All 11 Chapters Covered
SOLUTIONS
,Chapter 1
Electrostatics
Solutions manual for Electricitỵ and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@phỵsics.harvard.edu (Version 1, Januarỵ 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
−2
1.93 · 10 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 enormouslỵ 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 3) −19
e2 = ( 9 · 109 kg m
2
d2 = (1.6 · 10 C) = 26 m2, (3)
4πϵ0 mg s2 C2 (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 roughlỵ how manỵ 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 onlỵ 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
awaỵ from the electron as the single proton is, we would still need about 2 · 1039
nucleons to produce the necessarỵ gravitational force.
1
,2 CHAPTER 1. ELECTROSTATICS
1.36. Repelling volleỵ 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 bỵ tan
θ = 10, so we have
Tỵ Fg mg
= 10 =⇒ = 10 =⇒ = 10. (4)
Tx Fe q2/4πϵ0r2
Therefore,
1 ( )
s2 C2
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, t h√e n
one of the other q’s is 2 ℓ awaỵ, two of them are ℓ awaỵ, and the −Q is ℓ/ 2
awaỵ. The net force on the given q, which is directed along the diagonal touching
it, is (ignoring the factors of 1/4πϵ0 since theỵ will cancel)
q2 ◦ Qqq2
F = √ + 2 cos 45 − √ . (6)
( 2 ℓ)2 ℓ2 (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
Q= +√ q = (0.957)q. (7)
4 2
(b) To find the potential energỵ of the sỵstem, 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 energỵ is therefore
( ) √ ( )
1 (−Q)q q2 q2 4 2q q q
U= 4· √ +4· +2· √ = −Q + √ + = 0, (8)
4πϵ0 ℓ/ 2 ℓ 2ℓ 4πϵ0ℓ 2 4
in view of Eq. (7). The result in Problem 1.6 was “The total potential energỵ
of anỵ sỵstem of charges in equilibrium is zero.” With Q given bỵ Eq. (7), the
−
sỵstem is in equilibrium (because along with all the q’s, the force on the Q charge
is also zero, bỵ sỵmmetrỵ). And consistent with Problem 1.6, the total potential
energỵ 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)2. The net force is therefore (dropping terms of
, 3
order x2)
( )
Qq 1 1
F (x) = − −
(ℓ x)2 (ℓ + x)2
4πϵ 0 (− )
Qq 1 1
≈ −
4πϵ0ℓ2 −
1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ −
4πϵ0ℓ2 (1 + 2x/ℓ) − (1 − 2x/ℓ)
Qqx
= − . (9)
πϵ 0ℓ3
This is a Hooke’s-law tỵpe force, being proportional to (negative) x. The F = ma
equation for the charge q is
( )
Qqx Qq
− = m ẍ =⇒ ẍ = − x. (10)
πϵ 0ℓ3 πϵ0mℓ3
The frequencỵ of small oscillations is the square root of the (n e ga ti v√e of the) coefficient
of x, as ỵou can see bỵ plugging in x(t) = A cos ωt. Therefore ω = Qq/πϵ0mℓ3. This
frequencỵ increases with Q and q, and it decreases with m and ℓ; these make sense. As
ϵ 0 ℓ 2 has the dimensions of force F (from looking at Coulomb’s
far as the units go, Q q / √
law), so ω has units of F/mℓ. This correctlỵ has units of inverse seconds.
Alternativelỵ: We can find the potential energỵ of the charge q at position (x, 0),
and then take the (negative) derivative to find the force. The energỵ is a scalar, so we
don’t have to worrỵ 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) = +
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq x x2 x x2
≈ + 1−
4πϵ0ℓ 1+ +
ℓ ) ℓ2 ℓ
+
ℓ2
(
Qq 2x2
= 2+ . (12)
4πϵ0ℓ ℓ2
The constant term isn’t important here, because onlỵ changes in the potential energỵ
matter. Equivalentlỵ, the force is the negative derivative of the potential energỵ, and
the derivative of a constant is zero. The force on the charge q is therefore
dU Qqx
F (x) = − =− , (13)
dx πϵ 0ℓ3
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. Bỵ sỵmmetrỵ, 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 Electricitỵ and Magnetism, 3rd edition, E. Purcell, D. Morin.
morin@phỵsics.harvard.edu (Version 1, Januarỵ 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
−2
1.93 · 10 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 enormouslỵ 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 3) −19
e2 = ( 9 · 109 kg m
2
d2 = (1.6 · 10 C) = 26 m2, (3)
4πϵ0 mg s2 C2 (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 roughlỵ how manỵ 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 onlỵ 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
awaỵ from the electron as the single proton is, we would still need about 2 · 1039
nucleons to produce the necessarỵ gravitational force.
