ˆ (0.10 m) E ( N m /C ) 36 ˆj ( j C m)


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1 7.. = = 3 = 4 = 5. The electrc feld s constant everywhere between the plates. Ths s ndcated by the electrc feld vectors, whch are all the same length and n the same drecton Model: The dstances to the observaton ponts are large compared to the sze of the dpole, so model the feld as that of a dpole moment. The dpole conssts of charges q along the yaxs. The electrc feld n (a) ponts down. The feld n (b) ponts up. Solve: (a) The dpole moment s p ( qs, from to +) (. 9 C). m j. j C m The electrc feld at ( cm, cm), whch s at dstance r. m n the plane perpendcular to the electrc dpole, s p 9. j C m (9. N m /C ) 8. j N/C r (. m) The feld strength, whch s all we re asked for, s 8. N/C. (b) The electrc feld at ( cm, cm), whch s at r. m along the axs of the dpole, s p 9 (. j C m) (9. N m /C ) 36 j N/C r (. m) The feld strength at ths pont s 36 N/C.
2 7.9. Model: The rods are thn. Assume that the charge les along a lne. Because both the rods are postvely charged, the electrc feld from each rod ponts away from the rod. Because the electrc felds from the two rods are n opposte drectons at P, P, and P 3, the net feld strength at each pont s the dfference of the feld strengths from the two rods. Solve: xample 7.3 gves the electrc feld strength n the plane that bsects a charged rod: rod Q 4 r r / The electrc feld from the rod on the rght at a dstance of cm from the rod s rght C 9. m. m.5 m N m /C.765 N/C 5 The electrc feld from the rod on the rght at dstances cm and 3 cm from the rod are.835 N/C and 5.54 N/C. The electrc felds produced by the rod on the left at the same dstances are the same. Pont P s. cm from the rod on the left and s 3. cm from the rod on the rght. Because the electrc felds at P have opposte drectons, the net electrc feld strengths are At. cm At. cm At 3. cm N/C.54 N/C.5 N/C N/C.835 N/C N/C N/C.54 N/C.5 N/C
3 7.4. Model: Model each dsk as a unformly charged dsk. When the dsk s postvely charged, the onaxs electrc feld of the dsk ponts away from the dsk. Solve: (a) The surface charge densty on the dsk s 9 Q Q 5 C C/m A R.5 m From quaton 7.3, the electrc feld of the left dsk at z. m s C/m z 8.85 C /N m 38, N/C R z.5 m /. m Hence, (38, N/C, rght). Smlarly, the electrc feld of the rght dsk at z. m (to ts left) s (38, N/C, left). The net feld at the mdpont between the two dsks s N/C. (b) The electrc feld of the left dsk at z.5 m s.5 N/C.5 N/C, rght 8.85 C /N m.5 m /. m 4.85 N/C, left. The net feld C/m 5 5 z Smlarly, the electrc feld of the rght dsk at z.5 m (to ts left) s s thus The feld strength s N/C N/C, rght 7.9. Model: The electrc feld n a regon of space between the plates of a parallelplate capactor s unform. Solve: The electrc feld nsde a capactor s Q A. Thus, the charge needed to produce a feld of strength s 6 Q A 8.85 C /N m. m 3. N/C 33.4 nc The number of electrons transferred from one plate to the other s C 9..6 C
4 7.4. Model: The nfnte negatvely charged plane produces a unform electrc feld that s drected toward the plane. Solve: From the knematc equaton of moton v v ax and F q ma, q v a x mv m x q Furthermore, the electrc feld of a plane of charge wth surface charge densty s. Thus, kg. m/s 8.85 C /N m C. C/m mv x.85 m q 7.7. Model: The sze of a molecule s. nm. The proton s. nm away, so r >> s and we can use quaton 7. for the electrc feld n the plane that bsects the dpole. Solve: You can see from the dagram that F dpole on proton s opposte to the drecton of p. The magntude of the dpole feld at the poston of the proton s The magntude of p 5. C m (9. N m /C ) 5.64 N/C dpole r (. m) F dpole on proton Includng the drecton, the force s F s dpole on proton edpole (.6 C)(5.64 N/C) 9. N F dpole on proton (9. 3 N, drecton opposte p ).
