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Ch.16Solutions

Ch.16Solutions - Electrical Energy and Capacitance 41(a...

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Unformatted text preview: Electrical Energy and Capacitance 41 (a) TheworkdoAeis I W=F.scosB=(qE).scosB,or m)cosOo=16.40XlO19 JI W=(1.60X10-19 C)(200 N/C)(2.00x10-2 (b) The change ill the electrical potential energy is II ~Ee =- W =1-6.40x10-19 :! i J I I I (c) The change ip the electrical potential is AV = ~Ee = _-6_.4_0_>< = -= .. 00 V 1_0_-19_J. 4 19 q" 1.60x10. C '------, Ii u Ii 16.2 (a) We follow t~e path from (0,0) to (20 cm,O) to (20 cm,50 em). The work done on the charge by the field is = (qE)[(0.20 m)cosOo+(0.50 m)cos90o] =(12)<10-6 C)(250 Vjm)[(0.20 m)+0]=6.0x10-4 Thus, (b) !:1V= ~Ee q &?Ee = -W = 1-6.0X10-4 J J 'I I ,--, --. V i: -6.0x10-4 Ii I 12x10-6 C J ::::50 - J/C =1-50 I 16.3 The work done by the agent moving the charge out of the cell is W;nput = ..iWjield = -( -~Ee) = +q(!:1 V) 16.4 ~E =q(!:1V)=q ( V -V ) so q= ~E e = -1.92 X10- .J _-'3.20x10-19 C e: f I' VI - V; +60.0 J/C .-----. 17 I I 42 CHAPrER16 16.5 E = I~ Vi d 25 QOO J/C 1.5~10-2 m " 11.7XI06 ,-,---- N/C I 16.6 Since potential difference is work per unit charge ~ V = W , the V\7ork done is q ~I W=q(?V)=(3.6X10 'i 5 C)(+12 J/C)=14.3X106 JI 16.7 (a) E ='~ d VI J II 600 J/C 5.33 X10-3 m I,-,--~",-----, N/C I 1.13x10 5 , (c) W = F .scosB II = (1.80 lO- r 'i 14 N)[(5.33-2.90)xlO-3 m Jcosoo=14.38XlO-17 J I 16.8 Fromconserva:tionofenergy'2mv:-0=lq(~V)1 1 or vf= 2Iq(~V)1 m (a) For the proton, 19 21(1.60XlO- C)( -120 V)/ I' " 27 = ,1.52x105 m/s, 1.67x 10- kg '------'-- (b) For the elettron, II ;! zl( -1.60 x 10-" C)( +1Z0 V)I I 9.11x10-31 kg = 6.49x10 6 m/s I Electrical Energy and Capacitance 43 (a) Use conservati9n of energy II Q (KE+PEs +PEe)f = (KE + PEs +PEe); I I or d(KE)+d(EEs)+d(PEe)=O I' x~O d(KE) =0 since the block is at rest at both beginning and end. where Xmax is the maximum stretch of the spring. max d(PE e ) = -w = -(QE)x II Thus, 0 + ~ kx~x - (QE) xmax = 0 , giving xmax _ 2QE ~.; (50.0XlO~ C)(5.00X105 Vim) 2 -- _I k , 100 Nlm or 0.500 m . ~-- .1 (b) At equilibrium~ ~ = -J:: + F: =0, Therefore, -kxeq +QE =0 Xeq = ~E =~Xmax =10.250 m , Note that when the block is released from rest,it'overshoots the eqUilibrium position and oJcillates with siinple harmonic motion inJhe electric field. 16.10 Using dy = voi + ~Jye for the full flight _ 1 2 gives O-voi+-ai 2, ,or ay--- _ -2voy t Then, using v~ = v~y,+ 2ay ( D.y) for the upward part of the flight gives ---- -V~y 2(-2voy /t) V t (20.1 m/s)( 4.10 s) = 20.6 m =- Oy = ....0..-__ 4 4 II 'I 44 CHAPTER 16 . I,F . From Newton's s.econd law, ay = -y = ---=-~- = - g+ . !i , -mg-qE m m (qE) m .. Equating . this to the earliei result gives " = -( g + ~) = -ZtVOy , so the electric I field strength is E=(m]n.!.2Vt -g]=( q U OY 2.00kg )[2(20.1 m/s) -9.80 m/s2.] = 1.95X103 N/C 5.00x10-6 C 4.10 s N/C) = 4.02xl04 v=140.2kv Thus, (AV)max 7(AYmax)E=(20.6m)(1.95x103 l 16.11 (a) 9 2 2 kq (S!199x10 N.m /C )(1.60xIO-19 V =_e = .