Chapte 9 lectic Chages, Foces, an Fiels 6 9. One in a million (0 ) ogen molecules in a containe has lost an electon. We assume that the lost electons have been emove fom the gas altogethe. Fin the numbe of electons fom the infomation given in the poblem, an then multipl b the chage pe electon of.60 0 9 C. Multipl the numbe of electons b e Q f nn e A 6 9 0.85 mol 6.0 0 mol.60 0 C 0.8 C. A lage numbe of electons an potons ae collecte togethe into a single sstem. Use the total numbe of chages an the electical chage of the sstem to etemine the numbe of potons an electons. Sum the pouct of the paticle numbes an thei masses to fin the total mass.. (a) Wite out euations fo the total numbe of N + N 55 chages an thei chage ( p e). Substitute the fist euation into the secon an solve fo e p e N N Q ( 55 e) N. (b) Detemine N N p p 55 Ne 55 9 59 e e N Ne Q 5.56 0 C Ne ( 55 Qe) 55 9 9.60 0 C. Fin the total mass M Nm + Nm total e e p p 5 9 9.09 0 kg 59.6 0 kg 9.9 0 kg + 0. Two chages of uneual magnitue eet an electostatic foce on each othe. Use Coulomb s law (euation 9-5) to fin the magnitue of the foce between the two chages.. (a) Appl euation 9-5 iectl ( 6 )( 6 ). 0 C. 0 C 9 F k 8.99 0 N m / C.9 N ( 0.55 m). (b) The magnitue of the electostatic foce epens upon the pouct of the chages of both paticles, so the negative chage epeiences a foce magnitue that is the same as that epeience b the positive chage. 6. Thee chages ae aange as inicate in the figue an eet electostatic foces on each othe. Let the -ais be along the line of the thee chages with the positive iection pointing to the ight. Use Coulomb s law (euation 9-5) an the supeposition of foces to fin the electostatic foce (magnitue an iection) on. The foce fom will be attactive an to the left, an the foce fom will be attactive an to the ight.. (a) Wite Coulomb s law using vecto notation F F + F k + k
. Substitute the chage magnitues given in the figue k k F [ + ] (.0) + (.0)(.0) [ ] [ ] ( 9 )( 6 8.99 0 N m / C 0 C ) k.0.0 ( 0 N) ( 0.9 m). The electostatic foce on is 0 N 0. kn towa. (b) If the istance wee tiple, the magnitue woul be cut to a ninth an the iection woul be unchange. 9. A poton is situate below a 0.5 nc chage such that the upwa electical foce on the poton just balances the ownwa gavitational foce. The positivel chage poton must be positione below the negative chage so that the upwa foce of electical attaction will balance the ownwa foce of gavit on the poton. Set the magnitues of the two foces eual to each othe an solve fo. +e 0.5 nc Set the magnitues F F an solve fo e g ke ke ke 9 9 9 ( 8.99 0 N m / C )( 0.5 0 C)(.60 0 C) (.6 0 kg)( 9.8 m/s ) 5500 m 5.5 km below. Fou chages ae situate at the cones of a suae as shown in the iagam at the ight. The foce on chage is a vecto sum of the foces fom the othe thee chages. Let be at the oigin an be on the negative -ais. Use Coulomb s law (euation 9-5) to fin the vecto sum of the thee foces, fom which we can fin the magnitue an iection of the electostatic foce on.. (a) Fin F. Fin F. Fin F (.0 )(.0 ) 6.0 k k k F (.0 )(.0 ) k k k F (.0 ) k k.0 k F ( ). Fin the vecto sum of the thee foces 5. Fin the iection of F fom the + ais 6.0k.0 k k.0 k F + (.0 ) ( ).0k + 9 6 ( )( ).0 8.99 0 N m / C. 0 C + F + ( 0. m) (.5 N) (.8 N) F,.8 N θ tan tan 66 F.5 N, (.0 ) ( )
6. Fin the magnitue of F.5 N +.8 N 8.5 N F. (b) If the istance wee ouble, the magnitue of the foce woul be cut to one-fouth an the iection woul be unchange.. Thee chages ae aange along the -ais as inicate in the figue an eet electostatic foces on each othe. Let the -ais be along the line of the thee chages with the positive iection pointing to the ight. Let epesent the istance between an. Use Coulomb s law (euation 9-5) an the supeposition of foces to fin the electostatic foce (magnitue an iection) on an set it eual to zeo. Supposing to be a positive chage, the foce fom will be epulsive an to the ight, an the foce fom will be attactive an to the left. Fin the appopiate value of b fining the oots of the esulting uaatic epession.. (a) Use Coulomb s law to set 0 set F F + F k k 0 F ( D). Set the magnitues of the two tems eual to each othe, ivie both sies b k an simplif b iviing both sies b an then taking the suae oot D ( ) D D ± D 0 cm 5.8 cm, 5 cm ± ± 5. µ C 9.9 µ C. (b) No, the answe to pat (a) oes not epen on whethe is positive o negative. If wee negative, it woul be pulle to the left b an pushe to the ight b, so the foces woul still balance at 5 cm. 5. Two chages ae place on the -ais as shown at ight an ceate an electic fiel in the space aoun them. Use euation 9-8 to fin the magnitue an iection of the electic fiels ceate b each of the two chages at the specifie locations, then fin the vecto sum of those fiels to fin the electic fiel. At.0 cm the fiel fom will point in the iection an the fiel fom will point in the + iection..0 cm 0.0 cm 0.0 cm. (a) Sum the fiels pouce b the two chages at.0 cm. (b) Repeat fo.0 cm k k ( ) + k + 6 6 9 6. 0 C 9.5 0 C ( 8.99 0 N m / C ) + (.0 0 N/C ) ( 0.00 m) ( 0.0 m) k k + k + 6 6 9 6. 0 C 9.5 0 C ( 8.99 0 N m / C ) + ( 5.9 0 N/C ) ( 0.00 m) ( 0.060 m). An electic chage epeiences an upwa foce ue to the pesence of an electic fiel. The negative chage epeiences an upwa electic foce, so we can conclue the electic fiel points ownwa. Set the magnitue of the electic foce (euation 9-9) eual to the magnitue of the weight in oe to fin the magnitue of the electic fiel. Then use the known electic fiel to fin the foce an hence the acceleation of the object when its chage is ouble. Let upwa be the positive iection.
