Notes on Electric Fields of Continuous Charge Distributions


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1 Notes on Electic Fields of Continuous Chage Distibutions Fo discete pointlike electic chages, the net electic field is a vecto sum of the fields due to individual chages. Fo a continuous chage distibution along 1, 2, o 3 dimensions of space, dq(x) = λ dx, o dq(x,y) = σ dxdy, o dq(x,y,z) = dxdydz, (1) the sum becomes an integal E() = k( ) 3 dq( ) (2) In my notations, = (x,y,z) is the point whee we measue the electic field E while = (x,y,z ) the location of the chage. Also, k is the Coulomb constant: in MKSA units k = 1/(4πǫ ) Nm 2 /C 2, while in Gaussian units k = 1. In these notes, I am going to calculate the electic field fo a few unifom examples of continuous chages: (1) Infinite thin od; (2) finite thin od; (3) infinite 2D plate; (4) a pai of paallel plates; (5) a finite thin 2D disk. Cylindical Coodinates All the examples in these notes have axial symmety otating the whole wold though any angle aound a fixed axis (which I ll use as the z axis) will not change the locations of the chages o thei density, so the electic field should also emain unchanged by such a otation. To make use of this symmety, I am going to use the cylindical coodinates (,φ,z), whee (,φ) act as pola coodinates in the hoizontal (x,y) plane while z is the vetical elevation. In tems of Catesian (x, y, z) coodinates = x 2 +y 2, φ = actan y x, z = z. x = cosφ, y = sinφ, z = z. (3) The unit vecto ı in the diection point hoizontally away fom the z axis while the unit 1
2 vecto ı φ point along the cicle suounding that axis, y ı φ ı x ı = cosφ ı x + sinφ ı y, ı φ = sinφ ı x + cosφ ı y. (4) Finally, theunitvecto ı z inthez diectionpointveticallyup, justasitdoesinthecatesian coodinate system. In tems of these 3 unit vectos, E = E ı + E φ ı φ + E z ı z (5) which tanslate to Catesian components as E x = E cosφ E φ sinφ, E y = E sinφ + E φ cosφ, (6) E z = E z. Example 1: Unifomly chaged infinite thin od. In this example the electic chages ae unifomly distibuted along an infinitely thin, infinitely long od. We use coodinates whee this od lies along the z axis, so the chages ae at (x,y,z ) = (,,anyz ). The chage density is unifom, which means λ = dq(z ) dz = const. This chage system has many of symmeties, including otations aound the z axis, tanslations along that axis, and flipping the vetical axis upside down aound any point we like. The electic field E(x,y,z) must have simila symmeties, so its vetical component E z must 2
3 vanish eveywhee, while the hoizontal components E x and E y do not depend on z. Futhemoe, by otational symmety, the hoizontal components must in the adial diection ı : diectly away fom z axis fo λ >, o diectly towads the z axis fo λ < : In cylindical coodinates this means E φ as well as E z. Futhe moe, the adial component E depends only on the hoizontal distance fom the chaged od. In light of these symmeties, the only component we need to calculate is E as a function of a single vaiable. To do that, we need to evaluate the integal E () = od 3 kdq = + 3 kλdz (7) whee = 2 +(z z ) 2 (8) is the distance fom the chage at ( =,z ) to the point (,φ,z) whee we measue the field. To claify the geomety, hee is the pictue of the vetical plane though the chaged 3
4 od and the place whee the field is measued: z z chage θ (,z) pobe E = cosθ z z = sinθ 2 = 2 + (z z) 2 tanθ = z z (9) To simplify the integal (7), let s change the integation vaiable fom z to the θ angle on the diagam (9): and hence z = z + tanθ, dz = dtanθ = dθ cos 2 θ, = cosθ, 3 dz = cos3 θ 3 dθ cos 2 θ = cosθ dθ = d ( sinθ (1) ). (11) Also, while z anges fom to +, the angle θ uns fom π/2 to +π/2. Plugging all these data into the integal (7), we obtain 2 E () = kλ +π dsinθ π/2 = kλ ) (sin(+π/2) sin( π/2) = 2 (12) In Gaussian units this means = 2kλ, E = 2λ ı (13) 4
