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Field distribution of a uniformly charged circular arc
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Journal of Electrostatics 63 (2005) 1035–1047 www.elsevier.com/locate/elstat
Field distribution of a uniformly charged circular arc Ping Zhu Department of Physics, Simao Teacher’s College, Yunan 665000, China Received 16 September 2004; received in revised form 24 January 2005; accepted 7 February 2005 Available online 17 March 2005
Abstract Because the field of a uniformly charged circular arc involves elliptic integrals, not generally possessing symmetry of a uniformly charged ring, it is difficult to solve and discuss this problem. This paper investigates the problem and presents the potential function and the field function of a uniformly charged circular arc together with discussions about them by using the theory of elliptic integrals and the analytic method. r 2005 Elsevier B.V. All rights reserved. Keywords: Arbitrary circular arc; Uniform linear charge; Potential distribution; Field distribution; Elliptic integrals
1. Introduction The field distribution of a uniformly charged circular arc is a typical and important problem in electrostatics. In the research of electromagnetism it is significant for us to investigate this problem [1–3]. However, because the field of a uniformly charged circular arc involves elliptic integrals, not generally possessing symmetry of a uniformly charged ring, it is difficult for us to discuss and solve the problem. Several researchers [4–8] have discussed some cases of the typical circular ring of uniform charge. Applying Fourier Bessel transform pair [9,10], Ball [11] presented the potential on the uniformly charged circular ring. Using computers and E-mail address: [email protected]. 0304-3886/$ - see front matter r 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.elstat.2005.02.001
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the superposition principle of electric field, Zhang and Jiang [12] discussed the electric field distribution of annular linear electric charge. According to the superposition theorem of the field and the electric potential of a point charge, Cheng et al. [13] derived a series solution of the potential and the field of a uniformly charged ring. In the rectangular coordinate system, Zhou and Chen [14] directly calculated the spatial distribution of the electrical field generated by a uniformly changed ring. Making use of the analogous method, Trinh and Maruvada [15] investigated the electric field distribution of an arbitrary circular-arc shaped uniform current filament embedded in a conductive medium, the solution to which has been derived in terms of elliptic integrals and used in the numerical evaluation of the resistance of complex ground electrodes. Their work has not only further promoted the research on the physics problems of elliptic integrals, but also helped me to solve the case of a uniformly charged circular arc in analytic methods. In this paper, by using the theory of elliptic integrals, important properties of elliptic integrals and some special tricks, we can turn difficulty into simplicity, and gain the analytic potential function and the analytic field function of a uniformly charged circular arc, so that this problem is solved satisfactorily. In Section 2, we prove useful properties of elliptic integral functions and present the potential function of the uniformly charged circular arc. In Section 3, we present the spatial field distribution of the uniformly charged circular arc and discuss electric field distributions on some special positions. Finally, in Section 4, we present the summary. Discussions on the problem may help us understand methods to deal with this kind of physics problem of elliptic integrals.
2. Proof of properties and potential function There is a section of a uniformly charged rigid circular arc whose radius is R; the central angle is a ð0oap2pÞ; and the linear charge density is t; and we will discuss its potential and field distribution, as depicted in Fig. 1. In a cylindrical coordinate system, we place the circular arc on the polar coordinate plane, whose initial coordinates are AðR; 0; 0Þ and final coordinates are BðR; a; 0Þ: At a point on the arc we take a line element dl with electric quantity dq ¼ t dl; which generates that the potential at the point Pðr; y; zÞ in space is dj ¼ t dl=4p0 r where r2 ¼ z2 þ ðr þ RÞ2 2rR½1 þ cosða0 yÞ. The potential at the point Pðr; y; zÞ that the charged circular arc generates is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tR 2rR=ðz2 þ ðr þ RÞ2 Þ pffiffiffiffiffiffiffiffiffi j¼ 4p0 2rR Z a da0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi .
0 1 2rR½1 þ cosða0 yÞ=ðz2 þ ðr þ RÞ2 Þ
ð1Þ
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Fig. 1. A uniformly charged circular arc.
Defining a0 y ¼ p 2b; namely b ¼ ðy þ p a0 Þ=2; da0 ¼ 2 db; and 1 þ cosða0 yÞ ¼ 2 sin2 b: When a0 ¼ 0; b0 ¼ ðy þ pÞ=2: When a0 ¼ a; ba ¼ ðy þ p aÞ=2: We let k2 ¼ 4rR=½z2 þ ðr þ RÞ2 and get Z ðpþyÞ=2 tRk db pffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi j¼ 4p0 rR ðpþyaÞ=2 1 k2 sin2 b pffiffiffiffi Z b0 t Rk db qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . ¼ ð2Þ pffiffiffi
4p0 r ba 1 k2 sin2 b Defining Z
b
F ðb; kÞ ¼ 0
db qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 1 k2 sin2 b
from Eq. (2) we get pffiffiffiffi t Rk j¼ pffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ, 4p0 r
(3)
(4)
where F ðb; kÞ is the elliptic integral function of the first kind. When 0pbpp=2; we may directly get the corresponding results of the potential. When b4p=2 or bo0; we cannot directly get the results of Eq. (4). If we use the general elliptic integral method, it is very difficult to solve Eq. (2). However, if we use the important property of the elliptic integral function, we may turn difficulty into simplicity, so that we conveniently get the specific results of the problem. When the amplitude b4p=2 or bo0; we define b ¼ mp a0 and 0pa0 pp=2; where m is an integer. Then we have [16] F ðb; kÞ ¼ F ðmp a0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ, (5) q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R p=2 where KðkÞ ¼ 0 db= 1 k2 sin2 b: KðkÞ is the complete elliptic integral function of the first kind.
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Since Z
mp a0
dy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 k2 sin2 y Z mp Z mp a0 dy dy ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ þ 2 2 0 mp 1 k sin y 1 k2 sin2 y 2 3 Z Z m ð2n1Þp=2 np X6 dy dy 7 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ 4 2 2 2 2 ðn1Þp ð2n1Þp=2 n¼1 1 k sin b 1 k sin b
F ðmp a0 Þ ¼
F ða0 ; kÞ m X ½KðkÞ þ KðkÞ F ða0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ, ¼ n¼1
F ðb; kÞ ¼ F ðmp a0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ. Using the same method, we can prove that the elliptic integral function of the second kind satisfies Eðb; kÞ ¼ Eðmp a0 ; kÞ ¼ 2mEðkÞ Eða0 ; kÞ,
(6)
where Z
b
Eðb; kÞ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 b db.
