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Transcription

1 THE ELECTRIC FIELD 6 Conceptual Questons 6.. A tny, postve test charge wll be placed at the pont n space and the force wll be measured. From the force F measurement and the charge the electrc feld wll be calculated usng E =. The drecton of the feld wll be the q same as the drecton of the force snce q s postve E = E > E > E. The electrc feld strength s larger n the regon where the feld lnes are closer together ( E and E ) and smaller where the feld lnes are farther apart. λf ( Qf/ Lf ) λf L 6.. (a). But, so = Q = Qf = = λ ( QL / ) λ Lf Ff (b) F λ, so = F (c) tmes the orgnal amount of charge would gve a constant lnear charge densty. So the amount of charge to add to the orgnal s 9 tmes the orgnal charge. λ 6.5. F = ee = e. If the charge densty λ s doubled, then the dstance r from the wre must also be doubled π E r for the force to be the same. Thus r = cm The electrc feld at the center s zero. We can thnk of the straw as beng made up of many rngs of postve charge. At the center of the rng addng all feld vectors gves a resultant electrc feld equal to zero, as shown n the fgure below. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. 6-

2 6- Chapter (a) A f A η f ( QA / ) f A =, so = = =.6 =.6 η ( QA / ) A f Ff (b) =. The total charge on the object does not change as t shrnks. An electron very far from the object F experences t as a pont charge. By very far, we mean much farther away than the sze of the object. Q Q nc Q Q Q Q nc 6.8. N = = = 8, and N A = = = = = N =. π r cm A πr π( r) πr cm Q E f 6.9. (a) E =. If Q s halved, then E s also halved. Thus π E r E = (b) The feld outsde a sphere s the same as that of a pont charge Q located at the center of the sphere. So f the E f radus of the sphere changes, the feld remans the same outsde the sphere at the dstance r = R. So. E = (c) The feld outsde a unformly charged sphere s the same as that of a pont charge Q located at the center of the sphere. Snce r s stll greater than R after beng halved, Q E f Ef = = E = r E π E 6... Dschargng the ball wll cause the restorng force F = ( mg + qe)sn 5 to decrease. Therefore the perod of the pendulum wll ncrease. Also, one can thnk of ths as extendng the equaton for the perod of a pendulum by changng L L the acceleraton from smply g to g + qe/ m. Thus T = π π, and as q, T ncreases. g ( g + qe/ m) 6.. E = E = E = E = E 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. 6.. (a) E f Nf/ E Nf Qf/ Af Ef Qf = = =. If Q s doubled (A = constant), = =. E N / E N Q/ A E Q (b) If L s doubled then A = A (Q = constant), so E f (c) E does not depend on d.. E = f E f A A = = =. E A A f Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

3 The Electrc Feld (a) Accelerates to the rght. (b) Remans n place. (c) Accelerates to the left. Fp 6.. (a) The forces on the two charges are equal, so =, F = qe, and they each have the same amount of F e charge and are placed n the same feld. F ap me (b) The acceleraton of the electron s larger because the electron has smaller mass. Snce a =, = <. m a m 6.5. (a) No, the trangle s not n equlbrum. The electrc feld s unform between the plates, so the force on the two postve charges to the left s the same as the total force on the two negatve charges to the rght. The trangle wll rotate counterclockwse as shown. e p (b) The trangle wll reman n place, snce the total force on t s zero. Exercses and Problems Secton 6. The Electrc Feld of Multple Pont Charges 6.. Model: The electrc feld s that of the two charges placed on the y-axs. We denote the upper charge by q and the lower charge by q. Because both the charges are postve, ther electrc felds at P are drected away from the charges. Solve: The electrc feld from q s 9 9 q (9. Nm /C )(. C) E =, θ below x-axs = (cosθ sn θj) πε r (.5 m) + (.5 m) Because tanθ = 5 cm/5 cm =, the angle θ = 5. Hence, Smlarly, the electrc feld from q s E (5 N/C) = j q E =, θ above x-axs = (5 N/C) + j r E net at P = E + E = (5 N/C) = 7.6 N/C Thus, the strength of the electrc feld s 7.6 N/C and ts drecton s along the +x-axs. Assess: Because the charges are located symmetrcally on ether sde of the x-axs and are of equal value, the y-components of ther felds wll cancel when added. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

4 6- Chapter Model: The electrc feld s that due to superposton of the felds of the two. nc charges located on the y-axs. We denote the top. nc charge by q and the bottom. nc charge by q. The electrc felds ( E and E) of both the postve charges are drected away from ther respectve charges. Wth vector addton, they yeld the net electrc feld E net at the pont P ndcated by the dot. Solve: The electrc felds from q and q are 9 9 q (9. N m /C )(. C) E =, along + x-axs = =,8 N/C πε r (.5 m) Because tanθ = cm/5 cm, θ = tan () = 6.. So, q E =, θ above + x-axs r 9 9 (9. N m /C )(. C) E = (cos6. + sn 6. j) = ( j) N/C (. m) + (.5 m) The net electrc feld s thus E net at P = E + E = (, j) N/C To fnd the angle ths net vector makes wth the x-axs, we calculate 9 N/C tan φ = φ = 9.,766 N/C Thus, the strength of the electrc feld at P s E net = (,766 N/C ) + (9 N/C) =,9 N/C = N/C and E net makes an angle of 9. below the +x-axs. Assess: Because of the nverse square dependence on dstance, E < E. Addtonally, because the pont P has no specal symmetry relatve to the charges, we expected the net feld to be at an angle relatve to the x-axs. 6.. Model: The electrc feld s that due to superposton of the felds of the two. nc charges located on the y-axs. We denote the top. nc charge by q and the bottom. nc charge by q. The electrc felds ( E and E) of both the postve charges are drected away from ther respectve charges. Wth vector addton, they yeld the net electrc feld E net at the pont P ndcated by the dot. Solve: The electrc felds from q and q are 9 9 q (9. N m /C )(. C) E =, along + x-axs = =,8 N/C πε r (.5 m) Because tanθ = cm/5 cm, θ = tan () = 6.. So, q E =, θ below y-axs r 9 9 (9. N m /C )(. C) E = ( cos6. sn 6. j) = ( j) N/C (. m) + (.5 m) The net electrc feld s thus E net at P = E + E = (98 9 j) N/C To fnd the angle ths net vector makes wth the x-axs, we calculate 9 N/C tan φ = φ =. 98 N/C Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

5 The Electrc Feld 6-5 Thus, the strength of the electrc feld at P s E net = (98 N/C) + (9 N/C) = N/C = N/C and E net makes an angle of. below the +x-axs. Assess: Because of the nverse square dependence on dstance, E < E. Addtonally, because the pont P has no specal symmetry relatve to the charges, we expected the net feld to be at an angle relatve to the x-axs. 6.. Model: The electrc feld at the pont s found by superposton of the felds due to the two charges located on the y-axs. The electrc feld due to the postve charge q at the pont s away from q. On the other hand, the electrc feld due to the negatve charge q at the pont s toward q. These two electrc felds are then added vertcally to obtan the net electrc feld at the pont. Solve: The electrc feld from q s 9 9 q (9. N m /C )(. C) E =, θ below x-axs = ( cosθ sn θj) πε r (.5 m) + (.5 m) Because tanθ = 5 cm/5 cm, θ = 5. So, Smlarly, the electrc feld from q s E (5 N/C) = j q E =, θ below + x-axs = (5 N/C) + j r E net = E + E = (5 N/C) j = 767 j N/C = 7.6 j N/C Thus, the strength of the electrc feld s 7.6 N/C and ts drecton s vertcally downward. Assess: A quck vsualzaton of the components of the two electrc felds shows that the horzontal components cancel 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 x-axs. The electrc feld n (a) ponts rght. The feld n (b) ponts left. Solve: (a) The dpole moment s 9 p= ( qs, from to + ) = (. C)(. m) =. C m Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