~ . 7 1.00xlo-2 m C) = 1.44x10 I '-7 _ V I =(8.99xl0 'i 9 N.m2/C2)(1.60XlO-19 C)( 1 0.020 a m 1) 0.010 a m =1-7.19~10-8 V I where 71=0.60 16.12 V = v;, + V2 =ke(q ~ ..1 + q2) ~ m-O =0;,60 m, and ' 72 = 0.60 m - 0:i30m = 0.30 m. Thus, 2 9 V =(S.99X109:; N .m )(3.0XlOC + 6.0x10-9 C)=1.2.2X102 2 'C 0.60 m 0.30 m '-.---. Vi Electrical Energy and Capacitance 45 Calling the 2.00,uC charge q3' I! , 9:,1 ' .m2 )(s.00X10-6 C 4.00x10-6 C N = ( S.99x10' -- 1 2 ----+----+ C 0.0600 m 0.0300 m 6 V =12.67 X10 '1V _'==2.=00=X=1=0-6=C==---] ~(0.0600)2 +(0.0300)2 m I by -2.00x10-6 C in part (a) yields (b) Replacing 2.00x10-6C V =12.13X!06 V 'i I and 16.14 W = q(aV) = q(Vf:'" 11 v;), VI = 0 sinJe the S.OO ,uC is infinite distance from other charges. Thus, W =(S.00~10-6 C)(0-1.135X106 V)=1-9.0S II 16.15, (a) V = L ke qj , i 1i 2 9 =(8.99XIO~ N .m )(5.00XIO- C C2 0.175 m 3.00xl0-9 C) =1103 V 0.175 m ~- I / 46 CHAPTER 16 Ii (b) PE = keq;?2 '12' 9 N .m = 8.99xlO C2 II. ( 2 J -'---. --0-.3--S--0-m-----(S.OOXlO-9 C)(-3.00Xl0-9 I C) = ._-_3._8S_x_lO_ .... J, I .. -7 I The nega:~ve sign means that positive work must be donel to separate the charges (that is, bring them up to a state of zero potential energy). ! 16.16 The potential ft distance, k (Q v=_e_= (8.99xlO , =0.300 m from a charge Q = +9.00 XlO-9 C is 9 N .m2/C2)(9.00xlO-9 C) 0.300m . =+270 V IS Thus, the work'I required to carry a charge q = 3.00 XlO-9 C from infinity to this location . " W=q~=(3.00Xl0-9 ) C) (+270 V)=I 8.09xlO-7 JI 16.17 The Pythagor~fln theorem gives the distance from the midpoint' of the base to the charge at the apex of the triangle as '3 =~(1.00 " :1 cm)2 -(1.00 cm)2 =M cm=.J!5xlO-2 m Then, the potetltial at the midpoint of the base is V = Ikeq;/r; , or , i V = (8.9,:9Xl09 _N_.m_ : =-1.l0xl04 C 2 2 J(( ~7.00xl0-9 C) + (-_7._00_X_lO_-9_C )+ (+_7._00_X_lO_-9_C 0.0100 m O.OlO Om MxlO-2 m )J V =1-11.0 kV I .,' , - Electrical Energy and Capacitance 47 Outside the spheri~~l charge distribution, the potential is the same as for a point charge at the center of the sphere, V=keQjr,where Q=l.OOxlO-9 C Thus, ~(PEe)=q(~V)=-ekeQ " " ii 'I (1 1) --- rf f; " II and from conservaron of energy ~(KE) = -~( PEe) , v= 2 (S'9?,1 xl09 N ~~2 )V.00X10-9 C)(1.60X10, ' 9.11x10-31 kg , 19 C)(, ,0.0200 1 m J 0.03~0 m ' v=17.25X 06 i " mjs I = (KE +PEe )j' which gives 16.19 From conservation of energy, (KE +PEe) f 2(S.9pXI09 N ~~2 r f )(15S) (1.60 X10- 19 ct =!2.74X10- " 14 (6.64 X 10-" kg)(2.00x10' mlS)' . m I 16.20 / By definition, the: work required to move a charge from one point to any other point on an equipotential surface is zero. From the definition of work, W = (F cos8). s , the work is zero only if 0 or F cos 8 = 0'. The displacement s cannot be assumed to be zero ~ all cases. Thus, one inust require that F cos 8 = O. The force F is given by F = qE and neither II the charge q nor the field strength E can be assumed to be zero in all cases. Therefore, the only way the work can be zero in all cases is if cos8 = 0 . But if cos 8 = 0, then 8 = 90 or the force (and hence the