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
−2
1.93 · 10 g
·
6.02 10·23 = 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
9 kg
q2 = 9 · 10 (9.4 C)2
(
m3 11
F =k
r2 s2 C2 = 8 · 10 N. (2)
(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 e2/4πϵ0d2 to equal the downward
gravitational force mg. Hence,
(
1 e2 kg m 3) (1.6 ·10 −19 C)2 = 26 m2, (3)
d2 = = 9 · 109
−31 2
4πϵ0 mg s C (9 · 10 kg)(9.8 m/s )
2 2
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 q2/4πϵ0r2
Therefore,
(
2 1 2 s C
q = (4πϵ )mgr = (0.4)π .85 · 10−12 2 2 ) (0.3 kg)(9.8 m/s2)(0.5 m)2
0 8
10 kg m3
−12 2 −6
= 8.17 · 10 C =⇒ q = 2.9 · 10 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)
2 q 2 Qq
F = √ q + 2 cos 45◦ − √ . (6)
( 2 ℓ)2 ℓ2 (ℓ/ 2)2
Setting this equal to zero gives
( )
Q= 1 + 1 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
( )
1 (−Q)q q2 2 )
√ ( q q
U= 4 +4 q = 4 2q = 0, (8)
4πϵ0 ℓ/ 2 +2·√ 4πϵ0ℓ −Q + √ + 4
2ℓ 2
· √ ·
ℓ
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) = − 0 (ℓ − x)2 (ℓ + x)2
4πϵ ( )
Qq 1 1
≈ − −
4πϵ0ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ − 2
(1 + 2x/ℓ) − (1 − 2x/ℓ)
4πϵ0ℓ
Qqx
= − . (9)
πϵ ℓ03
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)
πϵ0ℓ3 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 far as the units
go, Qq/ϵ0ℓ2 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 +
)
U (x) = 1 . (11)
4πϵ0 ℓ −x ℓ+x
We’ll need to expand things to order x2 because the order x terms will cancel:
Qq ( 1 +
)
U (x) = 1
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq 1+ x x2 x + x22
≈ + + 1− ℓ ℓ
4πϵ0ℓ ( ℓ ) ℓ2
Qq 2x 2
= 2+ 2 . (12)
4πϵ0ℓ 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) = − =− , 0 (13)
dx πϵ 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
, 4 CHAPTER 1. ELECTROSTATICS
forces is equal and opposite to the sum of the vertical components of the tensions. This gives
(
qQ ) q2 q2 qQ . (14)
2 sin θ+ 4πϵ (2ℓ sin θ)2 = 2T sin θ =⇒ 3 = 2T ℓ2−
4πϵ ℓ 2 0 16πϵ 0 sin θ 2πϵ 0
0
Similarly, each charge Q is in equilibrium if
(
qQ ) Q2 Q2 qQ
2 cos θ + = 2T cos θ =⇒ = 2T ℓ2 −
.
4πϵ0ℓ2 4πϵ0(2ℓ cos θ)2 16πϵ0 cos3 θ
2πϵ0
(15)
The righthand sides of the two preceding equations are equal, so the same must be true of the
lefthand sides. This yields q 2/ sin3 θ = Q2/ cos3 θ, or q2/Q2 = tan3 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.
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 the rhombus
which are of fixed length. The variable terms are q2/4πϵ0(2ℓ sin θ) and Q2/4πϵ0(2ℓ cos θ). Minimizing
this as a function of θ yields
(
d q2 2 ) cos θ sin θ q2
0= + Q = −q2 + Q2 =⇒ = tan3 θ. (16)
sin θ cos θ 2 cos 2θ Q 2
y dθ sin θ
e
e
1.40. Zero potential energy
r2 Let’s first consider the general case where the three charges don’t necessarily lie on the same line.
2 r1 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
x assume the proton lies in the xy plane. For an arbitrary location of the proton in this
-2 -1 -e -e 1 2
plane, let the distances from the electrons be r1 and r2. Then setting the potential energy of the
Figure 1 system equal to zero gives
1 ( e2 ) 1 1
U= e2 e2 + = 1. (17)
=⇒
4πϵ0 1 — — r1 r2
r1 r2
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± 5
+ = 1 =⇒ x2 − 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
, 5
distance to the distance between the electrons (which is just 1) is therefore the golden ratio. If we
assume x < −1/2, then the mirror image at x = −2.118 works equally well. You can quickly check
that there is no solution for x between the electrons, that is, in the region −1/2 < x < 1/2. There are
therefore two solutions with all three charges on the same line.