6 6-6 Chapter 6 The electrc feld at ( cm, cm), whch s at dstance r =. m along the axs of the dpole, s p 9 (. C m) E = = (9. N m /C ) = 6 N/C πε r (. m) The feld strength, whch s all we re asked for, s 6 N/C. (b) The electrc feld at ( cm, cm), whch s at r =. m n the plane perpendcular to the electrc dpole, s p 9. Cm E = = (9. N m /C ) = 8. N/C πε r (. m) The feld strength at ths pont s 8 N/C 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 y-axs. The electrc feld n (a) ponts up. The feld n (b) ponts down. Solve: (a) The electrc feld at ( cm, cm), whch s at r =. m along the axs of the dpole, s p E = 6 j N/C = πε r re (. m) (6 j N/C) p= = = (. jc m) (/ πε 9 ) (9. N m /C ) By defnton, the dpole moment s p =. j C m = ( qs, from to + ) = q(. m) j. Thus. C m. 9 C. nc q = = =. m (b) Pont ( cm, cm) s n the plane perpendcular to the dpole. The electrc feld E = x(/ πε) p/ r s half the strength of the feld at an equal dstance r on the axs of the dpole. Hence the feld strength at ths pont s 8 N/C. Secton 6. The Electrc Feld of a Contnuous Charge Dstrbuton 6.7. Model: We wll assume that the wre s thn and that the charge les on the wre along a lne. Solve: From Equaton 6.5, the electrc feld strength of an nfntely long lne of charge havng lnear charge densty λ s λ Elne = r E lne( r λ 5. cm) (. cm) 5. m E r λ = = = = πε. m lne πε Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

7 O The Electrc Feld 6-7 Dvdng the above two equatons gves E. m lne( r = 5. cm) = lne(. cm) ( N/C). N/C N/C 5. m E r = = = = 6.8. Model: The rods are thn. Assume that the charge les along a lne. The electrc feld of the postvely charged glass rod ponts away from the glass rod, whereas the electrc feld of the negatvely charged plastc rod ponts toward the plastc rod. The electrc feld strength s the magntude of the electrc feld and s always postve. Solve: Example 6. shows that the electrc feld strength n the plane that bsects a charged rod s Q Erod = πε + The electrc feld from the glass rod at r = cm from the glass rod s E glass r r ( L/) 9 9 C 5 = (9. N m /C ) =.765 N/C (. m) (. m) + (.5 m) The electrc felds from the glass rod at r = cm and r = cm are.85 N/C and.5 N/C. The electrc feld from the plastc rod at dstances cm, cm, and cm from the plastc rod are the same as for the glass rod. Pont P s. cm from the glass rod and s. cm from the plastc rod, pont P s cm from both rods, and pont P s cm from the glass rod and cm from the plastc rod. Because the drecton of the electrc felds at P s the same, the net electrc feld strength cm from the glass rod s the sum of the felds from the glass rod at cm and the plastc rod at cm. Thus At. cm E =.765 N/C +.5 N/C =. N/C At. cm E =.85 N/C +.85 N/C =.67 N/C At. cm E =.5 N/C N/C =. N/C Assess: The electrc feld strength n the space between the two rods goes through a mnmum. Ths pont s exactly n the mddle of the lne connectng the two rods. Also, note that the arrows shown n the fgure are not to scale Model: The rods are thn. Assume that the charge les along a lne. 5 5 Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

8 6-8 Chapter 6 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, the net feld strength at each pont s the dfference of the feld strengths from the two rods. Solve: Example 6. gves the electrc feld strength n the plane that bsects a charged rod: E rod = πε Q r r + ( L/) The electrc feld from the rod on the rght at a dstance of cm from the rod s E rght 9 9 C 5 = (9. N m /C ) =.765 N/C (. m) (. m) + (.5 m) The electrc feld from the rod on the rght at dstances cm and cm from the rod are 5.85 N/C and 5.5 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. cm from the rod on the rght. Because the electrc felds at P have opposte drectons, the net electrc feld strengths are At. cm E =.765 N/C.5 N/C =. N/ C 5 5 At. cm E =.85 N/C.85 N/C = N/C At. cm E =.765 N/C.5 N/C =. N/C 6.. Model: Assume the glass bead s a pont charge; also assume the bead s n the mdplane of the rod so we Q can use Erod = K. r r + ( L/) We seek Q. We are gven L =. m. Solve: Use F = qe. Q Fon bead = qbeaderod = qbeadk r r + ( L/) Fon bead 8 μn Q = r r + ( L/) = (. m) (. m) + (.5 m) = nc q 9 9 beadk (6. C)(9. N m /C ) The bead s repelled by the rod, whch means ther charges are the same sgn. So the rod has a charge of + nc. Assess: The answer s n the ballpark of the charges we have seen so far. Secton 6. The Electrc Feld of Rngs, Dsks, Planes, and Spheres 6.. Model: Assume that the rngs are thn and that the charge les along crcle of radus R. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

9 The Electrc Feld 6-9 The rngs are centered on the z-axs. Solve: (a) Accordng to Example 6., the feld of the left (negatve) rng at z = cm s 9 9 zq z / / πε( z + R ) [(. m) + (.5 m) ] (9. N m /C )(. m)( C) ( E ) = = =.88 N/C That s, the feld s E = (.88 N/C, left). Rng has the same quantty of charge and s at the same dstance, so t wll produce a feld of the same strength. Because Q s postve, E wll also pont to the left. The net feld at the mdpont s E = E+ E = (.6 N/C, left) (b) The force s 9 5 F = qe = (. C)(.6 N/C, left) = (.6 N, rght) 6.. Model: Assume that the rngs are thn and that the charge les along crcle of radus R. Solve: (a) Let the rngs be centered on the z-axs. Accordng to Example 6., the feld of the left rng at z = cm s 9 9 zq z / / πε( z + R ) [(. m) + (.5 m) ] (9. N m /C )(. m)( C) ( E ) = = =.9 N/C That s, E = (.9 N/C, rght). Rng has the same quantty of charge and s at the same dstance, so t wll produce a feld of the same strength. Because Q s postve, E wll pont to the left. The net feld at the mdpont between the two rngs s E = E+ E = N/C. (b) The feld of the left rng at z = cm s ( E ) = N/C. The feld of the rght rng at z = cm to ts left s z 9 9 (9. N m /C )(. m)( C) ( E) z = =. N/C / [(. m) + (.5) ] E = E+ E = N/C + (. N/C, left) So the electrc feld strength s. N/C. 6.. Model: Each dsk s a unformly charged dsk. When the dsk s charged negatvely, the on-axs electrc feld of the dsk ponts toward the dsk. The electrc feld ponts away from the dsk for a postvely charged dsk. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

