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Unformatted text preview: 23.T. ‘ TDENTIFY: Apply Eq.(23.2) to'caiéﬁiét’é the work. The electric potential energy orﬁa‘ﬁaﬁ’rbf by Eq.(23.9).
SET UP: Let the initial position of q2 be point a and the final position be point [7, as shown in Figure 23.1. r“ = 0.150 m
rb = «0.250 my + (0.250 m)2
rb = 0.3536 m
0.250 m
Figure 23.1
EXECUTE: Wagh = Ua 4U},
—6 __ —6
U“ :_1_q1q2 =(8988X109 N‘mZ/CZ)(+2.40><10 C)( 4.30><10 C)
47M0 ra 0.150 m
Ua = —0.6184 J
1 —6 _ 6
Uh = _q1q2 = (8,988x109 N . m2 /C2)W
47:60 rb 0.3536 in
Uh = —0.2623 J W‘Hb = Ua ~Ub z—0.6184 J —(—0.2623 J) = —0.356 J EVALUATE: The attractive force on q2 is toward the origin, so it dees negative work on qz when qz moves to larger r.
23.5. (a) IDENTIFY: Use conservation of energy: Ka+Ua+W other = Kb + Ub
U for the pair of point charges is given by Eq.(23.9). SET UP: a O / \ I 0 Let point a be where q; is 0.800 m from
qz q‘ q; and point b be where qz is 0.400 m r = 0,300 m from ql, as shown in Figure 23.5a. Figure 23.5a =0 andU=~———1 ‘1qu.
47:60 r K =imv: =§(1.50x10‘3 kg)(22.0 m/s)2 = 0.3630 J a 2 EXECUTE: Only the electric force does work, so W0 the: = +0.2454 J — . *5 — . 1 *5
U“: 1 qlq2=(8.988X109N.m2/C2)( 280x10 C)( 780><0 C)
4,560 r 0.800m a _1 2
Kb—Emvb _ —6 _ —6
Uh: 1 qlq2=(8.988x109Nlm2/C2)( 2.80x10 C)( 7.80x10 C)
4,;60 rb 0.400m = +0.4907 J The conservation of energy equation then gives Kb = Ka + (UH — U 1,) §mvf = +0.3630 J + (0.2454 J —0.4907 J) = 0.1177 J . J MEWS
1.50x10 kg EVALUATE: The potential energy increases when the two positively charged spheres get closer together, so the
kinetic energy and speed decrease.
(b) IDENTIFY: Let point e be where qz has its speed momentarily reduced to zero. Apply conservation of energy to points a and c: Ka +Ua +W =Kc +Uc. other SET UP; Points 0 and c are shown Figure 23.5b. va = 22.0 m/s
==—B>~ v1. 3 0
an2 of)” Oql EXECUTE: Kg =+0.3630 J (from part (a))
U“ : +0.2454 J (from part (a))
rd = 0.800 m
rc = ? Figure 23.5b
K. = 0 (at distance of closest approach the speed is zero) U5: 1 51192
47550 n Thus conservation of energy Ka +Ua 2 U6 gives ZLq‘cb = +0.3630 J +0.2454 J = 0.6084 J
7560 re (—2.80><10J6 C)(—7.80><10~6 C) rc =L—ﬂ— = (8.988x109 N  mZ/Cz) = 0.323 m.
47:60 0.6084 J +0.6084 J
EVALUATE: U —> of: as r ——> 0 so qz will stop no matter what its initial speed is.
23.7. (a) IDENTIFY and SET UP: U is given by Eq.(23.9). EXECUTE: U = 1 W
’ 467% r '6 —6 U =(8_988x109 N.m2/C2)W=+o.198 J 0.250 m EVALUATE: The two charges are both of the same sign so their electric potential energy is positive.
(b) IDENTIFY: Use conservation of energy: Kg +Ua +er = K b + U b SET UP: Let point a be where q is released and point b be at its final position, as shown in Figure 23.7. EXECUTE: K“ = 0 (released from rest) Ua = +0.198 J (from part (a))
2 Kb 3 ime
Figure 23.7
Only the electric force does work, so Wother = 0 and U = 471's0 r (i) rb = 0.500 m
(+4.60x10‘6 C)(+1.20x10”5 C) _ Uh : L512 = (8938 x109 N . macaw—~— _ +0.0992 J
47:60 r 0.500 in Then KG +Ua +Wmher = Kb +U,, gives Kb = Ua —U,, and §mv§ = Ua —Ub and v ___ MUG—Uh): 2(+0.198J—0.0992J)=26‘6 m.
