r = (0.250 m) + (0.250 m) r = m = = ( N m / C )

Transcription

1 ELECTIC POTENTIAL IDENTIFY: Apply Eq() to clculte the wok The electic potentil enegy of pi of point chges is given y Eq(9) SET UP: Let the initil position of q e point nd the finl position e point, s shown in Figue 5 m (5 m) + (5 m) 56 m Figue EXECUTE: W U U 6 6 qq 9 ( + 4 C)( 4 C) U (8988 N m / C ) P 5 m U 684 J 6 6 qq 9 ( + 4 C)( 4 C) U (8988 N m / C ) P 56 m U 6 J W U U 684 J ( 6 J) 56 J EVALUATE: The ttctive foce on q is towd the oigin, so it does negtive wok on q when q moves to lge IDENTIFY: Apply W U U SET UP: 8 U + 54 J Solve fo U 8 EXECUTE: W 9 J U U U U W 9 J ( 54 J) 7 J EVALUATE: When the electic foce does negtive wok the electicl potentil enegy inceses IDENTIFY: The wok needed to ssemle the nucleus is the sum of the electicl potentil enegies of the potons in the nucleus, eltive to infinity SET UP: The totl potentil enegy is the scl sum of ll the individul potentil enegies, whee ech potentil enegy is U (/ 4 πp)( qq/ ) Ech chge is e nd the chges e equidistnt fom ech othe, so the totl e e e e potentil enegy is U + + P P EXECUTE: Adding the potentil enegies gives 9 9 e (6 C) (9 N m /C ) U 46 J 6 MeV 5 P m EVALUATE: lot of enegy This is smll mount of enegy on mcoscopic scle, ut on the scle of toms, MeV is quite -

2 - Chpte 4 IDENTIFY: The wok equied 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 (/ 4 πp)( qq/ ) we hve e e W Δ U U U P EXECUTE: Fctoing out the e nd sustituting numes gives 9 9 ( )( ) 4 W 9 N m /C 6 C 768 J 5 5 m m () SET UP: The potons hve equl momentum, nd since they hve equl msses, they will hve equl speeds nd hence equl kinetic enegy Δ U K+ K K mv mv 4 Δ U 768 J EXECUTE: Solving fo v gives v m/s 7 m 67 kg 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 5 () IDENTIFY: Use consevtion of enegy: K + U + Wothe K + U U fo the pi of point chges is given y Eq(9) SET UP: Let point e whee q is 8 m fom q nd point e whee q is 4 m fom q, s shown in Figue 5 Figue 5 qq EXECUTE: Only the electic foce does wok, so W othe nd U P U K mv (5 kg)( m/s) 6 J 6 6 qq 9 ( 8 C)( 78 C) (8988 N m /C ) J P 8 m K 6 6 qq 9 ( 8 C)( 78 C) U (8988 N m /C ) J P 4 m The consevtion of enegy eqution then gives K K + ( U U ) mv mv 6 J (454 J 497 J) 77 J + + v (77 J) 5 m/s 5 kg EVALUATE: The potentil enegy inceses when the two positively chged sphees get close togethe, so the kinetic enegy nd speed decese () IDENTIFY: Let point c e whee q hs its speed momentily educed to zeo Apply consevtion of enegy to points nd c: K + U + Wothe K + U c c

3 Electic Potentil - SET UP: Points nd c e shown in Figue 5 EXECUTE: K + 6 J (fom pt ()) U J (fom pt ()) Figue 5 K c (t distnce of closest ppoch the speed is zeo) qq Uc P c qq Thus consevtion of enegy K + U Uc gives + 6 J J 684 J P c 6 6 qq 9 ( 8 C)( 78 C) c (8988 N m /C ) m P 684 J J EVALUATE: U s so q will stop no mtte wht its initil speed is 6 IDENTIFY: qq Apply U k nd solve fo SET UP: 6 q 7 C 6, q + C EXECUTE: kqq (899 N m /C )( 7 C)( + C) 7 m U 4 J EVALUATE: The potentil enegy U is scl nd cn tke positive nd negtive vlues 7 () IDENTIFY nd SET UP: U is given y Eq(9) EXECUTE: qq U 4επ ( + 46 C)( + C) U (8988 N m /C ) + 98 J 5 m EVALUATE: The two chges e oth of the sme sign so thei electic potentil enegy is positive () IDENTIFY: Use consevtion of enegy: K + U + Wothe K + U SET UP: Let point e whee q is elesed nd point e t its finl position, s shown in Figue 7 Figue 7 Only the electic foce does wok, so W othe nd (i) 5 m EXECUTE: K (elesed fom est) U + 98 J (fom pt ()) K mv U P qq 6 6 qq 9 ( + 46 C)( + C) U (8988 N m /C ) + 99 J P 5 m Then K + U + Wothe K + U gives K U U nd mv U U nd ( U U) ( + 98 J 99 J) v 66 m/s 4 m 8 kg (ii) 5 m is now ten times lge thn in (i) so U is ten times smlle: U + 99 J / + 99 J ( U U) ( + 98 J 99 J) v 67 m/s 4 m 8 kg

4 -4 Chpte (iii) 5 m is now ten times lge thn in (ii) so U is ten times smlle: U + 99 J/ + 99 J ( U U) ( + 98 J 99 J) v 75 m/s 4 m 8 kg EVALUATE: The foce etween the two chges is epulsive nd povides n cceletion to q This cuses the speed of q to incese s it moves wy fom Q 8 IDENTIFY: Cll the thee chges, nd U U + U + U SET UP: U U U ecuse the chges e equl nd ech pi of chges hs the sme seption, 5 m 6 kq k( C) EXECUTE: U 78 J 5 m 5 m EVALUATE: When the thee chges e ought in fom infinity to the cones of the tingle, the epulsive electicl foces etween ech pi of chges do negtive wok nd electicl potentil enegy is stoed qq qq qq U k + + SET UP: In pt (), m, m nd m In pt () let pticle hve coodinte x, so 9 IDENTIFY: m, EXECUTE: () x nd x (4nC)( nc) (4nC)(nC) ( nc)(nc) U k + + ( m) ( m) ( m) 7 6 J qq qq qq () If U, then k + + Solving fo x we find: x x x 6x 6 x 74 m, 6 m x x + Theefoe, x 74 m since it is the only vlue etween the two chges EVALUATE: U is positive nd oth U nd U e negtive If U, then U U + U Fo x 74 m, U + 97 J, U 4 J nd U 54 J It is tue tht U t this x IDENTIFY: The wok done on the lph pticle is equl to the diffeence in its potentil enegy when it is moved fom the midpoint of the sque to the midpoint of one of the sides SET UP: We pply the fomul W U U In this cse, is the cente of the sque nd is the midpoint of one of the sides Theefoe W U U cente side cente side Thee e 4 electons, so the potentil enegy t the cente of the sque is 4 times the potentil enegy of single lph-electon pi At the cente of the sque, the lph pticle is distnce 5 nm fom ech electon At the midpoint of the side, the lph is distnce 5 nm fom the two neest electons nd distnce 5 nm fom the two most distnt electons Using the fomul fo the potentil enegy (eltive to infinity) of two point chges, U (/ 4 πp)( qq/ ), the totl wok is qq e qqe qq α α α e Wcente side Ucente Uside 4 + P P P Sustituting q e e nd q α e nd simplifying gives Wcente side 4e + P EXECUTE: Sustituting the numeicl vlues into the eqution fo the wok gives ( ) W + 5 m 5 nm 5 nm C 68 J EVALUATE: Since the wok is positive, the system hs moe potentil enegy with the lph pticle t the cente of the sque thn it does with it t the midpoint of side???