5 7.3. Model: The electrc feld s that of two charges q and q located at x a. Solve: q (a) At pont, the electrc feld from q s q q q 4 4 5a a a ponts toward q and makes an angle aa tan below the xaxs, hence q q q q j j j 4 5a 4 5a 4 a The electrc feld from the q s q cos sn q q q 4 a a 4 5a ponts away from q and makes an angle aa Addng these two vectors, tan above the xaxs. So, q j j q cos sn 4 q 4 5a 4 5 5a q 3 j net q q 4 5 5a At pont, the electrc feld from q ponts toward q, so The electrc feld from q ponts away from q, so Addng these two vectors, q q 4 a q q 4 9a
6 7.43. Model: The electrc feld s that of a lne charge of length. Please refer to Fgure P7.43. et the bottom end of the rod be the orgn of the coordnate system. Dvde the rod nto many small segments of charge q and length y. Segment creates a small electrc feld at the pont P that makes an angle wth the horzontal. The feld has both x and y components, but z N/C. The dstance to segment from pont P s Solve: x y The electrc feld created by segment at pont P s. q cos sn q x j y 4 x y j 4 x y x y x y The net feld s the sum of all the, whch gves. q s not a coordnate, so before convertng the sum to an ntegral we must relate charge q to length y. Ths s done through the lnear charge densty Q/, from whch we have the relatonshp Wth ths charge, the sum becomes Q q y y Q / x y y y j 3 / 3 / 4 x y x y Now we let y dy and replace the sum by an ntegral from y m to y. Thus, Q/ xdy ydy Q/ y j x j 3/ 3/ 4 x y x y 4 x x y x y Q Q x j 4 x x 4 x x
7 7.47. Model: Assume that the quartercrcle plastc rod s thn and that the charge les along the quartercrcle of radus R. The orgn of the coordnate system s at the center of the crcle. Dvde the rod nto many small segments of charge q and arc length s. Solve: (a) Segment creates a small electrc feld at the orgn wth two components: cos sn x y Note that the angle depends on the locaton of the segment. Now all segments are at dstance r R from the orgn, so q q 4 r 4 R The lnear charge densty of the rod s Q/, where s the rod s length ( quartercrcumference R/). Ths allows us to relate charge q to the arc length s through Q Q q s s s R Usng s R, the components of the electrc feld at the orgn are q Q Q cos cos cos R x 4 R 4 R R 4 R q Q Q sn sn sn R y 4 R 4 R R 4 R (b) The x and ycomponents of the electrc feld for the entre rod are the ntegrals of the expressons n part (a) from rad to /. We have (c) The ntegrals are Q / Q / x cos d sn d 4 4 y R R / snd cos / cos cos d The electrc feld s Q j 4 R / / cos sn sn sn
8 7.53. Model: Assume that the electrc feld nsde the capactor s constant, so constantacceleraton knematc equatons apply. Please refer to Fgure P7.53. Solve: (a) The force on the electron nsde the capactor s q F ma q a m Because s drected upward (from the postve plate to the negatve plate) and q 9.6 C, the acceleraton of the electron s downward. We can therefore wrte the above equaton as smply a y q/m. To determne, we must frst fnd a y. From knematcs, x x v xt t axt t.4 m m v cos45t t m.4 m t t 6 5. m/s cos 45 Usng the knematc equatons for the moton n the ydrecton, 8.34 s t t q t t v y v y ay m/s v sn45 m C.34 s m v sn kg 5. m/s sn 45 q t t N/C 3.6 N/C 6 (b) To determne the separaton between the two plates, we note that y m and v y (5. m/s)sn 45, but at y y, the electron s hghest pont, v y m/s. From knematcs, From part (a), v v a y y m /s v sn 45 a y y y y y y v sn 45 v y y a 4a.6 9 C355 N/C q 4 ay 6.4 m/s 3 m 9. kg 6 5. m/s y y. m. cm m/s Ths s the heght of the electron s trajectory, so the mnmum spacng s. cm. y y
9 7.58. Model: The electron orbtng the proton experences a force gven by Coulomb s law. Solve: The force that causes the crcular moton s where we used v r T rf. The frequency s qp qe mv m e e 4 r f F 4 r r r N m /C.6 C 3 qp qe 5 f 6.56 Hz mr e 4 9. kg 5.3 m = 4 > >. The electrc feld strength s larger n the regon where the feld lnes are closer together ( 3 and 4 ) and smaller where the feld lnes are farther apart (a) (b) A N QA / A Af 3.63 / f f, so 3.63 N Q A Af N f N F f Nf Ff e f e and F e e, so. F N 7.. = = 3 = 4 = 5. The electrc feld s constant everywhere between the plates. Ths s ndcated by the electrc feld vectors, whch are all the same length and n the same drecton.
