Lecture 17 - The Secrets we have Swept Under the Rug

Transcription

1 Lecture 17 - The Secrets we have Swept Under the Rug Today s ectures examines some of the uirky features of eectrostatics that we have negected up unti this point A Puzze... Let s go back to the basics and consider a point charge at the origin. We can appy Gauss s Law over a sphere of radius r centered at to confirm that the radia eectric fied E = E[r] r satisfies = E d a = 4 π r 2 E[r] and hence E = r, as expected. However, using the divergence theorem, we coud rewrite this integra over the surface 4 π r 2 of our sphere as the voume integra over the interior of the sphere (essentiay switching between the integra and differentia form of Gauss s Law), = E dv (1) The divergence of the vector A = A r r + A θ θ + A ϕ ϕ in spherica coordinates is given by A = 1 r 2 r 2 A r r + 1 r Sin[θ] [A θ Sin[θ]] θ + 1 r Sin[θ] A ϕ. But when we evauate the divergence for the point charge s ϕ eectric fied E = r, we get E = 0, impying that the right-hand side of Euation (1) euas zero instead of 4 π r 2. Expain what is going on? This probems highights the subty inherent in deaing with point charges, which are inherenty strange objects. For exampe, the eectric fied infinitey cose to a point charge diverges, and the energy reuired to make a point charge (which can be determined by integrating 2 E 2 d v over a space) simiary bows up. In this probem, we note that E = 0 everywhere in space...or rather, amost everywhere in space. The differentia form of Gauss s Law states that E = ρ, and because the ony charge is a point charge at the origin, it makes sense that E = 0 at a other points in space. Indeed, if we consider any sma cosed Gaussian surface at any point in space that does not contain the origin, we expect that a eectric fied ines entering it wi aso eave it, impying that there is no net fux through the surface and that E = 0. But at the origin, something specia happens. At this point, we know that E = ρ, but what is ρ (the charge per unit voume)? The charge density of a point charge is infinite, since any box that surrounds the point charge, regardess of how sma it is, wi contain net charge, so ρ diverges at the origin. It turns out that ρ bows up as a deta function (4 π δ 3 [r ]) which you wi earn about in junior-eve eectrodynamics in a way that precisey impies that when you integrate E dv over any voume that contains the origin, no matter how sma, the right-hand side of Euation (1) wi correcty evauate to.

2 2 Lecture nb Some Interesting Mathematics Non-Symmetric Gauss s Law Consider the cross section of three infinite sheets intersecting at eua anges. The sheets a have surface charge density σ. 1. By adding up the fieds from the sheets, find the eectric fied at a points in space. 2. Find the fied instead by using Gauss s aw. You shoud expain ceary why Gauss s aw is in fact usefu in this setup. 3. What is the fied in the anaogous setup where there are N sheets instead of three? What is your answer in the N imit? 1. The eectric fied form a given sheet points away from the sheet and has uniform magnitude σ 2. Therefore, in any of the 6 regions of space divided by the sheets, the eectric fied at every point in this region wi be the same. Considering a point on the x-axis (i.e. directy to the right of the intersection point in the diagram above), the eectric fied from the nearest two sheets wi be 2 σ Sin π = σ in the x-direction. The contribution of the third sheet wi aso be σ in the x-direction, eading to an eectric fied in this entire region of σ in the x-direction. By 2 radia symmetry, this is the magnitude of the eectric fied in a of the regions, with the direction of it appropriatey rotated in each region. 2. We use for our Gaussian surface a reguar hexagon centered on the intersection point of the sheet. Assume that the ength of the Gaussian surface between two sheets is and that its depth (going into and our of the page) is d. Because the eectric fied points perpendicuary outwards against our Gaussian surface, Gauss s Law yieds