(7)
0
R p=2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 b db is the complete elliptic integral function of the second EðkÞ ¼ 0 kind. For a linearly charged circular arc, 0pap2p: In Eq. (4) we substitute b0 ¼ mp f0 : When 0pb0 pp=2; m ¼ 0 and f0 ¼ b0 : If b0 4p=2 or b0 o0; m is selected as the corresponding integer except 0. Using the same method, ba ¼ m0 p fa : When 0pba pp=2; m0 ¼ 0 and fa ¼ ba : If ba 4p=2 or ba o0; m0 is selected as the corresponding integer except 0. Then, from Eq. (4) we have pffiffiffiffi t Rk j¼ pffiffiffi ½F ðmp f0 ; kÞ F ðm0 p fa ; kÞ 4p0 r pffiffiffiffi t Rk ¼ pffiffiffi f2mKðkÞ F ðf0 ; kÞ ½2m0 KðkÞ F ðfa ; kÞg. 4p0 r
ð8Þ
Here KðkÞ; F ðf0 ; kÞ and F ðfa ; kÞ satisfy the condition of consulting the table of elliptic integrals, so that the problem may conveniently be solved. Using Eq. (8), we may easily discuss the spatial potential distribution of a uniformly charged circular
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arc, such as the following: (1) The potential on the z-axis which the uniformly charged quadrant generates is given by tR tR pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½F ðb0 ; 0Þ F ðba ; 0Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba 2 2 2p0 z þ R 2p0 z2 þ R2 tR pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . ¼ 80 z2 þ R2
j¼
ð9Þ
When z ¼ 0; the potential at the origin is j0 ¼ t=80 : (2) The spatial potential distribution of the uniformly charged semicircular ring is given by
tR yþp y qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ;k F ;k . (10) j¼ 2 2 2p0 z2 þ ðr þ RÞ2 When r ¼ R; y ¼ 3p=2; z ¼ 0; b0 ¼ 5p=4; and ba ¼ 3p=4; we easily find the maximum of the potential in the line l passing the point CðR; 3p=2; 0Þ and paralleling the z-axis jlmax ¼ jl0 ¼ ðt=2p0 ÞF ðp=4; 1Þ ¼ 0:44t=p0 : (3) The spatial potential distribution of the uniformly charged circular ring is given by
tR yþp yp q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ;k F ;k j¼ F 2 2 2p0 z2 þ ðr þ RÞ2 ¼
tR qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi KðkÞ. p0 z2 þ ðr þ RÞ2
ð11Þ
In Fig. 2, we draw the spatial potential distribution diagram of the uniformly charged circular ring. From Fig. 2 we see that when the coordinates r ¼ R and z ¼ 0; the potential is indefinite; when raR and z ¼ 0; the potential gains corresponding maximum jmax ðrÞ: Eq. (8) is the spatial potential distribution function of a uniformly charged circular arc.
3. Electric field distribution of a uniformly charged circular arc Using gradient calculation, from Eqs. (4) and (8), we can get the field distribution of a uniformly charged circular arc. In a cylindrical coordinate system, the field E is given by E¼5j¼
qj 1 qj qj er ey ez . qr r qy qz
(12)
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ϕ [V]
1040
20 15 10 5 -100
100 80 60 40
-50 0
ρ [m]
20
z [m]
50 100 0
Fig. 2. The spatial potential distribution of a uniformly charged circular ring. Parameters chosen are t=4p0 ¼ 2½V and R ¼ 20½m:
From Eq. (4) we find that j depends on r; y and z: Then we have [16] qj tðr þ RÞk3 pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ ¼ qr 16p0 r3 R pffiffiffiffi tk R qF ðb0 ; kÞ qk qF ðba ; kÞ qk þ , pffiffiffi qk qr qk qr 4p0 r 2
qF ðb; kÞ Eðb; kÞ k0 F ðb; kÞ ¼ 2 qk kk0
k sin b cos b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 1 k2 sin2 b 02
and qk k ðr þ RÞk2 ¼ 1 , qr 2r 2R pffiffiffiffiffiffiffiffiffiffiffiffiffi where k0 ¼ 1 k2 : Then we obtain qj tðr þ RÞk3 tk4 4R pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ pffiffiffiffiffiffiffiffiffi ¼ Er ¼ 2ðr þ RÞ qr 16p0 r3 R 32p0 r3 R k2 82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 02 > 2 kk : k0 1 k2 sin2 b0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ 4 ð13Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . 2 > 2 kk0 k0 1 k2 sin2 ba ;
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Similarly we get 2 3 pffiffiffiffi qj tk R 6 1 1 7 pffiffiffiffiffi 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ Ey ¼ 3 rqy 2 2 2 2 8p0 r 1 k sin b0 1 k sin ba
(14)
and qj tk3 z tk4 z pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ þ pffiffiffiffiffiffiffiffiffi ¼ qz 16p0 r3 R 16p0 r3 R 82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 2 0 > 2 kk : k0 1 k2 sin2 b0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 5 . 2 > 02 2 2 kk0 1 k k sin ba ;
Ez ¼
ð15Þ
Eqs. (12)–(15) present the field distribution functions of a uniformly charged circular arc. We discuss some special cases in the following. A. The field distribution on the z-axis which a uniformly charged circular arc generates: Now y ¼ 0; r ! 0; k ! 0; and k0 ! 1; so we have tR2 tk4 4R qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba lim pffiffiffiffiffiffiffiffiffi 2 2ðr þ RÞ Er ¼ 3 r!0 32p 0 r R k 2p0 ðz2 þ R2 Þ3 ( 2 2 Eðb0 ; kÞ k0 F ðb0 ; kÞ Eðba ; kÞ k0 F ðba ; kÞ
2 2 kk0 kk0 0 19 > k B sin b0 cos b0 sin ba cos ba C= 0 2 @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA . ð16Þ > k 1 k2 sin2 b0 1 k2 sin2 ba ; In the expression of E r ; using L’Hospital’s method, we cannot get the value of the limit term. Now adopting the method of the series expansion and reserving the term on k to the second power of k; we have k2 ½b sin b cos b þ Oðk4 Þ, 4 k2 F ðb; kÞ ¼ b þ ½b sin b cos b þ Oðk4 Þ, 4
