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Unformatted text preview: nxysms H3 Helmets/or k 1‘ 7 CD 24.10. IDENTIFY: Capacitance depends on the geometry of the object. 27t€oL
1n(r./r.) 21’ng C (a) SET UP: The capacitance of a cylindrical capacitor is C = . Solving for n, gives r,, = r,e EXECUTE: Substituting in the numbers for the exponent gives
27:(8.85x10"2 c’mm‘)(o.120 m) 3.67x10'" F Now use this value to calculate r,,: n, = r, 20"” = (0.250 cm)e°‘m = 0.300 cm
(b) SET UP: For any capacitor, C = Q/V and 11 = Q/L. Combining these equations and substituting the numbers
gives 2. = Q/L = CV/L
EXECUTE: Numerically we get
CV (3.67x10"' F)(125V) 11 =—— =————— = 3.8ZXI04 C/m = 38.2 nC/rn
L 0.120m EVALUATE: The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just
0.50 mm. These cylinders would have to be carefully constructed. =0.182 IDENTIFY: Simplify the network by replacing series and parallel combinations of capacitors by their equivalents. SET UP: For capacitors in series the voltages add and the charges are the same; J =—1+—+ For capacitors eq q C2
Q in parallel the voltages are the same and the charges add; Cml = Cl + C2 +~ = V. EXECUTE: (a) The equivalent capacitance of the 5.0 [1F and 8.0 ,uF capacitors in parallel is 13.0 #F. When these two capacitors are replaced by their equivalent we get.the network sketched in Figure 24.22. The equivalent
capacitance of these three capacitors in series is 3.47 ,uF. (b) Q,“ = Cﬂv = (3.47 yF)(50.0 V) = 174 ,uC (c) Qmt is the same as Q for each of the capacitors in the series combination shown in Figure 24.22, so Q for each of
the capacitors is 174 pC. EVALUATE: 'Ihe voltages across each capacitor in Figure 24.22 are V", = %£‘ = 17.4 V , VI3 = %—”‘ = 13.4 V and
lo I! V, =gl=193 V. Vl0 +Vu +V9 =17.4 V+l3.4 V+19.3 V =50.l V . The sum of the voltages equals the applied
9 voltage, apart from a small difference due to rounding.
10.0 in: 9.0 pr “—ll—ll——ll——'b 13.0 uF
Figure 24.22 IDENTIFY: After the two capacitors are connected they must have equal potential difference, and their combined charge must add up to the original charge.
2
SET UP: C = Q/V . The stored energy is U = ‘2QC =;CV1 EXECUTE: (a) Q = CV0. =_Qi=_Q_1 = '= = =2 ghi =g =2
(b)V C, C, andalsle+Q2 Q CV,,.Cl 'CandC2 280C (C/2) ansz 2.Q 2Q1.
2 _Q=ZQ=Z
Q"3Qa“dv C 3c 3"0
2 2 2 2 2
(c, v1[9_1.a.]=1[§9>[email protected]]=152_=1wg
2 c1 c2 2 c c 3c 3 (a) The original U was U =12CV02 , so AU = —%cvoz. (e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation.
EVALUATE: The original charge of the charged capacitor must distribute between the two capacitors to make the
potential the same across each capacitor. The voltage V for each after they are connected is less than the original voltage Vo of the charged capacitor. IDENTIFY: Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance.
SET UP: For a parallelplate capacitor, C = K q,A/d . EXECUTE: (8) Solving for d gives
_ KeoA _ (3.0)(8.85x10"I2 C2/N ~ In2 )(0.22 m)(0.28 m) d———— =1.64x10‘3 m: 1.64 mm. C l.0><10"J F
Dividing this result by the thickness of a sheet of paper gives Amt; z 8 sheets .
0.20 mm/sheet
.9
(b) Solving for the area of the plates gives A = Ed— : W = 0.45 m2 . Kso (3.0)(8.85x10‘" C’IN  m2)
(c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same capacitance.