electric field) must be perpendicular to the displacement s (which is tangent to the surface). That is, the field must be perpendicular to the equipotential surface at all points on that surface. s~: 48 CHAPTER16 16.21 kQ: V=_e_ r ' Ii so 9 r = k,;Q V:: t (8.99xl0 , I' N .m2/C2)(8.00xlO-9 C) _ 71.9 V.m V . V For V = 100 V, 50.0 V, and 25.0 V, ,I I r = 0.719 m, 1.44 m, and 2.88 m I I to the potential. I The radii are ijwersely proportional I 16.22 " (a) ,(b) Q =C(LlY) = (4.00XI0-6 F)(12.0 V) = 48.0xlO-6 C =148.0,uC Q=C(LlY) = (4.00xl0-6 F)(1.50V):::6.00xl0-6 C=16.00,uCI 16.23 (a) A C=E -= o d ( 8.85xl0-12 , --' C )(1 Ox10 '. m ) =I1.1XlO-8FI N'm2 (800m) _ 2 6 2 (b) QIIIIIX =C(:~V)max =C(Emaxd) =(1.llXI0-8 F)(3.0X106 N/C)(800 m) =127 C I 16.24 Fora paralJelpkle capacitor, " av= Q = C Eo (Q ) = Qd A/d EoA' (a) Doubling~iwhileholdingQandAconstantdoubles , '-(b) LlV to 1800VI. Q (EoA)AV d :: Ii Thus, doubling d while holding Ll V and A constant will cut the charge in h~l1f,or Qf = Qi /21 I 16.25 (a) E=dV d = I: 20.0~ -1.11x104 V/m=ll1.1 1:80xlO m " kV/ml directed toward the negative . plate Eo . (b) C=T=, A (8.85xlO-12 C2/N .m2)(7.60XlO-4 m2) 1.80xl0-3 m i =3.74xlO-12 F =13.74 pF I Electrical Energy and Capacitance 49 (c) Q =C(AV)=(3.74X10-12 F)(20.0 V) = 7.47 X10-11 C ='174.7 pC Ion one plate and 1-74.7 pC ortiithe other plate. I C=_o- E A .d ' so riA (B.B5X10d=_o-= E 12 C C2/N .m2)(21.0xIO. 60.0x10.15 F 12 m) . 2 =3.10x10-9 m 16.27 (a) IAVI (b) E=--= d 16.28 90.4 V . = 9.04x104 V/~ 1.00x10-3 m -----'-. I I LFy= 0 ~ ::rcos 15.0 = mg !i or r = mg cos15.0 LFx = 0 ~ 'I qE = T sin 15.0 = mg tan 15.0 or E = mg tan:~5.00 q:! AV = Ed = mgd tan 15.0 ![ q (350X10-6 k!k)(9.BOm/s ).(0.0400m)tan15.00 A.v=. . 30.0x10~ C '_ 2 mg =1.~3X103 V= 1.23kV I I 16.29 () a . 1 1 1 For senes coqnection, - = - + ~ :1 . C eq I! Ceq C1 C2 C1C2 = -----""-=-- C1 +C2 :)u CHAPTER 16 , I' (b) Q1=C1(.1V) = (0.050,uF)( 400V) = [20.0,uC] \ I I Q2=C2 (.1V) = (0.100,uF)( 400 V) =[40.0 pC] 16.30 (a) For parallel connection, CeQ='C1+C2 +C3 = (5.00+ 4.00+ 9.00) ,uF = /18.0 ,uF] , .., . (b) For senes connection, _ 1 '! . Ceq 1 = _1 +_1 +_ C1 C2 C3 _1 =1,_1_+_1 Ceq ,I IIS.OO,uF __ +_1_ giving C =[178 ,uF] 4.00,uF 9.00 ,uF ' . eq' 16.31 (a) Using the rules for combining capacitors in series and in parallel, the circuit is reduced in steps as shown below. The equivalent capacitor is shown to be a [2.00,uF] capacitor. 4.00fJ-F a 2.00 jtF ) 12.0 V Figure 1 12.0V Figure 2 12.0V Figure 3 Electrical Energy and Capacitance ,I 51 From Figure 3: ,: Qae=Cae (a vte From Figure 2:' = (2.00 ,uF)(12.0 V) = 24.0 ,uC Qab=Qbe = Qae = 24.0 ,uC capacitor is Q3 =124~0 ,uC . Thus, the charg~ on the 3.00,uF ~ ., Continumg to se 't"F' J :1 I 19ure 2 (V) , a ab Qab 24.0 ,uC 4 00 V =-C = 6 00 F - . ~ ,u and (a V)" = (a V) . = Qbe = 24.0,uC 3 be Cbe 3.00 ,uF 18.00 V ,--~. I From Figure 1,il(aV)4 = (aV)2 and = (avtb =14.00 V I I . (1) Q4=~4 (a V)4 = (4.00 ,uF) (4.00 V) = 116.0 ,uC I il Q2=C2 (avj~ = (2.00 ,uF)( 4.00 V) =18.00,uc Cparallel 1 +C2 =9.00 pF => C1 =9.00 pF-C2 =C -C-=-C