Lectue 8 - Gauss s Law A Puzzle... Example Calculate the potential enegy, pe ion, fo an infinite 1D ionic cystal with sepaation a; that is, a ow of equally spaced chages of magnitude e and altenating sign. Hint: The powe-seies expansion of Log[1 + x] may be of use. Solution Suppose the aay is built inwads fom the left (that is, fom negative infinity) as fa as a paticula ion. To add the next positive ion on the ight, the amount of extenal wok equied equals - k e2 a + k e2 2 a - k e2 3 a + = - k e2 a 1-1 2 + 1 3 + (1) The tems in paenthesis look emakably simila to the Taylo seies of Log[1 + x] when x = 1. (What is the Taylo seies of Log[1 + x]? d 1 Log[1 + x] = dx 1+x = 1 - x + x2 -, and integating both sides yields Log[1 + x] = x - x2 2 + x3 -, since the ight hand side is a polynomial expansion, it is also the Taylo seies of 3 Log[1 + x]. This Taylo seies is conveging fo -1 < x 1 (with convegence on -1 < x < 1 assued by the altenating seies test)). Theefoe the wok equied to bing this chage in equals - k e2 Log[2], which equals the a enegy of the infinite chain pe ion. The addition of futhe paticles on the ight doesn t affect the enegy involved in assembling the pevious ones, so this esult is indeed the enegy pe ion in the entie infinite (in both diections) chain. The esult is negative, which means that it equies enegy to move the ions away fom each othe. This makes sense, because the two neaest neighbos ae of opposite sign. Note that this is an exact esult! It does not assume that a is small. Getting such a nice closed fom is much moe difficult in 2 and 3 dimensions. Theoy Visualizing the Electic Field (Extended) Electic Field Lines Advanced Section: Escaping Field Lines Math Backgound: What is a Suface Integal? We ae aleady familia with suface integals of scala functions. Fo example, the suface aea of a sphee with adius R centeed at the oigin is given by suface da = 0 2 π 0 π R 2 Sin[θ] dθ dϕ = 4 π R 2 (4) Although we use spheical coodinates in this paticula poblem, we could also have used Catesian coodinates
2 Lectue 8-02-02-2017.nb Although spheical paticula poblem, (with d a = dx dy) o any othe coodinate system. Moe geneally, suppose we ae given a function f[θ, ϕ] defined eveywhee on the suface of the spheical shell of adius R. What is the integal of f[θ, ϕ] acoss the entie spheical shell? This equies a vey mino modification to the fomula above, namely, suface f [θ, ϕ] da = 0 2 π 0 π f [θ, ϕ] R 2 Sin[θ] dθ dϕ (5) We would need to know the specific function f[θ, ϕ] to explicitly evaluate this integal, but we obseve that the suface aea is meely a specific case of Equation (5) with f[θ, ϕ] = 1 eveywhee on the sphee. Lastly, athe than being given a scala function f[θ, ϕ], we could be given a vecto function F[θ, ϕ] defined eveywhee on the sphee. In such a case, we define the suface integal to be suface F[θ, ϕ] da = suface F[θ, ϕ] da da (6) No need to panic! The ight hand side is simply the same suface integal fom Equation (5), except that now the scala function is the dot poduct of two vectos. The fist vecto F[θ, ϕ] is the function that you ae integating ove the sphee. The second vecto da is a unit vecto of an infinitesimal patch in the diection nomal to the suface. On the sphee, the unit nomal vecto at any point is given by d a =, so that we can wite the suface integal as suface F[θ, ϕ] da da = 0 2 π 0 π F[θ, ϕ] R 2 Sin[θ] dθ dϕ (7) Fo closed sufaces, we take the unit nomal to be pointing outwads by convention. Fo open sufaces, you can abitaily choose between the two possible diections that the unit vecto can point (if you switch, it will flip the sign of the suface integal). Usually, we will be woking in setups whee the unit nomal vecto is obvious. Fo example, if we have a sheet in the x-y plane, then the unit nomal vecto will be z (o -z, you can choose eithe as long as you ae consistent thoughout you suface integal) at all points. Similaly, if we have a cylinde whose axis lies on the z-axis, then the unit nomal vecto at its top cap will be z (emembeing that unit nomal vectos point outwads fo closed sufaces) and -z at its bottom cap. Finally, in cylindical coodinates (ρ, θ, z), the unit vecto will be ρ fo all points along the cuved suface of the cylinde. Guass s Law An incedibly useful and beautiful esult, Gauss s Law is definitely woth memoizing! Hee we wite it fo a discete and continuous chage distibution. The integal E da ove the suface, equals 1 times the total chage enclosed by the suface, E da = 1 j q j = 1 ρ dv (8) Fo a combination of both (fo example, a point chage nea an infinite sheet), the Pinciple of Supeposition tells us that we sum ove the discete chages and integate ove the chage distibutions within ou suface. Example Find the electic field due to an infinite line of chage with unifom chage density λ.
