ollege Physics Student s Mnul hpte 8 HAPTR 8: LTRI HARG AD LTRI ILD 8. STATI LTRIITY AD HARG: OSRVATIO O HARG. ommon sttic electicity involves chges nging fom nnocoulombs to micocoulombs. () How mny electons e needed to fom chge of.00 n? (b) How mny electons must be emoved fom neutl object to leve net chge of 0.00 µ? () Since one electon hs chge of.6 0, we cn detemine the e numbe of electons necessy to genete totl chge of -.00 n by using the Q.00 0 0 eution e Q, so tht. 0 electons..60 0 e (b) Similly we cn detemine the numbe of electons emoved fom neutl object to leve chge of 0.00 µ of chge so tht: 0.00 0.60 0. 0 electons 8. ODUTORS AD ISULATORS 7. A 0.0 g bll of coppe hs net chge of.00 µ. Wht fction of the coppe s electons hs been emoved? (ch coppe tom hs 9 potons, nd coppe hs n tomic mss of 6..) Recll tht Avogdo s numbe is A 6.0 0 toms/mole. ow we need to detemine the numbe of moles of coppe tht e pesent. We do this using the m 0.0 g mss nd the tomic mss: n A 6. g/mol So since thee e 9 potons pe tom we cn detemine the numbe of potons, 4
ollege Physics Student s Mnul hpte 8 p, fom: p n A 9 potons/tom 0. 0 gm 6. 0 0 toms 9 potons 7 0 6 g/mol mol.. cu tom potons Since thee is the sme numbe of electons s potons in neutl tom, befoe we emove the electons to give the coppe net chge, we hve.7 0 electons. et we need to detemine the numbe of electons we emoved to leve net chge of.00 µ. We need to emove.00 µ of chge, so the numbe of electons to be emoved is given by Q.00 0 e, emoved. 0 e.60 0 electons emoved. inlly we cn clculte the fction of coppe s electon by tking the tio of the numbe of electons emoved to the numbe of electons oiginlly pesent: e,emoved e,initilly. 0 9.09 0.7 0 8. OULOMB S LAW. Two point chges e bought close togethe, incesing the foce between them by fcto of. By wht fcto ws thei seption decesed? Using the eution k, we see tht the foce is invesely popotionl to the seption distnce sued, so tht K nd K Since we know the tio of the foces we cn detemine the tio of the seption distnces: so tht The seption decesed by fcto of.
ollege Physics Student s Mnul hpte 8 0. () ommon tnspent tpe becomes chged when pulled fom dispense. If one piece is plced bove nothe, the epulsive foce cn be get enough to suppot the top piece s weight. Assuming eul point chges (only n ppoimtion), clculte the mgnitude of the chge if electosttic foce is get enough to suppot the weight of 0.0 mg piece of tpe held.00 cm bove nothe. (b) Discuss whethe the mgnitude of this chge is consistent with wht is typicl of sttic electicity. () If the electosttic foce is to suppot the weight of 0.0 mg piece of tpe, it must be foce eul to the gvittionl foce on the tpe, so using the eution k nd the ssumption tht the point chges e eul, we cn set electosttic foce eul to gvittionl foce. k k mg mg k / (.000 m) ( 0.0 0 kg)( 9.80 m/s ) 9 ( 9.00 0 m / ) / 0.04 0. (b) This chge is ppoimtely n, which is consistent with the mgnitude of the chge of typicl sttic electicity.. Point chges of.00 µ nd.00 µ e plced 0.0 m pt. () Whee cn thid chge be plced so tht the net foce on it is zeo? (b) Wht if both chges e positive? () We know tht since the negtive chge is smlle, the thid chge should be plced to the ight of the negtive chge if the net foce on it to be zeo. So if we wnt net 0, we cn use the eution k to wite the foces in tems of distnces: K K 0 d K 0 ( 0.0 m d ), 0, o since 0.0m d nd d, so tht o d ( 0.0 m) d 6
ollege Physics Student s Mnul hpte 8 ( 0.0 m) d ( 0.0 m) nd finlly, d 0.89 m The chge must be plced t distnce of 0.89 m beyond the negtive chge on the line connecting the two chges. (b) This time we know tht the chge must be plced between the two positive chges nd close to the µ chge fo the net foce to be zeo. So if we wnt net 0, we cn gin use k to wite the foces in tems of K K distnces: K 0 O since 0 0 0.0m, o ( 0.0 ) ( 0.0 m ), o ( 0.0 ), nd finlly ( 0.0 m) 0.09 m m The chge must be plced t distnce of 0.09 m fom the lesse chge on the line connecting the two chges. 8.4 LTRI ILD: OPT O A ILD RVISITD. 7 () ind the diection nd mgnitude of n electic field tht eets 4.80 0 westwd foce on n electon. (b) Wht mgnitude nd diection foce does this field eet on poton? () Using the eution we cn find the electic field cused by given foce on given chge (tking estwd diection to be positive): 7
ollege Physics Student s Mnul hpte 8 4.80 0.60 0 7 00 / (est) (b) The foce should be eul to the foce on the electon only in the opposite diection. Using we get 7 (.60 0 )( 00 /) 4.80 0 (est), s we epected. 8. LTRI ILD LIS: MULTIPL HARGS. () Sketch the electic field lines ne point chge. (b) Do the sme fo point chge.00. () (b) 4. Sketch the electic field lines long distnce fom the chge distibutions shown in igue 8.6() nd (b). () 8