10 6- Chapter 6 Solve: (a) The surface charge densty on the dsk s 9 Q Q 5 C 6 η = = = = 6.66 C/m A πr π(.5 m) From Equaton 6., the electrc feld of the left dsk at z =. m s 6 η 6.66 C/m ( E) z = = = 8, N/C ε + R / z (8.85 C /N m ) + (.5 m/. m) In other words, E = (8, N/C, left). Smlarly, the electrc feld of the rght dsk at z =. m (to ts left) s E = (8, N/C, left). The net feld at the mdpont between the two rngs s E = E+ E = (7.6 N/C, left). (b) The force on the charge s 9 5 F = qe = (. C)(7.6 N/C, left) = (7.6 N, rght) Assess: Note that the force on the negatve charge s to the rght because the electrc feld s to the left. 6.. Model: Model each dsk as a unformly charged dsk. When the dsk s postvely charged, the on-axs 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 6 η = = = = 6.66 C/m A πr π(.5 m) From Equaton 6., the electrc feld of the left dsk at z =. m s 6 η 6.66 C/m ( E) z = = = 8, N/C ε + R / z (8.85 C /N m ) + (.5 m/. m) Hence, E = (8, N/C, rght). Smlarly, the electrc feld of the rght dsk at z =. m (to ts left) s E = (8, N/C, left). The net feld at the mdpont between the two dsks s E = E+ E = N/C. (b) The electrc feld of the left dsk at z =.5 m s C/m 5 5 ( E) z = =.5 N/C E = (.5 N/C, rght) (8.85 C /N m ) + (.5 m/. m) Smlarly, the electrc feld of the rght dsk at z =.5 m (to ts left) s E = (.85 N/C, left). The net feld s thus E = E+ E = (8.7 N/C, rght) The feld strength s 8.7 N/C Model: A sphercal shell of charge Q and radus R has an electrc feld outsde the sphere that s exactly the same as that of a pont charge Q located at the center of the sphere. In the case of a metal ball, the charge resdes on ts surface. Ths can then be vsualzed as a charged sphercal shell of radus R. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

11 The Electrc Feld 6- Solve: Equaton 6.8 gves the electrc feld of a charged sphercal shell at a dstance r > R: Q Eball = r r In the present case, E ball = 5, N/C at r = ( cm) +. cm = 7. cm =.7 m. So, (.7 m) 5, N/C 8 r Eball 9 Q= = =.7 C= 7nC 9. N m /C 6.6. Model: The dstance. mm s very small n comparson to the sze of the electrode, so we can model the electrode as a plane of charge. Solve: From Equaton 6.6, the electrc feld of a plane of charge s 9 η Q 8 C 5 Eplane = = = =. N/C ε ε A (8.85 C /N m )(.. m) Secton 6.5 The Parallel-Plate Capactor 6.7. Model: The electrc feld n a regon of space between two charged crcular dsks s unform. Solve: The electrc feld strength nsde the capactor s E = Q/ εa. Thus, the area s 9 9 Q (. )(.6 C) D A = = =.7 m = π ε 5 E (8.85 C /N m )(. N/C) A D = =.9 cm π Assess: As long as the spacng s much less than the plate dmensons, the electrc feld s ndependent of the spacng and depends only on the dameter of the plates Model: The electrc feld s unform n a regon of space between closely spaced capactor plates. Solve: The electrc feld nsde a capactor s E = Q/ εa. Thus, the charge needed to produce a feld of strength E s 6 Q= ε AE = (8.85 C /N m ) π(. m) (. N/C) = 5 nc Thus, one plate has a charge of 5 nc and the other has a charge of 5 nc. Assess: Note that the capactor as a whole has no net charge Model: The electrc feld n a regon of space between the plates of a parallel-plate capactor s unform. Solve: The electrc feld nsde a capactor s E = Q/ εa. Thus, the charge needed to produce a feld of strength E s 6 Q= ε AE = (8.85 C /N m )(. m. m)(. N/C) = nc The number of electrons transferred from one plate to the other s 9 C = C Secton 6.6 Moton of a Charged Partcle n an Electrc Feld 6.. Model: The charged glass bead s a pont charge. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

12 6- Chapter 6 The bead hangs suspended n the ar when the net force actng on the bead s zero. The two forces that act on the bead are the electrc force and the gravtatonal force. Because the bead s postvely charged, the electrc feld must be ponted upward to cause an upward force, whch wll balance the gravtatonal force. Solve: For the bead to be n statc equlbrum, mg (. kg)(9.8 N/kg) 5 ( Fnet ) y = qe mg = N E = = = 6. N/C q 9 (. )(.6 C) 5 Thus the requred feld s E = (6. N/C, up). 6.. Model: The dsks form a parallel-plate capactor. The electrc feld nsde a parallel-plate capactor s a unform feld, so the proton wll have a constant acceleraton. Solve: (a) The two dsks form a parallel-plate capactor wth surface charge densty 9 Q Q C 5 η = = = =.8 C/m A πr π(. m) From Equaton 6.9, the feld strength nsde a capactor s 5.8 C/m 6 E η = = =.6 N/C ε 8.85 C /N m 6 (b) The electrc feld ponts toward the negatve plate, so n the coordnate system of the fgure E =.6 j N/C. The feld exerts forces F = q E = ee on the proton, causng an acceleraton wth a y-component that s proton ee 9 6 F y (.6 C)(.6 N/C) ay = = = =.5 m/s m m 7.67 kg After the proton s launched, ths acceleraton wll cause t to lose speed. To just barely reach the postve plate, t should reach v = m/s at y= mm. The knematc equaton of moton s m /s v = = v + a Δ y y 5 v = a Δ y = (.5 m/s )(. m) = 8. m/s y Assess: The acceleraton of the proton n the electrc feld s enormous n comparson to the gravtaton acceleraton g. That s why we dd not explctly consder g n our calculatons. 6.. Model: A unform electrc feld causes a charge to undergo constant acceleraton. Solve: Knematcs yelds the acceleraton of the electron. 7 7 v v (. m/s) (. m/s) 6 = + Δ = = = 5. m/s v v a x a Δx (. m) The magntude of the electrc feld requred to obtan ths acceleraton s 6 Fnet mea (9. kg)(5. m/s ) 5 E = = = =.8 N/C. e e 9.6 C Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

13 The Electrc Feld 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 + aδ x and F = qe = ma, qe a= = v Δ x= mv m Δx qe Furthermore, the electrc feld of a plane of charge wth surface charge densty η s E = η/ ε. Thus, 7 6 ε 9 6 mv (.67 kg)(. m/s) (8.85 C /N m ) Δ x = = =.8 m qη (.6 C)(. C/m ) 6.. Model: Assume the ol drop s a pont charge. Model the plane as nfnte. The negatve drop wll be attracted to the postve plane. Ignore gravty. Solve: We wll combne a number of equatons together. Solve the equaton for the feld due to an nfnte plane of charge for the surface charge densty: η = εe. The only force s the electrc force, so t s the net force: ma vf F = qe = ma E =. Also solve vf v = aδ x for a where v = : a=. We also need the mass of the q drop Δ x drop: m= ρv = ρ πr. Combne all these together. ma m vf m v ρ πr f vf E qdrop qdrop Δx qdrop Δx qdrop Δx (9 kg/m ) π(.5 μm) (.5 m/s) μ 9 η = ε = ε = ε = ε = ε = (8.85 C /Nm ) = 6. C/m 5(.6 C). m Assess: The answer s n the ballpark of the surface charge densty n Example 6.6. Careful analyss wll confrm that the unts work out correctly. Secton 6.7 Moton of a Dpole n an Electrc Feld 6.5. Model: The external electrc feld exerts a torque on the dpole moment of the water molecule. Solve: From Equaton 6., the torque exerted on a dpole moment by an electrc feld s τ = pe sn θ. The maxmum torque s exerted when snθ = or θ = 9. Thus, 8 τ max = pe = (6. C m)(5. N/C) =. N m 6.6. Model: Because r >> s, we can use Equaton 6. for the electrc feld n the plane perpendcular to the dpole. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