Lectue 8-02-02-2017.nb 3 Solution By symmety, the electic field must point adially outwads fom the line of chage. Cylinde Show E Show a Out[21]= We can use a cylinde whose axis lies on the line of chage and calculate E d a along this cylinde. Since E points adially, E da = 0 along the (flat) top and bottom of the cylinde, and by adial symmety, E d a = E[] da will be a constant value eveywhee along the cone (whee E[] is the magnitude of the electic field at a distance fom the wie). Theefoe, Gauss's Law yields E da = E[] da = E[] (2 π L) = λ L (9) which yields E[] = λ 2 π (10) Altenatively, we could have computed this value by staight-up integation, choosing the adial component of the electic field
4 Lectue 8-02-02-2017.nb E[] = - k λ dx x 2 + 2 = x 2 + 2 1/2 k λ x x= x 2 + 2 1/2 x=- = 2 k λ = λ 2 π This yields the same esult as above, but afte significantly moe wok. A Note about Symmety The key to the last poblem was poving that fo a line of chage lying on the x-axis, the electic field E = E[] points adially away fom the wie. Let s pove this explicitly. If the infinitely long wie lies along the x-axis, then the points on the wie ae given by (x, 0, 0) whee x (-, ). Let us conside the electic field at a point (0, 0, z), and let us call the electic field E = E x, E y, E z. Fist, let us flip the setup along the x = 0 as shown below, which means that evey point (and evey vecto) goes fom (x, y, z) (-x, y, z). Theefoe the electic field will go fom E x, E y, E z -E x, E y, E z, shown by the solid aow dashed aow. Howeve, since evey point (x, 0, 0) on the line of chage goes to anothe point (-x, 0, 0) on the line of chage, and the point of inteest (0, 0, z) (0, 0, z) does not change, the physical setup of the poblem emains exactly the same. Theefoe, the electic field befoe and afte the eflection must be the same, and this is only possible if E x = 0. Similaly, conside a eflection about the y = 0 plane, so that evey point goes fom (x, y, z) (x, -y, z). Theefoe the electic field will go fom 0, E y, E z 0, -E y, E z. Each point on the line of chage will not be changed
Lectue 8-02-02-2017.nb 5 (x, 0, 0) (x, 0, 0), and similaly the point of inteest will not change (0, 0, z) (0, 0, z), so that once again the physical setup is the same and hence the electic field must be the same. This can only happen in E y = 0. We conclude that at the point (0, 0, z), the electic field point staight up (i.e. adially away) fom the line of chage. We can genealize this esult to find the diection of the electic field at an abitay point (x, y, z) using symmety. Fist, we note that because the wie is infinitely long, tanslating along the x-diection cannot change the electic field, since we could equally well have defined ou oigin at some othe point (X, 0, 0) and the physical setup would be identical to that found above, so that the electic field at (X, 0, z) must point in the z diection. Finally, we use the otational symmety of the setup (i.e. otating ou coodinate system about the x-axis) to find that the electic field at any point (x, y, z) points adially outwad fom the wie in the diection 0, y, z. Poblems Field fom a Cylindical Shell Example Conside a chage distibution in the fom of an infinitely long hollow cicula cylinde (like a long chaged pipe) of adius R. If the cylinde has unifom chage pe unit aea σ, what is the electic field inside and outside the cylinde? Solution The electic field inside the cylinde must be pointing adially about the axis of symmety. Theefoe, using a Gaussian suface of a small cylinde with adius < R and length l placed along the axis of the lage cylinde, we find that because thee is no chage inside the cylindical shell. Thus, E = 0 fo all < R. E (2 π l) = q in = 0 (12) The electic field outside the cylinde is found using a the same Gaussian suface but with > R so that o equivalently E = R E (2 π l) = q in = 2 π R l σ (13) σ (2 π R σ). Since the chage pe unit length on the cylindical shell equals 2 π R σ, E = so 2 π that the electic field outside the cylinde is the same as if all of the chage was concentated at on the axis of the cylinde. These esults ae vey inteesting, but let me point out one subtlety in this esult. If you ask fo the electic field
6 Lectue 8-02-02-2017.nb vey inteesting, point subtlety you E[] as a function of the adial distance fom the cylindical shell, then stating at and coming towads the axis the answe will be E[] = R σ > R 0 < R This implies that at = R, thee is a discontinuity; the value of E[] σ fom the outside of the cylindical shell and E[] 0 fom the inside of the shell. This begs the question, what exactly is the electic field at = R? As we will see fom the electic field of an infinite plane, the answe will be E[R] = σ, and you ae encouaged to do the explicit calculation youself! Fo now, we note that E[R] is the aveage of the values two limits of E[] fom inside and outside the cylindical shell. Exta Poblem: Field fom a Cylindical Shell, Right and Wong Example Find the electic field outside a unifomly chaged hollow cylindical shell with adius R and chage density σ, an infinitesimal distance away fom it. Do this in the following two ways: 1. Slice the shell into paallel infinite ods, and integate the field contibutions fom all the ods. You should obtain the incoect esult of σ. 2. Why isn't the esult coect? Explain how to modify it to obtain the coect esult of σ. Hint: You could vey well have pefomed the above integal in an effot to obtain the electic field an infinitesimal distance inside the cylinde, whee we know the field is zeo. Does the above integation povide a good desciption of what's going on fo points on the shell that ae vey close to the point in question? 3. Confim that the esult equals σ by staight up integation assuming that the point is a finite distance z away fom the cente of the cylinde. What happens when z < R, z = R, and z > R? Solution 1. Let the ods be paameteized by the angle θ as shown in the diagam below. (14) The width of a od is R dθ, so its effective chage pe unit length is λ = σ (R d θ). The od is a distance 2 R Sin θ 2 fom the point P in question, which is infinitesimally close to the top of the cylinde. Only the vetical component of the field fom the od suvives, and this bings in a facto of Sin θ. Using the fact that the field fom a od is 2 λ, we find that the field at the top of the cylinde is (incoectly) 2 π 2 0 π σ R dθ 2 π 2 R Sin θ 2 Sin θ 2 = σ 2 π 0 π dθ = σ (15) Inteestingly, we see that fo a given angula width of a od, all ods yield the same contibution to the vetical electic field at P (since the ones futhe away fom it contibute at a bette angle).
