The Electric Potential Part 2, Solution Manual

E for a sheet, from a cutoff potential Let’s slice the finite disk into concentric rings. √ The charge in a given ring is σ(2πr dr), and all points in the ring are a distance r2 + z 2 from a point that is a distance z from the center of the disk, on the axis. So the potential (relative to infinity) at this point is R ∫ R )  2πσr dr 1 σ (√ 2 σ √ 2 r + z 2  = √ ϕ(z) = R + z2 − z . (141) = 2ϵ0 2ϵ 4πϵ0 0 r2 + z 2 0 0 In the R ≫ z limit we can write √ ( ) √ z2 z2 z2 R2 + z 2 = R 1 + 2 ≈ R 1 + . =R+ 2 2R R 2R (142) We can ignore the second term because it goes to zero as R → ∞ (so the end result of the Taylor series was a trivial one). We therefore obtain ϕ(z) = ) σ ( R−z . 2ϵ0 (143) The constant R term here is irrelevant because it simply introduces a constant additive term to ϕ(z), which yields zero when we take the derivative to find E(z). (The σR/2ϵ0 term is the potential at the center of the disk, where z = 0.) So ϕ(z) is effectively equal to −σz/2ϵ0 . The field (which is perpendicular to the disk) is therefore E(z) = − dϕ σ , = 2ϵ0 dz (144) in agreement with the result obtained via the standard (and quicker) method involving Gauss’s law. This procedure of truncating the sheet into a disk is valid for the following reason. Since the field from a point charge falls off like 1/r2 , we know that the field from a very large disk is essentially equal to the field from an infinite sheet. The extra annulus that extends out to infinity gives a negligible contribution because the process of taking the perpendicular component adds another factor of ∼ r in the More the perpendicular field component from a ring equals (denominator. ) precisely, √ 2πσr dr/(r2 + z 2 ) (z/ r2 + z 2 ), and for large r this behaves like dr/r 2 , the integral of which converges. Therefore, the field we found for the above finite disk is essentially the same as the field for an infinite sheet, provided that R is large enough to make the approximation in Eq. (142) valid. Mathematically, the field in Eq. (144) is independent of R, so when we finally take the R → ∞ limit, nothing changes. Equivalently, we showed that although the potential in Eq. (143) depends linearly on R, the field in Eq. (144) is independent of R. As far as the field is concerned, it doesn’t matter that increasing the size of the disk changes the potential at every point, because the potential everywhere changes by the same amount (assuming R ≫ z). The variation with z remains the same, so E = −dϕ/dz doesn’t change. As an analogy, we can measure the gravitational potential energy mgy with respect to the floor. If we shift the origin to instead be the ceiling, then the potential energy at every point changes by the same amount, but the gravitational force everywhere is still mg . Dividing the charge The potential is constant over the surface of a given sphere, so we can pull the ϕ outside the integral in Eq. (2.32) and write the potential energy of a sphere as U = ∫ (ϕ/2) ρ dv = ϕq/2. So if the spheres of radii R1 and R2 have charge q and Q − q , respectively, the sum of the two potential energies is ) ( 2 q q (Q − q )2 q 1 Q−q Q−q U= . (145) + · + = · 2R2 4πϵ0 2R1 4πϵ0 R1 2 4πϵ0 R2 2 Minimizing this by setting the derivative with respect to q equal to zero yields ) ( 1 (Q − q) q dU = − 0= . (146) dq R2 4πϵ0 R1 Solving for q gives q = QR1 /(R1 + R2 ). So there is charge QR1 /(R1 + R2 ) on the first sphere and charge QR2 /(R1 + R2 ) on the second sphere. The two terms in Eq. (146) (without the minus sign in front of the second term) are simply the potentials of the two spheres. So the condition of minimum energy is equivalent to the condition of equal potentials. Note that the second derivative, d2 U/d2 q = 1/R1 + 1/R2 , is positive, so the extremum is indeed a minimum of U , not a maximum. This is consistent with the special case where R1 = R2 ; equal division of the charge involves half as much total energy as piling all of Q on one sphere, from Eq. (145). Potentials on the axis We’ll slice the cylinder into rings and then integrate over these rings. The charge in a ring with width dx is dQ = Q(dx/b). The ring is a convenient charge element to use in computing the potential at axial points, because all of the charge in the ring is equidistant from a given point on the axis, We’ll find the potential at a general axial point, a distance x0 from the midpoint. With x = 0 chosen to be the midpoint of the axis, we have (the integral in given in Appendix K) ϕ= 1 4πϵ0 ∫ dQ r = = = 1 4πϵ0 ∫ b/2 −b/2 √ Q dx/b a2 + (x − x0 )2 ) b/2 1 Q (√ 2 ln a + (x − x0 )2 + (x − x0 )  4πϵ0 b −b/2 ( √ ) a2 + (b/2 − x0 )2 + (b/2 − x0 ) 1 Q ln √ . (147) 2 2 4πϵ0 b a + (−b/2 − x0 ) + (−b/2 − x0 ) At the midpoint we have x0 = 0, so ϕmid = √ a2 + b2 /4 + b/2 1 Q ln √ . 4πϵ0 b a2 + b2 /4 − b/2 (148) At an endpoint we have x0 = b/2, so ϕend = a 1 Q 1 Q ln √ = ln 4πϵ0 b 4πϵ0 b a 2 + b2 − b √ a 2 + b2 + b . a (149) The difference is therefore ϕmid − ϕend = (√ ) a a2 + b2 /4 + b/2 1 Q ). )(√ ln (√ 4πϵ0 b a 2 + b2 + b a2 + b2 /4 − b/2 (150) This is a rather messy answer, so let’s look at how ϕmid and ϕend behave in some limits, to feel better about the result. Consider the limit b → 0, in which case the cylinder reduces to a thin ring. To first order in b, we can ignore the b2 terms under the square roots in the above expressions. We obtain ( ) Q Q Q b b Q a + b/2 · = ≈ ln 1 + . (151) 4πϵ0 ϕmid ≈ ln ≈ a b a b a − b/2 b a Likewise, 4πϵ0 ϕend ≈ ( ) Q b Q . ln 1 + ≈ a b a (152) Both of these results make sense, because all of the charge on the ring is essentially a distance a from the midpoint of the axis, which is essentially the same as an endpoint. The difference in the potentials is therefore zero. In the limit b → ∞, both ϕmid and ϕend go to zero because of the Q/b coefficient, and because we are holding the total charge Q constant. If we instead let the charge per unit length be constant, so that Q = λb, then you can show that 4πϵ0 ϕmid ≈ 2λ ln(b/a) and 4πϵ0 ϕend ≈ λ ln(2b/a). For large b, the “2” inside the log doesn’t matter much, so we see that ϕmid is approximately twice as large as ϕmid. The reason for this basically comes down to the fact that if you’re in the middle of a long cylinder, you see two long cylinders on either side of you (and also the fact that the potential due to a stick grows only like the log of the length, so that a long stick yields roughly the same potential as one that has twice the length). A rough sketch of the field lines is shown in Fig. 42. The field is zero at the center, by symmetry. We haven’t worried about drawing the density of field lines correctly; for a very long cylinder, you can show that the field at the surface is twice the field at the midpoint of an end face. Spherical cavity in a slab (a) The given setup can be considered to be the superposition of the given slab and a sphere with charge density −ρ. Our goal is to show that the potential at a point on the surface of the slab at infinity equals the potential at the center of the cavity. From the standard Gauss’s-law argument with a pillbox extending from −x to x, the field due to the slab in the interior is x̂ρx/ϵ0 . (Alternatively, the relevant part of the slab is effectively a sheet with surface charge density ρ(2x)). And outside the slab the field is x̂ρR/ϵ0 (the whole slab is effectively a sheet with surface charge density ρ(2R)). Similar Gauss’s-law arguments give the field due to the (negative) sphere in the interior as −r̂ρr/3ϵ0 , and outside as −r̂ρR3 /3ϵ0 r2 . Integrating the slab and sphere fields to find the total potential relative to the center of the cavity, we find that the potential inside the cavity equals −ρx2 /2ϵ0 + ρr2 /6ϵ0 . At the rightmost point of the cavity, we have x = r = R, so the potential of this point (relative to the center) is (ρR2 /ϵ0 )(−1/2 + 1/6) = −ρR2 /3ϵ0 . If we now march along the surface of the slab to infinity, only the field from the sphere matters. The change in potential as we march out is ∫ ∞ ρR3 ρR2 −ρR3 . (153) dr = = ∆ϕ = − 2 3ϵ 3ϵ r R 3ϵ0 0 0 R (Alternatively, the change in potential from R to ∞ is just −Q/4πϵ0 R, with Q = −4πR3 ρ/3.) This is correctly positive, because the potential increases as Figure 42 we move away from the negative sphere. This ∆ϕ exactly cancels the negative potential at the rightmost point of the cavity, so we end up with zero potential on the surface of the slab at infinity, as desired. The algebra here basically boils down to 1/2 = 1/6 + 1/3. The 1/2 is the (magnitude of the) decrease in potential due to the slab. The 1/6 + 1/3 is the increase in potential due to the sphere, inside and outside. (b) The potential inside the cavity is ϕ = −ρx2 /2ϵ0 + ρr2 /6ϵ0 . In the plane of the paper, we have r2 = x2 + y 2 , so ϕ becomes√ρ(y 2 − 2x2 )/6ϵ0 . Points on the ϕ = 0 curve therefore satisfy y 2 = 2x2 =⇒ y = ± 2x. This is consistent with Fig. 2.49, where the slope of the straight line looks to be a little larger than 1. (c) We found in part (a) that as we move from the rightmost point of the cavity out to infinity along the surface of the slab, the potential increases by ρR2 /3ϵ0 . If we want to end up at the same potential as at the rightmost point of the cavity, we must then move away from the slab by a distance that causes the potential to decrease by ρR2 /3ϵ0 . Since the field outside the slab equals x̂ρR/ϵ0 (far away, the sphere can be ignored), the change in potential as we move away from the slab is −(ρR/ϵ0 )∆x. This equals −ρR2 /3ϵ0 when ∆x = R/3, as desired. -Q Q Field from two shells By superposition, the electric field outside both shells is that of two point charges located at the centers. And also by superposition, the field inside each shell is that of a point charge at the center of the other shell (because a given spherical shell with uniform charge distribution produces no field in its interior). We therefore obtain the fields shown in Fig. 43. Note that the field lines inside each shell are straight. The two shells are equivalent (as far as their external fields go) to two point charges Q and −Q at their centers. Therefore we may replace the spheres with these point charges. Since the centers are initially 2a apart, the amount of work required to move them to infinity is (1/4πϵ0 )(Q2 /2a). Figure 43 In more detail: Let the positive shell be labeled A, and the negative shell B. The external field of A alone is that of a point charge Q at its center. So the work required to move B to infinity in the given setup is the same as the work required in an alternative setup where A is replaced by a point charge. But the work in this case is the same as if we instead held B fixed and moved the point charge to infinity. But since the external field of B alone is that of a point charge −Q at its center, we may replace B with a point charge. The work is therefore the same as in the case where we have two charges Q and −Q that are initially 2a apart. An equipotential for a disk σa φ = ___ πε0 disk Figure 44 From Eq. (2.30) the potential on the rim (with ϕ = 0 (√ at r = ∞) is)σa/πϵ0 . From Eq. (2.25) the potential on the symmetry axis is (σ/2ϵ0 ) y 2 + a2 − y . Setting these equal yields ( ) √ 2a 2 2a π2 − 4 =⇒ y 2 +a2 = y + y 2 + a2 −y = =⇒ y = a ≈ (0.467)a. (154) π π 4π The equipotential surface is (roughly) represented by the curve in Fig.44. The direction of the curve near the edge of the disk happens to be perpendicular to the plane of the disk. This follows from the result of Exercise 2.57(b) below; the tangential component of the field diverges near the edge of the disk, so the field is essentially parallel to the disk. The equipotential curve is therefore is perpendicular to the disk. Hole in a disk (a) Slicing the disk into concentric rings, we find the potential at the center to be (with ℓ = 1 cm) ∫ 3ℓ ∫ 2πrσ dr 1 σℓ dq 1 ϕ= . (155) = = 4πϵ0 ℓ ϵ0 r r 4πϵ0 Plugging in the various quantities gives ) ( − 10−5 mC2 (0.01 m) = −11, 300 V. ϕ= s2 C 2 8.85 · 10−12 kg m3 (156) (b) The electron’s final kinetic energy at infinity equals the loss in potential energy. This loss has magnitude (−e)ϕ = (−1.6 · 10−19 C)(−11, 300 V) = 1.81 · 10−15 J. (157) 2 Since this is only about 2% of the electron’s rest energy, namely mc = 8.2 · 10−14 J, a nonrelativistic calculation will suffice: )1/2 ( 1 2(1.81 · 10−15 J) mv2 = 1.81 · 10−15 J =⇒ v = = 6.3 · 107 m/s, (158) −31 9.1 · 10 2 kg which is about 20% of the speed of light. This answer is very close to the answer obtained via the correct relativistic calculation: Conservation of energy gives γmc2 = mc2 + |∆U | =⇒ γ = 1 + 1.81 · 10−15 J = 1.022. 