5 Electic Potentil -5 IDENTIFY: Apply Eq() The net wok to ing the chges in fom infinity is equl to the chnge in potentil enegy The totl potentil enegy is the sum of the potentil enegies of ech pi of chges, clculted fom Eq(9) SET UP: Let 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 Let q c e the thid, unknown chge Figue W Δ U ( U U ) EXECUTE: Wnt W, so W ( U U) gives U U U U + U + U q + qq ( ) c c P c d ( q qq c ) P d + q + qq c nd qc q/ EVALUATE: The potentil enegy fo the two chges q is positive nd fo ech q with q c it is negtive Thee e two of the q, q c tems so must hve qc < q IDENTIFY: Use consevtion of enegy U + K U + K to find the distnce of closest ppoch The qq mximum foce is t the distnce of closest ppoch, F k 7 SET UP: K Initilly the two potons e f pt, so U A poton hs mss 67 kg nd chge 9 q+ e+ 6 C EXECUTE: K U qq e ( mv ) k mv k nd 9 9 ke (899 N m /C )(6 C) 8 m 7 6 mv (67 kg)( m/s) 9 e 9 (6 C) F k (899 N m /C ) N (8 C) EVALUATE: The cceletion F/ mof ech poton poduced y this foce is extemely lge IDENTIFY: E W points fom high potentil to low potentil V V q SET UP: The foce on positive test chge is in the diection of E EXECUTE: V deceses in the estwd diection A is est of B, so VB > VA C is est of A, so VC < VA The foce on positive test chge is est, so no wok is done on it y the electic foce when it moves due south (the foce nd displcement e pependicul), nd VD VA EVALUATE: The electic potentil is constnt in diection pependicul to the electic field 4 IDENTIFY: W V V q kq V SET UP: Ech vcnt cone is the sme distnce, m, fom ech point chge EXECUTE: Tking the oigin t the cente of the sque, 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 q moves long digonl of the sque, 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

6 -6 Chpte 5 IDENTIFY nd SET UP: Apply consevtion of enegy to points A nd B EXECUTE: K A + UA KB + UB U qv, so K A + qva KB + qvb K K q V V 6 B A + ( A B) 5 J + ( 5 C)( V 8 V) 55 J v K / m 74 m/s B B EVALUATE: It is fste t B; negtive chge gins speed when it moves to highe potentil 6 IDENTIFY: The wok-enegy theoem sys W K W K V V q SET UP: Point is the stting nd point is the ending point Since the field is unifom, W Fscosφ E q scos φ The field is to the left so the foce on the positive chge is to the left The pticle moves to the left so φ nd the wok W is positive 6 6 EXECUTE: () W K K 5 J 5 J 6 W 5 J () V V 57 V Point is t highe potentil thn point 9 q 4 C W V V 57 V (c) Eqs W, so E 595 V/m qs s 6 m EVALUATE: A positive chge gins kinetic enegy when it moves to lowe potentil; V < V 7 IDENTIFY: Apply the eqution tht pecedes Eq(7): W q E d l SET UP: Use coodintes whee + y is upwd nd + x is to the ight Then E Eˆ 4 j with E 4 N/C () The pth is sketched in Figue 7 Figue 7 EXECUTE: E dl ( Eˆj) ( dxiˆ) so W q E d l EVALUATE: The electic foce on the positive chge is upwd (in the diection of the electic field) nd does no wok fo hoizontl displcement of the chge () SET UP: The pth is sketched in Figue 7 d l dyˆj Figue 7 EXECUTE: E dl ( Eˆj) ( dyˆj) Edy W q d qe dy qe ( y y E l ) y y + 67 m, positive since the displcement is upwd nd we hve tken + y to e upwd W qe y y ( ) ( + 8 C)(4 N/C)( + 67 m) + 75 J EVALUATE: The electic foce on the positive chge is upwd so it does positive wok fo n upwd displcement of the chge

7 Electic Potentil -7 (c) SET UP: The pth is sketched in Figue 7c y y sin θ (6 m)sin m The veticl component of the 6 m displcement is 88 m downwd Figue 7c EXECUTE: dl dxi+ ˆ dyˆj (The displcement hs oth hoizontl nd veticl components) E dl ( Eˆj) ( dxi ˆ+ dyˆj) Edy (Only the veticl component of the displcement contiutes to the wok) W q d qe dy qe ( y y E l ) 9 4 ( ) ( + 8 C)(4 N/C)( 88 m) 6 J W qe y y EVALUATE: The electic foce on the positive chge is upwd so it does negtive wok fo displcement of the chge tht hs downwd component 8 IDENTIFY: Apply K + U K + U SET UP: Let q + nc nd q + nc At point, 5 m At point, m nd 4 m The electon hs q end m e 9 kg K since the electon is elesed fom est keq keq keq keq EXECUTE: + mv e ( C) ( C) 7 E K + U k( 6 C) + 88 J 5 m 5 m 9 9 ( C) ( C E K + U k( 6 C) + + mv e 54 J + mv e m 4 m Setting E Egives v (54 J 88 J) 689 m s 9 kg 9 7 EVALUATE: V V + V 8 V V V + V 5 V V > V The negtively chged electon gins kinetic enegy when it moves to highe potentil kq 9 IDENTIFY nd SET UP: Fo point chge V Solve fo 9 kq (899 N m /C )(5 C) EXECUTE: () 5 m 5 mm V 9 V V 9 V () V kq constnt so V V (5 mm) 75 mm V V EVALUATE: The potentil of positive chge is positive nd deceses s the distnce fom the point chge inceses IDENTIFY: The totl potentil is the scl sum of the individul potentils, ut the net electic field is the vecto sum of the two fields SET UP: The net potentil cn only e zeo if one chge is positive nd the othe is negtive, since it is scl The electic field cn only e zeo if the two fields point in opposite diections EXECUTE: () (i) Since oth chges hve the sme sign, thee e no points fo which the potentil is zeo (ii) The two electic fields e in opposite diections only etween the two chges, nd midwy etween them the fields hve equl mgnitudes So E midwy etween the chges, ut V is neve zeo () (i) The two potentils hve equl mgnitude ut opposite sign midwy etween the chges, so V midwy etween the chges, ut E thee since the fields point in the sme diection (ii) Between the two chges, the fields point in the sme diection, so E cnnot e zeo thee In the othe two egions, the field due to the nee chge is lwys gete thn the field due to the moe distnt chge, so they cnnot cncel Hence E is not zeo nywhee EVALUATE: It does not follow tht the electic field is zeo whee the potentil is zeo, o tht the potentil is zeo whee the electic field is zeo

