Chpite 3 Potentiel électiue [18 u 3 mi] DEVOIR : 31, 316, 354, 361, 35 Le potentiel électiue est le tvil p unité de chge (en J/C, ou volt) Ce concept est donc utile dns les polèmes de consevtion d énegie Il est ussi tès utile dns les cicuits électiues, vec les lois de Kichhoff, etc) Nottion: énegie cinétiue K, énegie potentielle U, potentiel V 31 Énegie potentielle électiue De fçon généle, si une chge test se déplce du point u point dns un chmp électiue E, los l difféence d énegie potentielle (due à E) ente ces deux points est U U = ΔW = E dl Au esoin, évise les concepts de tvil et énegie potentielle Si on connit l foce, los une intégle (ci-dessus) pemet d oteni l énegie potentielle P conséuent, si on connit l énegie potentielle, une déivée pemett d oteni l foce: l E (718) ui fit inteveni le gdient, ui eviend à l section 35 Lie l section Attention : ne ps oulie ue l E (39) n est vlide ue pou des chges ponctuelles! Si l souce du chmp n est ps ponctuelle, los il fut utilise l fomule ci-dessus Exemples: 34 IDENTIFY: The wok euied is the chnge in electicl potentil enegy The potons gin speed fte eing elesed ecuse thei potentil enegy is conveted into kinetic enegy () SET UP: Using the potentil enegy of pi of point chges eltive to infinity, U = (1 / )( / ) we hve W = ΔU = U U 1 = 1 e e 1 Fctoing out the e nd sustituting numes gives W = ( 9 1 9 N m /C )( 16 1 19 C) 1 3 1 15 m 1 = 768 1 14 J 1 15 m () SET UP: The potons hve eul momentum, nd since they hve eul msses, they will hve eul speeds nd hence eul kinetic enegy ΔU = K 1 + K = K = 1 mv = mv Solving fo v gives v = ΔU m = 768 1 14 J 167 1 7 kg = 678 16 m/s EVALUATE: The potentil enegy my seem smll (comped to mcoscopic enegies), ut it is enough to give ech poton speed of nely 7 million m/s
311 IDENTIFY: Apply E(3) The net wok to ing the chges in fom infinity is eul to the chnge in potentil enegy The totl potentil enegy is the sum of the potentil enegies of ech pi of chges, clculted fom E(39) SET UP: Let 1 e whee ll the chges e infinitely f pt Let e whee the chges e t the cones of the tingle, s shown in Figue 311 Let c e the thid, unknown chge Figue 311 W = ΔU = (U U 1 ) U 1 = U = U +U c +U c = 1 d ( + c ) Wnt W =, so W = (U U 1 ) gives = U + c = nd c = / = 1 d ( + c ) EVALUATE: The potentil enegy fo the two chges is positive nd fo ech with c it is negtive Thee e two of the, c tems so must hve c < 3 Potentiel électiue Le potentiel électiue est l énegie potentielle p unité de chge Autement dit, l énegie potentielle U d une chge test en un point est otenue en multiplint l chge p l vleu du potentiel V en ce point En ce sens, le V est à U ce ue E est à F E L E (317) est l plus généle Attention: Les Es (314) à (315) ne sont vlides ue pou des chges souces ponctuelles Lie les exemples ux pp 791 à 794 Autes exemples : W 314 IDENTIFY: = V Fo point chge, V = k SET UP: Ech vcnt cone is the sme distnce, m, fom ech point chge Tking the oigin t the cente of the sue, the symmety mens tht the potentil is the sme t the two cones not occupied y the +5 µc chges This mens tht no net wok is done is moving fom one cone to the othe EVALUATE: If the chge moves long digonl of the sue, the electicl foce does positive wok fo pt of the pth nd negtive wok fo nothe pt of the pth, ut the net wok done is zeo
i 31 IDENTIFY: V = 1 i i SET UP: The loctions of the chnges nd points A nd B e sketched in Figue 31 () V A = 1 1 + A1 () V B = 1 1 + B1 A Figue 31 V A = (8988 1 9 N m /C ) +4 1 9 C 5 m + 65 1 9 C = 737 V 5 m B V B = (8988 1 9 N m /C ) +4 1 9 C 8 m + 65 1 9 C = 74 V 6 m (c) IDENTIFY nd SET UP: Use E(313) nd the esults of pts () nd () to clculte W W B A = ʹ (V B V A ) = (5 1 9 C)( 74 V ( 737 V)) = +8 1 8 J EVALUATE: The electic foce does positive wok on the positive chge when it moves fom highe potentil (point B) to lowe potentil (point A) 33 IDENTIFY: Fo point chge,v = k The totl potentil t ny point is the lgeic sum of the potentils of the two chges SET UP: () The positions of the two chges e shown in Figue 33 () V = k k( ) + = (c) The potentil long the x-xis is lwys zeo, so gph would e flt (d) If the two chges e intechnged, then the esults of () nd (c) still hold The potentil is zeo EVALUATE: The potentil is zeo t ny point on the x-xis ecuse ny point on the x-xis is euidistnt fom the two chges Figue 33
