All 11 Chapters Covered
SOLUTIONS
,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
The volume of a cube 1 mm on a side is 10−3 cm3 . So the mass of this 1 mm cube is
1.93 · 10−2 g. The number of atoms in the cube is therefore
1.93 · 10−2 g
6.02 · 1023 · = 5.9 · 1019 . (1)
197 g
Each atom has a positive charge of 1 e = 1.6 · 10−19 C, so the total charge in the cube
is (5.9 · 1019 )(1.6 · 10−19 C) = 9.4 C. The repulsive force between two such cubes 1 m
apart is therefore
q2 kg m3 (9.4 C)2
( )
F = k 2 = 9 · 109 2 2 = 8 · 1011 N. (2)
r s C (1 m)2
The weight of an aircraft carrier is mg = (108 kg)(9.8 m/s2 ) ≈ 109 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
Let the desired distance be d. We want the upward electric force e2 /4πϵ0 d2 to equal
the downward gravitational force mg. Hence,
1 e2 ( kg m3 ) (1.6 · 10−19 C)2
d2 = = 9 · 109 2 2 = 26 m2 , (3)
4πϵ0 mg s C (9 · 10−31 kg)(9.8 m/s2 )
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
3.6·1051 nucleons (that’s roughly how many are in the earth) to produce a gravitational
force at an effective distance of 6.4 · 106 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
accounts for a factor of only 1.6 · 1012 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
away from the electron as the single proton is, we would still need about 2 · 1039
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 q 2 /4πϵ 0r
2
Therefore,
s2 C 2
( )
1
q2 = (4πϵ0 )mgr2 = (0.4)π 8.85 · 10−12 (0.3 kg)(9.8 m/s2 )(0.5 m)2
10 kg m3
= 8.17 · 10−12 C2 =⇒ q = 2.9 · 10−6 C. (5)
1.37. Zero force at the corners
(a) Consider a charge q at √ a particular corner. If the square has side length ℓ, then
√
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)
q2 q2 Qq
F = √ + 2 cos 45◦ 2 − √ . (6)
( 2 ℓ) 2 ℓ (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
Q= +√ q = (0.957)q. (7)
4 2
(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
√ (
q2 q2
( ) )
1 (−Q)q 4 2q q q
U= 4· √ +4· +2· √ = −Q + √ + = 0, (8)
4πϵ0 ℓ/ 2 ℓ 2ℓ 4πϵ0 ℓ 2 4
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)2 , where the minus sign indicates leftward. And the force from the
left charge Q equals Qq/4πϵ0 (ℓ + x)2 . The net force is therefore (dropping terms of
, 3
order x2 )
( )
Qq 1 1
F (x) = − −
4πϵ0 (ℓ − x)2 (ℓ + x)2
( )
Qq 1 1
≈ − −
4πϵ0 ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ − (1 + 2x/ℓ) − (1 − 2x/ℓ)
4πϵ0 ℓ2
Qqx
= − . (9)
πϵ0 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
− = mẍ =⇒ ẍ = − x. (10)
πϵ0 ℓ3 πϵ0 mℓ3
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/πϵ0 mℓ3 . 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/ϵ√0 ℓ 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
We’ll need to expand things to order x2 because the order x terms will cancel:
( )
Qq 1 1
U (x) = +
4πϵ0 ℓ 1 − x/ℓ 1 + x/ℓ
x x2 x x2
(( ) ( ))
Qq
≈ 1+ + 2 + 1− + 2
4πϵ0 ℓ ℓ ℓ ℓ ℓ
2
( )
Qq 2x
= 2+ 2 . (12)
4πϵ0 ℓ ℓ
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 πϵ0 ℓ3
in agreement with the force in Eq. (9).
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
SOLUTIONS
,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
The volume of a cube 1 mm on a side is 10−3 cm3 . So the mass of this 1 mm cube is
1.93 · 10−2 g. The number of atoms in the cube is therefore
1.93 · 10−2 g
6.02 · 1023 · = 5.9 · 1019 . (1)
197 g
Each atom has a positive charge of 1 e = 1.6 · 10−19 C, so the total charge in the cube
is (5.9 · 1019 )(1.6 · 10−19 C) = 9.4 C. The repulsive force between two such cubes 1 m
apart is therefore
q2 kg m3 (9.4 C)2
( )
F = k 2 = 9 · 109 2 2 = 8 · 1011 N. (2)
r s C (1 m)2
The weight of an aircraft carrier is mg = (108 kg)(9.8 m/s2 ) ≈ 109 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
Let the desired distance be d. We want the upward electric force e2 /4πϵ0 d2 to equal
the downward gravitational force mg. Hence,
1 e2 ( kg m3 ) (1.6 · 10−19 C)2
d2 = = 9 · 109 2 2 = 26 m2 , (3)
4πϵ0 mg s C (9 · 10−31 kg)(9.8 m/s2 )
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
3.6·1051 nucleons (that’s roughly how many are in the earth) to produce a gravitational
force at an effective distance of 6.4 · 106 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
accounts for a factor of only 1.6 · 1012 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
away from the electron as the single proton is, we would still need about 2 · 1039
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 q 2 /4πϵ 0r
2
Therefore,
s2 C 2
( )
1
q2 = (4πϵ0 )mgr2 = (0.4)π 8.85 · 10−12 (0.3 kg)(9.8 m/s2 )(0.5 m)2
10 kg m3
= 8.17 · 10−12 C2 =⇒ q = 2.9 · 10−6 C. (5)
1.37. Zero force at the corners
(a) Consider a charge q at √ a particular corner. If the square has side length ℓ, then
√
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)
q2 q2 Qq
F = √ + 2 cos 45◦ 2 − √ . (6)
( 2 ℓ) 2 ℓ (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
Q= +√ q = (0.957)q. (7)
4 2
(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
√ (
q2 q2
( ) )
1 (−Q)q 4 2q q q
U= 4· √ +4· +2· √ = −Q + √ + = 0, (8)
4πϵ0 ℓ/ 2 ℓ 2ℓ 4πϵ0 ℓ 2 4
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)2 , where the minus sign indicates leftward. And the force from the
left charge Q equals Qq/4πϵ0 (ℓ + x)2 . The net force is therefore (dropping terms of
, 3
order x2 )
( )
Qq 1 1
F (x) = − −
4πϵ0 (ℓ − x)2 (ℓ + x)2
( )
Qq 1 1
≈ − −
4πϵ0 ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ − (1 + 2x/ℓ) − (1 − 2x/ℓ)
4πϵ0 ℓ2
Qqx
= − . (9)
πϵ0 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
− = mẍ =⇒ ẍ = − x. (10)
πϵ0 ℓ3 πϵ0 mℓ3
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/πϵ0 mℓ3 . 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/ϵ√0 ℓ 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
We’ll need to expand things to order x2 because the order x terms will cancel:
( )
Qq 1 1
U (x) = +
4πϵ0 ℓ 1 − x/ℓ 1 + x/ℓ
x x2 x x2
(( ) ( ))
≈ 1+ + 2 + 1− + 2
4πϵ0 ℓ ℓ ℓ ℓ ℓ
2
( )
Qq 2x
= 2+ 2 . (12)
4πϵ0 ℓ ℓ
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 πϵ0 ℓ3
in agreement with the force in Eq. (9).
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