8.2 · 10−14 J Hence (with β ≡ v/c), √ β = 1 − 1/γ 2 = 0.206 =⇒ v = βc = 6.2 · 107 m/s. (159) (160) Energy of a disk From Eq. (2.30) the potential at a point on the rim of a disk with radius r is ϕr = σr/πϵ0 . Adding on a ring with charge dq = σ2πr dr requires an energy of dU = ϕr dq = 2σ 2 r2 dr/ϵ0 . The total amount of energy required to assemble the disk of charge is therefore ∫ ∫ a 2σ 2 a 2 2σ 2 a3 . (161) U= dU = r dr = ϵ0 0 3ϵ0 0 But Q = πa2 σ =⇒ σ = Q/πa2 , so we can write U= 2 Q2 0.0675 Q2 . ≈ a 3π 2 ϵ0 a ϵ0 (162) From Problem 1.32 or Exercise 2.58, the energy required to build up a hollow spherical shell with radius a and charge Q is U= 0.0398 Q2 1 Q2 ≈ . ϵ0 8πϵ0 a a (163) As expected, this is smaller than the result for the disk, because the charges are generally closer to each other in the case of the disk. Field near a disk (a) A little area element within a wedge in Fig. 2.50 has area r dr dθ. So at the given point P , the magnitude of the field due to this little area is σ(r dr dθ)/4πϵ0 r2 = σ dr dθ/4πϵ0 r. But only the vertical component (in the plane of the page) survives, which brings in a factor of cos θ, yielding σ cos θ dr dθ/4πϵ0 r. If we integrate this from r = 0 up to either the r1 or r2 in Fig. 2.50, we obtain an infinite result. However, the divergence from one wedge cancels the divergence from the other, because the field due to the short wedge cancels the field due to the part of the long wedge out to a distance r1 . The uncanceled part of the long wedge comes from r values ranging from r1 out to the end at r2 .1 The net vertical component of the field from the two opposite wedges in Fig. 2.50 is therefore ∫ r2 ( ) σ σ cos θ dr dθ r2 E∥ = = cos θ dθ. (164) ln 4πϵ0 r r1 4πϵ0 r1 r1 P θ We must now integrate this result over θ. We can integrate from 0 to π/2 and then double the result; this will end up covering the whole disk. The task now is to find r1 and r2 in terms of θ. We can find r2 by using the law of cosines in the triangle involving r2 in Fig. 45. This gives R2 = (ηR)2 + r22 − 2(ηR)r2 cos θ. Solving this quadratic equation for r2 and choosing the positive root gives (√ ) r2 = R 1 − η 2 sin2 θ + η cos θ . (165) R θ r2 ηR R Figure 45 The process involving r1 (with the triangle containing the angle π − θ) is the same except for the replacement of cos θ with − cos θ. So twice the integration of the result in Eq. (164) from 0 to π/2 gives ) ∫ π/2 ( √ σ 1 − η 2 sin2 θ + η cos θ E∥ = cos θ dθ. (166) ln √ 2πϵ0 0 1 − η 2 sin2 θ − η cos θ (b) Let’s look at small values of η first. We can ignore terms of order η 2 , so the log term in Eq. (166) becomes ) ) ( ( ( ) (1 + η cos θ)2 1 + η cos θ ≈ ln 1 + 2η cos θ ≈ 2η cos θ, (167) = ln ln 1 − η 2 cos2 θ 1 − η cos θ where we have used ln(1 + x) ≈ x. Substituting this into Eq. (166) gives E∥ ≈ P disk ηR unbalanced part of disk circle centered at P Figure 46 2ηR ση πϵ0 ∫ π/2 cos2 θ dθ = 0 ση ση ( π ) = , 4ϵ0 πϵ0 4 (168) where we have used the fact that the average value of cos2 θ equals 1/2. It is reasonable that this result is linear in η, because as the given point P is moved away from the center, the unbalanced part of the disk (the shaded part in Fig. 46) grows linearly with η (for small η). Intuitively, the ση/4ϵ0 field from the unbalanced part should take the same form, up to a numerical factor, as the field from an effective line of charge with small width 2ηR (indicated by the dotted 1 If the given point doesn’t lie exactly in the plane of the disk, then there is actually no divergence. But it is still worth mentioning the canceling divergences, because E∥ is well defined even if the point does lie in the plane (whereas E⊥ isn’t, for an infinitesimally thin disk). lines in Fig. 46). And indeed, you can show that the former is π/4 times the latter. Now let’s look at values of η very close to 1. With η ≡ 1 − ϵ, the square root term in Eq. (166) becomes (dropping terms of order ϵ2 ) √ √ √ 2ϵ sin2 θ 2 2 2 2 1 − (1 − ϵ) sin θ ≈ cos θ + 2ϵ sin θ = cos θ 1 + cos2 θ ) ( ϵ sin2 θ ϵ sin2 θ = cos θ + , (169) ≈ cos θ 1 + cos2 θ cos θ √ where we have used 1 + x ≈ 1 + x/2. In the numerator in the log term in Eq. (166), the leading-order term (which happens to be zeroth order in ϵ, in this case) is simply 2 cos θ. But in the denominator, the cos θ terms cancel, so the leading-order term is ϵ sin2 θ/ cos θ + ϵ cos θ = ϵ/ cos θ. The argument of the log term is therefore (2 cos θ)/(ϵ/ cos θ) = 2 cos2 θ/ϵ. So Eq. (166) becomes (we’ll drop the ln(2 cos2 θ) term, because it is small compared with the leading-order ln ϵ term; so all that matters is the 1/ϵ behavior of the argument) ) ∫ π/2 ( σ 2 cos2 θ cos θ dθ E∥ ≈ ln ϵ 2πϵ0 0 ∫ π/2 σ ln ϵ σ ln ϵ ≈ − cos θ dθ = − . (170) 2πϵ0 0 2πϵ0 Note that this result is positive since ϵ < 1. We see that the field diverges as we get very close to the edge of the disk, but it diverges slowly, like a log. The log behavior makes sense, because as we get close to the edge, the short wedge in Fig. 2.50 cancels only a small part of the long wedge. So we need to integrate the 1/r field (see Eq. (164)) due to the long wedge almost down to r = 0. And the integral of 1/r diverges like ln r near r = 0. You can check, for various values of η near 0 or 1, that Eqs. (168) and (170) correctly agree with a numerical evaluation of Eq. (166). Energy of a shell The relevant volume in the integral in Eq. (2.32) is all located right on the surface of the shell where the potential ϕ takes on the uniform value of Q/4πϵ0 R. We can therefore take ϕ outside the integral, yielding ) ( ∫ Q2 1 1 1 Q U = ϕ ρ dv = ϕ Q = Q= . (171) 4πϵ R 8πϵ0 R 2 2 2 0 Energy of a cylinder The electric field outside the cylinder is λ/2πϵ0 r, where λ = ρπa2 is the charge per unit length in the cylinder. So the field outside is Eout = ρa2 /2ϵ0 r. The field at radius r inside the cylinder is due to the charge within radius r, so the effective charge per unit length is λr = ρπr2 . The field inside is therefore Ein = λr /2πϵ0 r = ρr/2ϵ0 . The potential at radius r inside the cylinder, relative to a given radius R outside the cylinder, is the negative integral of the field from R down to r. We must break this integral into two pieces: ( ) ∫ r ∫ a ρ ρa2 ρr′ ′ R ρa2 + dr′ − ln ϕ(r) = − (172) dr = (a 2 − r 2 ). ′ a 2ϵ r 2ϵ 4ϵ0 2ϵ 0 0 0 R a Equation (2.32) then gives the energy (relative to the configuration where the charge is distributed over a cylinder with radius R) in a length ℓ of the cylinder as [ ( ) ] ∫ ∫ ρ 1 R 1 a ρa2 ρ U= + ln (a2 − r2 ) 2πrℓ dr ρϕ dv = a 2 0 2ϵ0 4ϵ0 2 ] ∫ [ ( ) πρ2 a 2 1 U R =⇒ + (a2 − r2 ) r dr = a ln 2 a 2ϵ0 0 ℓ )] ( ) [ ( πρ2 a4 R a4 1 a4 = ln + − 4 a 2ϵ0 2 2 2 [ ( ) ] ρ2 π 2 a 4 R 1 = ln + a 4 4πϵ0 [ ( ) ] 2 R 1 λ ln + , (173) = 4πϵ0 a 4 as desired. If R = a, so that all of the charge is initially distributed over the surface of the cylinder, then it takes an amount of work per unit length equal to λ2 /16πϵ0 to move the charge inward and distribute it uniformly throughout the volume of the cylinder. Horizontal field lines In terms of r and θ, the y coordinate of a point in the plane is y = r cos θ. (Remember that θ is measured down from the vertical.) Using the given expression for r, we can write y as a function of only θ as y = (r0 sin2 θ) cos θ. The curves are horizontal at points where y achieves a maximum value, so we want to maximize y as a function of θ: 0= √ ( ) dy = r0 2 sin θ cos θ · cos θ + sin2 θ(− sin θ) =⇒ tan θ = ± 2. dθ (174) √ The field is therefore horizontal everywhere on the two lines with slopes of ±1/ 2 passing through the origin. (Since θ is measured from the vertical, the slope of a line is 1/ tan θ.) More generally, the field is horizontal everywhere in space on the cones generated by rotating these two lines around the y axis. Dipole field on the axes With the charges q and −q located at z = ℓ/2 and −ℓ/2, consider a distant point on the positive z axis with z = r. The charge q is slightly closer than the charge −q is to this point, so the upward field due to the charge q is slightly stronger than the downward field due to the charge −q. The net field will