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Electric Field and Potential Calculations, Manuais, Projetos, Pesquisas de Física

Detailed calculations and derivations related to electric fields and potentials. It covers topics such as the net force on a charge, the electric potential, the electric field on the earth, the electric field and potential inside a slab, and various other electromagnetic phenomena. The descriptions and equations presented in the document could be useful for students studying electromagnetism, physics, or related fields at the university level. The document delves into the mathematical and theoretical aspects of these topics, making it potentially valuable as study notes, lecture notes, or for preparing assignments, university essays, or exams.

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SOLUTIONS MANUAL
Electricity and Magnetism
Third Edition
Edward M. Purcell and David J. Morin
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SOLUTIONS MANUAL

Electricity and Magnetism

Third Edition

Edward M. Purcell and David J. Morin

TO THE INSTRUCTOR: I have tried to pay as much attention to detail in these exercise solutions as I did in the problem solutions in the text. But despite working through each solution numerous times during the various stages of completion, there are bound to be errors. So please let me know if anything looks amiss.

Also, to keep this pdf file from escaping to the web, PLEASE don’t distribute it to anyone, with the exception of your teaching assistants. And please make sure they also agree to this. Once this file gets free, there’s no going back.

In addition to any comments you have on these solutions, I welcome any comments on the book in general. I hope you’re enjoying using it!

David Morin

morin@physics.harvard.edu

(Version 1, January 2013)

⃝c D. Morin, D. Purcell, and F. Purcell 2013

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 Tx

Fg Fe

mg q^2 / 4 πϵ 0 r^2

Therefore,

q^2 =

(4πϵ 0 )mgr^2 = (0.4)π

8. 85 · 10 −^12

s^2 C^2 kg m^3

(0.3 kg)(9.8 m/s^2 )(0.5 m)^2

= 8. 17 · 10 −^12 C^2 =⇒ 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/

away. The net force on the given q, which is directed along the diagonal touching it, is (ignoring the factors of 1/ 4 πϵ 0 since they will cancel)

F =

q^2 (

2 ℓ)^2

  • 2 cos 45◦^

q^2 ℓ^2

Qq (ℓ/

2)^2

Setting this equal to zero gives

Q =

q = (0.957)q. (7)

(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

U =

4 πϵ 0

(−Q)q ℓ/

q^2 ℓ

q^2 √ 2 ℓ

2 q 4 πϵ 0 ℓ

−Q +

q √ 2

q 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/ 4 πϵ 0 (ℓ − x)^2 , where the minus sign indicates leftward. And the force from the left charge Q equals Qq/ 4 πϵ 0 (ℓ + x)^2. The net force is therefore (dropping terms of

order x^2 )

F (x) = −

Qq 4 πϵ 0

(ℓ − x)^2

(ℓ + x)^2

Qq 4 πϵ 0 ℓ^2

1 − 2 x/ℓ

1 + 2x/ℓ

Qq 4 πϵ 0 ℓ^2

(1 + 2x/ℓ) − (1 − 2 x/ℓ)

Qqx πϵ 0 ℓ^3

This is a Hooke’s-law type force, being proportional to (negative) x. The F = ma equation for the charge q is

Qqx πϵ 0 ℓ^3

= m¨x =⇒ x¨ = −

Qq πϵ 0 mℓ^3

x. (10)

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/πϵ 0 mℓ^3. This frequency increases with Q and q, and it decreases with m and ℓ; these make sense. As far as the units go, Qq/ϵ 0 ℓ^2 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

U (x) =

Qq 4 πϵ 0

ℓ − x

ℓ + x

We’ll need to expand things to order x^2 because the order x terms will cancel:

U (x) =

Qq 4 πϵ 0 ℓ

1 − x/ℓ

1 + x/ℓ

Qq 4 πϵ 0 ℓ

x ℓ

x^2 ℓ^2

x ℓ

x^2 ℓ^2

Qq 4 πϵ 0 ℓ

2 x^2 ℓ^2

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

F (x) = −

dU dx

Qqx πϵ 0 ℓ^3

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

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 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.

a a 2

a a 2

Figure 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

comes from the internal diagonals)