series 1 1, l' +-C 1 2 => Cseries=-C-+---C--=2.00 pF 12 C1 C2 Thus, using equati?n (1), Cseries= ( (9.00 pF -C2)C2 ). = 2.00 pF whicll reduces to 9.00pF-C2 +C2 6.00 pF)( C2 - 3.00 pF) = 0 (1), C1 = 3.00 pF C; - (9.00 JF)C2 " + 18.0 (pF)2 = 0 , or (C2 - Therefore, or either C2 = 6.00 pF . and, . from equation C2 = 3.00 pF and C1 = 6.00 pF . are 13.00 pF and 6.00 pF We conclude that ~he two capacitances I. 52 CHAPTER 16 16.33 2.50 a /-eb 20,0 j.tF a j.tF 6.00 j.tF c /-eb 20.0 j.tF ae--H-Y /-e b . 8.50. j.tF 20.0 j.tF Figure 1 Figure 2 Figure 3 \ (a) The equivalent capacitance of the upper branch between points a and c in Figure 1 is II C. = (15.0 ,uF)(3.00 ,uF) 15.0 ,uF + 3.00 ,uF 2.50,uF Then, using':Figure 2, the total capacitance between points aand c is i " . Cae= 2.50'I ,uF+6.00 ,uF=8.50,uF 'I From Figur~ 3, the total capacitance is C eJ1 = ( 8.501,uF + 20.01,uF )-1 = I.,u .. 596 , :i F I (b) Qab= Qac:= Qcb=(~ V)abCeq =(15.0 V)(5.96 ,uF) =89.5,uC Thus, the ch~rge on the 20.0,uC is Q20 = Q = 189.5 ,uC I cb Then, Q6 = (~ vtc (6.00 ,uF) = 163.2 ,uC I and 1 ; I Q15=Q3 =(~V)ac (2.50 ,uF) =126.3,uc I ;"--' Electrical Energy and Capacitance 53 : i : , (a) The combinati<?,nreduces t~ an equivalent capacitance of 112.0 p,F in stages as II I '-- shown below. 'I -1 I II -112.001 1 ~wr2~!C 1__ 1 36.0 V 6J[ 36.0 V ~F_ _ Figure 1 Figure 2 il J Figure 3 (b) From Figure 2~ Q2 and Q4 = (4.00 p,F)(36.0 V) = 144 p,C V) = 172.0 fiC I I I. = (2.00 ~F)(36.0 Q6 = (6.00 p,F)(36.0 V) = 216p,C Then, from Figure 1, 16.35 II : a --+ a --+ 1.00' ~ . 24.0 V 8.00 I: I:' ~ , I 5.00 ~ . 4.00 ~ LJ12.0 '. . 6.00" ~ n 400 . ~F '. c c' Figure 1 Figure 2 Figure 3 The circuit may ~e reduced in steps as shown above. Using the Figurep, Then, in Figure 2, (fj. Qac = (4.00 p,F)(24.0 V) = 96.0 p,C vtb = Qac Cab 96.0 p,C = 16.0 V 6.00 pF Finally, using Figure 1, Ql = C1 (d V)ab = (1.00 pF)(16.0 V) = 116.0 pC I Qs =(5.00 fF)(dV)ab =180.0 pC and I, Q4 = (4.00 pF) (d V)bc = 132.0 pC I Qs =(8.00 pF)(dV)bc =164.0 pC I 16.36 The technicianco~bines two of the capacitors in parallel making a c::apacitorof capacitance 200 pF . Then she does it again with two more of the capacitors. Then the two resulting 200 pF capacitors are conAected in series to yield an equivalent capacitbce of 100 pF . Because of the symmetry of the solution, every capacitor in the combination has the same voltage across it, II A , ' dV = (dV)(jb/2 = (90.0 V)/2=145.0 V \ I 16.37 (a) From Q =C(4,V), Q25 =(25.0 pF)(50.0 V) = 1.25x 103 pC =11.25 mC and. Q40:i=(40.0pF)(50.0V)=2.00x103 pC=1 2.00 mCI I (b) When the two; capacitors are connected in parallel, the equivalent capacitance is . Ceq = C1 + C2 ='25.0 pF+40.0 pF = 65.0 pF . :I Since the negative plate of one was connected to the positive plate of the other, the total charge stpred in the parallel combination is , Q=Q40 -Q25 =2.00x103 pC-1.25x103 pC=750 pC The potential 'I'" ~ifference across each capacitor of the parallel combination is - ,I and the final charge stored in each capacitor is Q~ =C1 (dV) = (25.0 pF)(ll.5 and V) =1288 pC I I