therefore point upward, and it has magnitude (with k ≡ 1/4πϵ0 ) ) ( 1 kq kq 1 kq − − = E= 2 2 2 (r − ℓ/2) (r + ℓ/2) (1 + ℓ/2r)2 r2 (1 − ℓ/2r) ) ( 1 1 kq − ≈ , (175) 2 1 − ℓ/r 1 + ℓ/r r where we have dropped terms of order ℓ2 /r2 . Using 1/(1 ± ϵ) ≈ 1 ∓ ϵ, we obtain (( ) ( )) ℓ ℓ kq 2kqℓ E ≈ 2 1+ − 1− = . (176) r r r r3 This field points in the positive rˆ direction, so it agrees with the result in Eq. (2.36), ) kqℓ ( 2 cos θ r̂ + sin θ θ̂ , r3 when θ = 0. q l/2 (177) β r l/2 -q In the transverse direction, we have the situation shown in Fig. 47. The magnitudes of the two fields are equal. The horizontal components cancel, but the downward components add. The distances from the given point to the two charges are essentially equal to r, so the magnitudes of the fields are kq/r2 . The (negative) vertical components are obtained by multiplying by sin β, which is approximately equal to (ℓ/2)/r in the small-angle approximation. The vertical field is therefore directed downward with magnitude ( ) kq ℓ/2 kqℓ E≈2 = 3 . (178) r r r2 Figure 47 This agrees with the result in Eq. (2.36) when θ = π/2, because the θ̂ vector points downward at the given point (in the direction of increasing θ, which is measured down from the vertical). This field is half as large as the field on the vertical axis, for a given value of r. Square quadrupole θ θ By symmetry, the desired field is radial. Let the lines to the negative charges make an angle θ with respect to the line to the center of the square, and let the distance to the negative charges be r1 , as shown in Fig. √48. Then the total field at a distance r from the center is (with k ≡ 1/4πϵ0 , d ≡ ℓ/ 2, and ϵ ≡ d/r) Er = = = ≈ = kq kq 2kq + − 2 cos θ r1 (r − d)2 (r + d )2 kq kq 2kq r √ + − (r − d)2 (r + d )2 r 2 + d 2 r 2 + d 2 [ ] 1 kq 1 2 + − 2 3/2 2 r2 (1 − ϵ)2 (1 + ϵ ) (1 + ϵ) ) ( ) ( )] kq [( 2 1 + 2ϵ + 3ϵ + 1 − 2ϵ + 3ϵ2 − 2 1 − 3ϵ2 /2 r2 kq [ 2 ] 9kqℓ2 . 9ϵ = 2r4 r2 r1 r r1 d d l l Figure 48 (179) There are various ways of obtaining the above Taylor series, but perhaps the easiest is to note that, for example, 1/(1 − ϵ)2 equals the derivative of 1/(1 − ϵ), which itself is just the sum of the geometric series 1 + ϵ + ϵ2 + · · · . Our result for Er is positive, so the field points away from the square. Along the other diagonal, it points toward the square. This implies that if we traverse a large circle around the quadrupole, there are four locations where the radial component of the field is zero. This should be contrasted with the field of a dipole, which has only two such locations where the radial component is zero. P r1 Two-dimensional dipole Consider the point P show in Fig. 49. The field due to the positive wire takes the form λ/2πϵ0 r. Since the field is radial and has a magnitude that depends only on r, r λ r2 l/2 θ l/2 -λ Figure 49 the potential at P (due to the positive wire) relative to the point midway between the wires is ( ) ∫ r1 ∫ r1 ℓ/2 λ λ dr ln ϕ=− Er dr = − = . (180) 2πϵ0 r1 ℓ/2 ℓ/2 2πϵ0 r As a double check on the sign, if r1 is very small (although we’re not concerned with this case), then ϕ is a large positive quantity,( as it should be. Likewise, the ) potential due to the negative wire is −(λ/2πϵ0 ) ln (ℓ/2)/r2 . When we add these drops out, and we end up with a total potential of two potentials, the ( ℓ dependence ) ϕ = (λ/2πϵ0 ) ln r2 /r1 . Using the same approximate forms of r1 and r2 that we used in Eq. (2.35) in the r ≫ ℓ limit, we find r2 r1 = ≈ r + (ℓ/2) cos θ 1 + (ℓ/2r) cos θ = 1 − (ℓ/2r) cos θ r − (ℓ/2) cos θ ( )2 1 + (ℓ/2r) cos θ ≈ 1 + (ℓ/r) cos θ, (181) where we have used 1/(1 − ϵ) ≈ 1 + ϵ. We can now use the Taylor approximation ln(1 + ϵ) ≈ ϵ to write ( ) ( ) λ r2 λ ℓ cos θ λℓ cos θ ϕ(r, θ) = ln . (182) ≈ = 2πϵ0 2πϵ0 r r1 2πϵ0 r The 1/r dependence in ϕ is the same as the 1/r dependence in the individual fields from the wires. This is analogous to the fact that in the 3D dipole case in Section 2.7, the 1/r2 dependence in ϕ was the same as the 1/r2 dependence in the individual fields from the point charges. We can now find the electric field via E = −∇ϕ. In polar coordinates the gradient operator is given by ∇ = r̂(∂/∂r) + θ̂(1/r)(∂/∂θ). So the electric field equals ( ) ( ) ) 1 ∂ λℓ cos θ ∂ λℓ cos θ λℓ ( (183) E = −r̂ cos θ r̂ + sin θ θ̂ . − ˆθ = r ∂θ 2πϵ0 r2 ∂r 2πϵ0 r 2πϵ0 r The calculation of the shapes of the field-line curves and the constant-potential curves is nearly the same as in Section 2.7. The equation for the constant-ϕ curves is immediately obtained from Eq. (182). The set of points for which ϕ takes on the constant value ϕ0 is given by ( ) λℓ cos θ λℓ = ϕ0 =⇒ r = cos θ =⇒ r = r0 cos θ, (184) 2πϵ0 r 2πϵ0 ϕ0 where r0 ≡ λℓ/2πϵ0 ϕ0 is the radius associated with the angle θ = 0. In the lower half plane, both ϕ0 and cos θ are negative, so r is still positive, as it should be. As an exercise, you can show that r = r0 cos θ describes a circle with diameter r0 . So the equipotential curves are circles; see the solid lines in Fig. 50. As explained in Section 2.7, the slope of a given curve at a given point, relative to the r̂ and θˆ basis vectors at that point, is dr/rdθ. So the slope of the r = r0 cos θ curve is 1 sin θ 1 dr = (−r0 sin θ) = − . r0 cos θ cos θ r dθ (185) Remember that this is the slope with respect to the local r̂-θ̂ basis (which varies with position), and not the fixed x̂-ŷ basis. θ =0 θ= π/2 1 Field lines and constant-potential curves for a dipole. The two sets of curves are orthogonal at all intersections. The solid lines show constant-ϕ curves (r = r0 cos θ), and the dashed lines show E field lines (r = r0 sin θ). Figure 50 Now consider the E field. As in Section 2.7, we’ll do things in reverse order, first finding the slope of the tangent, and then using that to find the equation of the fieldline curves. The slope of the tangent is immediately obtained from the Er and Eθ components given in Eq. (183). We have Er cos θ . = sin θ Eθ (186) This slope is the negative reciprocal of the slope of the tangent to the constant-ϕ curves, given in Eq. (185), as it should be. To find the equation for the field-line curves, we can use the fact that the slope in Eq. (186) must be equal to dr/rdθ. We can then separate variables and integrate to obtain cos θ 1 dr = sin θ r dθ =⇒ ∫ dr = r ∫ cos θ dθ sin θ =⇒ ln r = ln sin θ + C. (187) Exponentiating both sides gives r = r0 sin θ. (188) where r0 ≡ eC is the radius associated with the angle θ = π/2. As an exercise, you can show that r = r0 sin θ describes a circle with diameter r0 . So the field-line curves are also circles; see the dashed lines in Fig. 50. Field lines near the equilibrium point (a) Setting a = 1 and ignoring the factor of q/4πϵ0 , the potential due to the two charges, at locations in the xy plane, is ϕ(x, y) = √ 4 (x + 2)2 + y 2 −√ 1 (x + 1)2 + y 2 . (189) √ Using the Taylor expansion 1/ 1 + ϵ ≈ 1 − ϵ/2 + 3ϵ2 /8, and keeping terms up to second order in x and y, we have 4 1 1 + (2x + x2 + y 2 ) 1 = √ −√ 1 + (x + x2 /4 + y 2 /4) 1 + (2x + x2 + y 2 ) ( ) ) 3( )2 1( ≈2 1− x + x2 /4 + y 2 /4 + x + · · · 2 8 ( ) ) 3( )2 1 ( 2 2 − 1− 2x + x + y + 2x + · · · 2 8 = 1 + (1/4)(y 2 − 2x2 ). ϕ(x, y) = √ 4 + (4x + x2 + y 2 ) 2 −√ (190) (Alternatively, you can obtain this from the Series operation in Mathematica.) If we had included the z dependence, then a z 2 term would appear in the same manner as the y 2 term. That is, the term in parentheses would be y 2 + z 2 − 2x2 . In terms of all the given parameters, you can show that ( ) y 2 − 2x2 q ϕ(x, y) ≈ 1+ . (191) 2 4πϵ0 a 4a y x Figure 51 Some level surfaces of the function y 2 − 2x2 are shown in Fig. 51. The origin is a saddle point; it is a maximum with respect to variations in the x direction, and a minimum with respect to variations in the y direction.√The constant-ϕ lines passing through the equilibrium point are given by y = ± 2x (near the origin). If we zoom in closer to the origin, the curves keep the same general shape; the picture looks the same, with the only change being the ϕ value associated with each curve. (b) The electric field is the negative gradient of the potential, so we have E = −∇ϕ = (192) The field lines are the curves whose tangents are the E field vectors, by definition. Equating the slope of a curve with the slope of the tangent E vector, and separating variables and integrating, gives ∫ ∫ dy dy y dy 1 dx Ey =⇒ =− =⇒ =− = x Ex 2x y dx 2 dx 1 B =⇒ ln y = − ln x + A =⇒ y = √ , (193) 2 x y x Figure 52 q (2x, −y). 