EVALUATE: The use of dielectric makes it possible to construct reasonable—sized capacitors since the dielectric increases the capacitance by a factor of K. IDENTIFY: C = KC = K —A— . V = Ed for a parallel plate capacitor; this equation applies whether or not a dielectric
0 60 d is present.
SET UP: A =1.0 ch =1.0x10" Inz .
(8.85x10"2 F/m)(1.0x10" m2) =l.18 F rcmz.
7.5x10'9m ” pe EXECUTE: (a) C =(10) (b) E=K=M—=l.l3x107 V/m. K 7.5x 10" m
EVALUATE: The dielectric material increases the capacitance. If the dielectric were not present, the same charge
density on the faces of the membrane would produce a larger potential difference across the membrane. 7'2 @ 24.53. IDENTIFY: P = E/t , where E is the total light energy output. The energy stored in the capacitor is U = %CV2. SET UP: E = 0.95U
EXECUTE: (a) The power output is 600 W, and 95% of the original energy is converted, so E=Pt=(2.70x105 W)(1.48x10 S): 4001. Eo_=49ﬂ=4211. 0.95 2U 2(4211)
U='CV’ c=—=
0’) 1 so V2 (125V)2 EVALUATE: For a given V, the stored energy increases linearly with C. = 0.054 F . IDENTIFY: Simplify the network by replacing series and parallel combinations by their equivalent. The stored
energy in a capacitor is U = gCVz. SET UP: For capacitors in series the voltages add and the charges are the same; i =l—+CL+~. For capacitors
at} l 2
in parallel the voltages are the same and the charges add; C“I = Cl +C2 + =%. U =—;—CV2 . EXECUTE: (a) Find Ceq for the network by replacing each series or parallel combination by its equivalent. The
successive simpliﬁed circuits are shown in Figure 24.57a—c. Um =%c‘,qv2 =§(2.19><10*5 F)(12.0 V)2 =1.ssx10" J =158 p]
(b) From Figure 24. 57c, Q“ = qu = (2.19x10'° F)(12.0 V)= 2.63x10" C. From Figure 24.57b, Q“ = 2.63x10" C. 5C
v.,=&=3m=5.48v. U“ ='Cv’=—(4.80x10'°F)(5.48V)’=7.=21x10"J 72.1w This one capacitor stores nearly half the total stored energy.
' 2 EVALUATE: U = E . For capacitors in series the capacitor with the smallest C stores the greatest amount of energy.
4. 06 pf
a.—.l I—o——:: 3—41; 8.60 pF 7.56 “F
8 60 8 b 2.19 uF
14F 4. 350M: ao——Il—4I!olﬂF—li—o a._ll—_.b
(a) (b) (C) Figure 24.57 ® 24.60. IDENTIFY: Apply the rules for combining capacitors in series and in parallel.
SET UP: With the switch open each pair of 3.00 ,uF and 6.00 ,uF capacitors are in series with each other and each pair is in parallel with the other pair. When the switch is closed each pair of 3.00 pF and 6.00 ,uF capacitors are in
parallel with each other and the two pairs are in series. 1 t
EXECUTE: (a) With the switch open C... = [(371135 + 31;?) + ($ + $15] ]= 4.00 ,uF. QM = CqV =(4.00;1F) (210 V): 8.40x10‘4 C . By symmetry, each capacitor carries 4.20x10" C. The
voltages are then calculated via V = Q/C. This gives Vd = Q/C3 = 140 V and V” = Q/C6 = 70 V .
Vd=Vd—Vu=70V. ' (b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is l
1 l
C =————+————————— :45 F. =CV= 4.50 210 =9.5X10"C, deach
°‘ [(3.00+6.00);1F (3.00+6.00)/1F] " 9w .. ( #F)( V) an capacitor has the same potential difference of 105 V (again, by symmetry).
