Electrostatics Part 1, Solution Manual

Electrostatics 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 6.02 · 1023 · 1.93 · 10−2 g = 5.9 · 1019 . 197 g (1) 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 (1 m)2 s C 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. 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, d2 = ( kg m3 ) (1.6 · 10−19 C)2 1 e2 = 9 · 109 2 2 = 26 m2 , s C (9 · 10−31 kg)(9.8 m/s2 ) 4πϵ0 mg (3) 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. 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 = 10 =⇒ Tx Fg = 10 =⇒ Fe mg = 10. q 2 /4πϵ0 r2 (4) Therefore, q2 ( ) s2 C2 1 (4πϵ0 )mgr 2 = (0.4)π 8.85 · 10−12 (0.3 kg)(9.8 m/s2 )(0.5 m)2 3 10 kg m −12 2 −6 C =⇒ q = 2.9 · 10 C. = 8.17 · 10 (5) = 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 + 2 cos 45◦ 2 − F = √ √ . ℓ ( 2 ℓ)2 (ℓ/ 2)2 (6) Setting this equal to zero gives Q= ( ) 1 1 + √ q = (0.957)q. 4 2 (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= √ ( ) ( ) q (−Q)q q2 1 q q2 4 2q √ +4· −Q + √ + = 0, (8) 4· +2· √ = 4πϵ0 ℓ 4πϵ0 ℓ 2 4 ℓ/ 2 2ℓ 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. 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 order x2 ) F (x) ) ( 1 1 Qq − 2 2 (ℓ + x) 4πϵ0 (ℓ − x) ) ( Qq 1 1 ≈ − − 2 1 − 2x/ℓ 1 + 2x/ℓ 4πϵ0 ℓ ) Qq ( ≈ − (1 + 2x/ℓ) − (1 − 2x/ℓ) 2 4πϵ0 ℓ Qqx = − . πϵ0 ℓ3 = − (9) 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) 3 3 πϵ0 ℓ πϵ0 mℓ 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 far as the units go, Qq/ϵ√0 ℓ2 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 ( ) 1 1 Qq + U (x) = . (11) 4πϵ0 ℓ − x ℓ+x We’ll need to expand things to order x2 because the order x terms will cancel: ) ( 1 1 Qq + U (x) = 4πϵ0 ℓ 1 − x/ℓ 1 + x/ℓ (( ) ( )) x x x2 x2 Qq ≈ 1+ + 2 + 1− + 2 4πϵ0 ℓ ℓ ℓ ℓ ℓ ) ( 2x2 Qq (12) = 2+ 2 . ℓ 4πϵ0 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 F (x) = − dU Qqx =− , πϵ0 ℓ3 dx (13) in agreement with the force in Eq. (9). 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 forces is equal and opposite to the sum of the vertical components of the tensions. This gives ) ( qQ q2 q2 qQ = 2T ℓ2 − = 2T sin θ =⇒ . (14) sin θ+ 2 2 4πϵ0 (2ℓ sin θ)2 4πϵ0 ℓ 2πϵ0 16πϵ0 sin3 θ Similarly, each charge Q is in equilibrium if ) ( Q2 qQ = 2T cos θ =⇒ cos θ + 2 4πϵ0 (2ℓ cos θ)2 4πϵ0 ℓ2 Q2 qQ . = 2T ℓ2 − 2πϵ0 16πϵ0 cos3 θ (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 q 2 /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. y 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 q 2 /4πϵ0 (2ℓ sin θ) and Q2 /4πϵ0 (2ℓ cos θ). Minimizing this as a function of θ yields ( 2 ) q Q2 q2 d cos θ sin θ =⇒ 0= + = tan3 θ. (16) + Q2 2 = −q 2 cos θ dθ sin θ Q2 cos θ sin2 θ e e Zero potential energy r2 2 -2 -1 -e r1 -e 1 Figure 1 x 2 Let’s first consider the general case where the three charges don’t necessarily lie on 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 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 energy of the system equal to zero gives ) ( 2 e e2 e2 1 1 1 + U= =⇒ = 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 2± 5 1 + (18) . = 1 =⇒ x2 − 2x − 1/4 = 0 =⇒ x = 2 x − 1/2 x + 1/2 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 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. Work for an octahedron 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) ) ( e2 e2 1 1 1 U= = −(2.121) −6· −3· √ . (19) 6· a a 4πϵ0 a 4πϵ0 2a And for the second configuration: ) ( e2 1 1 e2 1 1 U= = −(3.293) . 4· −8· +2·√ −1· √ a a 4πϵ 4πϵ0 2 a 0a 2a 2a a a 2a 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. (20) 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.) 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 e2 1 e e2 e2 1 1 1 (21) + ··· = − 1 − + − + ··· . − + − 4 4πϵ0 a 2a 3a 4πϵ0 a 2 3 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 /4πϵ0 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. Figure 2 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. 