U =

e^2 4 πϵ 0

a

a

2 a

e^2 4 πϵ 0 a

And for the second configuration:

U =

e^2 4 πϵ 0

a

a

2 a

2 a

e^2 4 πϵ 0 a

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 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

1 4 πϵ 0

e^2 a

e^2 2 a

e^2 3 a

4 πϵ 0

e^2 a

The expansion of ln(1 + x) is x − x^2 /2 + x^3 / 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)e^2 / 4 πϵ 0 a. 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

U =

e^2 4 πϵ 0 a

m=

(−1)m m

∑^ ∞

n=

∑^ ∞

m=−∞

(−1)m+n √ m^2 + n^2

∑^ ∞

p=

∑^ ∞

n=−∞

m=−∞

(−1)m+n+p √ m^2 + n^2 + p^2

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

U =

e^2 4 πϵ 0 a

(0.874)e^2 4 πϵ 0 a

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

Figure 3

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 the central charge to a position far away is (in units of e^2 / 4 πϵ 0 s)

W = 4

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;

4Sum[(-1)^(n+1)/n, {n,1,H}] + 4Sum[Sum[(-1)^(n+m+1)/(n^2+m^2)^(.5), {n,1,H}], {m,1,H}]

8 CHAPTER 1. ELECTROSTATICS

where we have used sin 2θ = 2 sin θ cos θ. Letting θ = 22. 5 ◦^ gives Q = (3.154)q. Some limits: If all three charges are equally spaced (with θ = 30◦) then Q = q, as expected. If θ → 0 then Q ≈ q/(4θ^3 ). (Two of these powers of θ come from the r^2 in Coulomb’s law, and one comes from the act of taking the tangential component of Q’s field.) If the q’s are diametrically opposite (with θ = 45◦) then Q = 0, as expected.

1.47. Field from a semicircle Choose the semicircle to be the top half of a circle with radius R centered at the origin. So the diameter of the semicircle lies along the x axis. Let the angle θ be measured relative to the positive x axis. A small piece of the semicircle subtending an angle dθ has charge dQ = Q(dθ/π). The magnitude of the field at the center due to this piece is dQ/ 4 πϵ 0 R^2. The x components of the field contributions from the various pieces will cancel in pairs, so only the y component survives, which brings in a factor of sin θ. The total (vertical) field therefore equals

Ey = −

∫ (^) π

0

Q(dθ/π) 4 πϵ 0 R^2

sin θ = −

Q

(4πϵ 0 )πR^2

∫ (^) π

0

sin θ dθ = −

2 Q

(4πϵ 0 )πR^2

where the minus sign indicates that the field points downward (if Q is positive). This result can be written as −λ/ 2 πϵ 0 R, where λ is the linear charge density. Interestingly, it can also be written as −Q/ 4 πϵ 0 A, where A = πR^2 /2 is the area of the semicircle.

1.48. Maximum field from a ring At (0, 0 , z) the field due to an element of charge dQ on the ring has magnitude dQ/ 4 πϵ 0 (b^2 + z^2 ). But only the z component survives, by symmetry, and this brings in a factor of z/

b^2 + z^2. Integrating over the entire ring simply turns the dQ into Q, so we have Ez = Qz/ 4 πϵ 0 (b^2 + z^2 )^3 /^2. Setting the derivative equal to zero to find the maximum gives

(b^2 + z^2 )^3 /^2 (1) − z(3/2)(b^2 + z^2 )^1 /^2 (2z) (b^2 + z^2 )^3

b^2 − 2 z^2 (b^2 + z^2 )^5 /^2

=⇒ z = ± b √ 2

Since we’re looking for a point on the positive z axis, we’re concerned with the positive root, z = b/

  1. Note that we know the field must have a local maximum somewhere between z = 0 and z = ∞, because the field is zero at both of these points.

1.49. Maximum field from a blob

(a) Label the points on the curve by their distance r from the origin, and by the angle θ that the line of this distance subtends with the y axis, as shown in Fig. 7.

y

x θ r

Figure 7

Then a point charge q on the curve provides a y component of the electric field at the origin equal to Ey =

q 4 πϵ 0 r^2 cos θ. (32)

If we want this to be independent of the charge’s location on the curve, we must have r^2 ∝ cos θ. The curve may therefore be described by the equation,

r^2 = a^2 cos θ =⇒ r = a

cos θ, (33)

where the constant a is the value of r at θ = 0, that is, the height of the curve along the y axis. We therefore have a family of curves indexed by a.