14 6- Chapter 6 Solve: (a) From Newton s thrd law, the force of Q on the dpole s equal and opposte to the force of the dpole on Q. You can see from the dagram that F dpole on Q s down and F Q on dpole s up, n the drecton of p. The magntude of the dpole feld at the poston of Q s p qs Edpole = = r r The magntude of F dpole on Q s qqs Fdpole on Q = QEdpole = r F Q on dpole has the same magntude, so F qqs =, drecton of p Q on dpole r (b) The electrc feld of charge Q exerts a torque on the dpole. The feld E s perpendcular to the dpole p, so θ = 9. The torque s gven by Equaton 6.: Q qqs τ = pe sn θ = ( qs) sn9 = πε r r 6.7. Model: The sze of a molecule s. nm. The proton s. nm away, so r >> s and we can use Equaton 6. 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 p 9 5. Cm 5 Edpole = = (9. N m /C ) = 5.6 N/C πε 9 r (. m) The magntude of F dpole on proton s 9 5 Fdpole on proton = eedpole = (.6 C)(5.6 N/C) = 9. N Includng the drecton, the force s Fdpole on proton = (9. N, drecton opposte p) Model: The electrc feld s that of three pont charges q, q, and q. Assume the charges are n the x-y plane. The 5. nc charge s q, the 5. nc charge s q, and the nc charge s q. The net electrc feld at the dot s E net = E + E + E. The procedure wll be to fnd the magntudes of the electrc felds, to wrte them n component form, and to add the components. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

15 The Electrc Feld 6-5 Solve: (a) The electrc feld produced by q s 9 9 q πε r (. m) (9. N m /C )(5. C) E = = =,5 N/C E ponts away from q, so n component form E =,5 N/C. The electrc feld produced by q s E = 56,5 N/C. E ponts away from q, so E Fnally, the electrc feld produced by q s q 9 9 πε r (. m) + (. m) (9. N m /C )(5. C) E = = =,5 N/C = 56,5 j N/C. E ponts toward q and makes an angle φ = tan (/) = 6. wth the x-axs. So, E = Ecosφ + Esn φj = ( 6 +, j) N/C Addng these three vectors gves 5 E net = E + E + E = (,6 6,6 j) N/C = (..6 j) N/C Ths s n component form. (b) The magntude of the feld s 5 x y Enet = E + E = (,6 N/C) + ( 6,6 N/C) = 8,69 N/C =. N/C 5 θ = tan E / E = 9.. We can also wrte E = (. N/C, 9. CW from and ts angle from the x-axs s ( y x ) the + x-axs). net 6.9. Model: The electrc feld s that of three pont charges q, q, and q. Assume the charges are n the x-y plane. The 5. nc charge s q, the bottom nc charge s q, and the top nc charge s q. The net electrc feld at the dot s E net = E + E + E. The procedure wll be to fnd the magntudes of the electrc felds, to wrte them n component form, and to add the components. Solve: (a) The electrc feld produced by q s 9 9 q πε r (. m) (9. N m /C )(5. C) E = = =,5 N/C E ponts toward q, so n component form E =,5 j N/C. The electrc feld produced by q s E = 56,5 N/C. E ponts away from q, so E Fnally, the electrc feld produced by q s q 9 9 πε r (. m) + (. m) (9. Nm /C )( C) E = = = 5, N/C = 56,5 N/C. E ponts away from q and makes an angle φ = tan (/) = 6.6 wth the x-axs. So, E = Ecosφ Esn φ j = (,5, j) N/C Addng these three vectors gves E net = E + E + E = ( 96,5 + 9, j) N/C = ( j) N/C Ths s n component form. (b) The magntude of the feld s 5 x y Enet = E + E = (96,5 N/C) + (9, N/C) =,6 N/C =. N/C 5 θ = tan E / E =. We can also wrte E = (. N/C, 6 CCW and ts angle below the x-axs s ( y x ) from the + x-axs). net Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

16 6-6 Chapter Model: The electrc feld s that of three pont charges q, q, and q. Assume the charges are n the x-y plane. The nc charge s q, the nc charge s q, and the 5. nc charge s q. The net electrc feld at the dot s E net = E + E + E. The procedure wll be to fnd the magntudes of the electrc felds, to wrte them n component form, and to add the components. Solve: (a) The electrc feld produced by q s 9 9 q πε r (. m) (9. Nm /C )( C) E = = =, N/C E ponts toward q, so n component form E =, j N/C. The electrc feld produced by q s E = 8, N/C. E ponts toward q, so E Fnally, the electrc feld produced by q s q 9 9 πε r (. m) + (.5 m) = 8, N/C. (9. Nm /C )( C) E = = = 6,7 N/C E ponts away from q and makes an angle φ = tan (/5) = wth the x-axs. So, E = E cosφ E sn φj = (,7,6 j) N/C Addng these three vectors gves E = E + E + E = ( ,66 j) N/C = ( j) N/C net Ths s n component form. (b) The magntude of the feld s x y Enet = E + E = (,7 N/C) + (86,66 N/C) = 86,9 N/C = 8.6 N/C and ts angle from the x-axs s θ ( Ey Ex ) = tan / = 87, but we see that we need to be n the second quadrant so we add 8. We can also wrte Enet = (8.6 N/C, 9 CCW from the + x-axs). 6.. Model: The electrc feld s that of three pont charges q= Q, q = Q, and q =+ Q. Assume the charges are n the x-y plane. The net electrc feld at pont P s E net = E + E + E. The procedure wll be to fnd the magntudes of the electrc felds, to wrte them n component form, and to add the components. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

17 The Electrc Feld 6-7 Solve: The electrc feld produced by q ponts toward q and s gven by E = Q L The electrc feld produced by q ponts toward q and s gven by The electrc feld produced by q s E ponts away from q and makes an angle E = L Q j Q Q E = = L + L L φ = tan ( LL / ) = 5 wth the x-axs. So Q Q E = (cos sn j) j φ + φ = πε L + L Addng these three vectors gves Q Q E net = E + E + E = j ( )( j) + = + L L 6.. Model: The electrc feld s that of two charges q and +q located at x =± a. Solve: At pont, the electrc feld from q s E q E q q = = ( a) + ( a) πε 5a ponts toward q and makes an angle φ = tan ( aa / ) = 6. below the x-axs, hence q q q E (cos sn ) ( q = φ φ j = j = j) 5a 5a a q Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

18 6-8 Chapter 6 The electrc feld from the +q s E +q q q E+ = = q a + ( a) 5a ponts away from +q and makes an angle φ = tan ( aa / ) = 6. above the x-axs. So, q q E + q = (cos φ + sn φj) = ( j) + 5a 5 5a Addng these two vectors, q E net = E q + E+ q = ( + j) 5 5a At pont, the electrc feld from q ponts toward q, so q E q = 9a The electrc feld from +q ponts away from +q, so q E + q = πε a Addng these two vectors, 7q E net = E q + E+ q = πε 9 a At pont, the electrc feld from q ponts toward q, so q E q = πε a The electrc feld from +q ponts away from +q, so q E + q = 9a Addng these vectors, 7q E net = E q + E+ q = 9a Pont s a mrror mage of pont. Snce ponts to the left and down. Thus, q E net = ( j) 5 5a 6.. Model: The electrc feld s that of two postve charges. E net ponts to the left and up, E net has a reversed y-component and Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