34 IDENTIFY: Fo point chge,v = k The totl potentil t ny point is the lgeic sum of the potentils of the two chges SET UP: Conside the distnces fom the point on the y-xis to ech chge fo the thee egions y (etween the two chges), y > (ove oth chges) nd y < (elow oth chges) () y < :V = y < :V = k ( + y) k ( y) = ky y y > :V = k ( + y) k y = k y k ( + y) k ( y + ) = k y A genel expession vlid fo ny y isv = k y + y + () The gph of V vesus y is sketched in Figue 34 (c) y >> :V = k y k y (d) If the chges e intechnged, then the potentil is of the opposite sign EVALUATE: V = t y = V + s the positive chge is ppoched nd V s the negtive chge is ppoched Figue 34 37 IDENTIFY: K + V = K + V SET UP: Let point e t the cthode nd let point e t the node K = V = 95 V An electon hs = e nd m = 911 1 31 kg K = (V ) = (16 1 19 C)( 95 V) = 47 1 17 J K = 1 mv, so v = (47 1 17 J) 911 1 31 kg = 11 17 m s EVALUATE: The negtively chged electon gins kinetic enegy when it moves to highe potentil 39 () IDENTIFY nd SET UP: The diection of E is lwys fom high potentil to low potentil so point is t highe potentil () Apply E(317) to elte V to E V = E dl = Edx = E(x x ) E = V +4 V = = 8 V/m x x 9 m 6 m (c) W = (V ) = ( 1 6 C)(+4 V) = 48 1 5 J
EVALUATE: The electic foce does negtive wok on negtive chge when the negtive chge moves fom high potentil (point ) to low potentil (point ) 331 IDENTIFY nd SET UP: Apply consevtion of enegy, E(33) Use E(31) to expess U in tems of V () K 1 + V 1 = K + V (V 1 V ) = K K 1 ; = 16 1 19 C K 1 = 1 m v = 499 1 18 J; K e 1 = 1 m v = 915 1 17 J e V 1 V = K K 1 = 156 V EVALUATE: The electon gins kinetic enegy when it moves to highe potentil () Now K 1 = 915 1 17 J, K = EVALUATE: V 1 V = K K 1 = +18 V The electon loses kinetic enegy when it moves to lowe potentil 33 Clculs de potentiels Section ptiue Lie l encdé du s de l p 794 et l exemple 38 Lie les exemples de pp 796 à 798 Autes exemples : 334 IDENTIFY: Exmple 31 shows tht fo line of chge,v = λ πp ln( / ) Apply consevtion of enegy to the motion of the poton SET UP: Let point e 18 cm fom the line nd let point e t the distnce of closest ppoch, whee K = () K = 1 mv = 1 (167 1 7 kg)(15 1 3 m/s) = 188 1 1 J () K + V = K + V V = K K ln( / ) = πp λ ( 1175 V) = exp πp ( 1175 V) λ = 188 1 1 J 16 1 19 C = 1175 V = (18 m)exp πp (1175 V) 5 1 1 C/m = 158 m EVALUATE: The potentil inceses with decesing distnce fom the line of chge As the positively chged poton ppoches the line of chge it gins electicl potentil enegy nd loses kinetic enegy 337 IDENTIFY: Fo points outside the cylinde, its electic field ehves like tht of line of chge Since voltmete eds potentil diffeence, tht is wht we need to clculte SET UP: The potentil diffeence is ΔV = λ ln( πp / ) () Sustituting numes gives ΔV = 1 cm ( )ln 6 cm λ ln( πp / ) = ( 85 1 6 C/m) 9 1 9 N m /C