Eðb; kÞ ¼ b
EðkÞ ¼
p pk2 þ Oðk4 Þ; 2 8
KðkÞ ¼
p pk2 þ þ Oðk4 Þ 2 8
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and 1 k2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1 þ sin2 b þ Oðk4 Þ. 2 2 2 1 k sin b Then Er ¼
tR2 tR qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba lim qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½4R 2ðr þ RÞk2 r!0 8p0 ½z2 þ ðr þ RÞ2 3 2p0 ðz2 þ R2 Þ3
½b0 ba ðsin b0 cos b0 sin ba cos ba Þ þ Oðk4 Þ
¼
tR2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin a. 4p0 ðz2 þ R2 Þ3
Using the same method, we get Ey ¼
tR2 ð1 cos aÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4p0 ðz2 þ R2 Þ3
and Ez ¼
tRza qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . 40 ðz2 þ R2 Þ3
That is E¼
tR2 tR2 ð1 cos aÞ tRza qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin aer qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez , 4p0 ðz2 þ R2 Þ3 4p0 ðz2 þ R2 Þ3 4p0 ðz2 þ R2 Þ3 (17)
where ey is perpendicular to the polar axis. Eq. (17) is a typical result, which can also be obtained by the integral using the superposition principle of field. The component qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi of the field E on the z-axis direction is E z ¼ tRza=4p0 ðz2 þ R2 Þ3 : In Fig. 3, we pffiffiffi draw the field distribution diagram and see that when z ¼ 2R=2; E z gains the pffiffiffi maximum E zmax ¼ ta=6 3p0 R: B. The field distribution on the polar coordinate plane which a uniformly charged circular arc generates: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Now z ¼ 0; k ¼ 4rR=½ðr þ RÞ2 ; so we have Er ¼
tðr þ RÞk3 tk4 4R pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ pffiffiffiffiffiffiffiffiffi 2 2ðr þ RÞ 16p0 r3 R 32p0 r3 R k
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Ez [N / C] 0.75 0.5 0.25 √2 / 2
-√2 / 2
0
z/R
-0.25 -0.5 -0.75 -3
-2
-1
0
1
2
3
Fig. 3. The component of the field E on the z-axis direction E z as a function of z for a uniformly charged arc. Parameters chosen are ta=4p0 R ¼ 2½N=C:
82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 02 > 2 : kk k0 1 k2 sin2 b 0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . 02 > 2 kk k0 1 k2 sin2 ba ;
ð18Þ
Similarly we get
2 3 pffiffiffiffi tk R 6 1 1 7 pffiffiffiffiffi 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 Ey ¼ 2 2 2 2 8p0 r3 1 k sin b0 1 k sin ba
(19)
E z ¼ 0.
(20)
E ¼ E r er þ E y ey .
(21)
and Then C. Field distributions of some uniformly charged special circular arcs: 1. The field distribution of the uniformly charged quadrant Now substituting a ¼ p=2; b0 ¼ ðy þ pÞ=2; ba ¼ ð2y þ pÞ=4; and k ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4rR=½z2 þ ðr þ RÞ2 into Eqs. (13)–(15), and (12), we get E r ; E y ; E z ; and E: (1) The field distribution on the z-axis is given by E¼
tR2 tR2 tRz qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi er qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez . 3 3 2 2 80 ðz2 þ R2 Þ3 4p0 ðz2 þ R Þ 4p0 ðz2 þ R Þ
(22)
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(2) The field distribution on the line l passing CðR; p; 0Þ and paralleling the z-axis is given by " pffiffiffi 2 # p p p tk3 tk 2k 02 F ;k E ; k k F ; k pffiffiffiffiffiffiffiffiffiffiffiffiffi , Er ¼ 4 8p0 R 4 4 8p0 R 2 2 k2 " pffiffiffi # tk 2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi Ey ¼ 8p0 R 2 k2 and
" pffiffiffi 2 # p p p tk3 z tk3 z 2k 02 Ez ¼ F ;k þ E ; k k F ; k pffiffiffiffiffiffiffiffiffiffiffiffiffi , 2 02 4 4 4 16p0 R2 16p0 R k 2 2 k2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where k ¼ 4R2 =ðz2 þ 4R2 Þ: Specially when z ¼ 0; we get the field at the point CðR; p; 0Þ: pffiffiffi pffiffiffi t tð 2 1Þ EC ¼ ½lnð1 þ 2Þer þ ey . (23) 8p0 R 8p0 R 2. The field distribution of the uniformly charged semicircular arc qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Now substituting a ¼ p; b0 ¼ ðy þ pÞ=2; ba ¼ y=2; and k ¼ 4rR=½z2 þ ðr þ RÞ2 into Eqs. (13)–(15), and (12), we obtain the field distribution of a uniformly charged semicircular arc. (1) The field distribution on the z-axis: Since a ¼ p; from Eq. (17) we have E¼
tR2 tRz qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez . 40 ðz2 þ R2 Þ3 2p0 ðz2 þ R2 Þ3
(24)
pffiffiffi When z ¼ 0; E ¼ ðt=2p0 RÞey : When z ¼ R; E ¼ ðt 2=8p0 RÞey p ffiffiffi ðt 2=160 RÞez : (2) The field distribution on the cylindrical surface whose radius is 2R and parallels qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the z-axis: Now r ¼ 2R; b0 ¼ ðy þ pÞ=2 and ba ¼ y=2; k ¼ 8R2 =ðz2 þ 9R2 Þ; k0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðz2 þ R2 Þ=ðz2 þ 9R2 Þ; so we have pffiffiffi
pffiffiffi 2tkð2 3k2 Þ 3 2tk3 yþp y ;k F ;k Er ¼ F 2 2 2 64p0 R 64p0 Rk0 8 >
< y þ p y yþp y 02 ;k E ;k k F ;k F ;k
E > 2 2 2 2 : 2 39 > = y y6 1 1 7 2 þk sin cos 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 , > 2 2 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ ;
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2 3 pffiffiffi 2tk 6 1 1 7 Ey ¼ 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 32p0 R 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ and pffiffiffi 3 pffiffiffi 3