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 ) ( ∞ ∞ ∞ ∑ ∞ ∞ ∞ ∑ ∑ ∑ ∑ ∑ (−1)m (−1)m+n (−1)m+n+p e2 √ √ . U= + + 4πϵ0 a m=1 m m2 + n2 p=1 n=−∞ m=−∞ m2 + n2 + p2 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 U= e2 (0.874)e2 , (−0.693 − 0.115 − 0.066) = − 4πϵ0 a 4πϵ0 a (23) 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. Chessboard Figure 3 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 the central charge to a position far away is (in units of e2 /4πϵ0 s) ) ) ( ) ( ( 1 1 1 1 1 1 1 1 1 ≈ 1.4146, − + +4 − √ − √ − √ +8 √ − √ + √ W =4 1 2 3 13 2 2 2 3 2 5 10 (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 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 /4πϵ0 s): N W 3 1.1716 7 1.4146 101 1.6015 1001 1.6141 10,001 1.6154 Figure 4 100,001 1.6155 The W for an infinite chessboard is apparently roughly equal to (1.6155)e2 /4πϵ0 s. 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. y Zero field? 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 the two outer charges. This implies that 2· 3a a q q 1 1 ·√ · √ =2· , 4πϵ0 y 2 + (3a)2 4πϵ0 y 2 + a2 y 2 + (3a)2 y 2 + a2 a q 2a q -q -q Figure 5 (25) where the second factor on each side of the equation comes from the act of taking the horizontal component. Simplifying this gives 3 1 = 2 (y 2 + a2 )3/2 (y + 9a2 )3/2 y 2 + 9a2 = 32/3 (y 2 + a2 ) √ 9 − 32/3 =⇒ y = a ≈ (2.53)a. 32/3 − 1 =⇒ (26) 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 being equal.) Therefore, by continuity, there must exist a point on the y axis where Ex equals zero. Q θ θ Rcosθ 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 Q sin θ, 4πϵ0 (2R cos θ)2 (27) θ 4θ R q q and the tangential field (in the opposite direction) at one q due to the other q is q cos 2θ. 4πϵ0 (2R sin 2θ)2 2θ (28) Equating these fields gives cos2 θ cos 2θ q cos 2θ Q cos 2θ =⇒ Q = q sin θ = =q , (sin 2θ)2 (cos θ)2 sin θ sin2 2θ 4 sin3 θ θ (29) Figure 6 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/(4θ3 ). (Two of these powers of θ come from the r2 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. 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/4πϵ0 R2 . 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 ∫ π ∫ π Q(dθ/π) 2Q Q sin θ dθ = − Ey = − sin θ = − , (30) 2 2 (4πϵ0 )πR 0 (4πϵ0 )πR2 0 4πϵ0 R where the minus sign indicates that the field points downward (if Q is positive). This result can be written as −λ/2πϵ0 R, where λ is the linear charge density. Interestingly, it can also be written as −Q/4πϵ0 A, where A = πR2 /2 is the area of the semicircle. Maximum field from a ring At (0, 0, z ) the field due to an element of charge dQ on the ring has magnitude ). But only the z component survives, by symmetry, and this brings dQ/4πϵ0 (b2 + z 2 √ in a factor of z/ b2 + z 2 . Integrating over the entire ring simply turns the dQ into Q, so we have Ez = Qz/4πϵ0 (b2 + z 2 )3/2 . Setting the derivative equal to zero to find the maximum gives 0= x Figure 7 b =⇒ z = ± √ . (31) 2 Since we’re looking for a point on the positive z axis, we’re concerned with the positive √ root, z = b/ 2. 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. y θ b2 − 2z 2 (b2 + z 2 )3/2 (1) − z(3/2)(b2 + z 2 )1/2 (2z ) = 2 (b2 + z 2 )3 (b + z 2 )5/2 Maximum field from a blob r (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. Then a point charge q on the curve provides a y component of the electric field at the origin equal to q cos θ. (32) Ey = 4πϵ0 r2 If we want this to be independent of the charge’s location on the curve, we must have r2 ∝ cos θ. The curve may therefore be described by the equation, √ r2 = a2 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 ≈ 1.24a. 