ΔV = 78 1 4 V = 78, V = 78 kv () E = inside the cylinde, so the potentil is constnt thee, mening tht the voltmete eds zeo EVALUATE: Cution! The fct tht the voltmete eds zeo in pt () does not men tht V = inside the cylinde The electic field is zeo, ut the potentil is constnt nd eul to the potentil t the sufce 341 IDENTIFY nd SET UP: Use the esult of Exmple 39 to elte the electic field etween the pltes to the potentil diffeence etween them nd thei seption The foce this field exets on the pticle is given y E(13) Use the eution tht pecedes E(317) to clculte the wok () Fom Exmple 39, E = V d = 36 V = 8 V/m 45 m () F = E = (4 1 9 C)(8 V/m) = +19 1 5 N (c) The electic field etween the pltes is shown in Figue 341 Figue 341 The plte with positive chge (plte ) is t highe potentil The electic field is diected fom high potentil towd low potentil (o, E is fom + chge towd chge), so E points fom to Hence the foce tht E exets on the positive chge is fom to, so it does positive wok W = F dl = Fd, whee d is the seption etween the pltes W = Fd = (19 1 5 N)(45 m) = +864 1 7 J (d) V = +36 V (plte is t highe potentil) ΔU = U U = (V ) = (4 1 9 C)( 36 V) = 864 1 7 J EVALUATE: We see tht W = (U U ) = U U 344 IDENTIFY: Exmple 38 shows tht the potentil of solid conducting sphee is the sme t evey point inside the sphee nd is eul to its vlue V = / πp R t the sufce Use the given vlue of E to find SET UP: Fo negtive chge the electic field is diected towd the chge Fo points outside this spheicl chge distiution the field is the sme s if ll the chge wee concentted t the cente E = nd = E (38 N/C)( m) = = 169 1 8 C 899 1 9 N m /C Since the field is diected inwd, the chge must e negtive The potentil of point chge, tking s zeo, is V = = (899 19 N m /C )( 169 1 8 C) = 76 V t the sufce of the m sphee Since the chge ll esides on the sufce of conducto, the field inside the sphee due to this symmeticl distiution is zeo No wok is theefoe done in moving test chge fom just inside the sufce to the cente, nd the potentil t the cente must lso e 76 V EVALUATE: Inside the sphee the electic field is zeo nd the potentil is constnt 34 Sufces éuipotentielles L nlogie des coues de contou (Fig 33) ide à voi une compéhension intuitive P exemples, des coues ppochées coespondent à une pente plus ide; dns le contexte électiue, celà signifie E plus élevé
Lie pidement l section, ui est plutôt de ntue conceptuelle 35 Chmp électiue comme gdient du potentiel Si V est connu, los E (3) donne E Si V (et E) est dil, los on utilise E (33) Lie les deux exemples Autes exemples: 347 IDENTIFY nd SET UP: Use E(319) to clculte the components of E () E x = V x V = Axy Bx + Cy = Ay + Bx Ey = V y = Ax C E z = V z = () E = euies tht E x = E y = E z = E z = eveywhee E y = t x = C/A And E x is lso eul zeo fo this x, ny vlue of z, nd y = Bx/A = (B/A)( C/A) = BC/A EVALUATE: 348 IDENTIFY: Apply E(119) V doesn t depend on z so E z = eveywhee SET UP: E(17) sys E = 1 Similly, E y = kqy 3 () E x = V x = x () Fom pt (), E = kq EVALUATE: nd E z = kqz 3 ˆ is the electic field due to point chge kq x + y + z = kqx (x + y + z ) = kqx 3 3 xî + yĵ + z ˆk = kq ˆ, which gees with Eution (17) V is scl E is vecto nd hs components 349 IDENTIFY nd SET UP: Fo solid metl sphee o fo spheicl shell, V = k outside the sphee nd V = k R t ll points inside the sphee, whee R is the dius of the sphee When the electic field is dil, E = V () (i) < : This egion is inside oth spheesv = k k (ii) < < : This egion is outside the inne shell nd inside the oute shell V = k k = k 1 1 = k 1 1 (iii) > : This egion is outside oth sphees nd V = since outside
sphee the potentil is the sme s fo point chge Theefoe the potentil is the sme s fo two oppositely chged point chges t the sme loction These potentils cncel () V = 1 ndv =, sov = 1 1 1 (c) Between the sphees < < ndv = k 1 1 E = V = 1 1 = + 1 = V 1 1 1 (d) Fom Eution (33): E =, since V is constnt (zeo) outside the sphees (e) If the oute chge is diffeent, then outside the oute sphee the potentil is no longe zeo ut is V = 1 1 Q = 1 ( Q) All potentils inside the oute shell e just shifted y n mount V = 1 Q Theefoe eltive potentils within the shells e not ffected Thus () nd (c) do not chnge Howeve, now tht the potentil does vy outside the sphees, thee is n electic field thee: E = V = k + kq = k 1 Q = k ( Q) EVALUATE: In pt () the potentil is gete thn zeo fo ll <