19 The Electrc Feld 6-9 The fgure shows E and E due to the ndvdual charges. The total feld s E = E + E. Solve: (a) From symmetry, the y-components of the two electrc felds cancel out. The x-components are equal and add. Thus, E = ( E ) x = Ecosθ The feld strength and angle due to q are qx Thus the magntude of the net feld s E = πε ( x + s /) q q x x E = = cosθ = = r πε( x + s /) r x + s / / (b) The electrc feld strength at poston x s 9 9 (9. N m /C )()(. C) x 8x E = = [ x + (. m) ] [ x + (. m) ] where x has to be n meters. We can now evaluate E for dfferent values of x:. / / x (mm) x (m) E (N/C).. 768,. 576, ,. 58, N/C 6.. Model: The electrc feld of a dpole s that of two opposte charges ±q that make the dpole. Please refer to Fgure 6.8. The fgure shows a dpole algned on the y-axs, so the x-axs s the bsectng axs. The feld at a pont on the x-axs s Edpole = E + E. + Solve: From the symmetry of the stuaton we can see that the x-components of the two contrbutons to the electrc feld wll cancel, that they have equal y-components, and that E ponts n the y-drecton. Thus, dpole ( Edpole ) x = N/C ( Edpole) y = E+ snθ where θ s the angle E + makes wth the x-axs. From the geometry of the fgure, For a pont charge +q, the feld s s/ snθ = [ x + ( s/) ] / q E+ = πε x + (/) s Combnng these peces gves the dpole feld at dstance x along the bsectng axs: / Edpole = () q s j = qs j πε / / x + ( s /) [ x + ( s /) ] πε ( x + s /) / If x >> s, then ( x + s /) x. Thus E qsj dpole πε x If we note that p= qsj and f we replace x wth a more general varable r to denote the dstance from the dpole, then p Edpole r Ths s Equaton 6.. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

20 6- Chapter Model: The electrc feld s that of three pont charges. The feld at ponts on the x-axs s E net = E + E + E. Solve: (a) We note from the symmetry that the y-component of E and E cancel. Snce E has no y-component, the net feld wll have only an x-component. The x-components of E and E are equal, so ( E ) = E E cos θ ( E ) = ( E ) = N/C net x net y net z Note that the sgns of q and q were used n wrtng ths equaton. The electrc feld strength due to q s The electrc feld strength due to q s From geometry, Assemblng these peces, the net feld s ( E ) E E q q = = r x q = = r x + d x x cosθ = = r x + d q q x q x = = x + d q x E net = πε / x ( x + d ) net x πε / x πε x + d πε x ( x + d ) (b) For x << d, the observaton pont s very close to q =+. q Furthermore, at x m the felds E and E are nearly opposte to each other and wll nearly cancel. So for x << d we expect the feld to be that of a pont charge +q at the orgn. To test ths predcton, we note that for x << d x x q E / net = ( x + d ) d πε x Ths s, ndeed, the feld on the x-axs of pont charge q at the orgn. For x >> d, the three charges appear as a sngle charge of value qnet = q + q + q =. So we expect E when x >> d. In ths lmt, x x = = / x ( x + d ) x x x x so the feld does rapdly become zero, as expected. q Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

21 The Electrc Feld Model: The electrc feld s that of two nfnte lnes of charge extendng out of the page. The lne charges le on the x-axs. Solve: From Equaton 6.5, the electrc feld strength due to an nfnte lne of charge at a dstance r from the lne charge s λ E = r For the left and rght lne charges, the electrc felds are λ λ Eleft = Erght = = y + ( d/) y + d E left makes an angle φ above the +x-axs and E rght makes the same angle φ above the x-axs. From the geometry of the fgure, d/ d y cosφ = = snφ = y + d / y + d y + d λ d y E left = + j y + d y + d y + d λ d y E rght = + j y + d y + d y + d We now add these two vectors to fnd E net = E left + E rght. The x-components cancel to gve λ y 6λy E net = () j = j y + d y + d πε ( y + d ) Thus the feld strength s 6λ y E = y + d 6.7. Model: The electrc feld s that of two nfnte lnes of charge extendng out of the page. The lne charges le on the x-axs. Solve: (a) From Equaton 6.5, the electrc feld strength due to an nfnte lne of charge at a dstance r from the lne charge s λ E = r For the left and rght lne charges, the electrc felds are λ λ Eleft = Erght = = y + ( d/) y + d d/ d y cosφ = = snφ = y + d / y + d y + d λ d y E left = + j y + d y + d y + d λ d y E rght = j y + d y + d y + d λ d 8λd E net = Eleft + Erght = () = y + d y + d πε ( y + d ) Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

22 6- Chapter 6 Thus the feld strength s 8λd E = y + d 6.8. Model: The felds are those of two nfnte lnes of charge wth lnear charge densty λ. Solve: The feld at dstance r from an nfnte lne of charge s λ E = r r It ponts straght away from the lne. Wth two perpendcular lnes, the feld due to the lne along the x-axs ponts n the y-drecton and depends nversely on dstance y. Smlarly, the feld due to the lne along the y-axs ponts n the x-drecton and depends nversely on dstance x. That s E x λ λ = E = y y x πε λ E = E x + Eyj= + j x y The feld strength at ths pont n space s E = Ex + Ey = λ / x + / y 6.9. Model: Assume that the wre s thn and that the charge les on the wre along a lne. Solve: From Equaton 6.5, the electrc feld for an nfnte unformly charged lne s λ Elne = r where r s the dstance from the lne n the plane that bsects the lne. Solvng for the lnear charge densty, relne (.5 m)( N/C) 9 λ = = = 5.56 C/m (/ πε 9 ) (9. Nm /C ) The charge n. cm s 9 Q = L λ = (. m)(5.56 C/m) = 5.56 C=.56nC Because the electrc feld s drected toward the lne, Q s negatve. Thus Q =.56 nc. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

23 The Electrc Feld Model: The electrc feld s that of a lne of charge of length L. The orgn of the coordnate system s at the center of the rod. Dvde the rod nto many small segments of charge Δq and length Δ x. Solve: (a) Segment creates a small electrc feld E at pont P that ponts to the rght. The net feld E wll pont to the rght and have E = Ez = N/C. The dstance to segment s x, so y Δq E = ( E ) = E = ( E ) = x x x πε( x x ) ( x x ) Δq s not a coordnate, so before convertng the sum to an ntegral we must relate the charge Δq to length Δ x. Ths s done through the lnear charge densty λ = Q/L, from whch Wth ths charge, the sum becomes Q Δ q= λδ x = Δ x L ( QL / ) Δx Ex = πε ( x x ) Now we let Δx dx and replace the sum by an ntegral from x = L to x =+ L. Thus, L/ L/ ( QL / ) dx ( QL / ) ( QL / ) L Q E x = E Ñ = ( ) = = / / The electrc feld strength at x s (b) For x >> L, Δq x x x x L/ L/ x L x L Q E = πε x L / Q E = x That s, the lne charge behaves lke a pont charge. (c) Substtutng nto the above formula 9 9. C E = (9. N m /C ) = 9.8 N/C (. m) 5. m ( ) 6.. Model: The electrc feld s that of a lne charge of length L. Let the bottom end of the rod be the orgn of the coordnate system. Dvde the rod nto many small segments of charge Δq and length Δ y. Segment creates a small electrc feld at the pont P that makes an angle θ wth the horzontal. The feld has both x and y components, but E = N/C. The dstance to segment from pont P s / ( x + y ). z Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