1.41. Work for an octahedron
Consider an edge that has two protons at its ends (you can quickly show that at least one such edge a
must exist). There are two options for where the third proton is. It can be at one of the two vertices
such that the triangle formed by the three protons is a face of the octahedron. Or it can be at one
of the other two vertices. These two possibilities are shown in Fig. 2.
There are 15 pairs of charges, namely the 12 edges and the 3 internal diagonals. Summing
over these pairs gives the potential energy. By examining the two cases √
shown, you can show that for the first configuration the sum is (the term with the 2
comes from the internal diagonals)
)
2 ( 1 1 1 e2
U= e 6· 3 = −(2.121) . (19)
− 6 · a − · √2 a
a
4πϵ0 4πϵ0a
a
And for the second configuration:
(
e2 1 1 1 ) e2
U= 4· + 2 ·√ − 1· √ 1 = −(3.293) . (20)
4πϵ0 a −8 · a 2a 2a 4πϵ0a
Both of these results are negative. This means that energy is released as the octahedron is assembled.
Equivalently, it takes work to separate the charges out to infinity. You should think about why the Figure 2
energy is more negative in the second case. (Hint: the two cases differ only in the locations of the
leftmost two charges.)
1.42. Potential energy in a 1-D crystal
Suppose the array has been built inward from the left (that is, from negative infinity) as far as a
particular negative ion. To add the next positive ion on the right, the amount of external work
required is
( 2 (
1 e e2 e2 ) e2 1 1 ). (21)
+ − 1− + − 1 + ···
+ ··· 1 =−
−
4πϵ0 a 2a 3a 4πϵ0 a 2 3 4
The expansion of ln(1 + x) is x − x2/2 + x3/3 − · · · , converging for −1 < x ≤ 1. Evidently the sum
in parentheses above is just ln 2, or 0.693. The energy of the infinite chain per ion is therefore
−(0.693)e2/4πϵ0a. Note that this is an exact result; it does not assume that a is small. After all, it
wouldn’t make any sense to say that
“a is small,” because there is no other length scale in the setup that we can compare
a with.
The addition of further particles on the right doesn’t affect the energy involved in assembling the
previous ones, so this result is indeed the energy per ion in the entire infinite (in both directions) chain.
The result is negative, which means that it requires energy to move the ions away from each other.
This makes sense, because the two nearest neighbors are of the opposite sign.
If the signs of all the ions were the same (instead of alternating), then the sum in Eq. (21) would
be (1 + 1/2 + 1/3 + 1/4 + · · · ), which diverges. It would take an infinite amount of energy to assemble
such a chain.
, 6 CHAPTER 1. ELECTROSTATICS
An alternative solution is to compute the potential energy of a given ion due to the full infinite (in both
directions) chain. This is essentially the same calculation as above, except with a factor of 2 due to
the ions on each side of the given one. If we then sum over all ions (or a very large number N ) to find
the total energy of the chain, we have counted each pair twice. So in finding the potential energy per
ion, we must divide by 2 (along with N ). The factors of 2 and N cancel, and we arrive at the above
result.
1.43. Potential energy in a 3-D crystal
The solution is the same as the solution to Problem 1.7, except that we have an additional
term. We now also need to consider the “half-space” on top of the ion, in addition to the half-plane
above it and the half-line to the right of it. In Fig. 12.4 the half-space of ions is on top of the plane of
the paper (from where you are viewing the page).
If we index the ions by the coordinates (m, n, p), then the potential energy of the ion at (0, 0, 0) due to
the half-line, half-plane, and half-space is
(∑ ∞ ∞ ∞ )
2 ∞
(−1) m ∞
∑
∞ (−1)m+n ∑ ∑ ( 1) m+n+p
U= e √ ∑ √ − .
4πϵ0a m=1 m +∑ m2 + n2 + p=1 n=−∞ m=−∞ m2 + n 2 + p 2
n=1 m=−∞
(22)
The triple sum takes more computer time than the other two sums. Taking the limits to be 300 instead
of ∞ in the triple sum, and 1000 in the other two, we obtain decent enough results via Mathematica.