. (a) Set F an solve fo. (b) Use Newton's Secon Law to fin a ( 0.0kg ))( 9.8 m/s ) 6.6 0 C ( ) ma ( ) (. 0 N/C) F F F ma ma g ma a g ( 9.8 m/s ). 0 N/C 8. Thee chages ae positione as shown at ight. ach of the thee chages pouces its own electic fiel that suouns it. The total electic fiel at an point is the vecto sum of the fiels fom each chage. Use euation 9-0 an the component metho of vecto aition to fin the magnitue electic fiel at the points inicate in the poblem statement. Let be at the oigin an be on the positive -ais.. (a) At a point halfwa between chages an the vectos an cancel one anothe. The emaining contibution comes fom. Fin the istance fom to the mipoint of the opposite sie. Appl euation 9-0 to fin + 0.095 m 0.0555 m 9 6 k 8.99 0 N m / C 5.00 0 C ( 0.0555 m) 6.89 0 N/C. (b) At this location, the electic fiels fom an a, an the esulting fiel points towa. The fiel ue to will have the same magnitue as foun in pat (a) an will be pepenicula to the combine fiels of an. The vecto sum of the electic fiels fom all thee chages will have a magnitue geate than that foun in pat (a).. (c) Fin the components of 5. Fin the components of 6. Fin the components of. Let an fin the vecto sum 8. Detemine the magnitue of k k + + k k cos 60 sin 60 ( cos0 sin 0 ) + + k + + k k cos 60 sin 60 ( ) 9 6 ( )( ) k + + 8.99 0 N m / C 5.00 0 C 8.0 ( 0.095 m) 8.9 0 N/C 9 MN/C
50. Thee ientical chages ae place as shown in the figue at the ight. ach of the thee chages pouces its own electic fiel that suouns it. The total electic fiel at an point is the vecto sum of the fiels fom each chage. As illustate in the figue, at the mipoint of an of the thee sies of the tiangle two of the thee vectos will cancel. Use euation 9-0 to fin the magnitue electic fiel at the mipoints b fining the magnitue of the thi, unbalance vecto.. (a) Fist fin the istance fom to the mipoint of the opposite sie + 0. m 0.8 m 9 ( Nm 6 8.99 0 )(. 0 C). Now appl euation 9-0 k. 0 N/C C 6 ( 0.8). (b) Due to smmet, the thee electic fiel vectos at the cente of the tiangle cancel out an the fiel thee is zeo. So, the magnitue thee is less than that at the mipoint of a sie. 95. A positivel chage sphee attache to a elae, hoizontal sping slies without fiction as it is attacte to a negative chage that is bought to a hoizontal istance 0. m fom the cente of the sphee, at which point the sphee is in euilibium. Set the foce of electostatic attaction between the two chages eual to the sping foce that hols the sphee in static euilibium, an solve fo. At that point the sphee has move 0. m towa the chage, so that the two ae a istance 0. m apat.. Set the foce kq + on the sphee eual to zeo F k ( ). Solve the esulting s epession fo ( ) s 0 kq k ± kq k s 9 6 6 ( 8.99 0 N m / C )( 8.55 0 C)(. 0 C) kq ± 0. m ± k s 0.5 m o 0.006 m (the negative oot is etaneous) ( 89. N/m)( 0. m) Insight If the point chage wee positivel chage an bought to the location, the sping woul compess until the sping foce balances the electostatic foce, an the sphee woul come to euilibium at 0.06 m. T veifing this solution ouself, but be foewane the solution involves fining the oots of a cubic euation in! 5