8πϵ0 a3 ≡ eA . Different values of B give where A is a constant of integration, and B √ different field lines. Technically, this y = B/ x result is valid only in the first quadrant. But since the setup is symmetric with respect to the yz plane (at least near the origin), and also with respect to rotations around the x axis, the general form of the field lines in the xy plane is shown in Fig. 52. If we zoom in closer to the origin, the lines keep the same general shape. If we include the z dependence, then the correct expression for the field near the origin has the (2x, −y) vector in Eq. (192) replaced with (2x, −y, −z). As a check on this, the divergence of this vector is zero, which is correct because ∇ ·E = ρ/ϵ0 and because there are no charges at the origin. Although the abbreviated vector in Eq. (192) is sufficient for making a picture of what the field lines look like, it has nonzero divergence, so its utility goes only so far. See Exercise 2.65 for an extension of this exercise. A theorem on field lines (a) First note that the equilibrium point must lie on the line containing the two charges, because otherwise the fields from the two charges won’t point along the same line, which means that there is no way for the fields to exactly cancel. There are various cases to consider. (1) If both charges are nonzero and have the same sign, then the equilibrium point must lie between them, and it is easy to see that there is only one such point. (2) If both charges are nonzero and have opposite sign, then the equilibrium point must lie outside them, closer to the smaller charge. The one exception to this case is when the charges are equal and opposite, in which case there is no equilibrium point (technically, it is located at infinity). (3) If one of the charges is zero, then there is no equilibrium point. (4) If both charges are zero (a trivial case which probably isn’t worth considering), then every point is an equilibrium point. (b) If we choose the origin of our coordinate system to be the equilibrium point, with the two charges lying on the x axis, and if we Taylor-expand the potential around the origin, then it can’t have any terms that are linear in the coordinates. This is true because if it did, the electric field (which is the negative gradient of the potential) would have constant nonzero terms, violating the fact that E = 0 at the equilibrium point. So the potential must take the form of ϕ = A + ax2 + by2 + cz 2 (plus higher order terms, which we can ignore close to the origin). But b = c because the system is symmetric under rotations around the x axis. So ϕ = A + ax2 + b(y 2 + z 2 ). The relation ∇2 ϕ = −ρ/ϵ0 tells us that ∇2 ϕ = 0, because ρ = 0 at the equilibrium point (and everywhere else, except at the locations of the charges). Therefore, a = −2b. So the potential must take the form of ϕ = A + b(−2x2 + y 2 + z 2 ). (194) In the√ xy plane (that is, for z = 0), the equipotential lines are therefore given by y = ± 2x, as desired. Note that all of the above reasoning is still valid even if we replace one (or both) of the point charges with a stick (as in Exercise 2.45, for example), as long as all of the charge in the setup lies on a single line. Remark: The above reasoning can also be applied to the equipotential lines that cross at the origin in Fig. 12.42 in√ the solution to Problem 2.19. We claim that the slopes of these lines are equal to ±1/ 2. This follows from Gauss’s law and symmetry around the z axis (instead of the x axis in the above setup with two points); the potential must look like (195) ϕ = A + b(−2z 2 + x2 + y 2 ) near the origin. In the xz plane (that is, for y = 0), the equipotential lines are therefore √ given by z = ±x/ 2. All of the above results can be summarized by saying that if a system is symmetric under rotations around a given axis, then at an equilibrium √ point the equipotential lines make an angle of tan−1 ( 2) ≈ 54.7◦ with respect to the symmetry axis.