We find
e2 (−0.693 − 0.115 − 0.066) = − (0.874)e2
U= , (23)
4πϵ0a 4πϵ0a
which agrees with Eq. (1.18) to three digits. This result is negative, which means that it requires
energy to move the ions away from each other. This makes sense, because the six nearest neighbors
are of the opposite sign.
1.44. Chessboard
W is probably going to be positive, because the four nearest neighbors are all of the opposite sign.
Fig. 3 shows a quarter (or actually slightly more than a quarter) of a 7 × 7 chessboard. Three
different groups of charges are circled. The full chessboard
consists of four of the horizontal group, four of the diagonal group, and eight of the
triangular group. Adding up the work associated with each group, the total work required to move
Figure 3 the central charge to a position far away is (in units of e2/4πϵ0s)
( ) ( ) ( )
1 1 1 1 1 1 1 1
1
W =4 − + +4 −√ − √ − √ +8 √ −√ +√ ≈ 1.4146,
1 2 3 2 2 2 3 2 5 10 13
(24)
which is positive, as we guessed.
For larger arrays we can use a Mathematica program to calculate W . If we have an N × N
chessboard, and if we define H by 2H + 1 = N (for example, H = 50 corresponds to N = 101),
then the following program gives the work W required to remove the central charge from a 101 × 101
chessboard.
H=50;
4*Sum[(-1)^(n+1)/n, {n,1,H}] +
4*Sum[Sum[(-1)^(n+m+1)/(n^2+m^2)^(.5), {n,1,H}], {m,1,H}]
, H
H
7
This program involves dividing the chessboard into the regions shown in Fig. 4; the sub-squares
have side length H. (If you want, you can reduce the computing time by about a factor of 2 by
dividing the chessboard as we did in Fig. 3.) The results for
various N × N chessboards are (in units of e2/4πϵ0s):
Figure 4
N 3 7 101 1001 10,001 100,001
W 1.1716 1.4146 1.6015 1.6141 1.6154 1.6155
The W for an infinite chessboard is apparently roughly equal to (1.6155)e2/4πϵ0s. The prefactor
here is double the 0.808 prefactor in the result for Problem 1.7, due to the fact that the latter is the
energy per ion, so there is the usual issue of double counting.
1.45. Zero field?
y
The setup is shown in Fig. 5. We know that Ey = 0 on the y axis, by symmetry, so we need only
worry about Ex. We want the leftward Ex from the two middle charges to cancel the rightward Ex from a 2a
the two outer charges. This implies that q -q q -q
Figure 5
1 q a 1 q 3a
2· ·√
4πϵ0 y2 + a2 y 2 + a2 =2· ·√ y2 + (3a)2 , (25)
4πϵ 0 y2 + (3a)2
where the second factor on each side of the equation comes from the act of taking the horizontal
component. Simplifying this gives
1 3 2 2 2/3 2 2
2 2 3/2
= 2 2 3/2
=⇒ y + 9a = 3 (y + a )
(y + a ) (y + 9a )
√
=⇒ y = a 9 − 3
2/3
≈(2.53)a. (26)
32/3 − 1
In retrospect, we know that there must exist a point on the y axis with Ex = 0, by a continuity
argument. For small y, the field points leftward, because the two middle charges dominate. But for
large y, the field points rightward, because the two outer charges dominate. (This is true because for
large y, the distances to the four charges are all essentially the same, but the slope of the lines to
the outer charges is smaller than the slope of the lines to the middle charges (it is 1/3 as large). So
the
x component of the field due to the outer charges is 3 times as large, all other things Q
being equal.) Therefore, by continuity, there must exist a point on the y axis where
Ex equals zero.