Lectue 8-02-02-2017.nb 7 2. As stated in the poblem, it is no supise that this answe is incoect, since the same calculation would supposedly yield the field just inside the cylinde too, whee it is zeo instead of σ. Howeve, this calculation does yield the aveage of these two values. The eason why the calculation is invalid is that it doesn t coectly descibe the field due to ods vey close to the given point, that is, fo ods with θ 0. It is incoect fo two easons. Fist, the distance fom a od to the given point is not equal to 2 R Sin θ. Additionally, the field does not point along the line fom the od to the top of the 2 cylinde. It points moe vetically, so the exta facto of Sin θ which we used to pick out the vetical component 2 isn t valid. What is tue is that if we emove a thin stip fom θ = - ϕ 2 to θ = ϕ (fo vey small ϕ 1) at the top of the cylinde, 2 then the above integal is valid fo the emaining pat of the cylinde. In othe wods, the electic field due to the emaining cylinde will be 2 π-ϕ σ σ since we can make 2 π-ϕ abitaily close to 1. By supeposition, the 2 π 2 π total field of the entie cylinde in this field of σ plus the field due to the thin stip at the top. But if the point in question is infinitesimally close to the cylinde, then the thin stip will look like an infinite plane, the field of which we know is σ. The desied total field is then E outside = E cylinde minus stip + E stip = σ + σ = σ (16) E inside = E cylinde minus stip - E stip = σ - σ = 0 (17) whee the minus sign in E inside comes fom the fact that E stip (like an infinite sheet) points in diffeent diections inside and outside the cylinde. Technical note: We have two infinitesimally small quantities in this poblem: the width ϕ of the stip that we ae emoving fom the cylinde and the infinitesimal distance of the point P fom the cylinde. You may (o at least should) be wondeing which of these two infinitesimally small quantities is smalle. This is an impotant point to keep tack of - if you ae blasé about the matte and simply send both quantities to 0 without egad, you could end up with wong esults! In this poblem, we want fo the distance fom P to the cylinde to be smalle, and indeed that is the case. Fo we fist picked a small ϕ (which we can make as small as we like) and then we consideed a point P which was essentially touching the cylinde but on the outside; you can think of this as once you fix ϕ, you bing P in extemely close to the cylinde (and theefoe extemely close to the thin stip). 3. Oient the cylinde s axis to lie on the y-axis and conside the electic field at the point (0, 0, z).
8 Lectue 8-02-02-2017.nb Define the distance fom the wie at angle θ to the point (0, 0, z) to be, which by the Law of Cosines equals 2 = R 2 + z 2 λ - 2 R z Cos[θ]. Using the chage density λ = σ (R dθ), the electic field fom an infinite wie, 2 π and the z-component z-r Cos[θ] of the electic field, E = 0 2 π R σ dθ 2 π z-r Cos[θ] = R σ 2 π 0 2 π z-r Cos[θ] R 2 +z 2-2 R z Cos[θ] dθ (18) This integal is not supe-difficult to evaluate, but cae must be taken as to whethe z < R, z = R, o z > R. The full esult is given by E = R σ (1-Sign[R-z]) z (19) Integate R σ z - R Cos[θ], {θ, 0, 2 π}, Assumptions 0 < ϵ0 && 0 < σ && 0 < R && 0 < z 2 π ϵ0 R 2 + z 2-2 R z Cos[θ] R σ (-1 + Sign[R - z]) - 2 z ϵ0 In othe wods, E = 0 z < R σ z = R R z σ z > R We see that when we appoach the suface of the cylinde fom the inside E = 0, wheeas when we appoach it fom the outside, E = R σ σ. The electic field on the actual suface equals the aveage of these two values, z E = σ. (20) Mathematica Initialization