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 . And since y = −r cos θ = −a(cos θ)3/2 , the height of x = r sin θ = a sin θ cos θ√ a disk is dy = (3/2)a sin θ cos θ dθ. So the volume is ∫ 0 ∫ π/2 √ √ ( )2 3 V = (πx2 ) dy = π a sin θ cos θ · a sin θ cos θ dθ 2 −a 0 ∫ 3 3 π/2 3 sin θ cos3/2 θ dθ. (34) = πa 2 0 Writing sin2 θ as 1 − cos2 θ yields integrals of sin θ cos3/2 θ and − sin θ cos7/2 θ, which give 2/5 and −2/9, respectively. The sum of these is 8/45, so the volume is V = (4/15)πa3 . 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 ≈ 1.17a. 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 0.85 d d Field from a hemisphere (a) Consider the ring shown in Fig. 9, defined by the angle θ and subtending an angle dθ. Its area is 2π(R cos θ)(R dθ), so its charge is σ(2πR2 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 σ(2πR2 cos θ dθ) σ sin θ cos θ dθ sin θ = dE = . 4πϵ0 R2 2ϵ0 d Figure 8 R θ (35) 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  π/2 ∫ π/2 σ sin θ cos θ dθ σ sin2 θ  σ E= = = . (36) 2ϵ0 2ϵ0 2  0 4ϵ0 0 Figure 9 This is half as large as the field at the center of the end face of a half-infinite hollow cylinder (see Problem 1.11). You should convince yourself why the cylinder’s field must indeed be larger (although the factor of 2 is by no means obvious). Hint: consider the field contributions from corresponding bits of the surfaces subtending the same solid angle. The cylinder’s surface is tilted with respect to the line to the center of the end face. In terms of the total charge Q = 2πR2 σ on the hemisphere, the result in Eq. (36) can be written as Q/8πϵ0 R2 . If you solved Exercise 1.47 you will note that the present field is smaller (by a factor of π/4) than the field at the center of a semicircle with the same charge Q. This is because the semicircle’s charge is generally higher up, so the act of taking the vertical component doesn’t reduce the field as much as for the hemisphere. Equivalently, building a hemispherical cage out of a large number of semicircular pieces of wire would effectively yield a hemisphere with a larger surface charge density near the top than near the base. (b) The field due to a solid hemisphere with radius R and uniform volume charge density ρ can be found by slicing up the solid hemisphere into concentric hemispherical shells with thickness dR. The effective surface charge density of each shell is ρ dR, so the result from part (a) tells us that the field from each shell is (ρ dR)/4ϵ0 . Integrating over R simply turns the dR into an R, so the total field is ρR/4ϵ0 . Again, this is half as large as the field from a half-infinite solid cylinder (see Problem 1.11). Q0 θ/2 1.51. N charges on a circle θ/2 θ/2 N =12 Q4 Figure 10 Let Q0 be the point charge at the top of the circle, and consider the nth point charge away from it (call it Qn ), as shown in Fig. 10 for n = 4 and N = 12. The angle θ equals n(2π/N ), so the distance from Q0 to Qn is rn = 2R sin(θ/2) = 2R sin(nπ/N ). The horizontal components of all the fields at Q0 cancel in pairs, so we’re concerned only with the vertical component, which brings in a factor of sin(θ/2). The vertical component of the field at Q0 due to Qn is therefore En = Q/N ( )2 sin(nπ/N ). 4πϵ0 2R sin(nπ/N ) (37) The total field at Q0 from all the Qn charges is then E= −1 N ∑ 1 Q ( ). 16πϵ0 R2 N sin(nπ/N ) n=1 (38) If you compute this sum numerically for various values of N , you will find that it grows with N , although very slowly (like a log). The sum does in fact diverge as N → ∞, due to the behavior of the terms with small n (and likewise for n close to N , because the terms are symmetric around n = N/2). If n ≪ N , (we can write ) sin(nπ/N ) ≈ nπ/N , so for small n the field from Qn behaves like 1/ N (nπ/N ) = 1/nπ. And since ∑M the sum n=1 1/n diverges (like ln M ), we see that the total field diverges. We are assuming that N is large enough so that n can become very large