b m 2.80x10" kg (ii) rb = 5.00 m rb is now ten times larger than in (i) so U b is ten times smaller: U b = +0.0992 J / 10 = +0.00992 J. vb: (2(Ua—Ub)= 2(+0.198J—(100992J)=36.7 m.
m 2.80x10 kg (iii) rb : 50.0 m
r}, is how ten times larger than in (ii) so Ub is ten times smaller: Uh ::+0.00992 J/l0 : +0.000992 .l. m 2U~U 2+0.18J~. 21 ;‘
Vi—J ( “ b): ( 9 030099 )2375 m/s. _
2.80x10 kg ‘ EVALUATE: The force between the two charges is repulsive and provides an acceleration to q. This causes the
speed of q to increase as it moves away from Q. 23.18. IDENTIFY: Apply K“ +Ua = Kb +Ub.
SET UP: Let (11: +300 nC and q2 = +2.00 nC. At point a, r1“ = rm 2 0.250 m . At point 17, r“, = 0.100 m and er = 0.400 In. The electron has q = —e and me = 9.11><10'31 kg . K“ = 0 since the electron is released from rest. EXECUTE: —keq‘ — keqz = — keq‘ — kqu + lmevj.
0. r2. rib r21, 2
—9 9
E = K +Ua = k(—1.60><10“19 C) W+w =—2.88x10’” J.
a “ 0.250 m 0.250 m
i —9 I —9 1
Eb = Kb +Ub = k(—1.60><10“° C) W + W + lmay; : —5.04><10’” J + —mevZ
0.100 m 0.400 m 2 2 Setting Ea = Eb gives vb = WGD4XIO'” J — 2.88X10‘17 J): 6.89><106 m/s.
\’ . x g EVALUATE: Va = Vla +V2a = 180 V. Vb = V“, + V2,] = 315 V. Vb > Va . The negatively charged electron gains kinetic energy when it moves to higher potential. 23.21. IDENTIFY: V2; $
47560 i r, SET UP: The locations of the changes and points A and B are sketched in Figure 23.21. ql = +2.40 nC A £12 = ~6.50 nC
Figure 23.21 EXECUTE: (a) VA = +11) +2.4Ox10'9 C + —6.50><10‘9 c VA =(8.988><109 Nm2/C2)
0.050 m 0.050 m
=—737 V (b) v. =_1_[2L+a]
4x60 r B} 732 = —704 v “9 — _9
VB =(8.988><109 Nmz/C2)[W+M£j 0.080 m 0.060 In (c) IDENTIFY and SET UP: Use Eq.(23.13) and the results of parts (a) and (b) to calculate W.
EXECUTE: WBM = q'(VB —VA) 2 (2.50X10‘9 C)(—704 V — (—737 V)) = +8.2x10"8 J EVALUATE: The electric force does positive work on the positive charge when it moves from higher potential
(point B) to lower potential (point A). 23.27. IDENTIFY: Kg + an 2 Kb +qu .
SET UP: Let point a be at the cathode and let point b be at the anode. Kg = 0 . Vb *Va = 295 V . An electron has
q=—e and m=9.11><10‘31 kg.
1 EXECUTE: Kb 2 q(Va ~Vb) = —(1.60><10‘19 C)(~ 295 V) = 4.72x10“” J . Kb =Emv; , so 17 vb: Mil—9:1.01x107m/s. _ ;
9.11x10" kg 1 ‘EVALUATE: The negatively charged electron gains kinetic energy when it moves to higher potential. 23.31. IDENTIFY and SET UP: Apply conservation of energy, Eq.(23.3). Use Eq.(23. 12) to express U in terms of V.
(a) EXECUTE: K1+ qu = K2 + qu q(v1 —V2)=K2 ~Kl; q=—1.602><10“9 C
Kl =gmevf = 4.099><10*18 J; K2 = %mev22 = 2.915><10'17 J
Vl—V2 = K2 "K1 =—156 v q EVALUATE: The electron gains kinetic energy when it moves to higher potential.
(b) EXECUTE: Now Kl = 2.915><10’17 J, K2 = 0 K2 — Kl q
EVALUATE: The electron loses kinetic energy when it moves to lower potential. Vl—sz =+182V we ...
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This homework help was uploaded on 04/09/2008 for the course PH 1120 taught by Professor Keil during the Spring '08 term at WPI.
 Spring '08
 Keil

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