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Unformatted text preview: 23.1. IDENTE‘I: Apply Eqﬂll} to calculate the work. The electric potential energy ofa pair of point charges is given
hf: Eqﬂlﬂ].
SET L'P: Let the initial position of g: be point a and the f'ntal position he point in, as shown in Figure 23.1. :; =0.13{'n1 Figure 2.3.1 I 1115“ "1 a; = ..II'{0.250 n1)J — {0.250 n1}2 ,. :; =0.353tiin 0.250 Ill
Encore: it’d = U, —L; L: L“: =ceeeex1u° NmJ.'C::I _ {+2.40x10" C']{—4_sox1u* C)
4n“ 1; 0.150 in U“ = 410134 J u L; = = [3933 X m; H_m:I.C,2:II:+140 x 103;);450 : 10"E C]
L; = —0.2623 J
02% = If, —1','Eh = 4.5134 J—{—0.2I523 I] = —0.35ti J
EVALUATE: The attractive force on g: is. toward the origin. to it does negative work on g: when q; moves to
larger r'. 23.5. tummy: apply t‘=e‘l'q= amisolve for r'.
i. SETUP: air;=—'.'.2>c:1[f"1 C'. q: =+2.3«:=11l“i C Encore: r' =————
U 41400 J E‘L’ALUATE: The potential energy L’ is. a scalar and can take positive and negative T.'aues. 23.11. IDENTITY: Apply Eqﬂil}. Ihe net work to bring the charges in from inﬁnity 1:. equal to the change in potential
energy. The total potential energy is the sun: of the potential energies of each pair of charges, calculated from Eq.{23.9].
SET UP: Let 1 he where all the charges are inﬁniter ﬁr apart. let I be where the charges are at the corners, of the
triangle. as shown in Figure 23.11. .1, q
/ r1 Let qt. he the third, unknmsn charge.
:I {Ili
I'i' :' { EELU‘IE:1T’=—nL'= —{t'. 4:.) Figure 35'" U, =0
L'.=L" +Lr +U = 1—2
. a .. 5. dado as.)
Want TF=0. so Ti" =—I:L71—L7,] gives 0 =—i..'1
1
0: 2+2 .
“ﬁts! are} 91+ lea =0 and a =e"2
E‘L’ALUA'I'E: The potential energy for the two charges 9 is positive and for each :3 with qt it is. negative. There are
two of the q. qt. terms so must have qt < q. 23.16. 23.19. 21.1. H: IDENTIFY: The workenergy theorem says Hfﬂ.‘ = Kb — K“. P“ = 1: — ‘3
SET UP: Point or is the starting and point it is the ending point. Since the ﬁeld is tuJifornJ. WP“ =Fscossl=Eq .5 cos o. The field is to the left so the force on the posids‘e charge is to the leﬁ. The particle
moves to the left so all = [37 and the worlt l'l"Hm is positive. Blunt: (:1) 112* =5; —1P.."r =1.5I:]><1EI'5 J— D =l.31lx1C—FJ arm _ 1.50:. 10* .1 all Fl—F; = q _—4 ﬁnled? C = 33? U Point :2 is at higher potential than point .13.
— —ﬁ;*—E_Fl_&_'  3 r.
{c} EIqlr—Waﬁjo 5— ME — 5 — ﬁanmﬂ III—193x10 "Urn. EVALUATE: A positive charge gains kinetic energy when it moves to lower potential; 1“; <: F,. For a point charge I' = ﬂ . Solve for r'. 1,. ltq = {399K109 Nm:.'C:]{l.iD=lﬂ"‘ C} IDENTIFY and Err L's: E : . '=— —=l.'D I'll—3 =1'D
ELU‘I‘E (1]r F PM) v :I X m 3 rant
. . . . "1"  renew .
fr: =constantso Er=I'.r.. n=r —' = lJDInnr' '=.'.5C
“’3 “I "   '._t;_, { listless; E‘L’ALUA'I'E: The potential ofa positive charge is positis‘e and decreases as the distance ﬂow the point charge
increases. IDENTITY and SET UP: Expressions for the electric potential inside and otttside a solid conducting sphere are deris‘ed in Example 23.3. kt; _ H3.ij 1o" (3] T ' o .430 m Hsjoxm“ C]
11240 or {c} This is inside the sphere. The potential has the same Value as at the surface, 131 V.
E‘L'ALUATE: All poinm ofa conductor are at tbe same potential. EELU‘I‘E: (a) This is ouwide the sphere. so F = = did ‘3. Kb] This is at the surface ofthe sphere. so I" = =131‘i.'. Inrnrsrr and SET UP:
difference between the plates is F = Eo‘ . For oppositele cbargedparallel plates, E = .'T."£t between the plates and the potential EICELUTE: (a) E =— — — = 5310 NFC.
5:: a
Ifh] F = 1351 = {3311} h'"C]{ﬂ.0220 In] =11? "if.
{c} The electric ﬁeld stays the same ifthe separation of the plates doubles. The potential difference between the
plates doubles. E‘L’ALUATE: The electric ﬁeld ofan inﬁnite sheet of charge is tuiiforln_ independent of distance from the sheet. The force on a test charge between the two plates is constant becattse the electric field is constant. The potential
difference is the worlc per tuiit charge on a test charge when it moves ﬁ'ona one plate to the other. 1ithen the distance doubles the work. which is force tiLnes distance. doubles and the potential difference doubles. Lona—111's: C =£
tﬁa
SET UP: 1,uIF=1I:I“i F
Erect—rt: 9: CF; = mesa10* Bait) ETJ=1_32><1D" (3:13] all: EI'ALIJATE: One plate has charge —Q and the other has charge —Q . 31.5. Lossm: c= Q . c=Eﬂ—A.