and still have the approximation sin(nπ/N ) ≈ nπ/N be valid. This is of course true in the N → ∞ limit. Note that the only possible cause for the divergence of the total field is the behavior of the fields from nearby Qn . There is a finite total charge on the ring, so the field from the non-infinitesimally-close charges must be finite, because those charges don’t 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 2R sin(nπ/N ) in the denominator. But the total field ends up diverging because there are so many charges that are very close to Q0 . Even though the field at Q0 diverges in the N → ∞ limit, the actual force on Q0 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 circumference 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. 6F0 An equilateral triangle 6F0 −6 (a) Let F0 be the force between two charges of q = 10 C each, at a distance of a = 0.2 m. Then F0 = q 2 /4πϵ0 a2 = 0.225 N, as you can verify. The force between B and C has magnitude (2)(2)F0 = 4F0 , and the force between A and either B or C has magnitude (3)(2)F0 = 6F0 . From Fig. 11, the magnitude of the force on A is (39) FA = 2 cos 30◦ · 6F0 = 2.34 N. 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 a B 2q 4F0 C 2q 6F0 Figure 11 F0 = (8.72)F0 = 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 ( ) kg m3 10−6 C q √ = 9 · 109 2 2 = 6.75 · 105 N/C. (41) E= 2 (0.2 m)2 /3 s C 4πϵ0 (a/ 3) Concurrent field lines Consider √ a point at height z above the semicircle. All points on the wire are a distance ℓ = R2 + z 2 from this point, so a small piece of the wire with charge dq = λR dθ yields a field with magnitude dE = 3q A dq λR dθ = . 4πϵ0 (R2 + z 2 ) 4πϵ0 ℓ2 (42) 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: 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 λRz dθ , 4πϵ0 (R2 + z 2 )3/2 λRx dθ λR(R cos θ) dθ = , 4πϵ0 (R2 + z 2 )3/2 4πϵ0 (R2 + z 2 )3/2 (43) 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 ∫ π/2 in a factor of −π/2 cos θ dθ = 2. Therefore, dEz = Ez = and λRz , 4ϵ0 (R2 + z 2 )3/2 dEx = and Ex = λR2 . 2πϵ0 (R2 + z 2 )3/2 (44) Hence Ez /Ex = πz/2R, 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. Semicircle and wires A b b θ θ C r B dy θ Figure 12 r dθ (a) The charge dq in the piece at A is λb dθ. The magnitude of the field due to this charge is λ dθ λb dθ = . (45) EA = 4πϵ0 b 4πϵ0 b2 The charge dq in the piece at B is λ dy. But from Fig. 12 we have dy = r dθ/ cos θ = r dθ/(b/r) = r2 dθ/b. The magnitude of the field due to this charge is therefore λ(r2 dθ/b) λ dθ . (46) EB = = 4πϵ0 b 4πϵ0 r2 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 r2 /b2 , 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/4πϵ0 b) and Exercise 1.47 (which says that the downward field at C due to the semicircle is λ/2πϵ0 b). (b) Imagine cutting the two-dimensional setup into thin strips defined by a large number of vertical planes, rotated at small angles with respect to each other, all passing through the axis of the cylinder. Each thin strip is similar to the onedimensional setup in part (a), except for the fact that the strip gets narrower as it approaches the top of the hemisphere (it is somewhat like a curved pie piece). The linear charge density is therefore effectively smaller at the top. We know from part (a) that a uniform linear charge density leads to the downward field from the circular part of a strip canceling the upward field from the straight part. A smaller density on the circular part therefore means that its downward field can’t fully cancel the upward field from the straight part. The net field therefore points upward. This result is consistent with the results from Problem 1.11 (which says that the upward field at C due to the cylinder is σ/2ϵ0 ) and either Problem 1.12 or Exercise 1.50 (which say that the downward field at C due to the hemisphere is σ/4ϵ0 ). Field from a finite rod In Fig. 13, define the distances: ℓ = 0.05 m, a = 0.03 m, and b = 0.05 m. The linear charge density of the rod is λ = (8 · 10−9 C)/(0.1 m) = 8 · 10−8 C/m. At point A the field points leftward and has magnitude ) ( ) ( ∫ a+2ℓ λ 2ℓ 1 λ dx 1 1 λ EA = = − = 2 4πϵ0 a(a + 2ℓ) 4πϵ0 a a + 2ℓ 4πϵ0 a x ( )( ) 3 )( 2(.05 m) 9 kg m −8 C = 9 · 10 8 · 10 2 2 s C m (.03 m)(.13 m) N = 1.85 · 104 . (47) C As a check, if a ≫ ℓ this result approaches (1/4πϵ0 )(2ℓλ/a2 ), which is correctly the field from a point charge 2ℓλ at a distance a. At point B, only the vertical component of the field survives, by symmetry. So the field points downward and has magnitude ∫ ℓ b λ dx 1 ·√ EB = , (48) 4πϵ0 −ℓ b2 + x2 b2 + x2 where the second factor gives the vertical component. This integral can be evaluated with a trig substitution, x = b tan θ =⇒ dx = b dθ/ cos2 θ (or you can just look it up), which yields −1 ∫ ∫ tan−1 (ℓ/b) 1 1 λ tan (ℓ/b) λb2 dθ/ cos2 θ = EB = cos θ dθ 2 3 3/2 4πϵ0 b − tan−1 (ℓ/b) 4πϵ0 − tan−1 (ℓ/b) b (1 + tan θ) tan−1 (ℓ/b)  1 λ 1 λ 2ℓ = = √ sin θ 4πϵ0 b 4πϵ0 b ℓ2 + b2 − tan−1 (ℓ/b) ( ) kg m3 (8 · 10−8 C/m) 2(.05 m) √ = 9 · 109 2 2 s C (.05 m) (.05 m)2 + (.05 m)2 N (49) = 2.04 · 104 . C ∫ The cos θ dθ integral here is just what you would obtain if you parameterized the rod in terms of θ; see Eq. (1.38). A a l l b B Figure 13 As a check, if b ≫ ℓ this result approaches (1/4πϵ0 )(2ℓλ/b2 ), which is correctly the field from a point charge 2ℓλ at a distance b. Flux through a cube (a) The total flux through the cube is q/ϵ0 , 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/6ϵ0 . (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/8ϵ0 , 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/8ϵ0 ) = q/24ϵ0 . 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. 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 =⇒ 2q q =⇒ 2(x − a)2 = x2 = 4πϵ0 (x − a)2 4πϵ0 x2 √ √ √ 2a = (2 + 2)a ≈ (3.414)a. (50) 2(x − a) = x =⇒ x = √ 2−1 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/ϵ0 pointing in to the surface. By construction, the only place where there is flux is the spherical cap, so all of the q/ϵ0 flux must occur there. But the total flux emanating from the charge 2q is 2q/ϵ0 , 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 3 2 1 0 2q -q 0 1 E =0 -1 -2 -3 -2 -1 2 3 4 5 (in units of a) 5 (in units of a) Figure 14 3 2 1 2q 0 E= 0 -q -1 -2 -3 -2 -1 0 1 2 3 4 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◦ . 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/4πϵ0 R2 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 /R2 , 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. field is Qz/4πϵ0 R3 . Alternatively, you can calculate the field exactly (as in the solution to Exercise 1.48) and then take the z ≪ R limit. (b) The solution to Problem 1.8 tells us that in the plane of the ring, the field near the center, at radius r, points radially inward (assuming Q is positive) with magnitude λr/4ϵ0 R2 . But since λ = Q/2πR, this can be written as Qr/8πϵ0 R3 . Consider a point near the center, with a nonzero r value, and also a nonzero z value. To leading order, the horizontal component of the field is still Qr/8πϵ0 R3 , and the vertical component is still Qz/4πϵ0 R3 , from part (a). That is, these results are actually valid for all points in space near the origin, not just in the plane of the ring or on the axis. You can check this by writing out the exact expressions for the fields. For example, in part (a) the effective values of R change slightly if the point is off the axis, but this doesn’t change the field, to leading order. Alternatively, note that due to symmetry, the horizontal component Er (r, z) is an even function of z. This means that Er (r, z) has no linear dependence on z. The variation with z therefore starts only at order z 2 , which is