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Lecture 9 - The Curious Case of Discontinuities A Puzzle... Electron Jelly Imagine a sphere of radius a filled with negative charge of uniform density, the total charge being equivalent to that of two electrons. Assume that two protons are embedded in this jelly, and further suppose that, in spite of their presence, the negative charge distribution remains uniform. Where must the protons be located so that the force on each of them is zero? (This is a surprisingly realistic caricature of a hydrogen molecule; the magic that keeps the electron cloud in the molecule from collapsing around the protons is explained by quantum mechanics!) Solution Recalling Gauss’s Law, the electric field at a distance r ≤ a from the center of the jelly sphere (which has charge density ρ = - 4 2 e 3 ) equals E 4 π r2 = ϵ1 43 π r3 ρ or equivalently 3 πa 0 E= A proton at a distance r would feel the force Fjelly = E e = rρ 3 ϵ0 (1) r ρ e 3 ϵ0 pulling it towards the center of the jelly. To have stable equilibrium, the other proton must also be at a distance r on the same diameter as the first proton, but in the 2 opposite direction. The repelling force between the two protons equals Fproton = (2k er)2 . Equilibrium occurs when Fjelly = Fproton , which allows us to solve for r, re 3 ϵ0 r ρ e 3 ϵ0 = 2e π a3 = r3 a3 = 4 3 r= In retrospect, this factor of 1 2 k e2 (2 r)2 k e2 (2 r)2 1 8 a 2 (2) is clear. If all of the -2 e electron charge were located in a point charge at the center, it would provide a force on one of the protons that is 8 times the force due to the other proton (because the other proton is twice as far away, and half as big). So the forces will balance if we reduce the effective electron charge by a factor of 8. This is accomplished by reducing the effective radius of the jelly by a factor of 2. □ Advanced Electrostatics Problems Building a Sheet from Rods Printed by Wolfram Mathematica Student Edition 2 Lecture 9 - 02-06-2017.nb Example An infinite uniform sheet of charge can be thought of as consisting of an infinite number of adjacent uniformly charged rods. Using the fact that the electric field from an infinite rod is 2 πλϵ r pointing away from the rod, 0 integrate over these rods to show that the field from an infinite sheet with charge density σ is σ 2 ϵ0 pointing away from the sheet. Solution Assume the infinite sheet lies at z = 0 in the x-y plane. What is the electric field at the point (0, 0, z)? By symmetry, it must point in the z-direction. If we split up the infinite sheet into thin rods of thickness ⅆ x running parallel to the y-axis, their charge density would equal λ = σ ⅆ x. The contribution of the electric field at our point in the zdirection would equal ⅆ Ez = 2 πλr ϵ zr where the last term picks out the z-component. Using r2 = x2 + z2 , the total 0 electric field equals ∞ σ ⅆx ∫-∞ 2 π ϵ0 z x2 +z2 x=∞ = 2 πσϵ0 ArcTan xz x=-∞ = σ 2 ϵ0 (3) as desired. □ This is one of the most astounding results you find in physics! No matter how far away you are from the shell, the electric field has the value 2σϵ pointing away from the plane. Additionally, no matter how close you come to the 0 plane, you still feel the electric field σ 2 ϵ0 pointing away from the plane. However, if you pass through the plane, the electric field will discontinuously change from σ 2 ϵ0 pointing in one direction to σ 2 ϵ0 pointing in the opposite direc- tion! Furthermore, if you are a piece of charge embedded within the sheet, you will feel an electric field of 0 (by symmetry), but if you move an infinitesimal distance off the sheet you will feel an electric field of 2σϵ . 0 This type of behavior is only exhibited by infinitely thin sheets of charge. Volume charge distributions create continuous electric fields (since the integration in E = ∫ k ρ ⅆx'r2ⅆy' ⅆz' r smooths out discontinuities in ρ). And lines of charges don’t seem as alarming because the electric field goes to ∞ near the line of charge (just like for a point charge). So only 2D charge configurations have this bizarre behavior where the electric field is discontinuous but stays finite everywhere. As stated in the "Field from a Cylindrical Shell, Right and Wrong" section in the previous lecture, a great rule to remember about discontinuities in nature is: With all discontinuities in nature, the actual value at the discontinuity equals the average of the discontinuous values! (4) This statement is worth its weight in gold; print it on bumper stickers and share it with the