I; or
SET UP: 1i5.."l:1en the capacitor is connected to the battery: enough charge ﬂows onto the plates to make Fm =11'3' V.
LsIcLTE: {a}: 110 ‘.' {b} {i} When u‘ is doubled C is halved. I; =% and Q is constant: so I’doubles. I" = 2433' V . {ii} When 1" is doubled A increases by a factor of '4. T’decreases by a factor of4 and I’ = 3.9 V _
ETALIJATE: The electric ﬁeld between the plates is E = Q"snsl . T; = Ed _ 'sl'hen 11‘ is doubled E is unchanged and
I’douhles. 'thenA is increased by a factor ofd, E decreases by a factor ofd so T’decreases by a factor ofd. 21.1ﬂ. IDES—I'II'Y: Capacitance depends onthe geometry ofthe object.
7— L
11 on him ."1; :1 "I r .l"
15.. .. (a) SET UP: The capacitance ofa cylindrical capacitor is C = . Solsing for 1;, gives 1;, =1;la Ecs'EctTE: Substituting in the numbers for the exponent gives
2.":{835 1: 1G": C't‘N m2 :1 ('3'. 12:} m}
3.5? a 113"" F
Now use this value to calculate r”: 1;, =1'_,a:"m = {9.154} cmja':' :5: = 0.3130 cm
{1}} SET "UP: For any capacitor, C = QIT’and A. = £2.31. Combining these equations and substituting the numbers gives xi. = QII = CITE.
EJICLTE: Numerically we get
r “3.57 5: 111“ F” [133131
= CI = 3.82 a: 10'“ C.'m= 33.2 nil"m
L Ullﬂm ETALIJATE: The distance between the surfaces of the two cylinders would he only CLUSGI cm which is just
{till mm. These cylinders would have to be carefully constructed =I'J.lE2 P. 24.22. IDES—I'lI'Y: Simplify the network by replacing series and parallel combinations of capacitors by their equisalents. SET UP: For capacitors in series the voltages add and the charges are the same; L = i+ l —   For capacitors Cars—s n. in parallel the voltages are the same and the charges add; C“. = C, + C: +   C =I—E. EJICLTE: {a} The equivalent capacitance ofthe 5.1} ,uF and 3.0 ,uT' capacitors inparallel is 13.0 ,MF. When these two capacitors are replaced their equivalent we get the network sketched in Figure 24.22. The equivalent
capacitance ofthese three capacitors in series is 3.4? ,uF. {b} om = cm!” = (3.4? FREDA} 1;} =1T4 IraC {c} Q.“ is the same as Q for each ofthe capacitors in the series combination shown in Figure 24.221 so Q for each of
the capacitors is 1H _1.1C. ETALUATE: The voltages across each capacitor in Figure 24.22 are PM = g“ = 114 V , F; = 4'3"" =13_d "LT and lil 11 r’ = =19.3 y. rgt +11; +1; =13.4 v —13.4 v — 19.3 v = 511.1 v . The sum ofthe voltages equals the applied '3'
5‘ voltage._ apart ﬁom a small difference due to rounding.
Inn .111: 9.1:: “F ~—1—1l—H—3 13.11 s '
Figure 24.12 2" TL. SETUP: d =1.iU=<lﬂ_" m. 1313:“? c ExEctTE: [3} EM =933K13‘” F: 90.13 pF
100 1 E _ (9.00::10'” 31(1ij 10'; 1.11] q. 3.35mi?“ c1 JENml} {c} r' = Ed = [3.03106 mum .50 5: 1n:r1 m) = 4.5 210‘ v {d} Energy=§gr'=L.[uc1scimt mam V}=1_3'3')tlﬂ_ﬁ J=1.3£I .qu g: 10.313031? cf
2c 1:9.Gﬂxlﬂ'” F) 24.15. 13mm: C= .C=%.IL=Ed.T1estaredeue1g§ria%§'l’. {b} (3:3?353 A: =u.r:+152m=. ETALUATE: ‘We could 313c calculate the stared energy as. = 1.3U pl . 24.39. LDEFI'II'Y and SET L'P: Q is ccuuataut 3: we can apply Eq.{2—1.14}. The chalge dv:=_1':sit;.r on each stuface cfthe
dielectic ia given by Eq.[24.llfij.
q j _'i Irl.
EIECLTE: E = 5 30K =5=M=122
K E 2.:ICIK1EI' ‘L‘rm
(a) a]. = ail—1.5K}
LI = 52,130 = [3.334 : 13"1 '33'51  mljﬁﬂﬂ x11}E NFC) = 2.333x13'" C‘s'm2
err. ={1.333><13I'" C"m:}[1—1’1.23}= 6.20 :~<1I:'T IlliiiJ
{In} As calculated above, K =1.23. EI'ALUATE: The aurface charges an the dielectic produce 21.1.1 electric ﬁeld that partially cancels the electric ﬁeld
produced by the charges, Dll the capacitcrplatea. ...
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