negligible for small z. So Er is essentially independent of z near the axis. Similar reasoning works with Ez as a function of r. For simplicity, let’s define A ≡ Q/8πϵ0 R3 . Then the horizontal and vertical field components have magnitudes Ar and 2Az, respectively. The top and bottom faces of the small cylinder have a combined area of 2πr02 . And the vertical cylindrical side has an area of (2πr0 )(2z0 ). There is outward flux through the top and bottom, and inward flux through the side, so the net outward flux equals (2πr20 )(2Az0 ) − (4πr0 z0 )(Ar0 ) = 0, (51) as desired. If we work backwards, this exercise actually provides a much quicker method, compared with the one in Problem 1.8, for finding the horizontal component of the field near the center of the ring, assuming that we know the vertical component. Zero field inside a cylindrical shell A A' P b B B' Figure 16 a The solution to this exercise is essentially the same as the solution to Problem 1.17. The main points are that areas are proportional to lengths squared, and that the relevant surfaces make the same angle with respect to the line to the point in question. As in the solution to Problem 1.17, let a be the distance from point P to patch A, and let b be the distance from P to patch B; see Fig. 16. (Since the cones are assumed to be thin, it doesn’t matter exactly which points in the patches we use to define these distances.) Draw the “perpendicular” bases of the cones, and call them A′ and B ′ . The ratio of the areas of A′ and B ′ is a2 /b2 , because areas are proportional to lengths squared. And the angle between the planes of A and A′ is the same as the angle between the planes of B and B ′ . The ratio of the areas of A and B is therefore also equal to a2 /b2 . So the charge on patch A is a2 /b2 times the charge on patch B . The magnitudes of the fields due to the two patches take the general form of q/4πϵ0 r2 . We just found that the q for A is a2 /b2 times the q for B. But we also know that the r2 for A is a2 /b2 times the r2 for B. So the values of q/4πϵ0 r2 for the two patches are equal. The fields at P due to A and B (which can be treated essentially like point charges, because the cones are assumed to be thin) are therefore equal in magnitude; and opposite in direction, of course. If we draw enough cones to cover the whole cylinder, the contributions to the field from the little patches over the whole cylinder cancel in pairs, so we are left with zero field at P . This holds for any point P inside the cylinder. E3 E4 17 E1 E2 Field from a hollow cylinder The cylindrical tube of charge (the bold circle) in Fig. 17 has perfect axial symmetry. So inside the tube, E1 and E2 must be radial and equal in magnitude. Applying Gauss’s law to a cylinder of radius a and length ℓ yields 2πaℓE1 = 0, because there is no charge inside the tube. Therefore E = 0 inside the tube. Outside the tube, symmetry also demands that E3 = E4 . Applying Gauss’s law to a cylinder of radius r and length ℓ yields 2πrℓE3 = λℓ/ϵ0 , where λ is the charge per unit length. This gives E = λ/2πϵ0 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/ϵ0 times the charge enclosed, but nothing requires that E1 = E2 or E3 = E4 in Fig. 18. The integral of E over the small inner circle vanishes (as it does for any cross-sectional shape), but it can do so with E1 = E2 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 E2 must point inward (if the charge is positive). a r Figure 17 E4 E3 E1 E2 a r Potential energy of a sphere The charge inside a sphere of radius r (with r < R) is q = (4πr3 /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 = 1 q · dq 1 (4πρ)2 4 r dr. = 3 4πϵ0 4πϵ0 r (52) Building up the whole sphere this way, from r = 0 to r = R, requires the work: ∫ R 1 (4πρ)2 R5 1 (4πρ)2 4 W = r dr = . (53) 4πϵ 3 4πϵ 5 3 0 0 0 The charge in the