24 6- Chapter 6 Solve: The electrc feld created by segment at pont P s Δq Δq x y E (cos sn ) = θ θ j = j πε ( x + y ) πε( x + y ) x + y x + y The net feld s the sum of all the E, whch gves E = E. Δq s not a coordnate, so before convertng the sum to an ntegral we must relate charge Δq to length Δ y. Ths s done through the lnear charge densty λ = Q/L, from whch we have the relatonshp Q Δ q= λ Δ y = Δ y L Wth ths charge, the sum becomes QL / xδy y Δy E = j πε / / ( x + y ) ( x + y ) Now we let Δy dy and replace the sum by an ntegral from y = m to y = L. Thus, L L L L ( QL / ) xdy ydy ( QL / ) y E = j = x j Ñ / / ( x y Ñ ) ( x y ) + + x x + y x + y Q Q x = j x x L Lx + x + L 6.. Model: Assume that the rng of charge s thn and that the charge les along crcle of radus R. Solve: From Example 6., the on-axs feld of a rng of charge Q and radus R s zq Ez = / πε( z + R ) When z << R, ths means we are near the center of the rng. At that pont, segments of charge and j that are 8 apart create felds E and E j that cancel each other. When the felds of all segments of charge around the rng are added, the net result s zero. Ths s ndcated by the above expresson because when z = m, the electrc feld s zero. When z >> R, then If ths s used n the expresson for E z, we get / / ( z + R ) = ( z ) = z zq Ez = Ths s the feld of a pont charge Q as seen along the z-axs. Q z z 6.. Model: Assume that the rng of charge s thn and that the charge les along crcle of radus R. Solve: (a) From Example 6., the on-axs feld of a rng of charge Q and radus R s zq Ez = πε / ( z + R ) For the feld to be maxmum at a partcular value of z, de/ dz =. Takng the dervatve, (/)( ) / 5/ / 5/ de Q = z z = = z dz πε ( z + R ) ( z + R ) ( z + R ) ( z + R ) z R = z =± z + R Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

25 The Electrc Feld 6-5 (b) The feld strength at the pont z = R/ s ( E ) Q ( R/ ) Q = = ( R/ ) + R z max / R 6.. Model: Assume that the semcrcular rod s thn and that the charge les along the semcrcle 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 Δq and arc length Δs. Segment creates a small electrc feld E at the orgn. The lne from the orgn to segment makes an angle θ wth the x-axs. Solve: Because every segment at an angle θ above the axs s matched by segment j at angle θ below the axs, the y-components of the electrc felds wll cancel when the feld s summed over all segments. Ths leads to a net feld pontng to the rght wth E = ( E ) = E cosθ E = N/C x x y Note that angle θ depends on the locaton of segment. Now all segments are at the same dstance r = R from the orgn, so Δq Δq E = = r R The lnear charge densty on the rod s λ = Q/L, where L s the rod s length. Ths allows us to relate charge Δq to the arc length Δs through Δ q= λ Δ s = ( Q/ L) Δ s Thus, the net feld at the orgn s ( QL / ) Δs Q Ex = cosθ cos = θδs R LR The sum s over all the segments on the rm of a semcrcle, so t wll be easer to use polar coordnates and ntegrate over θ rather than do a two-dmensonal ntegral n x and y. We note that the arc length Δs s related to the small angle Δθ by Δ s= RΔ θ, so Q Ex = cosθδθ LR Wth Δθ dθ, the sum becomes an ntegral over all angles formng the rod. θ vares from Δ θ = π/ to θ =+ π/. So we fnally arrve at E x = Q Q Q LR = LR = LR π / π / Ñ cosθdθ snθ π / π / Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

26 6-6 Chapter 6 Snce we re gven the rod s length L and not ts radus R, t wll be convenent to let R = L/π. So our fnal expresson for E, now ncludng the vector nformaton, s (b) Substtutng nto the above expresson, π Q E = L 9 9 (9. N m /C ) π ( C) 5 E = =.7 N/C (. m) 6.5. Model: Assume that the quarter-crcle plastc rod s thn and that the charge les along the quarter-crcle 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 Δq and arc length Δs. Solve: (a) Segment creates a small electrc feld E at the orgn wth two components: ( E ) = E cos θ ( E ) = E snθ x y Note that the angle θ depends on the locaton of the segment. Now all segments are at dstance r = R from the orgn, so Δq Δq E = = r R The lnear charge densty of the rod s λ = Q/L, where L s the rod s length (L = quarter-crcumference = πr/). Ths allows us to relate charge Δq to the arc length Δs through Q Q Δ q= λδ s= Δ s= Δs L π R Usng Δs = RΔθ, the components of the electrc feld at the orgn are Δ ( E ) q cos θ Q R θ cos θ Q = = Δ = Δ θ cos θ x πε R π R π R π R Δ ( E ) q sn θ Q R θ sn θ Q = = Δ = Δ θ sn θ y πε R R π R π R (b) The x- and y-components of the electrc feld for the entre rod are the ntegrals of the expressons n part (a) from θ = rad to θ = π/. We have Q Q Ex = d E = snθdθ π/ π/ Ñ cosθ θ y πr πε Ñ πr Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

27 The Electrc Feld 6-7 (c) The ntegrals are Ñ π/ sn / / / [ cos π ] cos π cos π cos [ sn π d d ] sn π θ θ = θ = =+ Ñ θ θ = θ = sn =+ The electrc feld s Q E = ( + j) π R 6.6. Model: Assume that the plastc sheets are planes of charge. Solve: At pont the electrc felds due to the left sheet and the rght sheet are η η η η E left =, toward rght = Erght =, toward left = ε ε ε ε η E net = Eleft + Erght = ε At pont, Eleft = ( η / ε), Erght = ( η/ ε), and Enet = ( η/ ε). At pont, Enet =+η ( / ε) Model: An nsulatng sphere of charge Q and radus R has an electrc feld outsde the sphere that s exactly the same as that of a pont charge Q located at the center of the sphere. Solve: The electrc feld of a charged sphere at a dstance r > R s gven by Equaton 6.8: Q Esphere = r r In the present case, E = E + E, where E and E are the felds of the ndvdual spheres. The dstance from the center of each sphere to the mdpont between s r = r = cm. Thus, 9 9 (9. N m /C )( C) E = = 5.65 N/C (. m) 9 9 (9. N m /C )(5 C) E = = 8.8 N/C (. m) The felds pont n the same drecton, so 5 E = (5.65 N/C N/C, rght) = (. N/C, rght). The electrc feld wll pont left when Q and Q are nterchanged. The electrc feld strength n both cases s 5. N/C Model: The parallel plates form a parallel-plate capactor. The electrc feld nsde a parallel-plate capactor s a unform feld, so the electron and proton wll have constant acceleraton. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

28 6-8 Chapter 6 The negatve plate s at x = m and the postve plate s at x= d = cm. Solve: Both partcles accelerate from rest ( v = m/s), so at tme t ther postons are xe = xe + aet = aet xp = xp + apt = d + apt At some nstant of tme t the electron and proton have the same poston: x e = x p = x. Ths s the pont where they pass each other. At ths nstant, x at x d at = e = + p These are two equatons n the two unknowns x and t. From the frst equaton, t = x / a e. Usng ths n the second equaton gves a x d x x p = + a = e + ap/ ae To fnsh, we need to fnd the acceleratons of the electron and proton. Both partcles are n a parallel-plate capactor wth Ecap = Q/ εa. The feld ponts to the left, so E = Q/ εa. The proton s acceleraton s a p x Fp qpex e( Q/ εa) eq/ εa = = = = m m m m p p p p The proton s acceleraton s negatve, as expected. For the electron, Fe qeex e( Q/ εa) eq/ εa ae = = = = m m m m e e e e Consequently, the acceleraton rato s ap/ ae = me/ mp. Usng ths, the pont where the two charges pass s d cm x = = =.9995 cm / 9. kg/.67 kg + m 7 e m p + Assess: Ths s very close to where the proton starts. Snce there s a factor of 8 dfference n the masses of the proton and electron, the electron accelerates much faster than the proton Model: The electrc feld s unform nsde the capactor, so constant-acceleraton knematc equatons apply to the moton of the proton. d Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