(b) Assume that the material has been shaped and positioned so that the electric field at the origin takes on the maximum possible value. Assume that the field points in the y direction. Then all the small elements of charge dq on the surface of the material must give equal contributions to the y component of the field at the origin. This is true because if it weren’t the case, then we could simply move a tiny piece of the material from one point on the surface to another, thereby increasing the field at the origin, in contradiction to our assumption that the field at the origin is maximum. From part (a), any vertical cross section (formed by the intersection of the surface with a plane containing the y axis) must therefore look like the r = a

cos θ curve we found. Equivalently, the desired shape of the material is obtained by forming the surface of revolution of the r = a

cos θ curve, around the y axis. Let’s try to get a sense of what the surface looks like. We know that the height is a. To find the width, note that x = r sin θ = a sin θ

cos θ. Taking the derivative of this function of θ, you can show that the maximum value of x is achieved when tan θ =

2; the maximum value is (4/27)^1 /^4 a. The width is twice this value, or 2(4/27)^1 /^4 a ≈ 1. 24 a. So the ratio of width to height is about 5 to 4. Let’s compare our shape with a sphere of the same volume. The volume of our shape can be obtained by slicing it into horizontal disks. The radius of a disk is x = r sin θ = a sin θ

cos θ. And since y = −r cos θ = −a(cos θ)^3 /^2 , the height of a disk is dy = (3/2)a sin θ

cos θ dθ. So the volume is

V =

−a

(πx^2 ) dy =

∫ (^) π/ 2

0

π

a sin θ

cos θ

a sin θ

cos θ dθ

πa^3

∫ (^) π/ 2

0

sin^3 θ cos^3 /^2 θ dθ. (34)

Writing sin^2 θ as 1 − cos^2 θ yields integrals of sin θ cos^3 /^2 θ and − sin θ cos^7 /^2 θ, which give 2/5 and − 2 /9, respectively. The sum of these is 8/45, so the volume is V = (4/15)πa^3. Since the diameter of a sphere with volume V is (6V /π)^1 /^3 , we see that a sphere with the same volume would have a diameter of (8/5)^1 /^3 a ≈ 1. 17 a. Compared with a sphere of the same volume, our shape is therefore stretched by a factor of (1.24)a/(1.17)a ≈ 1 .06 in the x direction, and squashed by a factor of a/(1.17)a ≈ 0 .85 along the y direction. Cross sections of our shape and a sphere with the same volume are shown in Fig. 8.

1.06 d

d

0.85 d d

Figure 8

1.50. Field from a hemisphere

(a) Consider the ring shown in Fig. 9, defined by the angle θ and subtending an angle

R

θ

Figure 9

dθ. Its area is 2π(R cos θ)(R dθ), so its charge is σ(2πR^2 cos θ dθ). The horizontal component of the field at the center of the hemisphere is zero, by symmetry. So we need only worry about the vertical component from each piece of the ring, which brings in a factor of sin θ. Adding up these components from all the pieces of the ring gives the magnitude of the field at the center of the hemisphere, due to the given ring, as

dE =

σ(2πR^2 cos θ dθ) 4 πϵ 0 R^2

sin θ =

σ sin θ cos θ dθ 2 ϵ 0

The field points downward if σ is positive. Integrating over all the rings (θ runs from 0 to π/2) gives the total field at the center as