(c) The only way for the sum of the positive charge on one plate of C2 and the negative charge on one plate of CI to change is for charge to ﬂow through the switch. That is, the quantity of charge that flows through the
switch is equal to the change in Q2 — Q1 . With the switch open, Ql = Q2 and Q,  Ql = 0. After the switch is closed, Q2 — Ql = 315 pC , so 315 ,uC of charge flowed through the switch. EVALUATE: When the switch is closed the charge must redistribute to make points c and d be at the same
potential. IDENTIFY: C=5;1.C=—3.V=Ed.U=%QV. SETUP: d=3.0><103 m. A=7tr2,with r=1.0x103 m. I2 1 _ 2 3 2
EXECUTE: (a) C=E=W=93xm¢ F,
d 3.0xro’m
Q 20c ,
V=—=————=2.2x10 v
(b) c 9.3x109F
9 (c) E=K=3£x—193—1=7.3x105 V/m d 3.0x10 m (d) U = §QV = goo C)(2.2x10’ V) = 2.2x10'° J
EVALUATE: Thunderclouds involve very large potential differences and large amounts of stored energy. IDENTIFY: This situation is analogous to having two capacitors CI in series, each with separation ~;—(d — a). SET UP: For capacitors in series, 1—=l+—1.
q Cl C2
cm .1. .1. " 6o? . 60A
a = =J— .2L =
EXE . (a)C (C,+C,) 2Cl ’(d—a)/2 d—a
(b) C=ﬂi=§ﬁi=c d . (c)As a—)0, C——)C0.Themetalslabhasnoeffectifitisverythin.Andas a—)d, C—No. V=Q/C. V=Ey is the potential difference between two points separated by a distance y parallel to a uniform electric ﬁeld. When the
distance is very small, it takes a very large ﬁeld and hence a large Q on the plates for a given potential difference. Since Q = CV this corresponds to a very large C. 7—H ® 24.70. IDENTIFY: The electric ﬁeld energy density is u = %§E2 . U = — 2 2c‘
2. 21teor SETUP: Forthis charge distribution, E=0for r<r;‘, E: for t;<r<r;, and E=0 for r>7;. Example 24.4 shows that Z = 27““
L ln(r,, / 7;.) for a cylindrical capacitor. ,1 2 1’
EXECUTE: a = i E2 = l =
( ) u 2 6° 2 6° [ 27!er 871361;2 L12 "’ dr U 2’ U= dV=2L d= ——d——= lnl.
(b) I“ n ur r 4m:0 ,a r an L 47% (I; r.)
Q2 Q2 22L
(c) Using Equation (24.9). U = 52‘— = 4”60L1n(r,, Ira) = 47% ln(1;,/I;,). This agrees with the result of part (b). 2
EVALUATE: We could have used the results of part (b) and U = E to calculate U IL and would obtain the same result as in Example 24.4. IDENTIFY: The system can be considered to be two capacitors in parallel, one with plate area L(L— x) and air
between the plates and one with area Lx and dielectric ﬁlling the space between the plates. SET UP: C = KjA for a parallelplate capacitor with plate area A.
EXECUTE: (a) C = %((L — x)L + xKL) = ‘%‘(L + (K — l)x) (b) dU =;(dC)v1 , where c=co+‘%‘(~dx+dxx) , with Co=%(L+(K—1)x).Thisgives dU =%[‘o_’~d_x(x_l)]v2 = W—W’idx.
D 2D . _ QLV _ l 1 _ l 1 C (c) If the charge 15 kept constant on the plates, then Q  T(L+ (K — l)x) and U ;CV — ECoV F .
0
2 _ 2
U “EL 1 E°L(K—1)dx and AU =u_uo .1de
2 DCo 2D
. (Kl)q)V2L . . . . . . . (d) Since dU = —Fdx = ——2—D~———dx , the force IS in the opposne direction to the motion dx, meaning that the slab feels a force pushing it out.
EVALUATE: (c) When the plates are connected to the battery, the plates plus slab are not an isolated system. In addition to the work done on the slab by the charges on the plates, energy is also transferred between the battery and
(K — DeoV’L the plates. Comparing the results for dU in part (c) to dU = —Fdx gives F = 2D 75 IDENTIFY: C = Q/ V . Apply Gauss‘s law and the relation between potential difference and electric ﬁeld.