Q~= Q - Q~ = 750 pC - 288 pC = 1462 pC , Electrical Energy and Capacitance 55 ',tom Q = C (!1V) , ~e initial charge of each capacitor is, QI0 = (10.0 kF)(12.0 V) = 120 p,C and Qx = Cx (0) = 0 After the capacitor~ are connected in parallel, the potential difference across each is !1v' = 3.00 V, and the total charge of Q = Q10 +Qx = 120 p,C is divided between the two capacitors as Q~o= (10.0pF)(3.00 V) =30.0 p,C and Q: = Q- Q~() = 120 p,C - 30.0 p,C = 90.0 p,C Th s C = = 90.0 p,C = 130.0 P,F I u, x !1V' 3.00 V Q: 16.39 From Q = C( !1V) , the initial charge of each capacitor is Ql = (1.00 ,UF)(10.0 V) =10.0 p,C and Q2 = (2.00 p,F)(O) = 0 After the capacitors are connected in parallel, the potential difference across one is the same as that across the other. This gives " . !1V= Q~ Q; or Q;=2Q~ 1.00 j.LF 2.00 p,F [I (1) From conservatio~ of charge, Q~+ Q; = Ql + Q2 = 10.0 p,C . Then, substituting from equation (1), this ~ecomes' Q~+ 2Q~=110.0p,C, giving Q~=11~ ,cl P Q; Finally, from equation (I), =12~P,cl _----'!-----------~-=----=---::::::-I 56 CHAPTER 16 1 16.40 The origind,l circuit reduces to a single equivalent capacitor in the steps shown below. il a a C2 y C C' C2 y 1 C2 b = ( -+-C -1.= ( 5.00 ,uF + ,uF.)-1 =3.33,uF s C1 10.0 1 1 2 J 1 Cp1'= Cs + C3 + Cs = 2(3.33 ,uF) + 2.00 ,uF = 8.66,uF Cp2=:1 C2 + C2 = 2(10.0 ,uF) = 20.0 ,uF C =(._1 +_1 eq Cp1 Cp2 J-1 =(_1_+_1 8.66,uF _)-1 20.0 ,uF =16.04,uF) -.... 16.41 Refer to the solution of Problem 16.40 given above. The total charge stored between points a and b is Qeq = ~eq (.1Vtb = (6.04 ,uF)( 60.0 V) = 362,uC Then, looking at the third figure, observe that the charges of the series capacitors of that figure are Qpl = Qp2 = Qeq = 362 ,uC. Thus, the potential difference across the upper parallel combination shown in the second figure is " (.1V) i= P, 'I Qp1 = 362,uC = 41.8 V Cp1 8.66,uF Finally, the charge on C3 is Q3 = C3 (.1V)P1 = (2.00 ,uF)( 41.8 V) = 183.6 ,uC I Electrical Energy and Capacitance 57 ecogniZe that the 7~00p,F and the 5.00 p,F of the center branch are cokected in series. The total " capacitance of that branch is ' '! :j C = ( --+-' 5 5.00 1 7.00 !i'1 )-1 =2.92p,F a Then recognize that this capacitor, the 4.00 p,F capacitor, and the 6;00 p,F capacitor are all connected in parallel between points a and b. Thus, the equivalent capacitance betwee4 points a and b is Ceq = 4.00 p,F+ 2.92 p,F+6.00 p,F = 112.9 p,F I 16.43 The capacitance is E A C=_o_= d, (8.85xl0-12 . C2/N .m2)(2.00XI0-4 5.00xl0-3 m m2) ", = 3.54 X10-13 F and the stored energy is :1 16.44 (a) When connectJd in parallel, the energy stored is , II =~[(25.q+5.00)XI0~ FJ(100 V)2 =10.150 JI (b) When con:nect~d in series, tne equivalent capacitance is " 1 1 Ceq = (+, -'25.0 !i 5.00 From W )-1 p,F = 4.17 p,F = tCeq I' (I:::. , the potential difference required to store the same energy as V)2 in part (a) above is I:::.V = 2W,!