29 The Electrc Feld 6-9 Solve: From Equaton 6.9, the electrc feld between the parallel plates E = ( ηε / ) j. The force on the proton s qe F ma qe a q η j a q η = = = = y = m mε mε Usng the knematc equaton = + y( ) + ( y ), y y v t t a t t qη Δ y = y y = (m/s)( t t) + ay( t t) = ( t t) mε To determne t t, we consder the horzontal moton of the proton. The proton travels a dstance of. cm at a 6 constant speed of. m/s. The velocty s constant because the only force actng on the proton s due to the feld between the plate along the y-drecton. Usng the same knematc equaton,. m 8 x 6 Δ x=. m = v ( t t ) + m ( t t ) = =. s. m/s (.6 C)(. C/m )(. s) Δ y = =. mm 7 (.67 kg)(8.85 C /Nm ) 6.5. Model: Assume that the electrc feld nsde the capactor s constant, so constant-acceleraton knematc equatons apply. Solve: (a) The force on the electron nsde the capactor s qe F = ma = qe a = m Because E 9 s drected upward (from the postve plate to the negatve plate) and q =.6 C, the acceleraton of the electron s downward. We can therefore wrte the above equaton as smply ay = qe/ m. To determne E, we must frst fnd a y. From knematcs, x= x + vx( t t) + ax( t t). m = m + vcos5 ( t t) + m (. m) 8 ( t t) = =. s 6 (5. m/s) cos 5 Usng the knematc equatons for the moton n the y-drecton, t t qe t t vy = vy + ay m/s= vsn5 + m 6 m v sn 5 (9. kg)(5. m/s)sn 5 E = = = 55 N/C =.6 N/C qt ( t) 9 8 (.6 C)(. s) Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

30 6- Chapter 6 (b) To determne the separaton between the two plates, we note that y y = y, the electron s hghest pont, v = m/s. From knematcs, From part (a), y = m and v = (5. m/s)sn 5, but at y y = y + y = + y v v a ( y y ) m /s v sn 5 a ( y y ) a y sn 5 v v ( y y) = = a a 9 qe (.6 C)(55 N/C) = = = 6. m/s m 9. kg 6 (5. m/s) y y y y = =. m =. cm ( 6. m/s ) Ths s the heght of the electron s trajectory, so the mnmum spacng s. cm Model: The electrc feld s unform nsde the capactor, so constant-acceleraton knematc equatons apply to the moton of the electron. The condton for the electron to not ht the negatve plate s that ts vertcal velocty should just become zero as the electron reaches the plate. Solve: The force on the electron nsde the capactor s qe F = ma = qe a = m 9 (.6 C)(. N/C) 5 a y = =.756 m/s 9. kg The ntal velocty v has two components: vx = v cos5 and vy = vsn 5. Because the electrc feld nsde the capactor s along the +y-drecton, the electron has a negatve acceleraton that reduces the vertcal velocty. We requre v = m/s f t s not to ht the plate. Usng knematcs, y y y y( ) (m/s) y y v = v + a y y = v + a Δ y 5 6 v = a Δ y = (.756 m/s )(. m) = 8.8 m/s y y m/s 7 v = =.9 m/s sn Model: The parallel plates form a parallel-plate capactor. The electrc feld nsde a parallel-plate capactor s a unform feld, so the electrons wll have a constant acceleraton. 6 Solve: (a) The bottom plate should be postve. The electron needs to be repelled by the top plate, so the top plate must be negatve and the bottom plate postve. In other words, the electrc feld needs to pont away from the bottom plate so the electron s acceleraton a s toward the bottom plate. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

31 The Electrc Feld 6- (b) Choose an xy-coordnate system wth the x-axs parallel to the bottom plate and the orgn at the pont of entry. Then the electron s acceleraton, whch s parallel to the electrc feld, s a = aj. Consequently, the problem looks just lke a 7 projectle problem. The knetc energy / 6 K = mv =. J gves an ntal speed v = ( K/ m) = 8.5 m/s. Thus the ntal components of the velocty are 6 6 x = = y = = v v cos5 5.7 m/s v v sn5 5.7 m/s What acceleraton a wll cause the electron to pass through the pont ( x, y ) = (. cm, cm)? The knematc equatons of moton are x= x + vxt+ axt = vxt=. m y = y + vyt+ ayt = vyt+ at = m 9 From the x-equaton, t= x/ v x =.7 s. Usng ths n the y-equaton gves vy t 5 a = = 6.59 m/s t But the acceleraton of an electron n an electrc feld s 5 Felec qelece ee ma (9. kg)( 6.59 m/s ) a = = = E = = = 7,5 N/C =.8 N/C m m m e 9.6 C (c) The mnmum separaton d mn must equal the heght y max of the electron s trajectory above the bottom plate. (If d were less than y max, the electron would collde wth the upper plate.) Maxmum heght occurs at t = t = 8.7 s. At ths nstant, Thus, d mn =.5 mm. ymax = vyt+ at =.5 m =.5 mm 6.5. Model: Assume the soot partcle s a pont charge n F = qe and a small sphere where the drag force s concerned. Also assume the partcle becomes negatvely charged. Draw a free body dagram wth the electrc force drected down because the E feld s down, but the actual drecton of that force depends on the sgn of q. Solve: (a) Wrte Newton s frst law usng the free body dagram. Note that m= Vρ = π r ρ. Solve for term. Σ F = 6πηrv mg qe = term v ( π ρ ) v term = r g + qe 6πηr Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

32 6- Chapter 6 (b) Wth the qe term equal to zero the π and an r wll cancel. v r ρg μ term 5 (.5 m) ( kg/m )(9.8 m/s ) = = =.67 mm/s 6η 6(.8 kg/m s) (c) The charge on the soot partcle s negatve, so the electrc force s up. 9 π(.5 μm) ( kg/m )(9.8 m/s ) + (5)(.6 C)(5 N/C) vterm = =.9 mm/s 5 6 π(.8 kg/m s)(.5 μm) Assess: The termnal speed s smaller when the electrc force s present, as we would expect Model: An orbtng electron experences a force that causes centrpetal acceleraton. Solve: The electron orbts at a radus r = R + h = (. mm) +. mm =. mm. The force that causes the crcular moton s smply the electrc force F gven by Coulomb s law: KQ sphere qelec KeQ sphere (9 N m /C )(.6 C)(. C) F = = = =.6 N r r (. m) For crcular moton, mv rf (. m)(.6 N) 7 F = macentrpetal = v= = =.8 m/s r m 9. kg 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= πr/ T = πrf. The frequency s qp qe mv e me( π r f ) F = = = r r r f q q (9. Nm /C )(.6 C) = = = 6.56 Hz π π (9. kg)(5. m) 9 9 p e 5 mr e Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

33 The Electrc Feld Model: The electron orbtng the proton experences a force gven by Coulomb s law. Solve: The force that causes the crcular moton s qp qe mv e me( π r f ) F = = = r r r where we used v= πr/ T = πrf. The radus s / qp qe me π f / (.6 C)(.6 C) 8 r = πε ( ) = (9. N m /C ) (9. kg) π (. s ) =.86 m 9 nm Model: The electrc feld at the dpole s locaton s that of the on wth charge q. Solve: (a) We have p = αe. The unts of α are the unts of p/e and are Cm C m C s = = N/C N kg (b) The electrc feld due to the on at the locaton of the dpole s q q E at dpole =, away from q = πε r r Because p= αe, the nduced dpole moment s q p= α πε r From Equaton 6., the electrc feld produced by the dpole at the locaton of the on s p q q α E dpole = = α πε = 5 r r r r The force the dpole exerts on the on s q α F dpole on on = qedpole = 5 r Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