E =

∫ (^) π/ 2

0

σ sin θ cos θ dθ 2 ϵ 0

σ 2 ϵ 0

sin^2 θ 2

π/ 2

0

σ 4 ϵ 0

involve any infinitesimal distances that would make the fields diverge. Equivalently, once the sin(nπ/N ) term in the denominator of the field becomes non-infinitesimal, the fields go like 1/N , so the sum (which involves fewer than N terms) is bounded. We saw above that the vertical field contribution from each of the nearby charges equals 1/nπ, which is finite. In short, in Eq. (37) the small charge Q/N and the small factor of sin(nπ/N ) from the vertical component cancel the square of the small distance 2 R sin(nπ/N ) in the denominator. But the total field ends up diverging because there are so many charges that are very close to Q 0. Even though the field at Q 0 diverges in the N → ∞ limit, the actual force on Q 0 goes to zero. The force equals the field times the charge Q/N , and since the field only diverges like ln N , the force behaves like (ln N )/N , which goes to zero for large N. A continuous circle of charge is equivalent to the N → ∞ limit. So if an additional point charge with finite (non-infinitesimal) charge q were placed exactly on the cir- cumference of the (infinitesimally thin) circle, the force on it would be infinite, due to the infinite field. However, in reality there are no true point charges or infinitesimally thin distributions of charge.

1.52. An equilateral triangle

(a) Let F 0 be the force between two charges of q = 10−^6 C each, at a distance of a = 0.2 m. Then F 0 = q^2 / 4 πϵ 0 a^2 = 0.225 N, as you can verify. The force between B and C has magnitude (2)(2)F 0 = 4F 0 , and the force between A and either B or C has magnitude (3)(2)F 0 = 6F 0. From Fig. 11, the magnitude of

A

B C

a

2 q 2 q

3 q

4 F 0

6 F 0

6 F 0 6 F 0

Figure 11

the force on A is FA = 2 cos 30◦^ · 6 F 0 = 2.34 N. (39) The magnitude of the force on C is (squaring and adding the horizontal and vertical components)

FC =

[

(4 + 6 cos 60◦)^2 + (6 sin 60◦)^2

] 1 / 2

F 0 = (8.72)F 0 = 1.96 N. (40)

And the force on B has the same magnitude. (b) Three equal charges of 2· 10 −^6 C would yield zero field at the center, by symmetry. So the field at the center is due to the excess charge of q = 10−^6 C at A. Since A is a distance a/

3 from the center, the magnitude of the field at the center of the triangle is

E =

q 4 πϵ 0 (a/

3)^2

kg m^3 s^2 C^2

10 −^6 C

(0.2 m)^2 / 3

= 6. 75 · 105 N/C. (41)

1.53. Concurrent field lines

Consider a point at height z above the semicircle. All points on the wire are a distance ℓ =

R^2 + z^2 from this point, so a small piece of the wire with charge dq = λR dθ yields a field with magnitude

dE = dq 4 πϵ 0 ℓ^2

λR dθ 4 πϵ 0 (R^2 + z^2 )

Let the x axis split the semicircle in half. Then the net Ey field is zero, by symmetry. The (magnitudes of the) z and x components of the dE field in Eq. (42) are obtained by multiplying it by z/ℓ and x/ℓ. (The latter of these can be obtained in two steps:

12 CHAPTER 1. ELECTROSTATICS

multiply by R/ℓ to get the component in the x-y plane, then multiply by x/R to get the x component.) So we have

dEz = λRz dθ 4 πϵ 0 (R^2 + z^2 )^3 /^2

, and dEx = λRx dθ 4 πϵ 0 (R^2 + z^2 )^3 /^2

λR(R cos θ) dθ 4 πϵ 0 (R^2 + z^2 )^3 /^2

where θ runs from −π/2 to π/2. The net Ez and Ex components are obtained by integrating over θ. In Ez the integration simply brings in a factor of π. In Ex it brings in a factor of

∫ (^) π/ 2 −π/ 2 cos^ θ dθ^ = 2. Therefore,

Ez =

λRz 4 ϵ 0 (R^2 + z^2 )^3 /^2

, and Ex =

λR^2 2 πϵ 0 (R^2 + z^2 )^3 /^2

Hence Ez /Ex = πz/ 2 R, which can be written a little more informatively as Ez /Ex = z/(2R/π). This is the slope of the E vector at a point at height z on the z axis. The slope covers a horizontal distance 2R/π while covering a vertical distance z. The straight line that points in the direction of the electric field at the point (0, 0 , z) therefore passes through the point (2R/π, 0 , 0) in the plane of the semicircle. This point is independent of z, as desired. This point also happens to be the “center of charge” of the semicircle, or equivalently the center of mass of a semicircle made out of a piece of wire (we’ll leave it to you to verify this). So the result of this exercise is consistent with the following fact (which you may want to try to prove): Far away from a distribution of charges, the electric field points approximately toward the center of charge of the distribution. For nearby points it generally doesn’t, although it happens to (exactly) point in that direction for points on the axis of the present setup.