complete sphere is Q = (4πR3 /3)ρ, which gives 4πρ = 3Q/R3 . Thus the potential energy U , which is the same as the work W , can be written as U = (3/5)Q2 /4πϵ0 R. Note that Q2 /4πϵ0 R has the proper energy dimensions of (charge)2 /(ϵ0 · 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. Electron self energy Setting the potential energy (3/5)e2 /4πϵ0 r0 from Exercise 1.61 equal to mc2 gives r0 = = 3 1 e2 5 4πϵ0 mc2 kg m3 ) (1.6 · 10−19 C)2 3( 9 · 109 2 2 = 1.69 · 10−15 m. 5 s C (9.1 · 10−31 kg)(3 · 108 m/s)2 (54) 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 23 = 8 times as much Figure 18 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 82 = 64. But there is also one power of r0 in the denominator, so this cuts the result down to 32. Equivalently, the result for the energy in Eq. (53) in the solution to Exercise 1.61 is proportional to R5 , and 25 = 32. 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 4πR2 σ, so the potential energy of the particle at the surface of the shell is V (R) = (4πR2 σ )(−q ) Rσq =− . 4πϵ0 R ϵ0 (55) The initial potential energy was zero, so this loss √ in potential energy shows up as kinetic energy. Hence, mv2 /2 = Rσq/ϵ0 =⇒ v = 2Rσq/ϵ0 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 = σ2πr 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 ( ) Rσq dx σ2π(xR/L) dx q dQ(−q) =− . (56) dU = =− 2ϵ0 L 4πϵ0 x 4πϵ0 x Integrating from x = 0 to x = L simply turns the dx into an L, so we have U = −Rσq/2ϵ0 . We need to double this because there are two cones, so we end up with the same potential √ energy of −Rσq/ϵ0 as in part (a). We therefore obtain the same speed of v = 2Rσq/ϵ0 m, independent of L. Field between two wires The electric field from a single wire is λ/2πϵ0 r. Between the wires the fields from the two wires point in the same direction, so we have 15, 000 N/C = 2 λ 2πϵ0 r =⇒ λ = = = P rod dx into and out of page Figure 19 ) 15, 000 N/C πϵ0 r ( ) ) s 2 C2 15, 000 N/C (3.14) 8.85 · 10−12 (1.5 m) kg m3 −7 (57) 6.3 · 10 C/m. ( The amount of excess charge on 1 km of the positive wire is then (1000 m)λ = 6.3 · 10−4 C. r y θ sheet ( Building a sheet from rods In Fig. 19, the horizontal line represents the sheet, which extends into and out of the 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/2πϵ0 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 definition) and σℓ dx (because ℓ dx is the relevant area). The horizontal component of the field cancels with the horizontal component of the field arising from the rod located symmetrically on the left side of P . So (as expected) we care only about the vertical component. This brings in a factor of cos θ. And since x = y tan θ, we have dx = y dθ/ cos2 θ. The (vertical) field at P therefore equals ∫ ∞ ∫ π/2 σ(y dθ/ cos2 θ) σ dx E = cos θ cos θ = −∞ 2πϵ0 r −π/2 2πϵ0 (y/ cos θ) ∫ π/2 σ σ = dθ = , (58) 2πϵ0 −π/2 2ϵ0 as desired. Alternatively, you can write the integral in terms of x. Since cos θ = y/r we have ∫ ∞ ∫ σy ∞ dx σ dx y E = · = 2 2 r 2πϵ r 2πϵ 0 0 −∞ x + y −∞ ( )  ∞ σy 1 σy π σ x  −1 = = . (59) · tan · = 2ϵ y  2πϵ y 2πϵ y 0 −∞ 0 dr 0 Force between two strips (a) Let’s slice one of the strips into narrow rods, and then integrate over the rods. Consider a rod with width dr at a distance r from a given point P , which itself is a distance x away from the edge of the strip; see Fig. 20. The electric field at P due to the rod is λ/2πϵ0 r, where λ = σ dr. The distance r runs from x to x + b. So the total field at P due to the strip is ( ) ∫ x+ b x+b σ σ dr ln E(x) = = . (60) 2πϵ0 2πϵ0 r x x For x → 0 this result diverges, but slowly like ln x. For x → ∞ we can use the Taylor series ln(1 + b/x) ≈ b/x, which gives E ≈ (σb)/2πϵ0 x. This makes sense, because from far away the strip looks essentially like a rod with linear charge density σb. (b) Consider a rod with finite height h