8 -8 Chpte IDENTIFY: qi V P i i SET UP: The loctions of the chnges nd points A nd B e sketched in Figue EXECUTE: () q q VA + P A A V A q q () VB + P B B V B Figue C 65 C (8988 N m /C ) + 77 V 5 m 5 m C 65 C (8988 N m /C ) + 74 V 8 m 6 m (c) IDENTIFY nd SET UP: Use Eq() nd the esults of pts () nd () to clculte W 9 8 EXECUTE: W q( V V ) (5 C)( 74 V ( 77 V)) + 8 J B A B A EVALUATE: The electic foce does positive wok on the positive chge when it moves fom highe potentil (point B) to lowe potentil (point A) kq IDENTIFY: Fo point chge, V 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 + x Figue q EXECUTE: () V P q q (c) V( x) P P + x

9 Electic Potentil -9 (d) The gph of V vesus x is sketched in Figue Figue q EVALUATE: (e) When x >>, V, just like point chge of chge + q At distnces fom the chges P x much gete thn thei seption, the two chges ct like single point chge kq IDENTIFY: Fo point chge, V 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 kq k( q) EXECUTE: () V + (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 equidistnt fom the two chges Figue 4 IDENTIFY: kq Fo point chge, V 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) kq kq kqy EXECUTE: () y < : V ( + y) ( y) y kq kq kq y< : V ( + y) ( y + ) y q q A genel expession vlid fo ny y is V k + y y + () The gph of V vesus y is sketched in Figue 4 kq kq (c) y >> : V y y (d) If the chges e intechnged, then the potentil is of the opposite sign kq kq kq y > : V ( + y) y y

10 - Chpte EVALUATE: ppoched V t y V + s the positive chge is ppoched nd V s the negtive chge is Figue 4 5 IDENTIFY: kq Fo point chge, V 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 5 Figue 5 kq kq kq( x + ) kq kq kq( x ) () x> : V < x< : V x x x( x ) x x x( x ) kq kq kq( x + ) q q x< : V + A genel expession vlid fo ny y is V k x x x( x ) x x (c) The potentil is zeo t x nd / (d) The gph of V vesus x is sketched in Figue 5 Figue 5 kqx kq EVALUATE: (e) Fo x>> : V, which is the sme s the potentil of point chge q F fom x x the two chges they ppe to e point chge with chge tht is the lgeic sum of thei two chges

11 Electic Potentil - kq 6 IDENTIFY: Fo point chge, V The totl potentil t ny point is the lgeic sum of the potentils of the two chges SET UP: The distnce of point with coodinte y fom the positive chge is y nd the distnce fom the negtive chge is + y kq kq EXECUTE: () V kq y y + y + y () V, when y y y ± 4 (c) The gph of V vesus y is sketched in Figue 6 V s the positive chge t the oigin is ppoched kq EVALUATE: (d) y : V kq >>, which is the potentil of point chge q F fom the two y y y chges they ppe to e point chge with chge tht is the lgeic sum of thei two chges Figue 6 7 IDENTIFY: K + qv K + qv SET UP: Let point e t the cthode nd let point e t the node K V V 95 V An electon hs q end m 9 kg EXECUTE: v K q V V K 7 (47 J) 9 7 ( ) (6 C)( 95 V) 47 J 9 kg 7 m s mv, so EVALUATE: The negtively chged electon gins kinetic enegy when it moves to highe potentil 8 IDENTIFY: kq kq Fo point chge, E nd V SET UP: The electic field is diected towd negtive chge nd wy fom positive chge V kq/ kq 498 V EXECUTE: () V > so q > 45 m E k q / kq V/m V (45 m)(498 V) () q C 9 k 899 N m /C (c) q >, so the electic field is diected wy fom the chge EVALUATE: The tio of V to E due to point chge inceses s the distnce fom the chge inceses, ecuse E flls off s / nd V flls off s / 9 () IDENTIFY nd SET UP: The diection of E is lwys fom high potentil to low potentil so point is t highe potentil () Apply Eq(7) to elte V V to E EXECUTE: V V E d l Edx E ( x x ) V V + 4 V E 8 V/m x x 9 m 6 m

12 - Chpte 6 5 (c) W q( V V ) ( C)( + 4 V) 48 J EVALUATE: The electic foce does negtive wok on negtive chge when the negtive chge moves fom high potentil (point ) to low potentil (point ) kq IDENTIFY: Fo point chge, V The totl potentil t ny point is the lgeic sum of the potentils of the kq two chges Fo point chge, E The net electic field is the vecto sum of the electic fields of the two chges SET UP: E poduced y point chge is diected wy fom the point chge if it is positive nd towd the chge if it is negtive EXECUTE: () V VQ + VQ >, so V is zeo nowhee except fo infinitely f fom the chges The fields cn cncel only etween the chges, ecuse only thee e the fields of the two chges in opposite diections Conside kq k( Q) point distnce x fom Q nd d xfom Q, s shown in Figue EQ EQ ( d x) x x ( d x) x d The othe oot, x d, does not lie etween the chges + () V cn e zeo in plces, A nd B, s shown in Figue Point A is distnce x fom Q nd d x fom Q B is distnce y fom Q nd d + y fom Q k( Q) k( Q) At A: + x d x d x k( Q) k( Q) At B: + y d y d + y The two electic fields e in opposite diections to the left of Q o to the ight of Q in Figue c But fo the mgnitudes to e equl, the point must e close to the chge with smlle mgnitude of chge This cn e the cse only in the egion to the left of Q EQ EQ gives kq k ( Q ) d nd x x ( d + x) EVALUATE: (d) E nd V e not zeo t the sme plces E is vecto nd V is scl E is popotionl to / nd V is popotionl to / E is elted to the foce on test chge nd ΔV is elted to the wok done on test chge when it moves fom one point to nothe Figue IDENTIFY nd SET UP: Apply consevtion of enegy, Eq() Use Eq() to expess U in tems of V () EXECUTE: K+ qv K + qv qv ( V) K K; q 9 6 C 8 7 K m ev 499 J; K m ev 95 J K K V V 56 V q EVALUATE: The electon gins kinetic enegy when it moves to highe potentil 7 () EXECUTE: Now K 95 J, K V K K q V + 8 V EVALUATE: The electon loses kinetic enegy when it moves to lowe potentil IDENTIFY nd SET UP: Expessions fo the electic potentil inside nd outside solid conducting sphee e deived in Exmple 8 9 kq k(5 C) EXECUTE: () This is outside the sphee, so V 656 V 48 m 9 k(5 C) () This is t the sufce of the sphee, so V V 4 m (c) This is inside the sphee The potentil hs the sme vlue s t the sufce, V EVALUATE: All points of conducto e t the sme potentil