5 while in MKSA units E = λ 2πǫ ı. (14) Fo example, at = 1 cm away fom a od of chage density λ = 1 µc/m, the electic field has a athe lage magnitude E = 2 4πǫ λ 2 ( Nm 2 /C 2 ) ( C/m) 1.1 m = N/C o 1.8 million volts pe mete. To complete this example, let me convet the electic field (12) fom the cylindical to the Catesian coodinates. In light of eqs. (4), ı = cosφ ı x + sinφ ı y = x 2 ı x + y 2 ii y = x x 2 +y 2 ı x + y x 2 +y 2 ii y, (15) hence ( x E = 2kλ x 2 +y 2 ı x + ) y x 2 +y 2 ii y, (16) o in components x E x (x,y,z) = 2kλ x 2 +y 2, E y (x,y,z) = 2kλ E z (x,y,z). y x 2 +y 2, (17) 5
6 : Example 2: Unifom thin od of finite length. Simila to the pevious example, the chages ae unifomly distibuted along an infinitely thin od, but this time the od has a finite length L. In othe wods, the z coodinate along the od uns fom a finite z 1 to a finite z 2 = z 1 +L instead of fom to +. The finite od has an axial symmety of otations aound the z axis. but it no longe has the tanslational symmety along that axis, o the flipping symmety z z. Consequently, the hoizontal components of E must point in the adial diection (away fom the z axis o towads it), but the vetical component E z does not need to vanish (and geneally does not). In cylindical coodinates this means E φ but both E and E z. Also, the E and E z components may depend on both and z coodinates (of the point whee we measue the field), although by axial symmety they do not depend on the φ. Thus, instead of a single integal (7), we now need to evaluate two integals E (,z) = kλ E z (,z) = kλ z 2 z 1 z 2 z 1 dz 3, (18) (z z )dz 3, (19) Let us stat with the fist integal (18). Poceeding similaly to the fist example, we change the integation vaiable fom the z to the angle θ, exactly like on the diagam (9). Consequently, we obtain but this time the θ angle uns dz 3 = 1 dsinθ (11) Consequently, fom θ 1 = actan z 1 z E = kλ θ 2 θ 1 dsinθ = to θ 2 = actan z 2 z kλ ( sinθ 2 sinθ 1 ). (2). 6
7 By tigonomety sinθ 1 = z 1 z 1 (21) whee 1 = (z 1 z) is the distance fom the lowe end of the od to the measuing point, and likewise sinθ 2 = z 2 z 2. Theefoe, the adial component of the electic field evaluates to E (,z) = kλ ( z2 z 2 z ) 1 z. (22) 1 have Now conside the vetical component of the electic field. Inside the integal (19), we (z z )dz 3 = z z dz 3 = tanθ cosθdθ = sinθdθ = 1 dcosθ. (23) Consequently, E = kλ θ 2 θ 1 dcosθ = kλ ( cosθ 2 cosθ 1 ). (24) By tigonomety, cosθ 1 = 1, cosθ 2 = 2, (25) and theefoe ( 1 E z (,z) = kλ 1 ). (26) 2 1 Altogethe, the electic field of a finite od can be descibed by the following fieldline 7
8 pictue: Finally, let s look at the neatheod and faawayfomtheod limits of the electic field. Neatheod limit means we ae much close to some point in the middle of the od tan to the od s ends, thus z 1 < z < z 2, L = z 2 z 1. In this limit θ 1 π/2, θ 2 +π/2, hence in eq. (2) sinθ 2 sinθ 1 2 (27) and theefoe E 2kλ = [infinite od]. 8
9 At the same time, in eq. (24) cosθ 2 cosθ 1 1 and hence E z E. In othe wods, in the neatheod limit, the electic field of a finite od is simila to the field of the infinite od: It s the neaby pat of the od which dominates the field, and it does not matte if the fa ends of the od ae infinitely fa away o just fa enough. On the othe hand, the fafomtheod limit means that we measue the field at some distance fom the od which is much lage than the od s length L = z 2 = z 1. In this limit 1 2 = 2 +z 2, θ 1 θ 2 θ = actan z, while to a bette appoximation 2 1 Lcosθ, θ 2 θ 1 + L sinθ. Consequently, in eqs. (2) and (24) sinθ 2 sinθ 1 (θ 2 θ 1 ) dsinθ dθ cosθ 2 cosθ 1 (θ 2 θ 1 ) dcosθ dθ = Lsinθ = Lsinθ cosθ, ( cosθ), (28) while kλ = k(q/l) sinθ. (29) Theefoe E k(q/l) sinθ Lsinθ cosθ = kq kq sinθ = 2 2, E z k(q/l) sinθ Lsinθ ( cosθ) = kq kq ( cosθ) = 2 2 z, (3) 9
10 which is pecisely the cylindical components of the electic field vecto E = + kq 2 ı +z ı z = + kq 2 (31) of a point chage Q. Physically, this means that fa away fom the od its size is too small to matte so it may be appoximated by a chaged point of the same net chage Q = λl. 1