10 7.5. Model: The dstances to the observaton ponts are large compared to the sze of the dpole, so model the feld as that of a dpole moment. The dpole conssts of charges q along the yaxs. The electrc feld n (a) ponts down. The feld n (b) ponts up. Solve: (a) The dpole moment s p ( qs, from to +) (. 9 C). m j. j C m The electrc feld at ( cm, cm), whch s at dstance r. m n the plane perpendcular to the electrc dpole, s p 9. j C m (9. N m /C ) 8. j N/C r (. m) The feld strength, whch s all we re asked for, s 8. N/C. (b) The electrc feld at ( cm, cm), whch s at r. m along the axs of the dpole, s p 9 (. j C m) (9. N m /C ) 36 j N/C r (. m) The feld strength at ths pont s 36 N/C.
11 7.9. Model: The rods are thn. Assume that the charge les along a lne. Because both the rods are postvely charged, the electrc feld from each rod ponts away from the rod. Because the electrc felds from the two rods are n opposte drectons at P, P, and P 3, the net feld strength at each pont s the dfference of the feld strengths from the two rods. Solve: xample 7.3 gves the electrc feld strength n the plane that bsects a charged rod: rod Q 4 r r / The electrc feld from the rod on the rght at a dstance of cm from the rod s rght C 9. m. m.5 m N m /C.765 N/C 5 The electrc feld from the rod on the rght at dstances cm and 3 cm from the rod are.835 N/C and 5.54 N/C. The electrc felds produced by the rod on the left at the same dstances are the same. Pont P s. cm from the rod on the left and s 3. cm from the rod on the rght. Because the electrc felds at P have opposte drectons, the net electrc feld strengths are At. cm At. cm At 3. cm N/C.54 N/C.5 N/C N/C.835 N/C N/C N/C.54 N/C.5 N/C
12 7.4. Model: Model each dsk as a unformly charged dsk. When the dsk s postvely charged, the onaxs electrc feld of the dsk ponts away from the dsk. Solve: (a) The surface charge densty on the dsk s 9 Q Q 5 C C/m A R.5 m From quaton 7.3, the electrc feld of the left dsk at z. m s C/m z 8.85 C /N m 38, N/C R z.5 m /. m Hence, (38, N/C, rght). Smlarly, the electrc feld of the rght dsk at z. m (to ts left) s (38, N/C, left). The net feld at the mdpont between the two dsks s N/C. (b) The electrc feld of the left dsk at z.5 m s.5 N/C.5 N/C, rght 8.85 C /N m.5 m /. m 4.85 N/C, left. The net feld C/m 5 5 z Smlarly, the electrc feld of the rght dsk at z.5 m (to ts left) s s thus The feld strength s N/C N/C, rght 7.9. Model: The electrc feld n a regon of space between the plates of a parallelplate capactor s unform. Solve: The electrc feld nsde a capactor s Q A. Thus, the charge needed to produce a feld of strength s 6 Q A 8.85 C /N m. m 3. N/C 33.4 nc The number of electrons transferred from one plate to the other s C 9..6 C
13 7.4. Model: The nfnte negatvely charged plane produces a unform electrc feld that s drected toward the plane. Solve: From the knematc equaton of moton v v ax and F q ma, q v a x mv m x q Furthermore, the electrc feld of a plane of charge wth surface charge densty s. Thus, kg. m/s 8.85 C /N m C. C/m mv x.85 m q 7.7. Model: The sze of a molecule s. nm. The proton s. nm away, so r >> s and we can use quaton 7. for the electrc feld n the plane that bsects the dpole. Solve: You can see from the dagram that F dpole on proton s opposte to the drecton of p. The magntude of the dpole feld at the poston of the proton s The magntude of p 5. C m (9. N m /C ) 5.64 N/C dpole r (. m) F dpole on proton Includng the drecton, the force s F s dpole on proton edpole (.6 C)(5.64 N/C) 9. N F dpole on proton (9. 3 N, drecton opposte p ).