13 Electic Potentil - () IDENTIFY nd SET UP: The electic field on the ing s xis is clculted in Exmple The foce on the electon exeted y this field is given y Eq() EXECUTE: When the electon is on eithe side of the cente of the ing, the ing exets n ttctive foce diected towd the cente of the ing This estoing foce poduces oscilltoy motion of the electon long the xis of the ing, with mplitude cm The foce on the electon is not of the fom F kx so the oscilltoy motion is not simple hmonic motion () IDENTIFY: Apply consevtion of enegy to the motion of the electon SET UP: K + U K + U with t the initil position of the electon nd t the cente of the ing Fom Q Exmple, V, whee is the dius of the ing P x + EXECUTE: x cm, x K (elesed fom est), K mv Thus mv U U ( ev V) And U qv ev so v m 9 Q 9 4 C V (8988 N m / C ) 4 πp x + ( m) + (5 m) V 64 V 9 Q 9 4 C V (8988 N m / C ) 48 V P x + 5 m 9 ev ( V) (6 C)(48 V 64 V) v m 99 kg 7 67 m/s EVALUATE: The positively chged ing ttcts the negtively chged electon nd cceletes it The electon hs its mximum speed t this point When the electon moves pst the cente of the ing the foce on it is opposite to its motion nd it slows down λ 4 IDENTIFY: Exmple shows tht fo line of chge, V V ln( / ) Apply consevtion of enegy π P to the motion of the poton SET UP: Let point e 8 cm fom the line nd let point e t the distnce of closest ppoch, whee K 7 EXECUTE: () K mv (67 kg)(5 m/s) 88 J K K 88 J π P () K + qv K + qv V V 75 V ln( / ) ( 75 V) 9 q 6 C λ πp( 75 V) πp(75 V) exp (8 m)exp 58 m λ 5 C/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 5 IDENTIFY: The voltmete mesues the potentil diffeence etween the two points We must elte this quntity to the line chge density on the wie λ SET UP: Fo vey long (infinite) wie, the potentil diffeence etween two points is Δ V ln ( / ) π P EXECUTE: () Solving fo λ gives ( ΔV )π P 575 V λ C/m ln ( / ) 9 5 cm ( 8 N m /C ) ln 5 cm () The mete will ed less thn 575 V ecuse the electic field is weke ove this -cm distnce thn it ws ove the -cm distnce in pt () (c) The potentil diffeence is zeo ecuse oth poes e t the sme distnce fom the wie, nd hence t the sme potentil EVALUATE: Since voltmete mesues potentil diffeence, we e ctully given ΔV, even though tht is not stted explicitly in the polem We must lso e ceful when using the fomul fo the potentil diffeence ecuse ech is the distnce fom the cente of the cylinde, not fom the sufce

14 -4 Chpte 6 IDENTIFY: The voltmete eds the potentil diffeence etween the two points whee the poes e plced Theefoe we must elte the potentil diffeence to the distnces of these points fom the cente of the cylinde Fo points outside the cylinde, its electic field ehves like tht of line of chge λ / SET UP: Using Δ V ln ( / ) nd solving fo, we hve V e π P Δ λ π P 9 (75 V) 9 N m /C EXECUTE: The exponent is 648, which gives 9 5 C/m (5 cm) e cm The distnce ove the sufce is 478 cm 5 cm 8 cm EVALUATE: Since voltmete mesues potentil diffeence, we e ctully given ΔV, even though tht is not stted explicitly in the polem We must lso e ceful when using the fomul fo the potentil diffeence ecuse ech is the distnce fom the cente of the cylinde, not fom the sufce 7 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 λ SET UP: The potentil diffeence is Δ V ln ( / ) π P EXECUTE: () Sustituting numes gives λ 6 9 cm Δ V ln ( / ) ( 85 C/m)( 9 N m /C ) ln π P 6 cm Δ V 78 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 equl to the potentil t the sufce 8 IDENTIFY: The wok equied is equl to the chnge in the electicl potentil enegy of the chge-ing system We need only look t the eginning nd ending points, since the potentil diffeence is independent of pth fo consevtive field Q SET UP: () W Δ U qδ V q( Vcente V ) q ε EXECUTE: Sustituting numes gives ΔU ( -6 C)(9 9 N m /C )(5 6 C)/(4 m) 8 J () We cn tke ny pth since the potentil is independent of pth (c) SET UP: The net foce is wy fom the ing, so the ll will ccelete wy Enegy consevtion gives U Kmx mv EXECUTE: Solving fo v gives v U (8 J) m 5 kg 67 m/s EVALUATE: Diect clcultion of the wok fom the electic field would e extemely difficult, nd we would need to know the pth followed y the chge But, since the electic field is consevtive, we cn ypss ll this clcultion just y looking t the end points (infinity nd the cente of the ing) using the potentil 9 IDENTIFY: The electic field is zeo eveywhee except etween the pltes, nd in this egion it is unifom nd points fom the positive to the negtive plte (to the left in Figue ) SET UP: Since the field is unifom etween the pltes, the potentil inceses linely s we go fom left to ight stting t EXECUTE: Since the potentil is tken to e zeo t the left sufce of the negtive plte ( in Figue ), it is zeo eveywhee to the left of It inceses linely fom to c, nd emins constnt (since E ) pst c The gph is sketched in Figue 9 EVALUATE: When the electic field is zeo, the potentil emins constnt ut not necessily zeo (s to the ight of c) When the electic field is constnt, the potentil is line Figue 9

15 Electic Potentil -5 4 IDENTIFY nd SET UP: Fo oppositely chged pllel pltes, E σ / P etween the pltes nd the potentil diffeence etween the pltes is V Ed 9 σ 47 C m EXECUTE: () E 5 N C P P () V Ed (5 N/C)( m) 7 V (c) The electic field stys the sme if the seption of the pltes doules The potentil diffeence etween the pltes doules EVALUATE: The electic field of n infinite sheet of chge is unifom, independent of distnce fom the sheet The foce on test chge etween the two pltes is constnt ecuse the electic field is constnt The potentil diffeence is the wok pe unit chge on test chge when it moves fom one plte to the othe When the distnce doules the wok, which is foce times distnce, doules nd the potentil diffeence doules 4 IDENTIFY nd SET UP: Use the esult of Exmple 9 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 Eq() Use the eqution tht pecedes Eq(7) to clculte the wok V 6 V EXECUTE: () Fom Exmple 9, E 8 V/m d 45 m 9 5 () F q E (4 C)(8 V/m) + 9 N (c) The electic field etween the pltes is shown in Figue 4 Figue 4 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 d l Fd, whee d is the seption etween the pltes W Fd + (d) V 5 7 (9 N)(45 m) 864 J V + 6 V (plte is t highe potentil) Δ U U U q V V 9 7 ( ) (4 C)( 6 V) 864 J EVALUATE: We see tht W ( U U) U U 4 IDENTIFY: The electic field is zeo inside the sphee, so the potentil is constnt thee Thus the potentil t the cente must e the sme s t the sufce, whee it is equivlent to tht of point-chge SET UP: At the sufce, nd hence lso t the cente of the sphee, the field is tht of point-chge, E Q/(4 πp ) EXECUTE: () Solving fo Q nd sustituting the numes gives Q P V (5 m)(5 V)/(9 9 N m /C ) 8-8 C 8 nc () Since the potentil is constnt inside the sphee, its vlue t the sufce must e the sme s t the cente, 5 kv EVALUATE: The electic field inside the sphee is zeo, so the potentil is constnt ut is not zeo 4 IDENTIFY nd SET UP: Conside the electic field outside nd inside the shell nd use tht to deduce the potentil EXECUTE: () The electic field outside the shell is the sme s fo point chge t the cente of the shell, so the potentil outside the shell is the sme s fo point chge: q V fo > P The electic field is zeo inside the shell, so no wok is done on test chge s it moves inside the shell nd ll q points inside the shell e t the sme potentil s the sufce of the shell: V fo P kq V (5 m)( V) () V so q nc k k (c) EVALUATE: No, the mount of chge on the sphee is vey smll Since U qv the totl mount of electic 5 enegy stoed on the lloon is only ( nc)( V) 4 J

16 -6 Chpte 44 IDENTIFY: Exmple 8 shows tht the potentil of solid conducting sphee is the sme t evey point inside the sphee nd is equl to its vlue V q/πp t the sufce Use the given vlue of E to find q 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 q (8 N/C)( m) 8 EXECUTE: E nd q P E 69 C 9 P 899 N m /C Since the field is diected inwd, the chge must e negtive The potentil of point chge, tking s zeo, is 9 8 q (899 N m /C )( 69 C) V 76 V t the sufce of the sphee Since the chge ll esides P m 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 45 IDENTIFY: Exmple 9 shows tht V( y) Ey, whee y is the distnce fom the negtively chged plte, whose potentil is zeo The electic field etween the pltes is unifom nd pependicul to the pltes SET UP: V inceses towd the positively chged plte E is diected fom the positively chged plted towd the negtively chged plte V 48 V 4 V EXECUTE: () E 8 V/m nd y V t y, V V t y 4 cm, d 7 m E V 4 V t y 85 cm, V 6 V t y 8 cm nd V 48 V t y 7 cm The equipotentil sufces e sketched in Figue 45 The sufces e plnes pllel to the pltes () The electic field lines e lso shown in Figue 45 The field lines e pependicul to the pltes nd the equipotentil lines e pllel to the pltes, so the electic field lines nd the equipotentil lines e mutully pependicul EVALUATE: Only diffeences in potentil hve physicl significnce Letting V t the negtive plte is choice we e fee to mke Figue IDENTIFY: By the definition of electic potentil, if positive chge gins potentil long pth, then the potentil long tht pth must hve incesed The electic field poduced y vey lge sheet of chge is unifom nd is independent of the distnce fom the sheet () SET UP: No mtte wht the efeence point, we must do wok on positive chge to move it wy fom the negtive sheet EXECUTE: Since we must do wok on the positive chge, it gins potentil enegy, so the potentil inceses () SET UP: σ Since the electic field is unifom nd is equl to σ /ε, we hve Δ V Ed d P EXECUTE: Solving fo d gives P 885 ( C/Nm )( V) ΔV d 9 σ 6 C/m 95 m 95 mm EVALUATE: Since the spcing of the equipotentil sufces (d 95 mm) is independent of the distnce fom the sheet, the equipotentil sufces e plnes pllel to the sheet nd spced 95 mm pt

17 Electic Potentil IDENTIFY nd SET UP: Use Eq(9) to clculte the components of E EXECUTE: V Axy Bx + Cy V () Ex Ay + Bx x V Ey Ax C y V Ez z () E equies tht Ex Ey Ez E z eveywhee E y t x CA / And E x is lso equl zeo fo this x, ny vlue of z, nd y Bx/ A ( B/ A)( C/ A) BC/ A EVALUATE: V doesn t depend on z so E z eveywhee 48 IDENTIFY: Apply Eq(9) q SET UP: Eq(7) sys E ˆ is the electic field due to point chge q P V kq kqx kqx EXECUTE: () Ex x x x + y + z ( x + y + z ) kqy kqz Similly, Ey nd E z kq xiˆ yˆj zkˆ kq () Fom pt (), E ˆ, + + which gees with Eqution (7) EVALUATE: V is scl E is vecto nd hs components kq kq 49 IDENTIFY nd SET UP: Fo solid metl sphee o fo spheicl shell, V outside the sphee nd V t V ll points inside the sphee, whee is the dius of the sphee When the electic field is dil, E kq kq EXECUTE: () (i) < : This egion is inside oth sphees V kq kq kq (ii) < < : This egion is outside the inne shell nd inside the oute shell V kq (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 q q () V nd V, so V q P 4 π P (c) Between the sphees < < nd V kq V q q V E + P P (d) Fom Eqution (): 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 q Q ( q Q) V All potentils inside the oute shell e just shifted y n mount P P P Q V Theefoe eltive potentils within the shells e not ffected Thus () nd (c) do not chnge P Howeve, now tht the potentil does vy outside the sphees, thee is n electic field thee: V kq kq kq Q k E + ( q Q) q EVALUATE: In pt () the potentil is gete thn zeo fo ll <

18 -8 Chpte 5 IDENTIFY: Execise 49 shows tht V kq fo <, V kq fo < < nd V kq kq SET UP: E, dilly outwd, fo 5 V EXECUTE: () V kq 5 V gives q 76 C k m 96 m () V so V 5 V The inne metl sphee is n equipotentil with V 5 V V + V 4 V t kq 45 cm, V V t 85 cm, V V t 5 cm, V V t 4 cm, V t 96 cm The equipotentil sufces e sketched in Figue 5 EVALUATE: (c) The equipotentil sufces e concentic sphees nd the electic field lines e dil, so the field lines nd equipotentil sufces e mutully pependicul The equipotentils e closest t smlle, whee the electic field is lgest Figue 5 5 IDENTIFY: Outside the cylinde it is equivlent to line of chge t its cente SET UP: The diffeence in potentil etween the sufce of the cylinde ( distnce fom the centl xis) nd λ genel point distnce fom the centl xis is given y Δ V ln( / ) π P EXECUTE: () The potentil diffeence depends only on, nd not diection Theefoe ll points t the sme vlue of will e t the sme potentil Thus the equipotentil sufces e cylindes coxil with the given cylinde λ V/ () Solving Δ V ln( / ) fo, gives e π P Δ λ π P Fo V, the exponent is ( V)/[( 9 9 N m /C )(5 9 C/m)] 7, which gives ( cm) e 7 9 cm Likewise, the othe dii e 4 cm (fo V) nd 68 cm (fo V) (c) Δ 9 cm cm 9 cm; Δ 4 cm 9 cm cm; Δ 68 cm 4 cm 88 cm EVALUATE: As we cn see, Δ inceses, so the sufces get fthe pt This is vey diffeent fom sheet of chge, whee the sufces e eqully spced plnes 5 IDENTIFY: The electic field is the negtive gdient of the potentil V SET UP: Ex, so E x is the negtive slope of the gph of V s function of x x EXECUTE: The gph is sketched in Figue 5 Up to, V is constnt, so E x Fom to, V is line with positive slope, so E x is negtive constnt Pst, the V-x gph hs decesing positive slope which ppoches zeo, so E x is negtive nd ppoches zeo