2tk z 2tk z yþp y ;k F ;k þ Ez ¼ F 2 2 2 2 64p0 R 64p0 R2 k0 8 >
< y þ p y yþp y 2 ; k E ; k k0 F ;k F ;k
E > 2 2 2 2 : 2
39 > = 1 1 6 7 2 þk sinðy=2Þ cosðy=2Þ4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . > 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ ;
Specially when z ¼ 0 and y ¼ 3p=2; we get the field at the point Cð2R; 3p=2; 0Þ: ( " #) pffiffiffi! pffiffiffi! pffiffiffi! t p 2 2 t p 2 2 1 p 2 2 4 F E EC ¼ ; þ ; F ; pffiffiffi er 9p0 R 4 3 4p0 R 4 3 9 4 3 3 5 ¼
9:66 102 t er . p0 R
ð25Þ
3. The field distribution of the uniformly charged circular ring: Now substituting a ¼ 2p; b0 ¼ ðy þ pÞ=2 ¼ p þ ðy pÞ=2; and ba ¼ ðy pÞ=2 into Eqs. (13)–(15) and (12), we get E y ¼ 0; and E¼
pffiffiffiffi & ' tk R EðkÞ ðr þ RÞk2 tk3 zEðkÞ pffiffiffiffiffi KðkÞ 0 2 1 er þ pffiffiffiffiffiffiffiffiffi 2 ez : 2R 4p0 r3 k 8p0 r3 Rk0
ð26Þ
(1) The field on the z-axis Since r ! 0 we get k ! 0 and k0 ! 1: Then we have E r ¼ 0 and E ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ½tRz=20 ðz2 þ R2 Þ3 ez : When z ¼ 0; we have E0 ¼ 0: When z ¼ 2R=2; E z gains pffiffiffi the maximum E zmax ¼ t=3 30 R: (2) The field on the polar coordinate plane Since z ¼ 0; we get E z ¼ 0 and pffiffiffiffi & ' tk R EðkÞ ðr þ RÞk2 p ffiffiffiffiffi E¼ KðkÞ 0 2 1 er . 2R 4p0 r3 k The direction of the field E is identical to the direction of the polar radius.
(27)
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4. Summary The study and discussion described above show the following: (1) From the above discussion, we see that using important properties of elliptic integrals Eqs. (5) and (6), we can turn complex physics problems into a simple problem and get the potential functions and the field functions j; E r ; E y ; and E z easily. According to Eqs. (8) and (13)–(15), we can specifically and systematically discuss the potential distribution and the field distribution of a uniformly charged circular arc and can conveniently draw corresponding distribution diagrams by using computers. Therefore this method provides an effective way of dealing with other similar physics problems of elliptic integrals in the analytic method and of solving this kind of physics problem by using computers. pffiffiffi (2) For a uniformly charged circular arc, when z ¼ 2 2R=2; the component of the field pffiffiffi E on the z-axis which it generates E z possesses the maximum E zmax ¼ ta=6 3p0 R; which depends on the central angle a: If the uniformly charged circular arc is not a closed ring, on the z-axis the electric field E not only possesses the component of the ez direction, but also possesses the components of the ey direction and the er direction; on the polar coordinate plane, the field E only possesses the components of the er direction and ey direction. For a uniformly charged circular ring, on the z-axis the field E is always along the z-axis direction and on the polar coordinate plane the field E is always along the radius direction.
Acknowledgements The author wishes to express his most sincere thanks to the referee and Professor Mark N. Horenstein, who read the manuscript carefully and gave valuable advice, comments and help. References [1] O.D. Jefimenko, Torque exerted by a moving electric charge on a stationary electric charge distribution, J. Phys. A: Math. Gen. 35 (2002) 5305–5314. [2] G.A. Vugalter, A.K. Das, V.A. Sorokin, A charged particle on a ring in a magnetic field: quantum revivals, Eur. J. Phys. 25 (2004) 157–170. [3] R.E. Kribel, K. Shinsky, D.A. Phelps, H.H. Fleischmann, Generation of field-reversing electron rings by neutralized-cusp injection into a magnetic mirror trap, Plasma Phys. 16 (1974) 113–115. [4] L.D. Landau, E.M. Lifshitz, Electrodynamics of Continuous Medium, People’s Education Press, Beijing, 1963, p. 12. [5] J.D. Jackson, Classical Electrodynamics, second ed., Wiley, New York, 1998, pp. 102–104. [6] B.B. Badijing, N.H. Tuopudijing, Problem in Electrodynamics, People’s Education Press, Beijing, 1964, p. 21. [7] X.Y. Lin, Z.X. Zhang, Key to Problem in Electrodynamics, Science Press, Beijing, 1999, pp. 111–118. [8] J.Z. Zhong, Field of a linear charged ring, Coll. Phys. 1 (7) (1982) 15. [9] C.J. Tranter, Integral Transforms in Mathematic and Physics, Wiley, New York, 1966. [10] I.N. Sneddon, Special Functions of Mathematical and Chemistry, Oliver and Boyd, Edinburg, UK, 1961, p. 145.
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[11] E. Ball, Potential from a ring of charge, Proc. IEE. 124 (7) (1977) 664. [12] L.S. Zhang, W.L. Jiang, The distribution of electric field of annular liner electric charge, Coll. Phys. 17 (8) (1998) 21–23. [13] C.L. Cheng, H. Wang, et al., Electric field of a uniformly charged ring, Coll. Phys. 22 (6) (2003) 15–17. [14] H.Y. Zhou, H. Chen, Space distribution of electrical field generated by a uniformly charged ring, Coll. Phys. 23 (9) (2004) 32. [15] N.G. Trinh, P.S. Maruvada, On the potential and field distribution around a ground electrode for HVDC transmission, IEEE Trans. Power Appar. Syst. PAS 90 (6) (1971) 2793–2801. [16] P. Byrd, M. Friedman, Handbook of Elliptic Integrals for Engineers and Physicists, Springer, Berlin, 1954, pp. 12–13, 282–283.
