Chpte 4. Enegy nd Ptentil Hyt; 0/5/009; 4-4. Enegy Expended in Mving Pint Chge in n Electic Field The electic field intensity is defined s the fce n unit test chge. The fce exeted y the electic field n chge Q, F = QE E The diffeentil wk dne y the field t mve the chge in the diectin dl, dw = F dl E E The diffeentil wk equied y us t mve the chge y dl. We shuld pply fce ginst the fce y the electic field. dw = F dl dw = Q E dl ( E ) The wk equied t mve the chge finite distnce, W = Q E dl pth 4. The Line Integl Assume E = cnstnt f simplicity. The pth is divided int six segments. The wk equied t mve chge Q fm B t A, W Q E Δ L +... + E ΔL ( 6 6) F unifm electic field, E = E = E =.. = E6, W QE Δ L +... +ΔL ( 6) L BA, vect fm B t A The exct mthemticl expessin A A W = Q E dl W = QE dl B B E = cnstnt W = QE L Dependent n the initil nd the finl pints ut nt the pth. BA
Hyt; 0/5/009; 4-
Hyt; 0/5/009; 4-3 Diffeentil length vect dl = dx ˆ ˆ ˆ x + dy y + dz z dl = dˆ d ˆ + + dz ˆz dl = d ˆ + dθ ˆ + sinθdˆ θ : Rectngul Cdintes : Cylindicl Cdintes : Spheicl Cdintes Exmple An infinite line chge It is well knwn y nw L E = Eˆ = ˆ πε The diffeentil length vect n cicle f dius. dl = dˆ ˆ + d + dz ˆz =0 =0 The wk needed t mve chge Q ut the cicul pth. finl finl L L W = Q ˆ ˆ ˆ ˆ d Q d 0 initil initil πε πε Mve the chge fm = t = dl = d ˆ finl = L L d Q L W = Q ˆ d ˆ Q ln initil πε = πε πε Minus sign mens tht we eceive enegy fm the electic field. Mve the chge fm = t = When chge is mved in the diectin f decesing cdinte vlue, this is tken ce f y the limits f the integl nt y the sign f dl. dl = d ˆ finl = L L d Q L W = Q ˆ d ˆ Q ln initil πε + = πε πε
4.3 Definitin f Ptentil Diffeence nd Ptentil The ptentil diffeence V is the wk dne y mving unit psitive chge fm ne pint t nthe in the electic field. The ptentil diffeence etween tw pints A nd B. A V = E dl : B is ften tken t infinity AB B The unit f ptentil is jules pe culm vlt. Exmple The wk dne in tking chge Q fm Q L W = ln πε The ptentil diffeence etween W L V = = ln Q πε = t = Hyt; 0/5/009; 4-4 = t = in the pesence f line chge. Exmple The ptentil diffeence etween = A t = B in the pesence f pint chge. The electic field is Q E = E ˆ ˆ = The diffeentil length vect dl = d ˆ A = A Q Q VAB = E dl d B B = A B Ptentil slute ptentil is used when ze-efeence pint is defined. The ze efeence pint is tken t infinity in cnventin. V = V V AB A B Ptentil diffeence Ptentil t A Ptentil t B 4.4 The Ptentil Field f Pint Chge The ptentil diffeence etween A nd B A = A Q Q VAB = E dl d B B = A B Only -cmpnent. Independent f θ nd, nd theefe f the pth. = d ˆ + dθ ˆ + sinθ dˆ θ
Hyt; 0/5/009; 4-5 Ze efeence pint t infinity Let B, then the ptentil t A is V A Q = A Since A is ity, the ptentil t ny pint in spce is Q V = πε 4 Equiptentil sufce The sme vlue f ptentil n the equiptentil sufce. N wk is needed t mve chge n the equiptentil sufce. The equiptentil sufces in the ptentil field f pint chge e sphees. 4.5 The Ptentil Field f System f Chges: Cnsevtive Ppety The ptentil is defined s wk needed t ing unit chge fm the infinity t the pint. The ptentil is independent f the pth. This is why the ptentil is vey useful cncept. Ptentil fm system f chges () A pint chge Q lcted t The ptentil t pint Q V ( ) = () Pint chges Q nd Q lcted t nd, espectively. The ptentil t Q Q V ( ) = + (3) Similly, the ptentil ising fm N pint chges is N Q Qn Qm V ( ) = +... + = n m = m (4) The ptentil fm vlume chge density v v ( ) Δv ( ) ( ) ( ) v n Δvn As Δv...... Δv n v dv V = + + (8) vlume 4 πε n (5) The ptentil fm line chge L ( ) dl V ( ) = (9) cuve 4 πε (6) The ptentil fm sufce chge S ( ) ds V ( ) = (0) sufce 4 πε
Exmple The ptentil fm line chge in the fm f ing dl = d ˆ, = zˆz, = ˆ, = + z Hyt; 0/5/009; 4-6 Using (9) π Ld L V = + z ε + z 0 Summy. The ptentil is the wk dne in cying unit psitive chge fm infinity t the pint. The wk is independent f the pth chsen.. The ptentil fm nume f pint chges is the sum f individul ptentils fm ech chge. The pth independence cn e stted s dl = 0 : Clsed pth integl f E is ze. E is cnsevtive field. C E P P Kichhff s vltge lw: The lgeic sum f vltge dps und ny clsed cicuit is ze.