= Ceq I f1: il 2(0.150 J) =1268 V 4.17xl0-6 F ~~ I 58 CHAPTER 16 !i 16.45 The capa&itance of this parallel plate capacitor is , ( C2 )(1 Ox10 m2) C =E A = 8.85x10-12 . = 1.1x10-8 F. 0 d N'm2 (800m) '0 6 With an electric field strength of E =3.0 x 106 N/C and a plate separation of d = 800 m, the potential difference between plates is :~ " SV=Ed=(3.0X106 V/m)(800m)=2.4x109 V Thus, the energy available for release in a lightning strike is 16.46 The energy transferred to the water is .'j 1 '[1 ] (50.0 C)(1.00X108 V) W=- -Q(LlV) = . =2.50xlO 100 2 200 II 7 J Thus, if m is the mass of water boiled away, " W =m[c(LlT)+4J becoines giving m= 2.50x10 " 7 2.55 J/kg J=19.79 k ., g I 16.47 The initial d:apacitance (with air between the plates) is Cj = Qf( Ll V)j , and the final capacitanceil(with the dielectric inserted) is Cf =Q/(LlV)f quantity of charge stored on the plates. , C (LlV). Thus, the dielectric constant is /( =.....L = ( ) ii Cj Ll V f I where Q is the constant 00 V ~ 25 V = 14.0 I 16.48 (a) E= LlV:_ d Ii 6.00 V =13.00X103 V/~ 2.00xlO-3 m . ~. I Electrical Energy and Capacitance 59 .. With air between, the plates, the capacitance is ii . A C. =Eo-= air d and with water (Ii 2 -12 C2 )(2.00X10-4 m ). . ""13 . 8.85x10 . 2 ( 3) =8.85x10 F N.m 2.00x10- m = 80) between the plates, the capacitance is n (K ! ., I C = KCair= (80)(8.S5X10-13 F) = 7.0Sx10- F The stored charge when water is between the plates is Q =C(aV) == (7.0SX10-n F)(6.00 V) = 4.25 X10-10 C =142.5 nC I (c) When air is the ~ielectric between the plates, the stored charge is Qair =Cair (aV)=(S.S5x10-13F)(6.00 K V) = 5.31X10- 12 c =15.31 pC I 16.49 (a) The dielectric constant for TeflonQll is = 2.1, so the capacitance is 2 m ) 2 C= KEoA = (2.1)(S.S5X10-12 C2/N.m )(175x10-4 d' 0.0400x10.3 II\ C=8.13X10r F=I,8.13nFl (b) For TeflonQll, dielectric strength is Emax = 60.0x106. Vim, so the maxiIIltlin voltage the is Vmax = 2.40~103 V = 12.40 kV I ii 16.50 Before the capacitor is rolled, the capacitance of this parallel plate capacitor is C= KEOA::i: d KEo(wxL) d where A is the surface area of one side of a foil strip. Thus, the required length is C.d . L=-= KEoW, (9.50 x 10-8 F) (0.025 Ox10-3 m) =\104m 2 (3.70)(S.S5X10-12 C2/N.m2)(7.00XIO- m) . l ~-~-------r-------------------+--' 60 CHAPTER 16 i6.51 (a) 12 V = m h 1.00 x iO- kg P:~ 1100 kg/m3 .41l'r 19.09X10-16 m31 ----.. ' [3VJ1I3 41l' I Smce V = -3 -, the radIUs IS r = , 3 ' and the surface area IS ~3 . . , A = 41l'r = 41l' . 2 2/ 3 ~~ = 41l' [3(9 09 X10'. 41l' 16 3 m )] = 14.54 X10-10 m21 (b) c= KEbA d (5 .. .0)(8.85x10-12 'C2/N .m2)(4.54XlO-1O m2) 0 = " =12.01X10-13 F-'. ' 100x10-9 m ~, 'I ( !; (c) Q = C(AV) = (2.01xlO-13 F)(100XlO-3 V) = 12.01xlO-14C and the number of electronic charges is I n= 9 e 2.01x 10- 14 1.60 X 10-19 C C = 11.26 X105 .----. I 16.52 Since the capacitors are in parallel, the equivalent capacitance is I or 1~.53 Since the cap~citors are in series, the equivalent capacitance is given by or Ceq = iEoA were '-d- h d = d1 + d2 + d3 I I!--' -------------------- Electrical Energy and Capacitance 61 For the parallel cqrrt,.bination: C" = C1 + C2 which gives (1) t . 