Journal of Electrostatics 63 (2005) 1035–1047 www.elsevier.com/locate/elstat
Field distribution of a uniformly charged circular arc Ping Zhu Department of Physics, Simao Teacher’s College, Yunan 665000, China Received 16 September 2004; received in revised form 24 January 2005; accepted 7 February 2005 Available online 17 March 2005
Abstract Because the field of a uniformly charged circular arc involves elliptic integrals, not generally possessing symmetry of a uniformly charged ring, it is difficult to solve and discuss this problem. This paper investigates the problem and presents the potential function and the field function of a uniformly charged circular arc together with discussions about them by using the theory of elliptic integrals and the analytic method. r 2005 Elsevier B.V. All rights reserved. Keywords: Arbitrary circular arc; Uniform linear charge; Potential distribution; Field distribution; Elliptic integrals
1. Introduction The field distribution of a uniformly charged circular arc is a typical and important problem in electrostatics. In the research of electromagnetism it is significant for us to investigate this problem [1–3]. However, because the field of a uniformly charged circular arc involves elliptic integrals, not generally possessing symmetry of a uniformly charged ring, it is difficult for us to discuss and solve the problem. Several researchers [4–8] have discussed some cases of the typical circular ring of uniform charge. Applying Fourier Bessel transform pair [9,10], Ball [11] presented the potential on the uniformly charged circular ring. Using computers and E-mail address: [email protected]. 0304-3886/$ - see front matter r 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.elstat.2005.02.001
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the superposition principle of electric field, Zhang and Jiang [12] discussed the electric field distribution of annular linear electric charge. According to the superposition theorem of the field and the electric potential of a point charge, Cheng et al. [13] derived a series solution of the potential and the field of a uniformly charged ring. In the rectangular coordinate system, Zhou and Chen [14] directly calculated the spatial distribution of the electrical field generated by a uniformly changed ring. Making use of the analogous method, Trinh and Maruvada [15] investigated the electric field distribution of an arbitrary circular-arc shaped uniform current filament embedded in a conductive medium, the solution to which has been derived in terms of elliptic integrals and used in the numerical evaluation of the resistance of complex ground electrodes. Their work has not only further promoted the research on the physics problems of elliptic integrals, but also helped me to solve the case of a uniformly charged circular arc in analytic methods. In this paper, by using the theory of elliptic integrals, important properties of elliptic integrals and some special tricks, we can turn difficulty into simplicity, and gain the analytic potential function and the analytic field function of a uniformly charged circular arc, so that this problem is solved satisfactorily. In Section 2, we prove useful properties of elliptic integral functions and present the potential function of the uniformly charged circular arc. In Section 3, we present the spatial field distribution of the uniformly charged circular arc and discuss electric field distributions on some special positions. Finally, in Section 4, we present the summary. Discussions on the problem may help us understand methods to deal with this kind of physics problem of elliptic integrals.
2. Proof of properties and potential function There is a section of a uniformly charged rigid circular arc whose radius is R; the central angle is a ð0oap2pÞ; and the linear charge density is t; and we will discuss its potential and field distribution, as depicted in Fig. 1. In a cylindrical coordinate system, we place the circular arc on the polar coordinate plane, whose initial coordinates are AðR; 0; 0Þ and final coordinates are BðR; a; 0Þ: At a point on the arc we take a line element dl with electric quantity dq ¼ t dl; which generates that the potential at the point Pðr; y; zÞ in space is dj ¼ t dl=4p0 r where r2 ¼ z2 þ ðr þ RÞ2 2rR½1 þ cosða0 yÞ. The potential at the point Pðr; y; zÞ that the charged circular arc generates is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi tR 2rR=ðz2 þ ðr þ RÞ2 Þ pffiffiffiffiffiffiffiffiffi j¼ 4p0 2rR Z a da0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi .
0 1 2rR½1 þ cosða0 yÞ=ðz2 þ ðr þ RÞ2 Þ
ð1Þ
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Fig. 1. A uniformly charged circular arc.
Defining a0 y ¼ p 2b; namely b ¼ ðy þ p a0 Þ=2; da0 ¼ 2 db; and 1 þ cosða0 yÞ ¼ 2 sin2 b: When a0 ¼ 0; b0 ¼ ðy þ pÞ=2: When a0 ¼ a; ba ¼ ðy þ p aÞ=2: We let k2 ¼ 4rR=½z2 þ ðr þ RÞ2 and get Z ðpþyÞ=2 tRk db pffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi j¼ 4p0 rR ðpþyaÞ=2 1 k2 sin2 b pffiffiffiffi Z b0 t Rk db qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . ¼ ð2Þ pffiffiffi
4p0 r ba 1 k2 sin2 b Defining Z
b
F ðb; kÞ ¼ 0
db qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 1 k2 sin2 b
from Eq. (2) we get pffiffiffiffi t Rk j¼ pffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ, 4p0 r
(3)
(4)
where F ðb; kÞ is the elliptic integral function of the first kind. When 0pbpp=2; we may directly get the corresponding results of the potential. When b4p=2 or bo0; we cannot directly get the results of Eq. (4). If we use the general elliptic integral method, it is very difficult to solve Eq. (2). However, if we use the important property of the elliptic integral function, we may turn difficulty into simplicity, so that we conveniently get the specific results of the problem. When the amplitude b4p=2 or bo0; we define b ¼ mp a0 and 0pa0 pp=2; where m is an integer. Then we have [16] F ðb; kÞ ¼ F ðmp a0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ, (5) q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi R p=2 where KðkÞ ¼ 0 db= 1 k2 sin2 b: KðkÞ is the complete elliptic integral function of the first kind.