11 Example 3: Unifomly chaged infinite 2D plate. In this example, the chages ae spead out in two dimensions, on a plate which we appoximate as infinite in x and y diections but infinitely thin in the z diection. The chages have unifom density in two dimensions, thus σ = dq daea = dq(x,y ) dx dy = const. The symmeties of this chage system include tanslations in x and y diections and otations aound the z axis. The otational symmety tells us that the electic field E(x,y,z) does not have any hoizontal E x o E y components, while the tanslational symmety makes the field independent on x and y, thus E x, E y y, E z ( x, y,z) = E z (zonly). Thanks to this symmety, all we need to calculate the E z along the z axis, the values at any othe point (x,y,z) would be the same as at (,,z). So let s stat with a pictue of the chages and the point whee we measue the field: z E chage at (x,y,) pobe at (,,z) y φ x = x 2 +y 2 +z 2 = 2 +z 2 (32) 11
12 In light of this diagam and the Coulomb Law de z = z 3 kdq = z 3 kσdx dy (33) and theefoe + E z (z) = kσ dx + dy z 3. (34) To evaluate this double integal, let s change the integation vaiables fom x and y to the pola coodinates (,φ ) on the chaged plate. The aea diffeential in these vaiables becomes daea = dx dy = d dφ (35) while the coodinates anges ae φ 2π, <. Consequently, the integal (34) fo the electic field of the chaged plate becomes E z = kσ d 2π dφ z 3. (36) Inside the inne integal, = 2 +z 2 depends on but not on the angle φ, and nothing else depends on the angle φ eithe. Consequently, the inne integal ove dφ is an integal ove a constant, so the integation amounts to simply multiplying the integand by 2π, thus 2π z d E z = kσ 3 = ( 2 +z 2. (37) ) 3/2 To evaluate this integal, let s change the integation vaiable once again, fom to = 2 +z 2 Since z does not change with, we have 2 = 2 + z 2 = 2d = 2 d = 2πz d = 2πzd. (38) 12
13 also, while anges fom zeo to infinity, anges fom to infinity, thus 2π z d 3 = 2πz d 3 = 2πz d 2 ( ) 1 = 2πz = 2πz +1 = 2π sign(z). (39) Consequently, E z = 2πkσ sign(z) (4) and hence E = 2πkσ sign(z) ı z. (41) In paticula, the electic field of an infinite chaged plate has the same magnitude E = 2πk σ eveywhee in space, but has opposite diections on the two sides of the plate: Fo σ > the field point vetically up above the plate but vetically down below the plate; fo σ < the diections ae opposite, down above the plate but up below the plate. (42) In coodinateindependent tems, the electic field of a chaged plate is always to the plate itself; fo a positivelychaged plate E points away fom the plate, while fo a negatively chaged plate E points towads the plate. 13
14 As to the magnitude of the electic field, in MKSA units E = σ 2ǫ. (43) Fo example, a plate of chage density of 1. micocoulomb pe squae mete ceates electic field of magnitude E = C/m 2 2 ( ) , V/m. C/(Vm) Example 4: Two paallel plates with opposite chages. In this example, we have two infinite plates paallel to each othe. One plate has positive unifom chage density +σ while the othe has negative unifom chage density σ, so the net electic chages of the two plates cancel each othe. σ (44) +σ Inthepeviousexamplewehaveobtainedtheelecticfieldofasingleplate, sointhisexample we do not face any integals; all we need to do is to add up the fields of the two plates, E net = E (1) + E (2). (45) Moeove, fo paallel plates both fields ae to the two plates fo the hoizontal plates as on figue (44) the fields ae vetical and have equal magnitudes eveywhee in space, E (1) E (2) 2πkσ, (46) so all we need is to check whee each field points up and whee it points down. Specifically, the field E (1) of the positive plate always points away fom that plate while the field E (2) of the negative plate always points towads that plate. So assuming the negative plate is above the positive plate as in the pictue (44), thee ae thee distinct situations: 14