19 Electic Potentil -9 EVALUATE: Notice tht n incesing potentil does not necessily men tht the electic field is incesing Figue 5 5 () IDENTIFY: Apply the wok-enegy theoem, Eq(66) SET UP: Points nd e shown in Figue 5 EXECUTE: The electic foce W K K K K 5 tot Δ 45 J E Figue 5 F nd the dditionl foce F oth do wok, so tht Wtot W + W W W W FE tot F 45 J 65 J 5 J EVALUATE: The foces on the chged pticle e shown in Figue 5 FE F Figue 5 The electic foce is to the left (in the diection of the electic field since the pticle hs positive chge) The displcement is to the ight, so the electic foce does negtive wok The dditionl foce F is in the diection of the displcement, so it does positive wok () IDENTIFY nd SET UP: Fo the wok done y the electic foce, W q( V V) EXECUTE: 5 W 5 J V V 8 V 9 q 76 C EVALUATE: The stting point (point ) is t 8 V lowe potentil thn the ending point (point ) We know tht V > V ecuse the electic field lwys points fom high potentil towd low potentil (c) IDENTIFY: Clculte E fom V V nd the seption d etween the two points SET UP: Since the electic field is unifom nd diected opposite to the displcement W FEd qed, whee d 8 cm is the displcement of the pticle EXECUTE: W V V 8 V 54 4 E V/m qd d 8 m EVALUATE: In pt (), W tot is the totl wok done y oth foces In pts () nd (c) W is the wok done just y the electic foce 54 IDENTIFY: ke The electic foce etween the electon nd poton is ttctive nd hs mgnitude F Fo e cicul motion the cceletion is d v / U k SET UP: 9 9 e 6 C ev 6 J mv ke ke EXECUTE: () nd v m ke () K mv U

20 - Chpte 9 ke k(6 C) 8 (c) E K + U U 7 J 6 ev 59 m EVALUATE: The totl enegy is negtive, so the electon is ound to the poton Wok must e done on the electon to tke it f fom the poton 55 IDENTIFY nd SET UP: Clculte the components of E fom Eq(9) Eq() gives F fom E 4/ EXECUTE: () V Cx 4/ 4/ 4 4/ C V / x 4 V /( m) 785 V/m V 4 / 5 4/ / () Ex Cx (5 V/m ) x x The minus sign mens tht E x is in the x -diection, which sys tht E points fom the positive node towd the negtive cthode (c) F qe so 4 / Fx eex ecx Hlfwy etween the electodes mens x 65 m / / 5 F x (6 C)(785 V/m )(65 m) N F x is positive, so the foce is diected towd the positive node EVALUATE: V depends only on x, so Ey Ez E is diected fom high potentil (node) to low potentil (cthode) The electon hs negtive chge, so the foce on it is diected opposite to the electic field 56 IDENTIFY: At ech point ( nd ), the potentil is the sum of the potentils due to oth sphees The voltmete eds the diffeence etween these two potentils The sphees ehve like point-chges since the mete is connected to the sufce of ech one SET UP: () Cll the point on the sufce of one sphee nd the point on the sufce of the othe sphee, cll the dius of ech sphee, nd cll d the cente-to-cente distnce etween the sphees The potentil diffeence V etween points nd is then q q q q q V V V + + P d d ε d EXECUTE: Sustituting the numes gives 9 V V (75 µc) ( 9 N m /C ) 75 m 5 m 84 6 V The mete eds 84 MV () Since V V is negtive, V > V, so point is t the highe potentil EVALUATE: An esy wy to see tht the potentil t is highe thn the potentil t is tht it would tke positive wok to move positive test chge fom to since this chge would e ttcted y the negtive sphee nd epelled y the positive sphee kqq 57 IDENTIFY: U SET UP: Eight chges mens thee e 8(8 ) / 8 pis Thee e pis of q nd q septed y d, pis of equl chges septed y d nd 4 pis of q nd q septed y d 4 kq EXECUTE: () U kq + 46 q / π d d d d d + P EVALUATE: () The fct tht the electic potentil enegy is less thn zeo mens tht it is enegeticlly fvole fo the cystl ions to e togethe kqq 58 IDENTIFY: Fo two smll sphees, U Fo pt () pply consevtion of enegy SET UP: Let q μc nd q 5 μc Let 5 m nd (899 N m /C )( C)( 5 C) EXECUTE: () U 5 J 5 m () K U U 5 J K + U K + U gives K 5 J K mv, so K (5 J) v 8 m/s m 5 kg

21 Electic Potentil - EVALUATE: As the sphee moves wy, the ttctive electicl foce exeted y the othe sphee does negtive wok nd emoves ll the kinetic enegy it initilly hd Note tht it doesn t mtte which sphee is held fixed nd which is shot wy; the nswe to pt () is unffected 59 () IDENTIFY: Use Eq() fo the electon nd ech poton SET UP: The positions of the pticles e shown in Figue 59 Figue 59 (7 m)/ 55 m EXECUTE: The potentil enegy of intection of the electon with ech poton is ( e ) U, so the totl potentil enegy is 4 πp 9 9 e (8988 N m /C )(6 C) U πp 4 55 m U J( ev /6 J) 57 ev 8 86 J EVALUATE: The electon nd poton hve chges of opposite signs, so the potentil enegy of the system is negtive () IDENTIFY nd SET UP: The positions of the potons nd points nd e shown in Figue 59 + d 55 m Figue 59 Apply K + U + Wothe K + U with point midwy etween the potons nd point whee the electon instntneously hs v (t its mximum displcement d fom point ) EXECUTE: Only the Coulom foce does wok, so W othe 8 U 86 J (fom pt ()) 6 8 K mv K (99 kg)(5 m/s) 5 J U ke / Then U K + U K 5 J 86 J 7575 J 9 9 ke (8988 N m /C )(6 C) 8 U 7575 J Then d EVALUATE: (675 m) (55 m) 88 m 675 m The foce on the electon pulls it ck towd the midpoint The tnsvese distnce the electon moves is out 7 times the seption of the potons 6 IDENTIFY: Apply F x nd F y to the sphee The electic foce on the sphee is Fe qe The potentil diffeence etween the pltes is V Ed SET UP: The fee-ody digm fo the sphee is given in Figue 56 EXECUTE: Tcosθ mg nd Tsinθ Fe gives Fe mgtn θ (5 kg)(98 m s )tn( ) 85 N Vq Fd (85 N)(5 m) Fe Eq nd V 478 V 6 d q 89 C

