CHAPTER 2 ELECTROSTATIC POTENTIAL

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1 1 CHAPTER ELECTROSTATIC POTENTIAL 1 Intoduction Imgine tht some egion of spce, such s the oom you e sitting in, is pemeted by n electic field (Pehps thee e ll sots of electiclly chged bodies outside the oom) If you plce smll positive test chge somewhee in the oom, it will expeience foce F E If you ty to move the chge fom point A to point B ginst the diection of the electic field, you will hve to do wok If wok is equied to move positive chge fom point A to point B, thee is sid to be n electicl potentil diffeence between A nd B, with point A being t the lowe potentil If one joule of wok is equied to move one coulomb of chge fom A to B, the potentil diffeence between A nd B is one volt () The dimensions of potentil diffeence e ML T 1 All we hve done so f is to define the potentil diffeence between two points We cnnot define the potentil t point unless we bitily ssign some efeence point s hving defined potentil It is not lwys necessy to do this, since we e often inteested only in the potentil diffeences between points, but in mny cicumstnces it is customy to define the potentil to be zeo t n infinite distnce fom ny chges of inteest We cn then sy wht the potentil is t some neby point Potentil nd potentil diffeence e scl quntities Suppose we hve n electic field E in the positive x-diection (towds the ight) This mens tht potentil is decesing to the ight You would hve to do wok to move positive test chge to the left, so tht potentil is incesing towds the left The foce on is E, so the wok you would hve to do to move it distnce dx to the ight is E dx, but by definition this is lso equl to d, whee d is the potentil diffeence between x nd x + dx d Theefoe E 11 dx In moe genel thee-dimensionl sitution, this is witten E gd i + j + k 1 x x x We see tht, s n ltentive to expessing electic field stength in newtons pe coulomb, we cn eqully well expess it in volts pe mete ( m 1 ) The invese of eqution 11 is, of couse,

2 E dx + constnt 1 Potentil Ne ious Chged Bodies 1 Point Chge Let us bitily ssign the vlue zeo to the potentil t n infinite distnce fom point chge The potentil t distnce fom this chge is then the wok equied to move unit positive chge fom infinity to distnce At distnce x fom the chge, the field stength is The wok equied to δx move unit chge fom x to x + δx is x The wok equied to move unit dx chge fom to infinity is x The wok equied to move unit chge fom infinity to is minus this Theefoe + 1 The mutul potentil enegy of two chges 1 nd septed by distnce is the wok equied to bing them to this distnce pt fom n oiginl infinite seption This is 1 PE + Befoe poceeding, little eview is in ode Field t distnce fom chge : E, N C 1 o m 1 o, in vecto fom, E ˆ N C 1 o m 1

3 Foce between two chges, 1 nd : F 1 N Potentil t distnce fom chge : Mutul potentil enegy between two chges: PE 1 J We couldn t possibly go wong with ny of these, could we? Spheicl Chge Distibutions Outside ny spheiclly-symmetic chge distibution, the field is the sme s if ll the chge wee concentted t point in the cente, nd so, then, is the potentil Thus Inside hollow spheicl shell of dius nd cying chge the field is zeo, nd theefoe the potentil is unifom thoughout the inteio, nd equl to the potentil on the sufce, which is 4 A solid sphee of dius being chge tht is unifomly distibuted thoughout the sphee is esie to imgine thn to chieve in pctice, but, fo ll we know, poton might be like this (it might be but it isn t!), so let s clculte the field t point P inside the sphee t distnce (< ) fom the cente See figue II1 We cn do this in two pts Fist the potentil fom the pt of the sphee below P If the chge is unifomly distibuted thoughout the sphee, this is just Hee is

4 4 the chge contined within dius, which, if the chge is unifomly distibuted thoughout the sphee, is ( / ) Thus, tht pt of the potentil is P FIGURE II1 Next, we clculte the contibution to the potentil fom the chge bove P Conside n elementl shell of dii x, x + δx The chge held by it is 4π x δx x δx δ The contibution to the potentil t P fom the chge 4 π δ x δx in this elementl shell is The contibution to the potentil fom ll x ( ) the chge bove P is x dx Adding togethe the two pts of the potentil, we obtin