and width dx within one of the strips. The force on this rod due to the other strip is (σh dx)E (x). Since the strips are right next to each other, the distance x runs from 0 to b. So the total force on a height h of one of the strips, due to the other strip, is ∫ b (σh dx)E Fh = 0 ) ( ∫ x+b σ2 h b dx ln = x 2πϵ0 0 (∫ ) ∫ b b σ2 h ln(x + b) dx − = ln x dx 2πϵ0 0 0 ) (∫ ∫ b 2b σ2 h ln y dy − ln x dx = 2πϵ0 b 0 (  2b b) σ2 h   (y ln y − y) − (x ln x − x) = 2πϵ0 b 0 ) σ2 h ( 2b ln 2 . (61) = 2πϵ0 b x P r σ Figure 20 rod (into and out of page) P You should verify the algebra leading to the last line. The force per unit height is therefore Fh /h = σ 2 b(ln 2)/πϵ0 , which is finite. The field diverges as x → 0, but only like a log. This isn’t large enough to outweigh the fact that there is only a small range of x that is very close to the strip. Basically, the area under the ln x curve near x = 0 is finite. Field from a cylindrical shell, right and wrong R θ/2 θ/2 dθ Figure 21 (a) Let the rods be parameterized by the angle θ shown in Fig. 21. The width of a rod is R dθ, so its effective charge per unit length is λ = σ(R dθ). The rod is a distance 2R sin(θ/2) from the point P in question, which is infinitesimally close to the top of the cylinder. Only the vertical component of the field from the rod survives, and this brings in a factor of sin(θ/2), as you can check. Using the fact that the field from a rod is λ/2πϵ0 r, we find that the field at the top of the cylinder is (apparently) ∫ π ∫ π σR dθ σ σ 2 ) sin(θ/2) = ( dθ = . (62) 2ϵ0 2πϵ0 0 0 2πϵ0 2R sin(θ/2) Interestingly, we see that for a given angular width of a rod, all rods yield the same contribution to the vertical electric field at P . correct E direction correct distance P incorrect E direction 2Rsin(θ/2) incorrect distance intersection of rod with page Figure 22 (b) As noted in the statement of the exercise, it is no surprise that the above result is incorrect, because the same calculation would supposedly yield the field just inside the cylinder too, where it is zero instead of σ/ϵ0 . The calculation does, however, give the next best thing, namely the average of these two values. We’ll see why shortly. The reason why the calculation is invalid is that it doesn’t correctly describe the field due to rods on the cylinder very close to the given point, that is, for rods characterized by θ ≈ 0. It is incorrect for two reasons. The closeup view in Fig. 22 shows that the distance from a rod to the given point is not equal to 2R sin(θ/2). Additionally, it shows that the field does not point along the line from the rod to the top of the cylinder. It points more vertically, so the extra factor of sin(θ/2) in Eq. (62) isn’t valid. What is true is that if we remove a thin strip from the top of the cylinder (so we now have a gap in the circle representing the cross sectional view), then the above integral is valid for ∫ the remaining part of the cylinder. The thin strip contributes negligibly to the dθ integral, so we can say that the field due to the remaining part of the cylinder is equal to the above result of σ/2π. By superposition, the total field due to the entire cylinder is this field of σ/2π plus the field due to the thin strip. But if the point in question is infinitesimally close to the cylinder, then the thin strip looks like an infinite plane, the field of which we know is σ/2ϵ0 . The desired total field is then σ σ σ . (63) Eoutside = Ecylinder minus strip + Estrip = + = 2ϵ0 2ϵ0 ϵ0 By superposition we also obtain the correct field just inside the shell: Einside = Ecylinder minus strip − Estrip = σ σ − = 0. 2ϵ0 2ϵ0 (64) The relative minus sign arises because the field from the cylinder-minus-strip is continuous across the gap, but the field from the strip is not; it points in different directions on either side of the strip.