22 - Chpte 9 EVALUATE: E V/ d 956 V/m E σ / P nd σ EP 846 C/m Figue 6 6 () IDENTIFY: The potentil t ny point is the sum of the potentils due to ech of the two chged conductos SET UP: Fom Exmple, fo conducting cylinde with chge pe unit length λ the potentil outside the cylinde is given y V ( λ/ πp )ln( / ) whee is the distnce fom the cylinde xis nd is the distnce fom the xis fo which we tke V Inside the cylinde the potentil hs the sme vlue s on the cylinde sufce The electic field is the sme fo solid conducting cylinde o fo hollow conducting tue so this expession fo V pplies to oth This polem sys to tke EXECUTE: Fo the hollow tue of dius nd chge pe unit length λ : outside V ( λ/ πp )ln( / ); inside V since V t Fo the metl cylinde of dius nd chge pe unit length λ : outside V ( λ/ πp )ln( / ), inside V ( λ/ πp )ln( / ), the vlue t (i) < ; inside oth V ( λ/ πp )ln( / ) (ii) < < ; outside cylinde, inside tue V ( λ/ πp )ln( / ) (iii) > ; outside oth the potentils e equl in mgnitude nd opposite in sign so V () Fo, V ( λ/ πp )ln( / ) Fo, V Thus V V V ( λ/ πp )ln( / ) (c) IDENTIFY nd SET UP: Use Eq() to clculte E V λ λ V EXECUTE: E ln πp πp ln( / ) (d) The electic field etween the cylindes is due only to the inne cylinde, so V is not chnged, V ( λ/ πp )ln( / ) EVALUATE: The electic field is not unifom etween the cylindes, so V E( ) 6 IDENTIFY: V The wie nd hollow cylinde fom coxil cylindes Polem 6 gives E () ln( / ) SET UP: 6 45 m, 8 m EXECUTE: V 4 6 E nd V E ln( /) ( N C)(ln(8 m 45 m)) m 57 V ln( ) EVALUATE: The electic field t ny is diectly popotionl to the potentil diffeence etween the wie nd the 6 cylinde IDENTIFY nd SET UP: Use Eq() to clculte F nd then F m gives EXECUTE: () FE qe Since q e is negtive F E nd E e in opposite diections; E is upwd so F E is downwd The mgnitude of F E is F 9 6 q E ee (6 C)( N/C) 76 N E () Clculte the cceletion of the electon poduced y the electic foce: 6 F 76 N 4 9 m/s m 99 kg EVALUATE: This is much lge thn g 98 m/s, so the gvity foce on the electon cn e neglected F E is downwd, so is downwd (c) IDENTIFY nd SET UP: The cceletion is constnt nd downwd, so the motion is like tht of pojectile Use the hoizontl motion to find the time nd then use the time to find the veticl displcement

23 EXECUTE: x-component 6 v x 65 m/s; x ; x x 6 m; t? x x vxt+ xt nd the x tem is zeo, so x x 6 m 9 t 9 s 6 v x 65 m/s y-component 4 9 v 9 m/s ; t 9 m/s; y y? y ; y y v t+ t y y y 4 9 (9 m/s )(9 s) 8 m 8 cm y y (d) The velocity nd its components s the electon leves the pltes e sketched in Figue 6 v x v (since ) 6 x 65 m/s v v + t y y y v y y (9 m/s )(9 s) v 6 78 m/s x Electic Potentil - Figue 6 6 vy 78 m/s tnα 74 so α 5 6 vx 65 m/s EVALUATE: The gete the electic field o the smlle the initil speed the gete the downwd deflection (e) IDENTIFY nd SET UP: Conside the motion of the electon fte it leves the egion etween the pltes Outside the pltes thee is no electic field, so (Gvity cn still e neglected since the electon is tveling t such high speed nd the times e smll) Use the hoizontl motion to find the time it tkes the electon to tvel m hoizontlly to the sceen Fom this time find the distnce downwd tht the electon tvels EXECUTE: x-component 6 v x 65 m/s; x ; x x m; t? x x vxt+ xt nd the x tem is tem is zeo, so x x m 8 t 846 s 6 v x 65 m/s y-component 6 8 v y 78 m/s (fom pt ()); y ; t 846 m/s; y y? 6 8 y y vyt+ (78 m/s)(846 s) 9 m 9 cm yt EVALUATE: The electon tvels downwd distnce 8 cm while it is etween the pltes nd distnce 9 cm while tveling fom the edge of the pltes to the sceen The totl downwd deflection is 8 cm + 9 cm 4 cm The hoizontl distnce etween the pltes is hlf the hoizontl distnce the electon tvels fte it leves the pltes And the veticl velocity of the electon inceses s it tvels etween the pltes, so it mkes sense fo it to hve gete downwd displcement duing the motion fte it leves the pltes 64 IDENTIFY: The chge on the pltes nd the electic field etween them depend on the potentil diffeence coss the pltes Since we do not know the numeicl potentil, we shll cll this potentil V nd find the nswes in tems of V σ Qd () SET UP: Fo two pllel pltes, the potentil diffeence etween them is V Ed d P PA EXECUTE: Solving fo Q gives Q P AV d (885 C /N m )( m) V/(5 m) / Q 59V C 59V pc, when V is in volts () E V/d V/(5 m) V V/m, with V in volts (c) SET UP: Enegy consevtion gives mv ev EXECUTE: Solving fo v gives 9 ev 6 ( C) V 5 / v 59 V m/s, with V in volts m 9 kg EVALUATE: Typicl voltges in student lotoy wok un up to ound 5 V, so the chge on the pltes is typiclly out ound 4 pc, the electic field is out 5 V/m, nd the electon speed would e out million m/s

24 -4 Chpte V 65 () IDENTIFY nd SET UP: Polem 6 deived tht E, whee is the dius of the inne cylinde ln( / ) (wie) nd is the dius of the oute hollow cylinde The potentil diffeence etween the two cylindes is V Use this expession to clculte E t the specified EXECUTE: Midwy etween the wie nd the cylinde wll is t dius of 6 ( + )/ (9 m + 4 m)/ 74 m V 5 V 6 E ln( / ) ln(4 m /9 m)(74 m) 4 97 V/m () IDENTIFY nd SET UP: The electic foce is given y Eq() Set this equl to ten times the weight of the pticle nd solve fo q, the mgnitude of the chge on the pticle F mg EXECUTE: E 9 mg ( kg)(98 m/s ) 4 C qe mg nd q E 97 V/m EVALUATE: It equies only this modest net chge fo the electic foce to e much lge thn the weight 66 () IDENTIFY: Clculte the potentil due to ech thin ing nd integte ove the disk to find the potentil V is scl so no components e involved SET UP: Conside thin ing of dius y nd width dy The ing hs e π ydy so the chge on the ing is dq σ ( π y dy) EXECUTE: The esult of Exmple then sys tht the potentil due to this thin ing t the point on the xis t distnce x fom the ing is dq πσ y dy dv P x + y P x + y σ ydy σ σ V dv x y + ( x + x) P x + y P P EVALUATE: Fo x this esult should educe to the potentil of point chge with Q σπ / x + x( + / x ) x( + / x ) so x + x /x σ σπ Q Then V, s expected P x Px Px () IDENTIFY nd SET UP: Use Eq(9) to clculte E x EXECUTE: V σ x σx Ex x P x + P x x + EVALUATE: Ou esult gees with Eq() in Exmple 67 () IDENTIFY: Use V V E d λ SET UP: Fom Polem 48, E () π P fo (inside the cylindicl chge distiution) nd λ E () π P fo Let V t (t the sufce of the cylinde) EXECUTE: > Tke point to e t nd point to e t, whee > Let dl d E nd d e oth dilly outwd, so E d Ed Thus V V Ed Then V gives V Ed In this intevl ( > ), E( ) λ/ π P, so d V λ d λ λ ln πp πp πp EVALUATE: This expession gives V when nd the potentil deceses (ecomes negtive nume of lge mgnitude) with incesing distnce fom the cylinde