1 For the series com bmation: Cs Thus, we have I : 1 1 = -C +-C 2 1 or 1 C -Cs - 1 = _. - - 1 =---'-----1 C2 Cs C1 CsCI :~C2 = CC~~ s and equating this to Equation (1) above gives or We write this result as: and use the quadr~tic formula to obtain Then, Equation (1~gives 1655 The charge stored on the capacitor by the battery is Q =C(~V)1 =C(100 V) L This is also the t~~al charge stored in the parallel combination when this charged capacitor is connected in parallel with an uncharged 10.0-,uF capacitor. Thus, if (L\V)2 is " the resulting voltage across the parallel combination, , . 9 = C" (~ V)2 gives C(100 V) =(C + 10.0 ,uF)(30.0 V) or (70.0 V)C = (30.0V)(10.0 ,uF) " and 16.56 C = (~~:~ ~ }10.0 ,uF) = 14.29 IJF I (a) _The 1.0-,uC I'islocated 0.50 m from point P, so its contribution to the potential at P is .V =ke 1 ql. f l = (8.99xl09 N .m2/c2)(1.OXI0-6 C)=ll.SXl0 0.50 m 4 V . I (b) The potential at P due to the -2.0-IJC charge located 0.50 m away is __ ~ __ JI 'I 62 CHAPTER 16 (c) The total potential at point P is Vp == V't+ V2 = (+1.8 - 3.6) x 104 V = 1-1.8 X 104 V] . (d) The worf required to move a charge q = 3.0 pC to point P from infinity is 16.57 The stages fo! the reduction of this circuit are shown below. 3.00 J-tF Ii 4.00 J-tF 3.00 J-tF 6.00 J-tF 2.25J-tF 6.25 J-tF 7.00 J-tF 48.0 V 48.0 V 48.0 V 48.0 V Thus, Ceq = )6.25 pF I 16.58 (a) Due to spherical symmetry, the charge on each of the concentric spherical shells will be unifoqnIy distribt~.ted over that shell. Inside a spherical surface having a uniform charge distribution, the electric field due to the charge on that surface is zero. Thus, in this re~on, the potential due to the charge-on that surface is constant and equal to the potential at the surface. Outside a spherical surface having a uniform charge distribution, the potential due to the charge on that surface is given by V = keq where r is the distance from the center of that surface and q is the charge on that surface. . . . In the remon between a pair of concentric spherical shells, with the inner shell having chkrge +Q and the outer shell having radius b and charge -Q, the total electric pdtential is given by r iBi ~;IMi\!1iilI_ " . om}" :1 Ii Electrical Energy and Capacitance 63 The potential difference between the two shells is therefore, I' " ilV=V\. r=a',r=bIi -.I.liV! =kQ (1 1) -kQ (1 1) =kQ - ab e --ab ebb --e (b-a) The capacitance of this device is given by (b) When b a , ~hen' b - a"" b . Thus, in the limit as b -7 00 , the capacitance found above become~ . 16.59 The energy store~:in a charged capacitor is W =~C(ilV)2. Hence, L\V=~2W.: = Cli 2(300J) = 4.47 x 103 V=14.47kV. 30.0x10-6 F . ---. -_. I 16.60 From Q = C ( il V) ; the capacitance of the capacitor with air between the plates is " C o = Qo ::i: 150 ,uC L\V: L\V 'I After the dielectr~c is inserted, the potential difference is held to the original value, but the charge chang~s to Q = Qo + 200 ,uC=350,uC . Thus, the capacitance with the dielectric slab in place is C II = _Q__ ='.i._35_O,"-,u_C ilV " ilV Ii The dielectric constant of the dielectric slab is therefore . K=~=~350 Co \. ilV ,uC)( L\V 150 ,uC )= 350 150 =12.331 64 CHAPTER 16 16.61 The charges initially stored on the capacitors are 'I Ql =C1 (~V)j =(6.0 ,uF)(250 V)=1.5x103 ,uC and Q2 =C2 (I~V)j = (2.0 ,uF)(250 V) = 5.0x102 pC 1 When the capacitors are connected in parallel, with the negative plate of one connected to the positive plate of the other, the net stored charge is The equivalent c,apacitance of the parall~l combination is Ceq the final potenti~l difference across each of the capacitors is = C1 + C2 = 8.0 ,uF . Thus, and the final charge on each capacitor is Q; = C and 1 (~V)' = (6.0 ,uF)(125 V) = 750 ,uC = 10.75 mC I I Q~= C2 (~V)' = (2.0 ,uF) (125 V) = 250 ,uC = 0.25 mC I 16.62 When,connected in series, the equivalent capacitance is and the charge stored on each capacitor is When the capacitors are reconnected in parallel, with the positive plate of one connected to the positive plate of the other, the new equivalent capacitance is C;q = C1 + C2 = 6.0 ,uF and the net stor~d charge is Q' = Ql + Q2 = 800/3 ,u~. Therefore, the final potential difference across each of the capacitors is (!:!V)' = R' = 800/3 ,uC = 44.4 V C2eq 6.0,uF ----_-:-~-~---~-----------------....,.-- Electrical Energy and Capacitance 65 The final charge on:!each of the capaCitors is ,) (a) V = V, + V 1 2 + V, =_e. 3 , kQ x+d 2kQ kQ e_+_e - x x-d _ - keQ [X(X:-d)-2(X2 .. ! X X (2 -d )+X(X+d)] - 2 d2 ) (b) When xd, " thenx2 _d2 =x2 and V 16.64 The energy requJed to melt the lead sample is ! w=m[cpb,(~T)+Lf 3 =(6.00'~10-6 ,kg)[(128 Jjkg.oC)(327.3C-20.0C)+245X10 =0.383 J J/kgJ The energy stored in a capacitor is W =.!.C(~V)2, so the requir~d potential difference is , .2. ., !i . ~V =~2tv = 2(0.383 J) =1121 V C 52.0x 10-6F ~.--. I _____---L------------------. 66 CHAPTER 16 16.65 The capacitance ofa parallel plate capacitor is C = II . K~ A Thus, KEoA ::h C .d, and the given force equation may be rewritten as Withthe giver data values, the force is F= C~l1V)2 =_(2_0_X1_O_- _F_)(_10_0_V_}2';;:_I50 6 NI ii2d !k " 2(2.0xlO-3 m) 16.66 The electric fi~ld between the plates is directed downward with magnitude IE y I~,d = 2.00x10-3 m -5.00x10 4V 100 V ,i Ii 4 N/m Since the graV:ftational force experienced by the electron is negligible in comparison to the electrical force acting on it, the vertical acceleration is I' a =-:: =-= y m;, me FJ,: qEy (-1.60X10- C)( -5.00 X104 9.11xlO-31 kg 19 N/, m), , =+8.78x10 15 , m s I 2 . (a) At the clo~est approach to the bottom plate, vy = O. Thus, the vertical displacement from point 0 is found from _ Of(vosin8oY fly - -" ---------. ii v~ = V~y + 2ay (l1y) as 2ay = -=--(------) 2 8.78x10 m/s2 15 -[ -(S.6X106 m/s)sin4SoJ -=- = -0.89 mm The minimum distance above the bottom plate is then i; d=D +:lflY=l.OOmm-0.89 2 , mm=1 0.11mml '-- _ rljn~ FEmwE Electrical Energy and Capacitance 67 (b) The time for the electron to go from point 0 to the upper plate is found from . 1 /iy = voyf+-aye as 2 Solving for t g~ves a positive solution of t = 1.11 x 10-9 s. The horizontal displacement from point 0 at this time is ...
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