ollege Physics Student s Mnul hpte 8 (b) field of two opposite chges o net chge is seen fom f wy. 8.7 ODUTORS AD LTRI ILDS I STATI QUILIBRIUM 7. Sketch the electic field lines in the vicinity of the conducto in igue 8.48, given the field ws oiginlly unifom nd pllel to the object s long is. Is the esulting field smll ne the long side of the object? The field lines devite fom thei oiginl hoizontl diection becuse the chges within the object enge. The field lines will come into the object pependicul to the sufce nd will leve the othe side of the object pependicul to the sufce. Yes, the field is smlle ne the long side of the object. This is evident becuse thee e fewe field lines ne the long side of the object nd thee e moe field lines ne the point of the object. 44. () ind the totl oulomb foce on chge of.00 n locted t 4.00 cm in igue 8.(b), given tht.00 µ. (b) ind the - position t which the electic field is zeo in igue 8.(b). () Accoding to igue 8., the point chges e given by.00 µ t.00 cm;.00 µ t.00 cm; 8.00 µ t 8.00 cm nd 4.00 µ t 4.0 cm If n chge is plced t 4.00 cm, the foce it feels fom othe chges is 9
ollege Physics Student s Mnul hpte 8 K found fom the eution. The net foce is the vecto ddition of the foce due to ech point chge, but since the point chges e ll long the - is, the foces dd like numbes; thus the net foce is given by K K K8 K4 8 4 K 8 4 8 4 To the ight, notice tht the tem involving the chge hs the opposite sign becuse it pulls in the opposite diection thn the othe thee chges. Substituting in the vlues given: 9.00 0 9.m (.00 0 ).00 0.00 0.00 0 ( 0.0400 m 0.000 m) ( 0.000 m 0.0400 m) ( 0.0800 m 0.0400 m).00 0 ( 0.40 m 0.0400 m) (b) The only possible loction whee the totl electic field could be zeo is between.00 -nd 0. 8.00 (ight) cm, since o 0. in tht nge to the the leftclosest chges cete foces on the test chge in opposite diections. So tht is the only egion we will conside. o the totl electic field to be zeo between.00 nd 8.00 cm, we know tht: K K K8 K4 0 8 4 Dividing by common fctos nd ignoing units (but emembeing hs unit of cm), we cn get simplified epession: ( ) ( ) ( ) ( ) y 8 4 We cn gph this function, using gphing clculto o gphing pogm, to detemine the vlues of tht yield y 0. Theefoe the totl electic field is zeo t 6.07cm. 40
ollege Physics Student s Mnul hpte 8 0. () ind the electic field t the cente of the tingul configution of chges in igue 8.4, given tht.0 n, b 8. 00 n, nd.0 n. (b) Is thee ny combintion of chges, othe thn b c, tht will poduce zeo stength electic field t the cente of the tingul configution? c () To detemine the electic field t the cente, we fist must detemine the distnce fom ech of the chges to the cente of the tingle. Since the tingle is euiltel, the cente of the tingle will be hlfwy coss the bse nd / of the wy up the height. To detemine the height use the Pythgoen theoem, o the height is given by h (.0 cm) (. cm).7 cm fom ech chge to the cente of the tingle is / of.7 cm, o Q (.7 cm) 4.4 cm.since k,. So the distnce 9.0 0 k ( 9.00 0 m / ) 08 / ( 0.44 m) t 90 ngle below the hoizontl, b 9 8.00 0 b k ( 9.00 0 m / ) 47 / t 0 ( 0.44 m) ngle below the hoizontl, nd c 9.0 0 c k ( 9.00 0 m / ) 68.0 / ( 0.44 m) t 0 ngle bove the hoizontl. Adding the vectos by components gives: y y cos 0 / sin ( 90 ) b cos( 0 ) c cos0 47 / ( 0.860) 68.0 / ( 0.8660) 97 / ( 90 ) b sin( 0 ) c sin0 ( 0.000) 68/ ( 0.000) 48 / 08 / - 47 / So tht the electic field is given by: y θ tn tn below the hoizontl. y ( 97 /) ( 48 /) 470 / 48 / 4.6, o 4.7 0 97 / nd /, 4.6 (b) o, thee e no combintions othe thn tht will poduce b c 4
ollege Physics Student s Mnul hpte 8 zeo stength field t the cente of the tingul configution becuse of the vecto ntue of the electic field. onside the two cses: () ll chges hve the sme sign nd () one chge hve diffeent sign thn the othe two. o cse (), symmety dicttes tht the chges must be ll the sme mgnitude, if test chge is not to feel foce t the cente of the tingle. o cse (), positive test chge would feel foce towds the negtive chge(s) nd wy fom the positive chge(s). Theefoe thee is no combintion tht would poduce zeo stength electic field t the cente of the tingle. 8.8 APPLIATIOS O LTROSTATIS 6. Wht cn you sy bout two chges nd, if the electic field one- fouth of the wy fom to is zeo? If the electic field is zeo /4 fom the wy of nd, then we know fom the eution Q k tht K The chge is 9 times lge thn. K so tht ( ) ( ) 9 6. Unesonble Results () Two 0.00 g indops in thundehed e.00 cm pt when they ech cuie.00 m chges. ind thei cceletion. (b) Wht is unesonble bout this esult? (c) Which pemise o ssumption is esponsible? () To detemine the cceletion, use ewton's Lws nd the eution k : k m k m 9 ( 9.00 0 m )(.00 0 ) ( 0.00 0 kg)( 0.000 m).80 0 m s (b) The esulting cceletion is unesonbly lge; the indops would not sty togethe. (c) The ssumed chge of.00 m the ode of µ o less. is much too get; typicl sttic electicity is on 4