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Since Z
mp a0
dy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 k2 sin2 y Z mp Z mp a0 dy dy ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ þ 2 2 0 mp 1 k sin y 1 k2 sin2 y 2 3 Z Z m ð2n1Þp=2 np X6 dy dy 7 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ 4 2 2 2 2 ðn1Þp ð2n1Þp=2 n¼1 1 k sin b 1 k sin b
F ðmp a0 Þ ¼
F ða0 ; kÞ m X ½KðkÞ þ KðkÞ F ða0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ, ¼ n¼1
F ðb; kÞ ¼ F ðmp a0 ; kÞ ¼ 2mKðkÞ F ða0 ; kÞ. Using the same method, we can prove that the elliptic integral function of the second kind satisfies Eðb; kÞ ¼ Eðmp a0 ; kÞ ¼ 2mEðkÞ Eða0 ; kÞ,
(6)
where Z
b
Eðb; kÞ ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 b db.
(7)
0
R p=2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 b db is the complete elliptic integral function of the second EðkÞ ¼ 0 kind. For a linearly charged circular arc, 0pap2p: In Eq. (4) we substitute b0 ¼ mp f0 : When 0pb0 pp=2; m ¼ 0 and f0 ¼ b0 : If b0 4p=2 or b0 o0; m is selected as the corresponding integer except 0. Using the same method, ba ¼ m0 p fa : When 0pba pp=2; m0 ¼ 0 and fa ¼ ba : If ba 4p=2 or ba o0; m0 is selected as the corresponding integer except 0. Then, from Eq. (4) we have pffiffiffiffi t Rk j¼ pffiffiffi ½F ðmp f0 ; kÞ F ðm0 p fa ; kÞ 4p0 r pffiffiffiffi t Rk ¼ pffiffiffi f2mKðkÞ F ðf0 ; kÞ ½2m0 KðkÞ F ðfa ; kÞg. 4p0 r
ð8Þ
Here KðkÞ; F ðf0 ; kÞ and F ðfa ; kÞ satisfy the condition of consulting the table of elliptic integrals, so that the problem may conveniently be solved. Using Eq. (8), we may easily discuss the spatial potential distribution of a uniformly charged circular
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arc, such as the following: (1) The potential on the z-axis which the uniformly charged quadrant generates is given by tR tR pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½F ðb0 ; 0Þ F ðba ; 0Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba 2 2 2p0 z þ R 2p0 z2 þ R2 tR pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . ¼ 80 z2 þ R2
j¼
ð9Þ
When z ¼ 0; the potential at the origin is j0 ¼ t=80 : (2) The spatial potential distribution of the uniformly charged semicircular ring is given by
tR yþp y qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ;k F ;k . (10) j¼ 2 2 2p0 z2 þ ðr þ RÞ2 When r ¼ R; y ¼ 3p=2; z ¼ 0; b0 ¼ 5p=4; and ba ¼ 3p=4; we easily find the maximum of the potential in the line l passing the point CðR; 3p=2; 0Þ and paralleling the z-axis jlmax ¼ jl0 ¼ ðt=2p0 ÞF ðp=4; 1Þ ¼ 0:44t=p0 : (3) The spatial potential distribution of the uniformly charged circular ring is given by
tR yþp yp q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ;k F ;k j¼ F 2 2 2p0 z2 þ ðr þ RÞ2 ¼
tR qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi KðkÞ. p0 z2 þ ðr þ RÞ2
ð11Þ
In Fig. 2, we draw the spatial potential distribution diagram of the uniformly charged circular ring. From Fig. 2 we see that when the coordinates r ¼ R and z ¼ 0; the potential is indefinite; when raR and z ¼ 0; the potential gains corresponding maximum jmax ðrÞ: Eq. (8) is the spatial potential distribution function of a uniformly charged circular arc.
3. Electric field distribution of a uniformly charged circular arc Using gradient calculation, from Eqs. (4) and (8), we can get the field distribution of a uniformly charged circular arc. In a cylindrical coordinate system, the field E is given by E¼5j¼
qj 1 qj qj er ey ez . qr r qy qz
(12)
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ϕ [V]
1040
20 15 10 5 -100
100 80 60 40
-50 0
ρ [m]
20
z [m]
50 100 0
Fig. 2. The spatial potential distribution of a uniformly charged circular ring. Parameters chosen are t=4p0 ¼ 2½V and R ¼ 20½m:
From Eq. (4) we find that j depends on r; y and z: Then we have [16] qj tðr þ RÞk3 pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ ¼ qr 16p0 r3 R pffiffiffiffi tk R qF ðb0 ; kÞ qk qF ðba ; kÞ qk þ , pffiffiffi qk qr qk qr 4p0 r 2
qF ðb; kÞ Eðb; kÞ k0 F ðb; kÞ ¼ 2 qk kk0
k sin b cos b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k 1 k2 sin2 b 02
and qk k ðr þ RÞk2 ¼ 1 , qr 2r 2R pffiffiffiffiffiffiffiffiffiffiffiffiffi where k0 ¼ 1 k2 : Then we obtain qj tðr þ RÞk3 tk4 4R pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ pffiffiffiffiffiffiffiffiffi ¼ Er ¼ 2ðr þ RÞ qr 16p0 r3 R 32p0 r3 R k2 82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 02 > 2 kk : k0 1 k2 sin2 b0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ 4 ð13Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . 2 > 2 kk0 k0 1 k2 sin2 ba ;
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Similarly we get 2 3 pffiffiffiffi qj tk R 6 1 1 7 pffiffiffiffiffi 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 ¼ Ey ¼ 3 rqy 2 2 2 2 8p0 r 1 k sin b0 1 k sin ba
(14)
and qj tk3 z tk4 z pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ þ pffiffiffiffiffiffiffiffiffi ¼ qz 16p0 r3 R 16p0 r3 R 82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 2 0 > 2 kk : k0 1 k2 sin2 b0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 5 . 2 > 02 2 2 kk0 1 k k sin ba ;
Ez ¼
ð15Þ