5 5 ( ) 5 8πε Long Chged Rod The field t distnce fom long chged od cying chge λ coulombs pe mete λ is πε Theefoe the potentil diffeence between two points t distnces nd b fom the od ( < b) is b λ πε b d λ â b ln( b / ) 6 πε 4 Lge Plne Chged Sheet The field t distnce fom lge chged sheet cying chge σ coulombs pe σ sque mete is Theefoe the potentil diffeence between two points t distnces ε nd b fom the sheet ( < b) is σ b ( b ) 7 ε 5 Potentil on the Axis of Chged Ring The field on the xis of chged ing is given in section 164 The ede is invited to show tht the potentil on the xis of the ing is 1/ ( + ) 8 x You cn do this eithe by integting the expession fo the field o just by thinking bout it fo few seconds nd elizing tht potentil is scl quntity

6 6 6 Potentil in the Plne of Chged Ring We suppose tht we hve ing of dius being chge We shll ty to find the potentil t point in the plne of the ing nd t distnce ( < ) fom the cente of the ing P θ A Conside n element δθ of the ing t P The chge on it is this element of chge is δθ The potentil t A due π 1 δθ 1 δθ π + cos θ π b c cos θ, 9 whee is / b 1 + nd c / The potentil due to the chge on the entie ing π dθ 1 π b c cos θ I cn t immeditely see n nlyticl solution to this integl, so I integted it numeiclly fom to 99 in steps of 1, with the esult shown in the following gph, in which is in units of, nd is in units of

7 The field is equl to the gdient of this nd is diected towds the cente of the ing It looks s though smll positive chge would be in stble equilibium t the cente of the ing, nd this would be so if the chge wee constined to emin in the plne of the ing But, without such constint, the chge would be pushed wy fom the ing if it styed t ll bove o below the plne of the ing Some computtionl notes Any ede who hs tied to epoduce these esults will hve discoveed tht the lot of hevy computtion is equied Since thee is no simple nlyticl expession fo the integtion, ech of the 1 points fom which the gph ws computed entiled numeicl integtion of the expession fo the potentil I found tht Simpson s Rule did not give vey stisfctoy esults, minly becuse of the steep ise in the function t lge, so I used Gussin qudtue, which poved much moe stisfctoy Cn we void the numeicl integtion? One possibility is to expess the integnd in eqution 1 s powe seies in cos θ, nd then integte tem by tem Thus c ( / ) b c cos θ b 1 ecos θ, whee e And then b ( / ) + 1

8 8 1 e cos θ ecos θ e 5 cos 5 8 e θ + cos 6 56 e θ + 6 cos e θ + cos 1 14 θ + e cos 7 e 4 cos θ + 4 θ e 8 cos 8 θ + 11 We cn then integte this tem by tem, using obviously zeo if n is odd We finlly get: ( n 1)!! π!! n π cos θdθ n if n is even, nd (1 + e + e + e + e ) Fo computtionl puposes, this is most efficiently endeed s (1 + e ( + e ( + e ( + e )))) I shll efe to this s Seies I It tuns out tht it is not vey efficient seies, s it conveges vey slowly This is becuse e is not smll fction, nd is lwys gete thn / Thus fo 1, e 8 / We cn do much bette if we cn obtin powe seies in / Conside the expession 1 1, which occus in eqution + cos θ 1 + ( / ) ( / ) cos θ 9 This expession, nd othes vey simil to it, occu quite fequently in vious physicl situtions It cn be expnded by the binomil theoem to give powe seies in / (Admittedly, it is tinomil expession, but do it in stges) The esult is (1+ ( / ) ( / )cos θ) 1/ P (cos θ) + P (cos θ)( 1 ) + P (cos θ)( ) + P (cos θ)( ) + 15 whee the coefficients of the powes of ( ) e polynomils in cos θ, which hve been extensively tbulted in mny plces, nd e clled Legende polynomils See, fo exmple my notes on Celestil Mechnics, Sections 114 nd 511 Ech tem in the Legende polynomils cn then be integted tem by tem, nd the esulting seies, fte bit of wok, is (1 + ( ) + ( ) + ( ) + ( ) )