14 7.3. Model: The electrc feld s that of two charges q and q located at x a. Solve: q (a) At pont, the electrc feld from q s q q q 4 4 5a a a ponts toward q and makes an angle aa tan below the xaxs, hence q q q q j j j 4 5a 4 5a 4 a The electrc feld from the q s q cos sn q q q 4 a a 4 5a ponts away from q and makes an angle aa Addng these two vectors, tan above the xaxs. So, q j j q cos sn 4 q 4 5a 4 5 5a q 3 j net q q 4 5 5a At pont, the electrc feld from q ponts toward q, so The electrc feld from q ponts away from q, so Addng these two vectors, q q 4 a q q 4 9a
15 7.43. Model: The electrc feld s that of a lne charge of length. Please refer to Fgure P7.43. et the bottom end of the rod be the orgn of the coordnate system. Dvde the rod nto many small segments of charge q and length y. Segment creates a small electrc feld at the pont P that makes an angle wth the horzontal. The feld has both x and y components, but z N/C. The dstance to segment from pont P s Solve: x y The electrc feld created by segment at pont P s. q cos sn q x j y 4 x y j 4 x y x y x y The net feld s the sum of all the, whch gves. q s not a coordnate, so before convertng the sum to an ntegral we must relate charge q to length y. Ths s done through the lnear charge densty Q/, from whch we have the relatonshp Wth ths charge, the sum becomes Q q y y Q / x y y y j 3 / 3 / 4 x y x y Now we let y dy and replace the sum by an ntegral from y m to y. Thus, Q/ xdy ydy Q/ y j x j 3/ 3/ 4 x y x y 4 x x y x y Q Q x j 4 x x 4 x x
16 7.47. Model: Assume that the quartercrcle plastc rod s thn and that the charge les along the quartercrcle of radus R. The orgn of the coordnate system s at the center of the crcle. Dvde the rod nto many small segments of charge q and arc length s. Solve: (a) Segment creates a small electrc feld at the orgn wth two components: cos sn x y Note that the angle depends on the locaton of the segment. Now all segments are at dstance r R from the orgn, so q q 4 r 4 R The lnear charge densty of the rod s Q/, where s the rod s length ( quartercrcumference R/). Ths allows us to relate charge q to the arc length s through Q Q q s s s R Usng s R, the components of the electrc feld at the orgn are q Q Q cos cos cos R x 4 R 4 R R 4 R q Q Q sn sn sn R y 4 R 4 R R 4 R (b) The x and ycomponents of the electrc feld for the entre rod are the ntegrals of the expressons n part (a) from rad to /. We have (c) The ntegrals are Q / Q / x cos d sn d 4 4 y R R / snd cos / cos cos d The electrc feld s Q j 4 R / / cos sn sn sn
17 7.53. Model: Assume that the electrc feld nsde the capactor s constant, so constantacceleraton knematc equatons apply. Please refer to Fgure P7.53. Solve: (a) The force on the electron nsde the capactor s q F ma q a m Because s drected upward (from the postve plate to the negatve plate) and q 9.6 C, the acceleraton of the electron s downward. We can therefore wrte the above equaton as smply a y q/m. To determne, we must frst fnd a y. From knematcs, x x v xt t axt t.4 m m v cos45t t m.4 m t t 6 5. m/s cos 45 Usng the knematc equatons for the moton n the ydrecton, 8.34 s t t q t t v y v y ay m/s v sn45 m C.34 s m v sn kg 5. m/s sn 45 q t t N/C 3.6 N/C 6 (b) To determne the separaton between the two plates, we note that y m and v y (5. m/s)sn 45, but at y y, the electron s hghest pont, v y m/s. From knematcs, From part (a), v v a y y m /s v sn 45 a y y y y y y v sn 45 v y y a 4a.6 9 C355 N/C q 4 ay 6.4 m/s 3 m 9. kg 6 5. m/s y y. m. cm m/s Ths s the heght of the electron s trajectory, so the mnmum spacng s. cm. y y
18 7.58. Model: The electron orbtng the proton experences a force gven by Coulomb s law. Solve: The force that causes the crcular moton s where we used v r T rf. The frequency s qp qe mv m e e 4 r f F 4 r r r N m /C.6 C 3 qp qe 5 f 6.56 Hz mr e 4 9. kg 5.3 m = 4 > >. The electrc feld strength s larger n the regon where the feld lnes are closer together ( 3 and 4 ) and smaller where the feld lnes are farther apart (a) (b) A N QA / A Af 3.63 / f f, so 3.63 N Q A Af N f N F f Nf Ff e f e and F e e, so. F N
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