SET UP: Each conductor is an equipotential surface. V, —V,, = :0 EU di" = 3’ EL 'di" , so EU = EL , where these
0 u are the ﬁelds between the upper and lower hemispheres. The electric ﬁeld is the same in the air space as in the
dielectric. EXECUTE: (a) For a normal Spherical capacitor with air between the plates, Co = 47:60 [A] . The capacitor in ’5”. this problem is equivalent to two parallel capacitors, CL and Cu , each with half the plate area of the normal capacitor. cL = “0 = 2nKe,[—i'2] and cU = 9°— : 2neo[—’e"—]. c = Cu + CL = 2mm K)[ '0'” J.
r.  r. 2 r.  r. r.  r. . . . . . 4m2 Q Q (h) Usrng a hemispherical Gaussran surface for each respective half, EL = —— , so EL = —L—z , and
2 K60 27rK£or
2 EU 172" = Q— , so EU = Q" 2 . But QL =VCL and QU = VCU . Also, Q + QU = Q . Therefore, Q = VC°K = KQu 2 en Znsor
and Q, =—Q— , QL =—K—Q— . This gives EL =—Q———1——2=—2— Q 2 and 1+K 1+K l+K27tK§r 1+K47tq,r EU=—Q— 1 2 =—2— Q 2 .Wedofmdthat EU=EL. 1+K ZnKeor 1+ K 47thor . . Q Q
c The free char e densrt on u er and lower herms heres are: a = ———" = —————— and
( ) g y pp p ( Ta)” 2”,}: 2nr.2(l+K)
Q, Q _ Q; KQ QL KQ and (01,")L = = ————— ; 0' —— = ——— ~— = ——"—“'.
2;": 27m} (1 + K) ( "" L 2m,2 27rraz(1+ K) 27",} 2m,’(1+ K) _ _ _(K1)Q _K_=1<__1_Q_
(d) din651.0 l/K)—( K )Zynf(K+1) [K‘FIJZIU‘Z _ _ =(K—1)_g_K =£:_IQ
Gin—atria UK) ( K )2m:(K+l) [KHJZmb’ (c) There is zero bound charge on the ﬂat surface of the dielectricair interface, or else that would imply a
circumferential electric ﬁeld, or that the electric field changed as we went around the sphere. EVALUATE: The charge is not equally distributed over the surface of each conductor. There must be more charge on
the lower half, by a factor of K, because the polarization of the dielectric means more free charge is needed on the
lower half to produce the same electric ﬁeld. 7'6 IDENTIFY: The object is equivalent to two identical capacitors in parallel, where each has the same area A, plate
separation d and dielectric with dielectric constant K. . . A
SET UP: For each capacitor in the parallel combination, = 52—. EXECUTE: (a) The charge distribution on the plates is shown in Figure 24.77.
60A 2(4.2)eo(0.120 m)2 _ _.,
=2 —— =———’——2.38x10 F.
(b) C ( d J 4.5x10" m
. 60A 2.38x10'° F
EVALUATE: If two of the plates are separated by both sheets of paper to form a capacrtor, C = ~27 = ——T— , smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77 IDENTIFY: As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has
plate separation d, plate area w(L — h) and air between the plates. The other has the same plate separation d, plate area wh and dielectric constant K.
K “160A SET UP: Deﬁne Keff by C“I = , where A = wL . For two capacitors in parallel, C“I = Cl + C1. EXECUTE: (a) The capacitors are in parallel, so C = Khh
K=l+——.
“'( L L] eow(L—h) + Keowh _ cowL( Kh h
L 1+ — — —) . This gives
(1 d d L (b) For gasoline, with K = 1.95 : % full: Keff (h = %) = 1.24; % full: Kc,f (h = g] = 1.48; 3m11:K,,,[h=—3£)=1.71.
4 4 (c) For methanol, with K = 33: :1; full: Keff [h = %) = 9; a full: Keff [h =%)=17 ; %full: Kg“ (h = 3:1) = 25. (d) This kind of fuel tank sensor will work best for methanol since it has the greater range of Ken values.
EVALUATE: When h=0, K“, =1.When h=L, Keff =K . 7»7 ...
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 Spring '11
 Corbin
 Physics

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