15 Above both plates, the field E (1) of the positive plate points up while the field E (2) of the negative plate points down. Consequently, the two fields cancel each othe, and the net electic field vanishes! Below both plates, the field E (1) of the positive plate points down while the field E (2) of the negative plate points up. Again, the two fields cancel each othe, and the net electic field vanishes! Between the two plates i.e., above the positive plate but below the negative plate the field E (1) of the positive plate points up while the field E (2) of the negative plate also points up. Consequently, the two fields add up, and the net electic field is E net = +4πkσ ı z. (47) To summaize, outside the two plates the electic field vanishes, while between the plates the field has magnitude E = 4πkσ and diection fom the positive plate towads the negative plate, as on the following figue: σ E = E = 4πkσ (48) +σ E = In MKSA units, the field between the plates has magnitude E = σ ǫ (49) Fo example, if one plate has chage density +1. µc/m 2 while the othe plate has chage density 1. µc/m 2, then the field between the plates is E = 113, V/m. 15
16 Example 5: Unifomly chaged disk. In this example, the chaged 2D suface is a finite disk o adius R. We use a coodinate system with anoiginat thecente of the disk, with the z axis to the disk, thus the chages lie at (x,y,z ) with z, finite x,y, x 2 +y 2 R 2. Fo simplicity, we limit this example to calculating the electic field on the z axis only alas, the offaxis field is much moe complicated thus z E pobe at (,,z) y = x 2 +y 2 +z 2 = 2 +z 2 chage at (x,y,) φ R x (5) Thanks to the otational symmety of the disk, on the z axis the electic field is puely vetical, E x = E y =, while the z components obtains fom a double integal simila two the one we had fo an infinite plate, E z = kσ dx dy z 3, (51) disk except that now the integation ange is limited to the disk, x 2 + y 2 R 2. In Catesian coodinates (x,y ) this is a athe complicated ange of a double integal, but it becomes 16
17 much simple in the pola coodinates (,φ ) whee uns fom zeo to R while φ uns fom zeo to 2π, thus E z = kσ R d 2π dφ z 3 (52) Thanks to the otational symmety, nothing inside the integal depends on the pola angle φ, so integating ove that angle poduces simply 2π the integand, thus R 2π z d E z = kσ 3 = ( 2 +z 2. (53) ) 3/2 At this point, we again poceed as in example#3 and change the emaining integation vaiable fom to, which now uns fom (fo ) to max = R 2 +z 2. (54) Consequently, E z = kσ max = 2πkσz = 2πkσz 2πz d 3 max d 2 = 2πkσz ( ) 1 1, R 2 +z 2 ( ) 1 max (55) o equivalently E(,,z) = 2πkσ sign(z) ı z ( 1 ). (56) R 2 +z 2 Note that the fist two factoson the ight hand side aeexactly as in the field of aninfinitely chaged plane, and only the thid facto depends on the disk s adius R. 17
18 To complete this example, lets conside the neathedisk and faawayfomthedisk limits of the electic field (56). In the neathedisk limit, we ae measuing the electic field much close to the disk than its adius, R, so in the last facto in eq. (56) 1 1. (57) R 2 +z2 Consequently, E 2πkσ sign(z) ı z (58) nea the disk, its electic field is simila to the field of an infinite chaged plane. In the opposite faawayfomthedisk limit we measue the field at distance much lage than the disk adius, so we expect the disk s field to be simila to the field of a point souce with the same net chage Q = πr 2 σ. And indeed, fo R, the last facto in eq. (56) becomes 1 hence R 2 +z = 1 1 (1 R2 2 z 2 +R 2 1 ) 12 R2 z 2 +R 2 = R 2 2(z 2 +R 2 ) R2 2z 2, (59) E(,,z) 2πkσ sign(z) ı z R2 2z 2 = k πr2 σ z 2 sign(z) ı z (6) Note that we ae measuing this field on the z axis, so z 2 in the denominato is the distance 2 fom the measuement point to the disk, while sign(x) ı z is the unit vecto in the diection of the measuement point. Finally, πr 2 σ is the total electic chage Q of the disk, so eq. (6) indeed agees with the Coulomb field of a point chage Q, E() = kq 2 ı. (61) 18
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