Eqs. (12)–(15) present the field distribution functions of a uniformly charged circular arc. We discuss some special cases in the following. A. The field distribution on the z-axis which a uniformly charged circular arc generates: Now y ¼ 0; r ! 0; k ! 0; and k0 ! 1; so we have tR2 tk4 4R qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba lim pffiffiffiffiffiffiffiffiffi 2 2ðr þ RÞ Er ¼ 3 r!0 32p 0 r R k 2p0 ðz2 þ R2 Þ3 ( 2 2 Eðb0 ; kÞ k0 F ðb0 ; kÞ Eðba ; kÞ k0 F ðba ; kÞ
2 2 kk0 kk0 0 19 > k B sin b0 cos b0 sin ba cos ba C= 0 2 @qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA . ð16Þ > k 1 k2 sin2 b0 1 k2 sin2 ba ; In the expression of E r ; using L’Hospital’s method, we cannot get the value of the limit term. Now adopting the method of the series expansion and reserving the term on k to the second power of k; we have k2 ½b sin b cos b þ Oðk4 Þ, 4 k2 F ðb; kÞ ¼ b þ ½b sin b cos b þ Oðk4 Þ, 4
Eðb; kÞ ¼ b
EðkÞ ¼
p pk2 þ Oðk4 Þ; 2 8
KðkÞ ¼
p pk2 þ þ Oðk4 Þ 2 8
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and 1 k2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1 þ sin2 b þ Oðk4 Þ. 2 2 2 1 k sin b Then Er ¼
tR2 tR qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½b0 ba lim qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½4R 2ðr þ RÞk2 r!0 8p0 ½z2 þ ðr þ RÞ2 3 2p0 ðz2 þ R2 Þ3
½b0 ba ðsin b0 cos b0 sin ba cos ba Þ þ Oðk4 Þ
¼
tR2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin a. 4p0 ðz2 þ R2 Þ3
Using the same method, we get Ey ¼
tR2 ð1 cos aÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4p0 ðz2 þ R2 Þ3
and Ez ¼
tRza qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . 40 ðz2 þ R2 Þ3
That is E¼
tR2 tR2 ð1 cos aÞ tRza qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin aer qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez , 4p0 ðz2 þ R2 Þ3 4p0 ðz2 þ R2 Þ3 4p0 ðz2 þ R2 Þ3 (17)
where ey is perpendicular to the polar axis. Eq. (17) is a typical result, which can also be obtained by the integral using the superposition principle of field. The component qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi of the field E on the z-axis direction is E z ¼ tRza=4p0 ðz2 þ R2 Þ3 : In Fig. 3, we pffiffiffi draw the field distribution diagram and see that when z ¼ 2R=2; E z gains the pffiffiffi maximum E zmax ¼ ta=6 3p0 R: B. The field distribution on the polar coordinate plane which a uniformly charged circular arc generates: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Now z ¼ 0; k ¼ 4rR=½ðr þ RÞ2 ; so we have Er ¼
tðr þ RÞk3 tk4 4R pffiffiffiffiffiffiffiffiffi ½F ðb0 ; kÞ F ðba ; kÞ pffiffiffiffiffiffiffiffiffi 2 2ðr þ RÞ 16p0 r3 R 32p0 r3 R k
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Ez [N / C] 0.75 0.5 0.25 √2 / 2
-√2 / 2
0
z/R
-0.25 -0.5 -0.75 -3
-2
-1
0
1
2
3
Fig. 3. The component of the field E on the z-axis direction E z as a function of z for a uniformly charged arc. Parameters chosen are ta=4p0 R ¼ 2½N=C:
82 3 > < Eðb ; kÞ k0 2 F ðb ; kÞ k sin b0 cos b0 7 6 0 0
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 02 > 2 : kk k0 1 k2 sin2 b 0 2 39 > 02 k sin ba cos ba 7= 6Eðba ; kÞ k F ðba ; kÞ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . 02 > 2 kk k0 1 k2 sin2 ba ;
ð18Þ
Similarly we get
2 3 pffiffiffiffi tk R 6 1 1 7 pffiffiffiffiffi 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 Ey ¼ 2 2 2 2 8p0 r3 1 k sin b0 1 k sin ba
(19)
E z ¼ 0.
(20)
E ¼ E r er þ E y ey .
(21)
and Then C. Field distributions of some uniformly charged special circular arcs: 1. The field distribution of the uniformly charged quadrant Now substituting a ¼ p=2; b0 ¼ ðy þ pÞ=2; ba ¼ ð2y þ pÞ=4; and k ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4rR=½z2 þ ðr þ RÞ2 into Eqs. (13)–(15), and (12), we get E r ; E y ; E z ; and E: (1) The field distribution on the z-axis is given by E¼
tR2 tR2 tRz qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi er qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez . 3 3 2 2 80 ðz2 þ R2 Þ3 4p0 ðz2 þ R Þ 4p0 ðz2 þ R Þ
(22)
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(2) The field distribution on the line l passing CðR; p; 0Þ and paralleling the z-axis is given by " pffiffiffi 2 # p p p tk3 tk 2k 02 F ;k E ; k k F ; k pffiffiffiffiffiffiffiffiffiffiffiffiffi , Er ¼ 4 8p0 R 4 4 8p0 R 2 2 k2 " pffiffiffi # tk 2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi Ey ¼ 8p0 R 2 k2 and
" pffiffiffi 2 # p p p tk3 z tk3 z 2k 02 Ez ¼ F ;k þ E ; k k F ; k pffiffiffiffiffiffiffiffiffiffiffiffiffi , 2 02 4 4 4 16p0 R2 16p0 R k 2 2 k2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where k ¼ 4R2 =ðz2 þ 4R2 Þ: Specially when z ¼ 0; we get the field at the point CðR; p; 0Þ: pffiffiffi pffiffiffi t tð 2 1Þ EC ¼ ½lnð1 þ 2Þer þ ey . (23) 8p0 R 8p0 R 2. The field distribution of the uniformly charged semicircular arc qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Now substituting a ¼ p; b0 ¼ ðy þ pÞ=2; ba ¼ y=2; and k ¼ 4rR=½z2 þ ðr þ RÞ2 into Eqs. (13)–(15), and (12), we obtain the field distribution of a uniformly charged semicircular arc. (1) The field distribution on the z-axis: Since a ¼ p; from Eq. (17) we have E¼
tR2 tRz qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ey þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ez . 40 ðz2 þ R2 Þ3 2p0 ðz2 þ R2 Þ3
(24)