1.54. Semicircle and wires

(a) The charge dq in the piece at A is λb dθ. The magnitude of the field due to this charge is EA =

λb dθ 4 πϵ 0 b^2

λ dθ 4 πϵ 0 b

The charge dq in the piece at B is λ dy. But from Fig. 12 we have dy =

A

B

C

θ θ

θ dy r^ d θ

r

b b

Figure 12

r dθ/ cos θ = r dθ/(b/r) = r^2 dθ/b. The magnitude of the field due to this charge is therefore EB =

λ(r^2 dθ/b) 4 πϵ 0 r^2

λ dθ 4 πϵ 0 b

Since the magnitudes EA and EB are equal, and since the fields are directed oppositely, the sum of the two fields is zero. The entire filament can be built up from these corresponding pairs, so the total field at C is zero. In short, the field contributions from equal point charges located at A and B would be in the ratio r^2 /b^2 , due to the inverse-square nature of the Coulomb field. But this effect is canceled by the fact that the actual charge at B is larger than at A by a factor (r/b)^2. One of these factors of r/b comes from the fact that B is farther from C, and the other comes from the fact that the B segment is tilted with respect to the line to C. This result is consistent with the results from Problem 1.10 (which says that the upward component of the field at C due to each of the straight segments is λ/ 4 πϵ 0 b) and Exercise 1.47 (which says that the downward field at C due to the semicircle is λ/ 2 πϵ 0 b).

14 CHAPTER 1. ELECTROSTATICS

As a check, if b ≫ ℓ this result approaches (1/ 4 πϵ 0 )(2ℓλ/b^2 ), which is correctly the field from a point charge 2ℓλ at a distance b.

1.56. Flux through a cube

(a) The total flux through the cube is q/ϵ 0 , by Gauss’s law. The flux through every face of the cube is the same, by symmetry. Therefore, over any one of the six faces we have

E · da = q/ 6 ϵ 0. (b) Because the field due to q is parallel to the surface of each of the three faces that touch q, the flux through these faces is zero. The total flux through the other three faces must therefore add up to q/ 8 ϵ 0 , because our cube is one of eight such cubes surrounding q. Since the three faces are symmetrically located with respect to q, the flux through each must be (1/3)(q/ 8 ϵ 0 ) = q/ 24 ϵ 0. Note: if the charge were a true point charge, and if it were located just inside or just outside the cube, then the field would not be parallel to each of the three faces that touch the given corner. The flux would depend critically on the exact location of the point charge. Replacing the point charge with a small sphere, whose center lies at the corner, eliminates this ambiguity.

1.57. Escaping field lines

(a) You can quickly show that the desired point with E = 0 must satisfy x > a. Equating the magnitudes of the fields from the two given charges then gives

2 q 4 πϵ 0 x^2

q 4 πϵ 0 (x − a)^2

=⇒ 2(x − a)^2 = x^2

2(x − a) = x =⇒ x =

2 a √ 2 − 1

2)a ≈ (3.414)a. (50)

A few field lines, are shown in Fig. 14.^1 Note that the field points in four different directions near the E = 0 point. This is consistent with the fact that the zero vector is the only vector that can simultaneously point in different directions. (b) Consider a field line that emerges from the 2q charge and ends up at the x = (3.414)a point where the field is zero. (There are actually no field lines that end up right at this point, but we can pick a line infinitesimally close.) Field lines that emerge at a smaller angle (with respect to the x axis) end up at the −q charge, and field lines that emerge at a larger angle end up at infinity. Consider the Gaussian surface indicated in Fig. 15; the surface is formed by rotating the black curve around the x axis. This surface follows the field lines except very close to the 2q charge, where it takes the form of a small spherical cap. The total charge enclosed within this surface is simply −q, so from Gauss’s law there must be an electric-field flux of q/ϵ 0 pointing in to the surface. By construction, the only place where there is flux is the spherical cap, so all of the q/ϵ 0 flux must occur there. But the total flux emanating from the charge 2q is 2q/ϵ 0 , so the spherical cap must represent half of the total area of a small sphere surrounding the charge 2q. (Very close to the 2q charge, that charge dominates the electric field, so the field is essentially spherically symmetric.) The cap must therefore be a hemisphere, so the desired angle is 180◦. (^1) This figure technically isn’t a plot of field lines, because you can see that some of the lines begin in empty space, whereas we know that field lines can begin and end only at charges or at infinity. So the density of the lines on the page doesn’t indicate the field strength. (Well, it fails to do that even if we