9 9 Since this is seies in ( ) the thn in e, it conveges much fste thn eqution 1 I shll efe to it s seies II Of couse, fo computtionl puposes it should be witten with nested pentheses, s we did fo seies I in eqution 14 Hee is tble of the esults using fou methods The fist column gives the vlue of / The next fou columns give the vlues of, in units of, clculted by fou methods Column, integtion by Gussin qudtue Column, integtion by Simpson s Rule Column 4, ppoximtion by Seies I Column 5, ppoximtion by seies II In ech cse I hve given the numbe of digits tht I believe to be elible It is seen tht Gussin qudtue gives by f the best esults Seies I is not vey good t ll, while Seies II is lmost s good s Simpson s Rule Of couse ny of these methods is completed lmost instntneously on moden compute, so one my wonde if it is wothwhile spending much time seeking the most efficient solution Tht will depend on whethe one wnts to do the clcultion just once, o whethe one wnts to do simil clcultions millions of times 7 Potentil on the Axis of Chged Disc The field on the xis of chged disc is given in section 165 The ede is invited to show tht the potentil on the xis of the disc is 1/ [( + x ) x] 9 Electon-volts The electon-volt is unit of enegy o wok An electon-volt (e) is the wok equied to move n electon though potentil diffeence of one volt Altentively, n electonvolt is equl to the kinetic enegy cquied by n electon when it is cceleted though potentil diffeence of one volt Since the mgnitude of the chge of n electon is bout C, it follows tht n electon-volt is bout J Note lso tht,

10 1 becuse the chge on n electon is negtive, it equies wok to move n electon fom point of high potentil to point of low potentil Execise If n electon is cceleted though potentil diffeence of million volts, its kinetic enegy is, of couse, 1 Me At wht speed is it then moving? Fist ttempt 1 m v e (Hee e, witten in itlics, is not intended to men the unit electon-volt, but e is the mgnitude of the electon chge, nd is the potentil diffeence (1 6 volts) though which it is cceleted) Thus v e / m With m kg, this comes to v m s 1 Oops! Tht looks wfully fst! We d bette do it popely this time Second ttempt ( γ 1) mc e Some edes will know exctly wht we e doing hee, without explntion Othes my be completely mystified Fo the ltte, the difficulty is tht the speed tht we hd clculted ws even gete thn the speed of light To do this popely we hve to use the fomuls of specil eltivity See, fo exmple, Chpte 15 of the Clssicl Mechnics section of these notes At ny te, this esults in γ 958, whence β 9411 nd v m s 1 4 A Point Chge nd n Infinite Conducting Plne An infinite plne metl plte is in the xy-plne A point chge + is plced on the z-xis t height h bove the plte Consequently, electons will be ttcted to the pt of the plte immeditely below the chge, so tht the plte will cy negtive chge density σ which is getest t the oigin nd which flls off with distnce ρ fom the oigin Cn we detemine σ(ρ)? See figue II

11 11 + h σ FIGURE II ρ Fist, note tht the metl sufce, being conducto, is n equipotentil sufce, s is ny metl sufce The potentil is unifom nywhee on the sufce Now suppose tht, insted of the metl sufce, we hd (in ddition to the chge + t height h bove the xy-plne), second point chge,, t distnce h below the xy-plne The potentil in the xy-plne would, by symmety, be unifom eveywhee Tht is to sy tht the potentil in the xy-plne is the sme s it ws in the cse of the single point chge nd the metl plte, nd indeed the potentil t ny point bove the plne is the sme in both cses Fo the pupose of clculting the potentil, we cn eplce the metl plte by n imge of the point chge It is esy to clculte the potentil t point (z, ρ) If we suppose tht the pemittivity bove the plte is ε, the potentil t (z, ρ) is / 1/ [ ( ) ] [ ( ) ] 41 πε ρ + h z ρ + h + z The field stength E in the xy-plne is / z evluted t z, nd this is E πε 4 h ( ρ + h 4 ) / The D-field is ε times this, nd since ll the lines of foce e bove the metl plte, Guss's theoem povides tht the chge density is σ D, nd hence the chge density is σ h 4 π ( ρ + h ) /

