. The force on the charge q is therefore
dU Qqx
F (x) = − =− , (13)
dx πϵ ℓ3
in agreement with the force in Eq. (9).
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams
typically have a more structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types
of law exams include:Essay/Problem-Based Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require
students to apply specific legal principles and statutes to the facts presented in the scenario.Multiple Ch
1.39. Rhombus of charges
We’ll do the balancing-the-forces solution first. Let the common length of the strings
be ℓ. By symmetry, the tension T is the same in all of the strings. Each of the two
charges q is in equilibrium if the sum of the vertical components of the electrostatic
, 4 CHAPTER 1. ELECTROSTATICS
forces is equal and opposite to the sum of the vertical components of the tensions.
This gives
( ) 2
qQ q q2 2 qQ
2
4πϵ0ℓ 2 16πϵ
Similarly, each charge Q is in equilibrium if
( ) 2
qQ Q Q2 2 qQ
2
4πϵ0ℓ2 16πϵ 2πϵ
(15)
The righthand sides of the two preceding equations are equal, so the same must be
2 3 2 3 2 2 3
true of the lefthand sides. This yields q / sin θ = Q / cos θ, or q /Q = tan θ, as
desired.
Some limits: If Q q, then θ 0. And if q Q, then θ π/2. Also, if q = Q, then
θ = 45◦. These all make intuitive sense.
prepares students for legal practice, with specializations that may include constitutional law, criminal law, corporate law, human rights law, or intellectual property law.2.2. Structure of Law ExamsLaw exams typically have a more
structured format compared to business exams, often requiring students to demonstrate knowledge of legal statutes, case law, and their ability to apply legal reasoning. Common types of law exams include:Essay/Problem-Based
Questions: In law exams, students are often given hypothetical scenarios (problem questions) and are asked to analyze the legal issues involved. These require students to apply specific legal principles and statutes to the facts presented
in the scenario.Multiple Ch
ALTERNATIVELY: To solve the exercise by minimizing the electrostatic energy, note
that the only variable terms in the sum-over-all-pairs expression for the energy are the
ones involving the diagonals of the rhombus. The other four pairs involve the sides of
2
the rhombus which are of fixed length. The variable terms are q /4πϵ0(2ℓ sin θ) and
2
Q /4πϵ0(2ℓ cos θ). Minimizing this as a function of θ yields
( 2 )
d q Q2 2 cos θ 2 sin θ q2 3
y sin θ cos θ
e
e
1.40. Zero potential energy
Let’s first consider the general case where the three charges don’t necessarily lie on
2
the same line. Without loss of generality, we can put the two electrons on the x axis a
unit distance apart (that is, at the values x = 1/2), as shown in Fig. 1. And we may
-2 -1 -e -e 1 2 x assume the proton lies in the xy plane. For an arbitrary location of the proton in this
plane, let the distances from the electrons be r1 and r2. Then setting the potential
Figure 1 energy of the system equal to zero gives
( 2 2 )
1 e e e2 1 1
4πϵ
One obvious location satisfying√this requirement has the proton on the y axis with
r1 = r2 = 2, that is, with y = 15/2 ≈ 1.94. In general, Eq. (17) defines a curve in
the xy plane, and a surface of revolution around the x axis in space. This surface is
the set of all points where the proton can be placed to give U = 0. The surface looks
something like a prolate ellipsoid, but it isn’t.
Let’s now consider the case where all three charges lie on the x axis. Assume that
the proton lies to the right of the right electron. We then have r1 = x − 1/2 and
r2 = x + 1/2, so Eq. (17) becomes
√
1 1 2 2± 5
+ = 1 =⇒ x − 2x − 1/4 = 0 =⇒ x = . (18)
x − 1/2 x + 1/2 2
The negative root must be thrown out because it violates our assumption that x > 1/2.
(With x < 1/2, the distance r1 isn’t represented by x − 1/2)√. So we find x = 2.118.
The distance from the right electron at x = 1/2 equals (1 + 5)/2. The ratio of this