Rcos
1.46. Charges on a circular track
Let’s work with the general angle θ shown in Fig. 6. In the problem at hand, 4θ = 90◦, so θ = 22.5◦. The
tangential electric field at one of the q’s due to Q is 4
Q R
sin θ, (27)
4πϵ0(2R cos θ)2 q q
and the tangential field (in the opposite direction) at one q due to the other q is
2
q
cos 2θ. (28) Figure 6
4πϵ0(2Rsin 2θ)2
Equating these fields gives
Q q cos2θ cos 2θ cos 2θ
sin θ = cos 2θ =⇒ Q = q =q , (29)
(cos θ)2 (sin 2θ)2 sin θ sin2 2θ 4 sin3 θ
, 8 CHAPTER 1. ELECTROSTATICS
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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
−2
1.93 · 10 g
·
6.02 10·23 = 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
9 kg
q2 = 9 · 10 (9.4 C)2
(
m3 11
F =k
r2 s2 C2 = 8 · 10 N. (2)
(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 e2/4πϵ0d2 to equal the downward
gravitational force mg. Hence,
(
1 e2 kg m 3) (1.6 ·10 −19 C)2 = 26 m2, (3)
d2 = = 9 · 109
−31 2
4πϵ0 mg s C (9 · 10 kg)(9.8 m/s )
2 2
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 q2/4πϵ0r2
Therefore,
(
2 1 2 s C
q = (4πϵ )mgr = (0.4)π .85 · 10−12 2 2 ) (0.3 kg)(9.8 m/s2)(0.5 m)2
0 8
10 kg m3
−12 2 −6
= 8.17 · 10 C =⇒ q = 2.9 · 10 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)
2 q 2 Qq
F = √ q + 2 cos 45◦ − √ . (6)
( 2 ℓ)2 ℓ2 (ℓ/ 2)2
Setting this equal to zero gives
( )
Q= 1 + 1 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
( )
1 (−Q)q q2 2 )
√ ( q q
U= 4 +4 q = 4 2q = 0, (8)
4πϵ0 ℓ/ 2 +2·√ 4πϵ0ℓ −Q + √ + 4
2ℓ 2
· √ ·
ℓ
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) = − 0 (ℓ − x)2 (ℓ + x)2
4πϵ ( )
Qq 1 1
≈ − −
4πϵ0ℓ2 1 − 2x/ℓ 1 + 2x/ℓ
Qq ( )
≈ − 2
(1 + 2x/ℓ) − (1 − 2x/ℓ)
4πϵ0ℓ
Qqx
= − . (9)
πϵ ℓ03
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)
πϵ0ℓ3 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 far as the units
go, Qq/ϵ0ℓ2 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 +
)
U (x) = 1 . (11)
4πϵ0 ℓ −x ℓ+x
We’ll need to expand things to order x2 because the order x terms will cancel:
Qq ( 1 +
)
U (x) = 1
4πϵ0ℓ 1 − x/ℓ 1 + x/ℓ
(( ) ( ))
Qq 1+ x x2 x + x22
≈ + + 1− ℓ ℓ
4πϵ0ℓ ( ℓ ) ℓ2
Qq 2x 2
= 2+ 2 . (12)
4πϵ0ℓ 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) = − =− , 0 (13)
dx πϵ 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
, 4 CHAPTER 1. ELECTROSTATICS
forces is equal and opposite to the sum of the vertical components of the tensions. This gives
(
qQ ) q2 q2 qQ . (14)
2 sin θ+ 4πϵ (2ℓ sin θ)2 = 2T sin θ =⇒ 3 = 2T ℓ2−
4πϵ ℓ 2 0 16πϵ 0 sin θ 2πϵ 0
0
Similarly, each charge Q is in equilibrium if
(
qQ ) Q2 Q2 qQ
2 cos θ + = 2T cos θ =⇒ = 2T ℓ2 −
.
4πϵ0ℓ2 4πϵ0(2ℓ cos θ)2 16πϵ0 cos3 θ
2πϵ0
(15)
The righthand sides of the two preceding equations are equal, so the same must be true of the
lefthand sides. This yields q 2/ sin3 θ = Q2/ cos3 θ, or q2/Q2 = tan3 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.
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 the rhombus
which are of fixed length. The variable terms are q2/4πϵ0(2ℓ sin θ) and Q2/4πϵ0(2ℓ cos θ). Minimizing
this as a function of θ yields
(
d q2 2 ) cos θ sin θ q2
0= + Q = −q2 + Q2 =⇒ = tan3 θ. (16)
sin θ cos θ 2 cos 2θ Q 2
y dθ sin θ
e
e
1.40. Zero potential energy
r2 Let’s first consider the general case where the three charges don’t necessarily lie on the same line.
2 r1 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
x assume the proton lies in the xy plane. For an arbitrary location of the proton in this
-2 -1 -e -e 1 2
plane, let the distances from the electrons be r1 and r2. Then setting the potential energy of the
Figure 1 system equal to zero gives
1 ( e2 ) 1 1
U= e2 e2 + = 1. (17)
=⇒
4πϵ0 1 — — r1 r2
r1 r2
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± 5
+ = 1 =⇒ x2 − 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
, 5
distance to the distance between the electrons (which is just 1) is therefore the golden ratio. If we
assume x < −1/2, then the mirror image at x = −2.118 works equally well. You can quickly check
that there is no solution for x between the electrons, that is, in the region −1/2 < x < 1/2. There are
therefore two solutions with all three charges on the same line.