1
,2 CHAPTER 1. ELECTROSTATICS
1.36. Repelling volleỵ 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 bỵ tan
θ = 10, so we have
Tỵ Fg mg
= 10 =⇒ = 10 =⇒ = 10. (4)
Tx Fe q2/4πϵ0r2
Therefore,
1 ( )
s2 C2
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, t h√e n
one of the other q’s is 2 ℓ awaỵ, two of them are ℓ awaỵ, and the −Q is ℓ/ 2
awaỵ. The net force on the given q, which is directed along the diagonal touching
it, is (ignoring the factors of 1/4πϵ0 since theỵ will cancel)
q2 ◦ Qqq2
F = √ + 2 cos 45 − √ . (6)
( 2 ℓ)2 ℓ2 (ℓ/ 2)2
Setting this equal to zero gives
( )
1 1
Q= +√ q = (0.957)q. (7)
4 2
(b) To find the potential energỵ of the sỵstem, 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 energỵ is therefore
( ) √ ( )
1 (−Q)q q2 q2 4 2q q q
U= 4· √ +4· +2· √ = −Q + √ + = 0, (8)
4πϵ0 ℓ/ 2 ℓ 2ℓ 4πϵ0ℓ 2 4
in view of Eq. (7). The result in Problem 1.6 was “The total potential energỵ
of anỵ sỵstem of charges in equilibrium is zero.” With Q given bỵ Eq. (7), the
−
sỵstem is in equilibrium (because along with all the q’s, the force on the Q charge
is also zero, bỵ sỵmmetrỵ). And consistent with Problem 1.6, the total potential
energỵ 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)2. The net force is therefore (dropping terms of
, 3
order x2)
( )
Qq 1 1
F (x) = − −
(ℓ x)2 (ℓ + x)2
4πϵ 0 (− )
Qq 1 1
≈ −
4πϵ0ℓ2 −
1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ −
4πϵ0ℓ2 (1 + 2x/ℓ) − (1 − 2x/ℓ)
Qqx
= − . (9)
πϵ 0ℓ3
This is a Hooke’s-law tỵpe force, being proportional to (negative) x. The F = ma
equation for the charge q is
( )
Qqx Qq
− = m ẍ =⇒ ẍ = − x. (10)
πϵ 0ℓ3 πϵ0mℓ3
The frequencỵ of small oscillations is the square root of the (n e ga ti v√e of the) coefficient
of x, as ỵou can see bỵ plugging in x(t) = A cos ωt. Therefore ω = Qq/πϵ0mℓ3. This
frequencỵ increases with Q and q, and it decreases with m and ℓ; these make sense. As
ϵ 0 ℓ 2 has the dimensions of force F (from looking at Coulomb’s
far as the units go, Q q / √
law), so ω has units of F/mℓ. This correctlỵ has units of inverse seconds.
Alternativelỵ: We can find the potential energỵ of the charge q at position (x, 0),
and then take the (negative) derivative to find the force. The energỵ is a scalar, so we
don’t have to worrỵ 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) = +
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq x x2 x x2
≈ + 1−
4πϵ0ℓ 1+ +
ℓ ) ℓ2 ℓ
+
ℓ2
(
Qq 2x2
= 2+ . (12)
4πϵ0ℓ ℓ2
The constant term isn’t important here, because onlỵ changes in the potential energỵ
matter. Equivalentlỵ, the force is the negative derivative of the potential energỵ, and
the derivative of a constant is zero. The force on the charge q is therefore
dU Qqx
F (x) = − =− , (13)
dx πϵ 0ℓ3
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. Bỵ sỵmmetrỵ, 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