3 Lecture nb 3 E (6 d) = (6 d) σ (2) or euivaenty E = σ, as found above. Note that whie Gauss s Law is aways vaid, it is ony usefu when you can find a simpe surface that is everywhere perpendicuar to the eectric fied and where the eectric fied has a uniform vaue. 3. For genera N, we use Gauss s Law (or, for the mathematicay minded, set up a summation over pairs of eectric sheets as we did in the first part of this probem) using a reguar 2 N-gon with side ength and depth d. The surface area of this 2 N-gon is (2 N) 2 Sin π d and the charge encosed is (2 N) d σ. So Gauss s Law 2 N yieds or euivaenty E = σ 2 Sin π 2 N N = 3. For arge N, Sin π 2 N π 2 N so E N π E (2 N d) = σ(2 N) d 2 Sin π 2 N (3). As expected, this is independent of. And it agrees with the above resut when σ. In the case of arge N, the sheets are very cose to each other, so we effectivey have a continuous voume charge distribution that depends on r. The separation between adjacent sheets grows ineary with r, so we have ρ[r] 1 N σ. More precisey, you can easiy show that ρ[r] =. As a simpe r π r exercise, you can use Gauss s Law to easiy show that for a voume charge distribution with cyinica symmetry and a constant eectric fied that is independent of r, we must have ρ[r] 1. r Eectric Fied at Corners 1. Consider an thin sheet of uniform charge density σ (shown beow) that extends infinitey in one direction and has a width b the other direction. What is the eectric fied at a distance x from the sheet? What happens as x 0? 2. A very ong tube has a suare cross section and uniform charge density σ. What wi be the eectric fied at a corner? What is the eectric fied if the tube is a conductor so that the charge is free to fow? E x r b Let s start with a brief survey of the second part of the uestion, that is, of the eectric fied at the corner of a conducting tube. By symmetry, we know that the eectric fied must point diagonay away from the centra axis of the tube. But this is probematic, since the eectric fied must point away from the surface of a conductor, and at the corner it seems that the eectric fied must point in two directions at once (to be norma to both faces it is σ

4 4 Lecture nb point (to touching). Your physics senses shoud be tinging. 1. To get a hande on the probem, et s first consider the eectric fied from just one face of the conducting tube, or euivaenty from an infinite conducting strip. Let s assume for now that the strip has a uniform surface charge density σ, tota width b and negigibe thickness as shown above. Treating each of the narrow strips as thin rods whose eectric fied (you wi reca from Gauss s Law) euas E rod = λ r, the tota eectric fied at a distance x from the side of the strip is given by 2 π r E = x x+b σ 2 π r = σ 2 π Log x+b x (4) As x 0, this resut diverges (sowy, ike Log[x]), and hence the eectric fied at the corner is infinite. 2. Now et s turn to the fu probem of a conducting tube. For the moment, we wi assume that the tube has a constant charge density. From above, we know that the eectric fied at the corner must diverge due to the two faces that the point touches. The eectric fied from the other two faces is finite and aso points in this same direction. Therefore, the eectric fied at the diverges at the corner of the tube. Next, et us turn to a conducting tube and reax the assumption that the charge density σ as constant. In a conducting tube, the density woud increase near the corners because of the sef-repusion of the charges. This has the effect of making the fied even arger than the above reasoning woud impy, so the above concusion of a diverging fied is sti vaid. In other words, this concusion is true for both a conducting and nonconducting tube. More generay, this type of divergence happens at any corner. For exampe, if you pace a charge Q on a conducting cube, the eectric fied wi diverge at the corners. As you can show, the divergence found above for the infinite strip does not occur as x 0 because the rods were infinitey ong, but rather because of the bits of charge right on the edge of the strip, which wi come infinitey cose to our point of interest as x 0. This divergence occurs even when the ange at the corner is not 90 - it woud aso be true for a poygon with 100 sides, in which the surface bends by ony a few degrees at each corner. This divergence aso happens at a kink in a wire. Conducting Disk There is a very sneaky way of finding the charge distribution on a conducting circuar disk with radius R and charge Q. Our goa is to find a charge distribution such that the eectric fied at any point in the disk has zero component parae to the disk. Reca from Probem 1.17 in the textbook (which you did in Homework 5) that the fied at any point P inside a spherica she with uniform surface charge density is zero, since the contributions due to both cones emerging from P cance.

5 Lecture nb 5 Consider the projection of this she onto the euatoria pane containing P. Expain why this setup is reevant, and use it to find the desired charge density on a conducting disk. Reca the argument in Probem 1.17 that showed why the fied inside a spherica she is zero. In the figure beow, the two cones that define the charges 1 and 2 on the surface of the she are simiar, so the areas of the end patches are in the ratio of r 1 r 2 2. This factor exacty cances the 1 r 2 effect in Couomb s aw, so the fieds from the two patches are eua and opposite at point P. The fied contributions from the entire she therefore cance in pairs. Now project the charges residing on the upper and ower hemispheres onto the euatoria pane containing P. The charges 1 and 2 in the patches mentioned above end up in the shaded patches shown in the figure above. The (horizonta) fieds at point P from these shaded patches have magnitudes k 1 x2 and k 2 1 x2. But due to the simiar trianges in the figure, x 1 and x 2 are in the same ratio as r 1 and r 2. Hence the two forces have eua magnitudes, just as they did in the case of the spherica she. The forces from a of the various parts of the disk therefore again cance in pairs, so the horizonta fied is zero at P. Since P was arbitrary, we see that the horizonta fied is zero everywhere in the disk formed by the projection of the charge from the origina she. We have therefore accompished our goa of finding a charge distribution that produces zero eectric fied component parae to the disk. Since the spherica she has a arger sope near the sides, more charge is above a given point in the euatoria pane near the edge, compared with at the center. The density of the conducting disk therefore grows with r, as shown in the figure beow. To be uantitative, we want to find the charge density σ disk [r] when we coapse the sphere with uniform charge density σ sphere onto the disk. The charge on an infinitesima patch of the disk a given point (r, ϕ) wi be