1.41. Work for an octahedron
Consider an edge that has two protons at its ends (you can quickly show that at least one such edge a
must exist). There are two options for where the third proton is. It can be at one of the two vertices
such that the triangle formed by the three protons is a face of the octahedron. Or it can be at one
of the other two vertices. These two possibilities are shown in Fig. 2.
There are 15 pairs of charges, namely the 12 edges and the 3 internal diagonals. Summing
over these pairs gives the potential energy. By examining the two cases √
shown, you can show that for the first configuration the sum is (the term with the 2
comes from the internal diagonals)
)
2 ( 1 1 1 e2
U= e 6· 3 = −(2.121) . (19)
− 6 · a − · √2 a
a
4πϵ0 4πϵ0a
a
And for the second configuration:
(
e2 1 1 1 ) e2
U= 4· + 2 ·√ − 1· √ 1 = −(3.293) . (20)
4πϵ0 a −8 · a 2a 2a 4πϵ0a
Both of these results are negative. This means that energy is released as the octahedron is assembled.
Equivalently, it takes work to separate the charges out to infinity. You should think about why the Figure 2
energy is more negative in the second case. (Hint: the two cases differ only in the locations of the
leftmost two charges.)
1.42. Potential energy in a 1-D crystal
Suppose the array has been built inward from the left (that is, from negative infinity) as far as a
particular negative ion. To add the next positive ion on the right, the amount of external work
required is
( 2 (
1 e e2 e2 ) e2 1 1 ). (21)
+ − 1− + − 1 + ···
+ ··· 1 =−
−
4πϵ0 a 2a 3a 4πϵ0 a 2 3 4
The expansion of ln(1 + x) is x − x2/2 + x3/3 − · · · , converging for −1 < x ≤ 1. Evidently the sum
in parentheses above is just ln 2, or 0.693. The energy of the infinite chain per ion is therefore
−(0.693)e2/4πϵ0a. Note that this is an exact result; it does not assume that a is small. After all, it
wouldn’t make any sense to say that
“a is small,” because there is no other length scale in the setup that we can compare
a with.
The addition of further particles on the right doesn’t affect the energy involved in assembling the
previous ones, so this result is indeed the energy per ion in the entire infinite (in both directions) chain.
The result is negative, which means that it requires energy to move the ions away from each other.
This makes sense, because the two nearest neighbors are of the opposite sign.
If the signs of all the ions were the same (instead of alternating), then the sum in Eq. (21) would
be (1 + 1/2 + 1/3 + 1/4 + · · · ), which diverges. It would take an infinite amount of energy to assemble
such a chain.
, 6 CHAPTER 1. ELECTROSTATICS
An alternative solution is to compute the potential energy of a given ion due to the full infinite (in both
directions) chain. This is essentially the same calculation as above, except with a factor of 2 due to
the ions on each side of the given one. If we then sum over all ions (or a very large number N ) to find
the total energy of the chain, we have counted each pair twice. So in finding the potential energy per
ion, we must divide by 2 (along with N ). The factors of 2 and N cancel, and we arrive at the above
result.
1.43. Potential energy in a 3-D crystal
The solution is the same as the solution to Problem 1.7, except that we have an additional
term. We now also need to consider the “half-space” on top of the ion, in addition to the half-plane
above it and the half-line to the right of it. In Fig. 12.4 the half-space of ions is on top of the plane of
the paper (from where you are viewing the page).
If we index the ions by the coordinates (m, n, p), then the potential energy of the ion at (0, 0, 0) due to
the half-line, half-plane, and half-space is
(∑ ∞ ∞ ∞ )
2 ∞
(−1) m ∞
∑
∞ (−1)m+n ∑ ∑ ( 1) m+n+p
U= e √ ∑ √ − .
4πϵ0a m=1 m +∑ m2 + n2 + p=1 n=−∞ m=−∞ m2 + n 2 + p 2
n=1 m=−∞
(22)
The triple sum takes more computer time than the other two sums. Taking the limits to be 300 instead
of ∞ in the triple sum, and 1000 in the other two, we obtain decent enough results via Mathematica.