pffiffiffi When z ¼ 0; E ¼ ðt=2p0 RÞey : When z ¼ R; E ¼ ðt 2=8p0 RÞey p ffiffiffi ðt 2=160 RÞez : (2) The field distribution on the cylindrical surface whose radius is 2R and parallels qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the z-axis: Now r ¼ 2R; b0 ¼ ðy þ pÞ=2 and ba ¼ y=2; k ¼ 8R2 =ðz2 þ 9R2 Þ; k0 ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðz2 þ R2 Þ=ðz2 þ 9R2 Þ; so we have pffiffiffi
pffiffiffi 2tkð2 3k2 Þ 3 2tk3 yþp y ;k F ;k Er ¼ F 2 2 2 64p0 R 64p0 Rk0 8 >
< y þ p y yþp y 02 ;k E ;k k F ;k F ;k
E > 2 2 2 2 : 2 39 > = y y6 1 1 7 2 þk sin cos 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 , > 2 2 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ ;
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2 3 pffiffiffi 2tk 6 1 1 7 Ey ¼ 4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 32p0 R 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ and pffiffiffi 3 pffiffiffi 3
2tk z 2tk z yþp y ;k F ;k þ Ez ¼ F 2 2 2 2 64p0 R 64p0 R2 k0 8 >
< y þ p y yþp y 2 ; k E ; k k0 F ;k F ;k
E > 2 2 2 2 : 2
39 > = 1 1 6 7 2 þk sinðy=2Þ cosðy=2Þ4qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 . > 1 k2 cos2 ðy=2Þ 1 k2 sin2 ðy=2Þ ;
Specially when z ¼ 0 and y ¼ 3p=2; we get the field at the point Cð2R; 3p=2; 0Þ: ( " #) pffiffiffi! pffiffiffi! pffiffiffi! t p 2 2 t p 2 2 1 p 2 2 4 F E EC ¼ ; þ ; F ; pffiffiffi er 9p0 R 4 3 4p0 R 4 3 9 4 3 3 5 ¼
9:66 102 t er . p0 R
ð25Þ
3. The field distribution of the uniformly charged circular ring: Now substituting a ¼ 2p; b0 ¼ ðy þ pÞ=2 ¼ p þ ðy pÞ=2; and ba ¼ ðy pÞ=2 into Eqs. (13)–(15) and (12), we get E y ¼ 0; and E¼
pffiffiffiffi & ' tk R EðkÞ ðr þ RÞk2 tk3 zEðkÞ pffiffiffiffiffi KðkÞ 0 2 1 er þ pffiffiffiffiffiffiffiffiffi 2 ez : 2R 4p0 r3 k 8p0 r3 Rk0
ð26Þ
(1) The field on the z-axis Since r ! 0 we get k ! 0 and k0 ! 1: Then we have E r ¼ 0 and E ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ½tRz=20 ðz2 þ R2 Þ3 ez : When z ¼ 0; we have E0 ¼ 0: When z ¼ 2R=2; E z gains pffiffiffi the maximum E zmax ¼ t=3 30 R: (2) The field on the polar coordinate plane Since z ¼ 0; we get E z ¼ 0 and pffiffiffiffi & ' tk R EðkÞ ðr þ RÞk2 p ffiffiffiffiffi E¼ KðkÞ 0 2 1 er . 2R 4p0 r3 k The direction of the field E is identical to the direction of the polar radius.
(27)
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4. Summary The study and discussion described above show the following: (1) From the above discussion, we see that using important properties of elliptic integrals Eqs. (5) and (6), we can turn complex physics problems into a simple problem and get the potential functions and the field functions j; E r ; E y ; and E z easily. According to Eqs. (8) and (13)–(15), we can specifically and systematically discuss the potential distribution and the field distribution of a uniformly charged circular arc and can conveniently draw corresponding distribution diagrams by using computers. Therefore this method provides an effective way of dealing with other similar physics problems of elliptic integrals in the analytic method and of solving this kind of physics problem by using computers. pffiffiffi (2) For a uniformly charged circular arc, when z ¼ 2 2R=2; the component of the field pffiffiffi E on the z-axis which it generates E z possesses the maximum E zmax ¼ ta=6 3p0 R; which depends on the central angle a: If the uniformly charged circular arc is not a closed ring, on the z-axis the electric field E not only possesses the component of the ez direction, but also possesses the components of the ey direction and the er direction; on the polar coordinate plane, the field E only possesses the components of the er direction and ey direction. For a uniformly charged circular ring, on the z-axis the field E is always along the z-axis direction and on the polar coordinate plane the field E is always along the radius direction.
Acknowledgements The author wishes to express his most sincere thanks to the referee and Professor Mark N. Horenstein, who read the manuscript carefully and gave valuable advice, comments and help. References [1] O.D. Jefimenko, Torque exerted by a moving electric charge on a stationary electric charge distribution, J. Phys. A: Math. Gen. 35 (2002) 5305–5314. [2] G.A. Vugalter, A.K. Das, V.A. Sorokin, A charged particle on a ring in a magnetic field: quantum revivals, Eur. J. Phys. 25 (2004) 157–170. [3] R.E. Kribel, K. Shinsky, D.A. Phelps, H.H. Fleischmann, Generation of field-reversing electron rings by neutralized-cusp injection into a magnetic mirror trap, Plasma Phys. 16 (1974) 113–115. [4] L.D. Landau, E.M. Lifshitz, Electrodynamics of Continuous Medium, People’s Education Press, Beijing, 1963, p. 12. [5] J.D. Jackson, Classical Electrodynamics, second ed., Wiley, New York, 1998, pp. 102–104. [6] B.B. Badijing, N.H. Tuopudijing, Problem in Electrodynamics, People’s Education Press, Beijing, 1964, p. 21. [7] X.Y. Lin, Z.X. Zhang, Key to Problem in Electrodynamics, Science Press, Beijing, 1999, pp. 111–118. [8] J.Z. Zhong, Field of a linear charged ring, Coll. Phys. 1 (7) (1982) 15. [9] C.J. Tranter, Integral Transforms in Mathematic and Physics, Wiley, New York, 1966. [10] I.N. Sneddon, Special Functions of Mathematical and Chemistry, Oliver and Boyd, Edinburg, UK, 1961, p. 145.
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