Solutions to Problems : Chapter 19 Problems appeared on the end of chapter 19 of the Textbook

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Rudolph Hall
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1 Solutions to Poblems Chapte 9 Poblems appeae on the en of chapte 9 of the Textbook 8. Pictue the Poblem Two point chages exet an electostatic foce on each othe. Stategy Solve Coulomb s law (equation 9-5) fo the sepaation istance. Solution Solve equation 9-5 fo q ( 9 )( N m / C. C)( 9. 6 C) F. N Insight Although. N is only about 6. ounces of foce, these micocoulomb-size chages can exet such a foce ove the substantial istance of.9 m o. ft..9 m 8. Pictue the Poblem The honeybee acquies an electostatic chage in active flight. Stategy Use the magnitue of an electon s chage e to fin the numbe of electons that coespon to the 9. pc total chage. Then use Coulomb s law to fin the magnitue of the foce between the two chage bees, an compae it with the weight of a single bee. Solution. (a) Fin the numbe of electons Q 9. C N 9 e.6 C/electon electons. (b) Use Coulomb s law to fin the ( ) q 9. C 9 F 8.99 N m / C foce between two chage bees 5. N. Detemine the weight of a bee W mg. Calculate the atio of the foces F mg (. m). kg 9.8 m/s. N 5. N.9. N Insight The electical foce between the bees is a tiny faction of thei weight because the amount of electical chage is quite small. It woul equie a chage of only.68 nc on each bee fo the electical foce to equal the weight! 5. Pictue the Poblem An electic fiel exists aoun a 5. µc chage at the oigin. Stategy Use the efinition of the electic fiel (equation 9-) to etemine its magnitue. Solution. (a) Apply equation 9- iectly. (b) Repeat fo the new istance 9 6 ( 8.99 N m / C )( 5. C) (.m) 9 6 ( 8.99 N m / C )( 5. C) (.m) Insight When the istance fom the chage is ouble, the fiel is cut to a fouth..5 N/C. N/C

2 . Pictue the Poblem Two chages ae place on the x-axis as shown at ight an ceate an electic fiel in the space aoun them. Stategy Use equation 9-8 to fin the magnitue an iection of the electic fiels ceate by each of the two chages at the specifie locations, then fin the vecto sum of those fiels to fin the electic fiel. At x. cm the fiel fom q will point in the xˆ iection an the fiel fom q will point in the +xˆ iection. q q. cm. cm. cm x Solution. (a) Sum the fiels pouce by the two chages at x. cm q q ( xˆ) + xˆ k + xˆ C 9.5 C ( 8.99 N m / C ) + x ˆ (. N/C ) x ˆ (.m) (.m). (b) Repeat fo x. cm q q xˆ + xˆ k + xˆ C 9.5 C ( 8.99 N m / C ) + x ˆ ( 5.9 N/C ) x ˆ (. m) (.6 m) Insight Although q has a lage magnitue than q, at x. cm the close istance to q means its contibution to the fiel is lage than the contibution fom q, an the fiel points in the xˆ iection. 6. Pictue the Poblem Thee chages ae positione as shown at ight. Stategy ach of the thee chages pouces its own electic fiel that suouns it. The total electic fiel at any point is the vecto sum of the fiels fom each chage. Use equation 9- an the component metho of vecto aition to fin the magnitue electic fiel at the points inicate in the poblem statement. Let q be at the oigin an q be on the positive x-axis. Solution. (a) At a point halfway between chages q an q the vectos an cancel one anothe. The emaining contibution comes fom q. Fist fin the istance fom q to the mipoint of the opposite sie. Apply equation 9- to fin +.5 m.8 m 9 6 ( 8.99 N m / C )( 5. C) (.8 m).9 N/C. (b) At this location, the electic fiels of q an q a, an the esulting fiel points towa q. The fiel ue to q will have the same magnitue as foun in pat (a) an will be pepenicula to the

3 combine fiels of q an q. 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 q q q q an fin the vecto sum ( cos xˆ + sin yˆ) ˆ ˆ + x y x y x y cos 6 ˆ sin 6 ˆ ( ˆ ˆ) x y x y cos 6 ˆ sin 6 ˆ ( ˆ ˆ) + + ˆ ˆ + + x y 8. Detemine the magnitue of ( )( ) 8.99 N m / C 5. C N/C (.5 m) Insight As expecte, the fiel is lage at the point miway between q an q, about six times lage in magnitue than at the point miway between q an q. 8. Pictue the Poblem Thee ientical chages ae place as shown in the figue at the ight. Stategy ach of the thee chages pouces its own electic fiel that suouns it. The total electic fiel at any point is the vecto sum of the fiels fom each chage. As illustate in the figue, at the mipoint of any of the thee sies of the tiangle two of the thee vectos will cancel. Use equation 9- to fin the magnitue electic fiel at the mipoints by fining the magnitue of the thi, unbalance vecto. Solution. (a) Fist fin the istance fom q to the mipoint of the opposite sie +. m.8 m 9 ( Nm )(. C). Now apply equation 9-. N/C C 6 (.8). (b) Due to symmety, 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. Insight Wheneve ientical chages ae aange in a symmetical fashion, the electic fiel at thei geometic cente is zeo.