12 1 This cn lso be witten σ h, 44 π ξ whee ξ ρ + h, with obvious geometic intepettion Execise: How much chge is thee on the sufce of the plte within n nnulus bounded by dii ρ nd ρ + dρ? Integte this fom zeo to infinity to show tht the totl chge induced on the plte is 5 A Point Chge nd Conducting Sphee P O I ζ R ξ + FIGURE II A point chge + is t distnce R fom metl sphee of dius We e going to ty to clculte the sufce chge density induced on the sufce of the sphee, s function of position on the sufce We shll be in mind tht the sufce of the sphee is n equipotentil sufce, nd we shll tke the potentil on the sufce to be zeo Let us fist constuct point I such tht the tingles OPI nd PO e simil, with the lengths shown in figue II The length OI is /R Then R / ξ / ζ, o 1 ξ / R ζ 51 This eltion between the vibles ξ nd ζ is in effect the eqution to the sphee expessed in these vibles Now suppose tht, insted of the metl sphee, we hd (in ddition to the chge + t distnce R fom O), second point chge (/R) t I The locus of points whee the potentil is zeo is whee

13 1 1 ξ / R ζ 5 Tht is, the sufce of ou sphee Thus, fo puposes of clculting the potentil, we cn eplce the metl sphee by n imge of t I, this imge cying chge of (/R) Let us tke the line O s the z-xis of coodinte system Let X be some point such tht OX nd the ngle XO θ The potentil t P fom chge + t nd chge (/R) t I is (see figue II4) X O I OI /R + FIGURE II4 1 / R 4 1/ 4 1/ ( cos ) ( / cos / ) 5 πε + R R θ + R θ R The E field on the sufce of the sphee is / evluted t The D field is ε times this, nd the sufce chge density is equl to D Afte some ptience nd lgeb, we obtin, fo point X on the sufce of the sphee R 1 σ 5 4π (X)

14 14 6 Two Semicylindicl Electodes This section equies tht the ede should be fmili with functions of complex vible nd confoml tnsfomtions Fo edes not fmili with these, this section cn be skipped without pejudice to undestnding following chptes Fo edes who e fmili, this is nice exmple of confoml tnsfomtions to solve physicl poblem y A B O x FIGURE II5 We hve two semicylindicl electodes s shown in figue II5 The potentil of the uppe one is nd the potentil of the lowe one is We'll suppose the dius of the cucle is 1; o, wht mounts to the sme thing, we'll expess coodintes x nd y in units of the dius Let us epesent the position of ny point whose coodintes e (x, y) by complex numbe z x + iy 1 z Now let w u + iv be complex numbe elted to z by w i ; 1 tht is, + z 1 + iw z Substitute w u + iv nd z x + iy in ech of these equtions, nd equte 1 iw el nd imginy pts, to obtin

15 15 u y ( 1+ x) + y ; 1 x y v ; 61 ( 1+ x) + y x 1 u v u y u + ( 1 + v) ; u + ( 1 + v) 6 In tht cse, the uppe semicicle ( ) in the xy-plne mps on to the positive u-xis in the uv-plne, nd the lowe semicicle ( ) in the xy-plne mps on to the negtive u-xis in the uv-plne (Figue II6) Points inside the cicle bounded by the electodes in the xy-plne mp on to points bove the u-xis in the uv-plne v B O u A FIGURE II6 In the uv-plne, the lines of foce e semicicles, such s the one shown The potentil goes fom t one end of the semicicle to t the othe, nd so eqution to the semicicul line of foce is w g π 6 o π tn ( v / u) 64 1 The equipotentils ( constnt) e stight lines in the uv-plne of the fom v fu 65

16 16 (You would pefe me to use the symbol m fo the slope of the equipotentils, but in moment you will be gld tht I chose the symbol f) If we now tnsfom bck to the xy-plne, we see tht the eqution to the lines of foce is nd the eqution to the equipotentils is 1 1 x y tn, 66 π y 1 x y fy, 67 o x + y + fy 1 68 Now en't you gld tht I chose f? Those who e hndy with conic sections (see Chpte of Celestil Mechnics) will undestnd tht the equipotentils in the xy-plne e cicles of dii f + 1, whose centes e t (,! f ), nd which ll pss though the points (!1, ) They e dwn s blue lines in figue II7 The lines of foce e the othogonl tjectoies to these, nd e of the fom x + y + gy These e cicles of dii g 1 nd hve thei centes t (,! g) They e shown s dshed ed lines in figue II7 FIGURE II7

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