6 6 Lecture nb σ disk [r] (r dϕ). Using the diagram above, the patch on the sphere above the point (r, ϕ) wi have ength width r d ϕ, and therefore a tota charge of 2 σ sphere (r d ϕ) Cos[θ]. Note that the factor of 2 emerges since we coapse both this patch and its symmetric counterpart at θ π - θ. Euating these two charges, we find Cos[θ], σ disk [r] = 2 σ sphere Cos[θ] (5) To remove the Cos[θ], we note that r = R Sin[θ] so that Cos[θ] = 1 - r R 2 1/2, and hence As a doube check, we can use σ sphere = 0 2 π 0 R 2 σ sphere σ disk [r] = 2 σ sphere 1- r R 2 1/2 (6) Q 4 π R 2 and ensure that the tota charge on the disk wi be Q, 1- r R 2 1/2 1- r R 2 1/2 r dϕ = 4 π σ sphere 0 R r r=r = 4 π σ sphere -R r R 2 1/2 r=0 = 4 π σ sphere R 2 = Q The density diverges at the edge of the conducting disk, but the tota charge has the finite vaue Q. It is fairy intuitive that the density shoud grow as r increases, because charges repe each other toward the edge of the disk. However, one shoud be carefu with this type of reasoning. In the ower-dimensiona anaog invoving a onedimensiona rod of charge, the density is actuay essentiay uniform, a the way out to the end; see the next probem. Conducting Stick The same strategy from the previous probem can be used to find the charge distribution on a conducting stick. Consider a stick of ength 2 R centered at the origin and ying on the x-axis. To find the charge λ[x] d x which ies between x and x + dx on the stick, consider a sphere of radius R centered at the origin. Coapse a of the charge between x and x + dx onto the x-axis, and find the charge distribution λ[x] of the conducting stick. Then expain why this answer is absoutey absurd (and yet sti correct!). We foow the same strategy as for the Conducting Disk, namey, to take a sphere of radius R and coapse it down onto a rod. We know that each patch on the two cones emerging from P wi cance each other, and this canceation continues after we project the two patches onto the x-axis (see beow, eft). Therefore, the charge λ[x] dx between x and x + dx on the rod wi be given by the tota charge on the ring of the charged sphere between x and x + d x (beow, right). (7)

7 Lecture nb 7 As we saw above for the Conducting Disk, the diagona ength of a ring at a distance x from the center of the dx sphere is, and the circumference of this ring (which has radius Cos[θ] R2 - x 2 1/2 ) is 2 π R 2 - x 2 1/2. Since x = R Sin[θ], Cos[θ] = 1 - x R 2 1/2 so that the tota charge on this ring is given by λ[x] dx = 2 π σ sphere (R 2 - x 2 ) 1/2 dx 1- x R 2 1/2 = 2 π R σ sphere dx As a uick doube check, the fu charge over the entire sphere assuming σ sphere = Q 4 π R 2 is -R R λ[x] dx = 4 π R 2 σ sphere = Q (9) as expected. Euation (8) impies that the charge density on a conducting rod is λ[x] = 2 π R σ sphere = Q 2 R (10) Shockingy, this is uniform! This is an absoutey astonishing resut!!! In case your jaw isn t hanging open, consider how unintuitive this resut is. A point at the center of a uniformy charged rod wi certainy fee zero force by symmetry, but any point not in the exact center wi fee a force pointing away from the center there wi be more charged rod on one of its sides than the other. As a particuary troubesome case, consider the force on the very ends of the rod - ceary these must point outwards. So how can this charge density be correct? As a fun historica aside, there was a very ivey debate over the charge density of a conducting stick. Part of the probem was the apparenty ridicuous nature of the uniform charge density soution. Another issue is that the charge density on a conducting neede seemed to be different depending on which mode you use (see Charge on a Conducting Neede by Griffiths and Li 1996 for an exceent discussion of this probem). This probem is discussed extensivey in Probem 3.5 of the textbook. Loosey speaking, it turns out that if you correct the uniform charge density by breaking up the stick into N sma segments and etting a of the segments (except the two end segments) move to their euiibrium positions, then in the imit N the movement reuired by each segment wi shrink to zero. In this imit, the two end segments can be ignored since they are singe points on a 1D rod. You are invited to think about this perpexing probem and read over the fantastic description in the text. Image Charge of a Conducting Cyinder Maximum Fied from a Bob Symmetry and Superposition Mathematica Initiaization (8)