4.6 Ptentil Gdient The ptentil cn e tined in tw methds. () Using line integl f the electic field. () Using vlume integl ve chge distiutin. (3) Find the ptentil fm the undy cnditin. Hyt; 0/5/009; 4-7 The mthemticl definitin f the ptentil V = E dl Fig. 4.6 Electic field stemlines Fig. 4.7 Equiptentil sufces F diffeentil length ΔL, ΔV E ΔL E ΔLcsθ ΔV ΔL =E csθ The mximum vlue is tined when csθ = ΔV = E ΔL mx Using vect nttin ΔV ΔV E = ˆ ˆ N ΔL ΔN mx N Since the mx. ccus lng ˆN ΔL is in ppsite diectin t E, ˆN : A unit vect nml t the equiptentil sufce in the diectin f highe ptentil Summy. E is the mximum vlue f the te f chnge f V with distnce.. E is ppsite t the diectin f the mst pidly incesing ptentil. E is pependicul t the equiptentil sufce. Δ V = E Δ L = 0 cnditin f equiptentil sufce Tw vects e pependicul
Hyt; 0/5/009; 4-8 Gdient f scl field The gdient f scl field T is defined s dt Gdient f T = gd T ˆN dn is unit vect nml the equiptentil sufce in the diectin f incesing T. ˆN The gdient f T is the mximum spce te f chnge f T in the diectin f incesing T. Reltin etween E nd V E =gdv Since V is functin f psitin, V V V dv = dx + dy + dz x y z Fm the definitin f the ptentil dv = E dl = E dx + E dy + E dz ( x y z ) = Eˆ ˆ ˆ x x + E y y + E z z = dx ˆ + dy ˆ + dz ˆ x y z Fm () nd (c) V x, V E = Ey =, E V z = x y z V V V E = ˆ ˆ ˆ x + y + z x y z () () (c) (d) Fm () nd (d) V ˆ V gd V = + ˆ + V ˆ x y z x y z del pet It is defined s = ˆ + ˆ + ˆ x y z T T T Its usge, T = ˆx + ˆy + ˆz x y z Theefe E = V x y z The gdient in the cdinte systems V ˆ V V = x + ˆ V y + ˆz x y z V V V V = ˆ ˆ + + ˆz z V V V V = ˆ ˆ ˆ + θ + θ sinθ gd T = T Rectngul cdintes Cylindicl cdintes Spheicl cdintes
Hyt; 0/5/009; 4-9 4.7 The Diple An electic diple is tw pint chges f equl mgnitude nd ppsite sign septed y smll distnce. The electic field cn e fund y using Culm s lw y using the negtive gdient f the ptentil. The ttl ptentil fm tw chges, Q Q R R V = = R R 4 πε RR Using the ppximtin Qdcsθ V = πε 4 f R R d csθ (e) Nte tht z=0 plne θ = 90 plne is t ze ptentil Tking gdient in spheicl cdintes Qd E = 3 ( csθ ˆ + sinθ ˆ θ ) πε 4
Plt f equiptentil sufces Qd / =, then V = cs θ /. : Eq. f the equiptentil sufces Let ( ) Hyt; 0/5/009; 4-0 Blck, electic field stemlines; Red, equiptentil lines. Eq. f the electic field stemlines. In = cnstnt plne, Eθ d sin E = θ θ d csθ d = ct θ dθ = C sin θ Diple mment The diple mment p is vect quntity. p = Qd : d is the chge septin diected fm Q t +Q. The ptentil fm the electic diple. Using the definitin f diple mment in (e) p ˆ V = Genelize V = p 4 πε is field pint, is the cente f the diple. 3 We nte E ~/. It flls ff fste thn tht f pint chge. At gete distnces the electic diple lks t hve n chge t ll.
4.8 Enegy Density in the Electsttic Field Binging chge t ne nthe chge equies wk. This wk is sted s the ptentil enegy in the system f tw chges. Hyt; 0/5/009; 4- Binging chge Q fm infinity t ny psitin. N wk needed since thee is n electic field. Binging the secnd chge Q in the field f Q The wk dne = Q V,. The ptentil t the psitin f Q pduced y Q. Binging the thid chge Q3 in the field pduced y Q nd Q. The wk dne = Q3 V 3, + Q3 V 3, The ttl ptentil enegy f the field W E= Q V, +(Q3 V 3, + Q3 V 3,)+ (Q4 V 4, + Q4 V 4, + Q4 V 4,3 )+ () Nte, f exmple, Q Rewite Q3 QV = 3 3, Q3 Q QV R,3 3 is scl R R = () 3 3 Using () in () W E = Q V, +(Q V,3 + Q V,3 )+ (Q V,4 + Q V,4 + Q3 V 3,4 )+ (c) ()+(c) W E = Q (V, + V,3 + V,4 + ) + Q (V, + V,3 + V,4 + ) + Q3 (V 3, + V 3, + V 3,4 + ) + = V 3, Ptentil due t ll chges t the psitin f Q3 N W QV Q V Q V Q V = ( + + 3 3 +...) E m m m = The enegy sted in egin f cntinuus chge distiutin WE = vvdv vl Use the mthemticl identity, ( VD) V( D) + D ( V) W ( D ) Vdv E ( VD ) D = ( V ) Vdv vl vl Using the divegence theem W E ( ) ( ) VD ds C D = V dv vl /, /, E W E D Edv = = ε vl E dv vl (45)
Exmple Enegy sted in the electic field f cxil cle f length L. Fm sectin 3.3 S S D = ˆ nd E = ˆ ε Hyt; 0/5/009; 4- The sted enegy is L π S πl S WE = ln 0 0 d d dz ε = ε ε Find the ptentil diffeence etween the inne nd the ute cnduct. Let the ute cnduct e the ze-ptentil efeence. S S V = E d d ln = = ε ε The enegy sted is L π ln ( / S S ) WE = SVdS = ln π S 0 0 S d dz L ε ε : The sme esult Using the ttl chge, Q = π LS, WE = QV : The sme s the enegy sted in cpcit. Vlume enegy density Fm (45) WE = D Edv vl Diffeentil vlue dwe = D Edv dw E D E dv =