4 . Pictue the Poblem The fiel pouce by an unknown chage at an unknown position is given at two points, as epicte in the figue at the ight. Stategy Because both measuements of the electic fiel point in the positive x-iection, the point chage must be on the x-axis. Also, since the magnitue of an electic fiel is lage the close it is measue to its souce, the position at the point chage must be geate than x. cm. Futhemoe, since the vectos point towa the ight, they must point towa the chage an the chage must be negative. Wite out equation 9- fo the two given fiels an combine them to get a quaatic equation in x. Solve the expession to fin the location of the chage. Then use eithe of the two fiel equations to fin the magnitue of the chage. Solution. (a) Use equation 9- to expess an the fiel magnitues ( xx) ( xx). Combine the equations to eliminate q, take the squae oot of both sies, an solve fo x. (b) Fin the magnitue of q fom ( ) ( ) ( ) ( ) x x x x x x x x x x x. N/C ( 5. cm) 5. N/C (. cm) cm. N/C 5. N/C ( ) (. N/C)(..5 m) 5. cm. cm x 9 x x q 8. C 8 pc k 8.99 N m / C. Because the fiel vecto points towa the chage it must be negative q 8 pc Insight If the chage wee positive, the fiel magnitues woul be the same but the vectos woul point towa x. The ules of subtaction limit the answes to only two significant figues. q x 56. Pictue the Poblem A poton is acceleate fom est in a unifom electic fiel an tavels along a staight line. Stategy Use Newton's Secon Law to fin the acceleation of the poton fom the electic foce F q exete by the electic fiel. Once the acceleation is known we can fin the spee of the paticle fom the equation fo velocity as a function of position, v v + aδ x (equation -). Solution. (a) Apply Newton's Secon Law to fin a. Solve equation - fo v when v F q ma a q m v v + aδ x + q m Δ x. Fin v fo Δ x. m. (b) Fin v fo Δ x. m 9 5 ( )( ).6 C.8 N/C. m v.6 kg 9 5 ( )( ).6 C.8 N/C. m v.6 kg 5.55 m/s 6. m/s Insight The lage chage-to-mass atio of elementay paticles pouces lage acceleations when they

ae immese in an electic fiel. In this case the acceleation of the poton is an astouning. m/s! 58. Pictue the Poblem A point chage situate at the oigin pouces an electic fiel that completely suouns it.

5 ae immese in an electic fiel. In this case the acceleation of the poton is an astouning. m/s! 58. Pictue the Poblem A point chage situate at the oigin pouces an electic fiel that completely suouns it. Stategy Use the efinition of the electic fiel (equation 9-) to fin the magnitue of the chage that ceates the stipulate electic fiel. Since the electic fiel points towa the oigin an theefoe towa the chage, the sign of the chage must be negative. Solution Solve equation 9- fo q, assuming the chage is negative ( 6, N/C)(.5 m) q 9 k 8.99 N m /C q. μc Insight If the sign of the chage ha been positive, the fiel woul have pointe in the 6. Pictue the Poblem A cylinical Gaussian suface is built aoun a long, thin wie that has a chage pe unit length λ on it. Stategy In poblem 55 we use Gauss s law to fin that λ πε whee is the istance fom the wie an λ is the wie s chage pe unit length. Use this expession an the given infomation to fin λ, an then use a atio to fin when is euce by a facto of. Solution. (a) Solve λ fo λ πε. (b) Use a atio to fin the new λ πε 6. C x ˆ iection. π 8.85 C / N m. m 5, N/C 6. C/m λ πε new new ol ol ol λ πε ol new ol new ol. m.95 m Insight Anothe way to solve pat (b) is to use the λ fom pat (a) to solve the given expession fo. Howeve, using a atio can often be a time-saving step that also avois the accumulation of ouning eos that can occu.

4. Compare the electric force holding the electron in orbit ( r = 0.53

4. Compare the electric force holding the electron in orbit ( r = 0.53 Electostatics WS Electic Foce an Fiel. Calculate the magnitue of the foce between two 3.60-µ C point chages 9.3 cm apat.. How many electons make up a chage of 30.0 µ C? 3. Two chage ust paticles exet a