We find
e2 (−0.693 − 0.115 − 0.066) = − (0.874)e2
U= , (23)
4πϵ0a 4πϵ0a
which agrees with Eq. (1.18) to three digits. This result is negative, which means that it requires
energy to move the ions away from each other. This makes sense, because the six nearest neighbors
are of the opposite sign.
1.44. Chessboard
W is probably going to be positive, because the four nearest neighbors are all of the opposite sign.
Fig. 3 shows a quarter (or actually slightly more than a quarter) of a 7 × 7 chessboard. Three
different groups of charges are circled. The full chessboard
consists of four of the horizontal group, four of the diagonal group, and eight of the
triangular group. Adding up the work associated with each group, the total work required to move
Figure 3 the central charge to a position far away is (in units of e2/4πϵ0s)
( ) ( ) ( )
1 1 1 1 1 1 1 1
1
W =4 − + +4 −√ − √ − √ +8 √ −√ +√ ≈ 1.4146,
1 2 3 2 2 2 3 2 5 10 13
(24)
which is positive, as we guessed.
For larger arrays we can use a Mathematica program to calculate W . If we have an N × N
chessboard, and if we define H by 2H + 1 = N (for example, H = 50 corresponds to N = 101),
then the following program gives the work W required to remove the central charge from a 101 × 101
chessboard.
H=50;
4*Sum[(-1)^(n+1)/n, {n,1,H}] +
4*Sum[Sum[(-1)^(n+m+1)/(n^2+m^2)^(.5), {n,1,H}], {m,1,H}]
, H
H
7
This program involves dividing the chessboard into the regions shown in Fig. 4; the sub-squares
have side length H. (If you want, you can reduce the computing time by about a factor of 2 by
dividing the chessboard as we did in Fig. 3.) The results for
various N × N chessboards are (in units of e2/4πϵ0s):
Figure 4
N 3 7 101 1001 10,001 100,001
W 1.1716 1.4146 1.6015 1.6141 1.6154 1.6155
The W for an infinite chessboard is apparently roughly equal to (1.6155)e2/4πϵ0s. The prefactor
here is double the 0.808 prefactor in the result for Problem 1.7, due to the fact that the latter is the
energy per ion, so there is the usual issue of double counting.
1.45. Zero field?
y
The setup is shown in Fig. 5. We know that Ey = 0 on the y axis, by symmetry, so we need only
worry about Ex. We want the leftward Ex from the two middle charges to cancel the rightward Ex from a 2a
the two outer charges. This implies that q -q q -q
Figure 5
1 q a 1 q 3a
2· ·√
4πϵ0 y2 + a2 y 2 + a2 =2· ·√ y2 + (3a)2 , (25)
4πϵ 0 y2 + (3a)2
where the second factor on each side of the equation comes from the act of taking the horizontal
component. Simplifying this gives
1 3 2 2 2/3 2 2
2 2 3/2
= 2 2 3/2
=⇒ y + 9a = 3 (y + a )
(y + a ) (y + 9a )
√
=⇒ y = a 9 − 3
2/3
≈(2.53)a. (26)
32/3 − 1
In retrospect, we know that there must exist a point on the y axis with Ex = 0, by a continuity
argument. For small y, the field points leftward, because the two middle charges dominate. But for
large y, the field points rightward, because the two outer charges dominate. (This is true because for
large y, the distances to the four charges are all essentially the same, but the slope of the lines to
the outer charges is smaller than the slope of the lines to the middle charges (it is 1/3 as large). So
the
x component of the field due to the outer charges is 3 times as large, all other things Q
being equal.) Therefore, by continuity, there must exist a point on the y axis where
Ex equals zero.
Rcos
1.46. Charges on a circular track
Let’s work with the general angle θ shown in Fig. 6. In the problem at hand, 4θ = 90◦, so θ = 22.5◦. The
tangential electric field at one of the q’s due to Q is 4
Q R
sin θ, (27)
4πϵ0(2R cos θ)2 q q
and the tangential field (in the opposite direction) at one q due to the other q is
2
q
cos 2θ. (28) Figure 6
4πϵ0(2Rsin 2θ)2
Equating these fields gives
Q q cos2θ cos 2θ cos 2θ
sin θ = cos 2θ =⇒ Q = q =q , (29)
(cos θ)2 (sin 2θ)2 sin θ sin2 2θ 4 sin3 θ
, 8 CHAPTER 1. ELECTROSTATICS
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