2 1 0 1 2 3 4 5

3

2

1

0

1

2

3

2 q - q E = 0

(in units of a )

Figure 14

2 1 0 1 2 3 4 5

3

2

1

0

1

2

3

E = 0

2 q

  • q

(in units of a ) Figure 15

If the charges take on the more general values of N q and −q, then the spherical cap represents 1/N of the complete sphere. So the task is to find the angle subtended by a cap with 1/N of the total area. This requires an integral (whereas in the above case we could simply say we had a hemisphere). But one nice case is N = 4, which leads to an angle of 60◦.

1.58. Gauss’s law at the center of a ring

(a) Let the ring lie in the horizontal plane. A small piece of the ring with charge dq produces a field dq/ 4 πϵ 0 R^2 at the center. At a small vertical distance z above the center, the magnitude of the field due to the dq piece is essentially the same (it differs only at order z^2 /R^2 , by the Pythagorean theorem ), so the vertical component is obtained by simply tacking on a sin θ type factor, which is z/R here. Integrating over the whole ring turns the dq into Q, so the desired vertical

don’t have any lines that abruptly end, because a 2D picture can’t mimic the actual density in 3D space.) However, every curve shown is at least part of a field line, so the figure is still helpful in visualizing the flow of the actual field lines.

1.60. Field from a hollow cylinder

The cylindrical tube of charge (the bold circle) in Fig. 17 has perfect axial symmetry.

E 1

E 3 E 4

E 2

a r

Figure 17

So inside the tube, E 1 and E 2 must be radial and equal in magnitude. Applying Gauss’s law to a cylinder of radius a and length ℓ yields 2πaℓE 1 = 0, because there is no charge inside the tube. Therefore E = 0 inside the tube. Outside the tube, symmetry also demands that E 3 = E 4. Applying Gauss’s law to a cylinder of radius r and length ℓ yields 2πrℓE 3 = λℓ/ϵ 0 , where λ is the charge per unit length. This gives E = λ/ 2 πϵ 0 r outside the tube, just as if the charge were concentrated on the axis. For a square tube of charge the integral over any cylinder must equal 1/ϵ 0 times the charge enclosed, but nothing requires that E 1 = E 2 or E 3 = E 4 in Fig. 18. The

E 1

E 3

E 4

E 2

a r

Figure 18

integral of E over the small inner circle vanishes (as it does for any cross-sectional shape), but it can do so with E 1 ̸= E 2 if, as is the case, these two fields point in opposite directions at the locations shown. By comparing this tube with a square charged conducting tube, within which the field is in fact zero (see Chapter 3), you can deduce that E 2 must point inward (if the charge is positive).

1.61. Potential energy of a sphere

The charge inside a sphere of radius r (with r < R) is q = (4πr^3 /3)ρ. The external field of this sphere is the same as if all of the charge were at the center. So the sphere acts like a point charge, as far as the potential energy of an external object is concerned. The next shell to be added, with thickness dr, contains charge dq = (4πr^2 dr)ρ. The work done in bringing in this dq (which is the same as the potential energy of the shell due to the sphere) is therefore

dW =

4 πϵ 0

q · dq r

4 πϵ 0

(4πρ)^2 3

r^4 dr. (52)

Building up the whole sphere this way, from r = 0 to r = R, requires the work:

W =

∫ R

0

4 πϵ 0

(4πρ)^2 3 r^4 dr =

4 πϵ 0

(4πρ)^2 3

R^5

The charge in the complete sphere is Q = (4πR^3 /3)ρ, which gives 4πρ = 3Q/R^3. Thus the potential energy U , which is the same as the work W , can be written as U = (3/5)Q^2 / 4 πϵ 0 R. Note that Q^2 / 4 πϵ 0 R has the proper energy dimensions of (charge)^2 /(ϵ 0 · distance). Indeed, we could have predicted that much of the result without any calculation. The only question is what the numerical factor out front is. It happens to be 3/5. Note that we don’t have to worry about the self energy of each infinitesimally thin shell, because by dimensional analysis this energy is proportional to (dq)^2. So it is a second-order small quantity and hence can be ignored.

1.62. Electron self energy

Setting the potential energy (3/5)e^2 / 4 πϵ 0 r 0 from Exercise 1.61 equal to mc^2 gives

r 0 =

4 πϵ 0

e^2 mc^2

kg m^3 s^2 C^2

) (1. 6 · 10 − 19 C) 2

(9. 1 · 10 −^31 kg)(3 · 108 m/s)^2

= 1. 69 · 10 −^15 m.

It is interesting to ask what happens to the mass if the charge density is kept constant but the radius is doubled. You might think that since there is 2^3 = 8 times as much

18 CHAPTER 1. ELECTROSTATICS

stuff (charge) present, the mass should be 8 times as large. However, the charge e is squared in the above formula for the potential energy, so this yields a factor of 8^2 = 64. But there is also one power of r 0 in the denominator, so this cuts the result down to

  1. Equivalently, the result for the energy in Eq. (53) in the solution to Exercise 1. is proportional to R^5 , and 2^5 = 32.

1.63. Sphere and cones

(a) There is no change in speed inside the shell, because the electric field is zero inside. So we just need to find the speed of the particle when it reaches the surface. The charge on the shell is 4πR^2 σ, so the potential energy of the particle at the surface of the shell is

V (R) =

(4πR^2 σ)(−q) 4 πϵ 0 R

Rσq ϵ 0

The initial potential energy was zero, so this loss in potential energy shows up as kinetic energy. Hence, mv^2 /2 = Rσq/ϵ 0 =⇒ v =

2 Rσq/ϵ 0 m. (b) Let’s find the potential energy U of the particle, due to one of the cones, when it is located at the tip of the cones. We’ll slice the cone into rings and then integrate over the rings. Consider a thin ring around the cone, located at a slant distance x away from the tip. The charge in this ring is dQ = σ 2 πr dx, where the radius r is given by r/x = R/L =⇒ r = xR/L. Every point in the ring is the same distance x from the tip, so

dU =

dQ(−q) 4 πϵ 0 x

σ 2 π(xR/L) dx

q 4 πϵ 0 x

Rσq dx 2 ϵ 0 L

Integrating from x = 0 to x = L simply turns the dx into an L, so we have U = −Rσq/ 2 ϵ 0. We need to double this because there are two cones, so we end up with the same potential energy of −Rσq/ϵ 0 as in part (a). We therefore obtain the same speed of v =

2 Rσq/ϵ 0 m, independent of L.

1.64. Field between two wires The electric field from a single wire is λ/ 2 πϵ 0 r. Between the wires the fields from the two wires point in the same direction, so we have

15 , 000 N/C = 2

λ 2 πϵ 0 r

=⇒ λ =

15 , 000 N/C

πϵ 0 r

15 , 000 N/C

8. 85 · 10 −^12

s^2 C^2 kg m^3

(1.5 m)

= 6. 3 · 10 −^7 C/m. (57)

The amount of excess charge on 1 km of the positive wire is then (1000 m)λ = 6. 3 · 10 −^4 C.

1.65. Building a sheet from rods In Fig. 19, the horizontal line represents the sheet, which extends into and out of the

θ

dx

P

sheet

rod

r y

into and out of page

Figure 19

page (and also to the left and right). The short segment represents a rod extending into and out of the page, with small width dx. The field at point P due to the rod is λ/ 2 πϵ 0 r, where the effective linear charge density of the rod is λ = σ dx. This is true because the amount of charge in a length ℓ of the rod can be written as both λℓ (by