25 Electic Potentil -5 EXECUTE: < Tke point t, whee <, nd point t E d Ed () λ / π, V Ed In this intevl ( < ), E P so s efoe Thus V V Ed Then V gives λ λ λ V d d πp πp πp λ V P EVALUATE: This expession lso gives V when The potentil is λ /P t nd deceses with incesing () EXECUTE: Gphs of V nd E s functions of e sketched in Figue 67 Figue 67 EVALUATE: E t ny is the negtive of the slope of V( ) t tht (Eq) 68 IDENTIFY: The lph pticles stt out with kinetic enegy nd wind up with electicl potentil enegy t closest ppoch to the nucleus SET UP: () The enegy of the system is conseved, with U (/ 4 πp )( qq/ ) eing the electic potentil enegy With the chge of the lph pticle eing e nd tht of the gold nucleus eing Ze, we hve Ze mv P EXECUTE: Solving fo v nd using Z 79 fo gold gives 9 9 4Ze 9 N m /C (4)(79) 6 C v P m ( 67 kg)( 56 m) 7 m/s We hve neglected ny eltivistic effects () Outside the tom, it is neutl Inside the tom, we cn model the 79 electons s unifom spheicl shell, which poduces no electic field inside of itself, so the only electic field is tht of the nucleus EVALUATE: Neglecting eltivistic effects ws not such good ide since the speed in pt () is ove % the speed of light Modeling 79 electons s unifom spheicl shell is esonle, ut we would not wnt to do this with smll toms V V E d l Qx SET UP: Fom Exmple, we hve: Ex E dl E / xdx Let so V 4 πp ( x + ) 69 IDENTIFY: EXECUTE: ( ) ( ) x u x + / / x + u x + Q x Q Q V dx u 4 πp ( ) P P EVALUATE: Ou esult gees with Eq(6) in Exmple 7 IDENTIFY: Divide the od into infinitesiml segments with chge dq The potentil dv due to the segment is dq dv Integte ove the od to find the totl potentil P SET UP: dq λdl, with λ Q/ π nd dl dθ EXECUTE: π dq λ dl Q dl Q dθ Qdθ Q dv V P P P π P π P π P

26 -6 Chpte EVALUATE: All the chge of the ing is the sme distnce fom the cente of cuvtue 7 IDENTIFY: We must integte to find the totl enegy ecuse the enegy to ing in moe chge depends on the chge ledy pesent SET UP: If ρ is the unifom volume chge density, the chge of spheicl shell o dius nd thickness d is dq ρ d, nd ρ Q/(4/ π ) The chge ledy pesent in sphee of dius is q ρ(4/ π ) The enegy to ing the chge dq to the sufce of the chge q is Vdq, whee V is the potentil due to q, which is q/4 πp EXECUTE: The totl enegy to ssemle the entie sphee of dius nd chge Q is sum (integl) of the tiny incements of enegy 4 ρ π q Q U Vdq dq ( ρ d) P P 5 P whee we hve sustituted ρ Q/(4/ π ) nd simplified the esult EVALUATE: Fo point-chge, so U, which mens tht point-chge should hve infinite selfenegy This suggests tht eithe point-chges e impossile, o tht ou pesent tetment of physics is not dequte t the extemely smll scle, o oth 7 IDENTIFY: V V E d l The electic field is dilly outwd, so E dl E d SET UP: Let, so V kq d kq EXECUTE: Fom Exmple 9, we hve the following Fo > : E nd V kq kq kq kq kq kq kq kq kq kq Fo < : E nd V d d d E E + () The gphs of V nd E vesus e sketched in Figue 7 EVALUATE: Fo < the potentil depends on the electic field in the egion to Figue 7 7 Q Q IDENTIFY: Polem 7 shows tht V ( ) fo nd V 8π P P fo SET UP: Q Q V, V 8πP P Q EXECUTE: () V V 8π P () If Q >, V is highe t the cente If Q <, V is highe t the sufce EVALUATE: Fo Q > the electic field is dilly outwd, E is diected towd lowe potentil, so V is highe t the cente If Q <, the electic field is diected dilly inwd nd V is highe t the sufce 74 IDENTIFY: Fo < c, E nd the potentil is constnt Fo > c, E is the sme s fo point chge nd SET UP: V EXECUTE: () Points,, nd c e ll t the sme potentil, so V V V Vc V Vc 9 6 kq (899 N m C )(5 C) 6 Vc V 5 V 6 m () They e ll t the sme potentil 6 (c) Only Vc V would chnge; it would e 5 V kq V

27 Electic Potentil -7 EVALUATE: The voltmete eds the potentil diffeence etween the two points to which it is connected 75 IDENTIFY nd SET UP: Apply F du / d nd Newton's thid lw EXECUTE: () The electicl potentil enegy fo spheicl shell with unifom sufce chge density nd point chge q outside the shell is the sme s if the shell is eplced y point chge t its cente Since F du d, this mens the foce the shell exets on the point chge is the sme s if the shell wee eplced y point chge t its cente But y Newton s d lw, the foce q exets on the shell is the sme s if the shell wee point chge But q cn e eplced y spheicl shell with unifom sufce chge nd the foce is the sme, so the foce etween the shells is the sme s if they wee oth eplced y point chges t thei centes And since the foce is the sme s fo point chges, the electicl potentil enegy fo the pi of sphees is the sme s fo pi of point chges () The potentil fo solid insulting sphees with unifom chge density is the sme outside of the sphee s fo spheicl shell, so the sme esult holds (c) The esult doesn t hold fo conducting sphees o shells ecuse when two chged conductos e ought close togethe, the foces etween them cuses the chges to edistiute nd the chges e no longe distiuted unifomly ove the sufces qq kqq EVALUATE: Fo the insulting shells o sphees, F k nd U, whee q nd q e the chges of the ojects nd is the distnce etween thei centes 76 IDENTIFY: Apply Newton's second lw to clculte the cceletion Apply consevtion of enegy nd consevtion of momentum to the motions of the sphees qq kqq SET UP: Polem 75 shows tht F k nd U, whee q nd q e the chges of the ojects nd is the distnce etween thei centes EXECUTE: Mximum speed occus when the sphees e vey f pt Enegy consevtion gives kqq m 5v5 + m5v5 Momentum consevtion gives m5v5 m5v5 nd v5 v5 5 m Solve fo v 5 nd v 5 : v5 7 m s, v5 44 m s Mximum cceletion occus just fte sphees e elesed F m kqq (9 N m C )( C)( C) gives m 55 (5 kg) m s nd (5 m) m s EVALUATE: The moe mssive sphee hs smlle cceletion nd smlle finl speed 77 IDENTIFY: Use Eq(7) to clculte V SET UP: Fom Polem 4, fo (etween the sphee nd the shell) E Q/P Tke t nd t Q d Q Q EXECUTE: V V V Ed P P P Q V 8π P EVALUATE: The electic field is dilly outwd nd points in the diection of decesing potentil, so the sphee is t highe potentil thn the shell 78 IDENTIFY: V V E d l SET UP: E is dilly outwd, so E dl E d Polem 4 shows tht E ( ) fo, E() kq/ fo < <, E ( ) fo < < c nd E() kq/ fo > c c kq kq EXECUTE: () At c: Vc d c c kq kq () At : V E d E d c c c c kq d (c) At : V E d E d E d kq kq + c c c (d) At : V kq + c since it is inside metl sphee, nd thus t the sme potentil s its sufce EVALUATE: The potentil diffeence etween the two conductos is V V kq