world! You will see it come up again and again in physics (from charge distributions) and mathematics (Fourier series). Extra Problem: Intersecting Sheets Hole in a Shell Consider a spherical shell of charge, of radius R and surface charge density σ, from which a small circular piece of radius b ≪ R has been removed. Printed by Wolfram Mathematica Student Edition Lecture 9 - 02-06-2017.nb 3 What is the direction and magnitude of the field at the midpoint of the aperture? Solution Method 1: Imagine that we have a complete spherical shell overlaid with a thin hole of charge density -σ. By the Principle of Superposition, this is identical to the above setup. For the spherical shell, Gauss’s Law easily yields the electric field E = 0 inside the sphere and E = ϵσ pointing radially outwards. Now we add the effect of the thin 0 hole, which at the midpoint looks like an infinite plane of charge with an electric field plane. Thus, inside the shell the electric field will be 0 + electric field will be σ ϵ0 - σ 2 ϵ0 = σ 2 ϵ0 σ 2 ϵ0 = σ 2 ϵ0 σ 2 ϵ0 pointing towards the pointing outwards and outside the sphere the pointing outwards. Hence, the electric field has magnitude σ 2 ϵ0 and points radially outwards at the midpoint. Note that the electric field is not discontinuous! This makes sense, because in the problem there is no sheet of charge at the point we are considering (and only sheets of charges can cause discontinuities). In our alternate setup, there are two sheets of charges, but they have charge densities σ and -σ and therefore cancel each other’s discontinuities. Method 2: We could also find the solution by straight up integration. Instead of integrating over the entire sphere. Note that if the aperture has radius b, then θ spans from Rb to π. θ R Note that the distance from the point at angle θ to the middle of the aperture equals 2 R Sin 2θ (which can either be found through the Law of Cosines or simple geometry). Therefore, the electric field in the middle of the aperture (which only has a component in the vertical direction) has magnitude 2 π π k R2 Sin[θ] σ θ 2 Sin 2 ⅆθ ⅆϕ 4 R2 Sin 2θ π k σ π Sin[θ] ⅆθ 2 ∫b/R Sin 2θ π π k σ ∫b/R Cos 2θ ⅆθ E = ∫0 ∫b/R = = θ=π = 2 π k σ Sin 2θ θ=b/R = 2 π k σ 1 - Sin 2bR Printed by Wolfram Mathematica Student Edition (7) 4 Lecture 9 - 02-06-2017.nb In the limit b ≪ R, Sin 2bR ≈ b 2R ≈ 0 so that E ≈ 2 π k σ = σ , 2 ϵ0 as desired. □ A Plane and a Slab Example An infinite plane has uniform surface charge density σ. Adjacent to it is an infinite parallel layer of charge of thickness d and uniform volume charge density ρ. All charges are fixed. Find E everywhere. Solution We will analyze the sheet and the slab separately, and then combine them using the Principle of Superposition. We already know that the sheet has an electric field σ 2 ϵ0 pointing away at all points in space. How can we deter- mine the electric field from the slab? There are many ways to compute the slab’s electric field: considering it as a conglomeration of thin sheets, straight up integration, and Gauss’s Law name three distinct methods. We will use Gauss’s Law (but you are invited to check the other methods). Assume that the slab stretches infinitely far in the y- and z-directions, and that it spans from x = - d2 to x = d2 . We use a Gaussian cylinder with length 2 x centered around x = 0 and cross sectional area A in the y-z plane. (Note that the Gaussian cylinder must be centered around the middle of the slab, although it does not need to by a cylinder and can instead be any prism shape.) Gauss’s Law yields 2 E A = ϵ1 (2 x A ρ) for x ≤ d2 and 2 E A = ϵ1 (d A ρ) for 0 x ≥ d . 2 0 In other words, E= x ρ ϵ0 dρ 2 ϵ0 x ≤ x > d 2 d 2 (8) with the direction +x for points x ≥ 0 and -x for points x < 0. All that remains is to add together the contribution of the electric field from the sheet and the slab (noting the directions of the fields!) to yield the final answer. The following Manipulate shows E as a function of x for various values of ρσd . Printed by Wolfram Mathematica Student Edition Lecture 9 - 02-06-2017.nb ρd σ 2 E Out[4]= ρd ϵ0 - d2 d 2 x σ ϵ0 Note that there is only a discontinuity in the electric field at x = - d2 where the sheet of charge resides. There is no discontinuity for the slab of charge. □ Advanced Section: Maximum Field from a Blob Recommended Problems This is a list of excellent problems (with solutions) in David Morin’s book. ◼ 1.6 Zero potential energy for equilibrium Mathematica Initialization Printed by Wolfram Mathematica Student Edition 5