34 6- Chapter 6 Accordng to Newton s thrd law, F dpole on on = F on on dpole. Therefore, F q α =, toward on on on dpole 5 r Model: The electrc feld at the dpole s locaton s that of the nfnte lne of charge of lnear charge densty λ. Solve: The feld at dstance r from an nfnte lne of charge s λ E = r r It ponts straght away from the lne. The dpole conssts of charge +q at dstance r + s/ and charge q at dstance r s/. The net force on the dpole due to the feld of the lne s qλ qλ ( r s/) ( r+ s/) F = qe+ qe = r= r πε r+ s/ r s/ πε ( r+ s/)( r s/) qsλ pλ = r= r πε ( ( /) ) ( ( /) ) r s πε r s where we used p = qs for the dpole moment. The mnus sgn shows that ths s an attractve force, toward the lne of charge. If r >> s, the (/) s term n the denomnator s neglgble and we fnd that the lne of charge exerts an attractve force of magntude pλ F = r Solve: (a) Charges ±. nc form an electrc dpole. The electrc feld strength.5 cm from the dpole n the plane perpendcular to the dpole s 5 N/C. What s the charge separaton? (b) Solvng for s, (5 N/C)(.5 m) s = = 9.98 m =. m =. mm 9 9 (9 N m /C )(. C) 6.6. Solve: (a) A very long charged wre has lnear charge densty. C/m. At what dstance from the wre s the feld strength 5, N/C? (b) Solvng for r, (9 N m /C )(. C/m) r = =. m cm 5, N/C Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

35 The Electrc Feld Solve: (a) At what dstance along the axs of a charged dsk s the electrc feld strength half ts value at the center of the dsk? Gve your answer n terms of the dsk s radus R. (b) Dvdng both sdes of the equaton by η/ ε, z R = z = z + R z = z + R z = z + R 6.6. Solve: (a) A proton s released from the postve plate of a parallel-plate capactor and accelerates toward the negatve plate at. m/s. If the capactor plates are. cm. cm squares, what s the magntude of the charge on each? (b) Solve the frst equaton for E and substtute nto the second equaton. The charge s 7 (.67 kg)(. m/s )(8.85 C /N m )(. m) Q = = 7. C 9.6 C 6.6. Model: The long charged wre s an nfnte lne of charge. The charges on the wre and the plastc rod are unform. The plastc strrer s located on the x-axs. Solve: The electrc feld of the nfnte lne of charge at a dstance x from ts axs s λ E x = x Because the electrc feld s a functon of x, the plastc strrer experences an electrc feld that vares along the length of the strrer. We handled such problems earler. Take a small segment of the charge on the strrer and calculate the electrc force due to the lne charge on ths charge segment Δq. A summaton of all such forces on the charge segments Δq wll yeld the net force on the strrer. Ths procedure s equvalent to ntegratng the electrc force on a small segment Δq over the length of the strrer. The force on charge Δq of length Δx at poston x due to the nfnte lne of charge s λλ Δx Δ F = EΔ q= Eλ Δ x= x In the above expresson, λ = QL / s the lnear charge densty of the strrer and λ s the lnear charge densty of the plastc rod. We have also used the relaton Δq/ Δ x= λ. Changng Δx dx and ntegratng x from x =. m to x =.8 m, the total force on the strrer s.8 m λλ.8 m.8 m F = Ñ dx= ( λλ )ln x = ( )ln. m λλ πε. m x. m C = (9. N m /C )(. C/m) ln() =. N.6 m 6.6. Model: The rods are lnes of charge wth a unform lnear charge densty. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

36 6-6 Chapter 6 Solve: From Example 6., the electrc feld strength E at dstance d from the center of a charged rod s E = πε Q d d + ( L/) Because pont P s equdstant from the center for each of the three rods, the electrc feld strength s the same for each rod. We have 9 9 C E = E= E = E = (9. N m /C ) d d + (.5 m) From the geometry n the fgure, d L = tan d = tan (.5 m)tan.887 m = = L 9 N m /C E = = 5, N/C (.887 m) (.887 m) + (.5 m) E s along the y-drecton, E makes an angle of below the x-axs, and E makes an angle of below the +x-axs. In component form, E = (5, N/C) j E = (5, N/C)( cos sn j) E = (5, N/C)(cos sn j) Addng these three vectors, 5 E = (5, N/C)sn j (5, N/C) j =.8 j N/C net Thus, the electrc feld strength at the center of the trangle s 5.8 N/C Model: The rod s thn and s assumed to be a lne of charge of length L. Solve: (a) The λ-versus-y graph over the length of the rod s shown n the fgure. (b) Consder a segment of charge Δq of length Δy at a dstance y from the center of the rod. The amount of charge n ths segment s Δ q= λδ y = a y Δ y Convertng Δq to dq, Δy to dy, and ntegratng from y = L/ to y =+ L/, the total charge s Thus the constant a s + L/ L/ L/ y al Q= dq= Ñ a y dy = Ñ aydy = a = L/ Q a = L Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

37 The Electrc Feld 6-7 (c) Wth the orgn of the coordnate system at the center of the rod, consder two symmetrcally located charge segments Δq wth length Δy. The electrc feld at P from the top charge segment makes an angle θ below the x-axs and the electrc feld of the bottom charge segment makes an angle θ above the x-axs. Because the charge densty s symmetrc about the orgn (.e., λ (at y) = λ (at y)), the y-components of the two contrbutons cancel out. Thus, we have to calculate only the x-component of the electrc feld at P. Because the electrc feld strength of the lower half of the rod s the same as that of the upper half, we only need to obtan the electrc feld strength of half the rod, then multply by two. The electrc feld along the x-drecton due to a charge segment Δq s Δq λδy x xa yδy Δ Ex = cosθ = = / ( x + y ) ( x + y ) x + y ( x + y ) Changng ΔE to de, Δy = dy, ntegratng y from y = m to y = L/, and multplyng by to take nto account the entre rod, the electrc feld s E x L/ L/ ax y dy ax y dy ax = = = Ñ / / ( x + y ) ( x + y ) x y + ax 8Q x = = πε x x + L / L x + L / The last step used the expresson for a from part (b) Model: Model the nfntely long sheet of charge wth wdth L as a unformly charged sheet. Ñ L/ Solve: (a) Dvde the sheet nto many long strps parallel to the y-axs and of wdth Δx. Each strp has a lnear charge densty λ = ηδx and acts lke a long, charged wre. At the pont of nterest, strp contrbutes a small feld E of strength λ ηδx E = = r r By symmetry, the x-components of all the strps wll add to zero and the net feld wll pont straght away from the sheet along the z-axs. The net feld s z-component wll be E = ( E ) = E cosθ / z z / zr z x z The dstance r = ( x + z ) and cos θ = / = /( + ). Thus, ηδx z ηz Δx Ez = = πε ( x z ) ( x z ) x z / / Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.

38 6-8 Chapter 6 If we now let Δx dx, the sum becomes an ntegral rangng from x = ΔL/ to x = L/. Ths gves L/ L/ Ñ x z L/ L/ ηz dx ηz x η L L Ez = = tan tan tan πε z z = πε z z + From trgonometry, tan ( φ) = tan ( φ). So fnally, η L η L E z = E = k πε z (b) As z m, tan tan tan z πε L π η π E k η k = z πε ε Ths s the electrc feld due to a plane as you can see from Equaton 6.6. We obtan ths result because n the lmt as z m, the dmenson L becomes extremely large. As z, L L η L λ tan E k k = z z πε z z where we have used ηl = λ as the charge per unt length of the sheet. Ths s the electrc feld due to a long, charged wre. We obtan ths result because for z >> L, the nfntely long sheet looks lke an nfnte lne charge. (c) The followng table shows the feld strength E z n unts of η/ ε for selected values of z n unts of L. A graph of E z s shown n the fgure above. zl / Ez η ε Model: Model the nfntely long sheet of charge wth wdth L as a unformly charged sheet. Copyrght Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst.