ch25 kho tài liệu bách khoa

1 Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery Thus, q = (25 × 10–6 F)(120 V) = 3.0 × 10–3 C (a) The capacitance of the system is C= q 70 pC = = 35 pF ∆V 20 V (b) The capacitance is independent of q; it is still 3.5 pF (c) The potential difference becomes ∆V = q 200 pC = = 57 V C 35 pF (a) The capacitance of a parallel-plate capacitor is given by C = ε0A/d, where A is the area of each plate and d is the plate separation Since the plates are circular, the plate area is A = πR2, where R is the radius of a plate Thus, C= ε 0π R d (8.85 ×10 = −12 F m ) π ( 8.2 ×10−2 m ) −3 1.3×10 m = 1.44 ×10−10 F = 144pF (b) The charge on the positive plate is given by q = CV, where V is the potential difference across the plates Thus, q = (1.44 × 10–10 F)(120 V) = 1.73 × 10–8 C = 17.3 nC We use C = Aε0/d (a) Thus, c hd 100 m2 8.85 × 10−12 Aε d= = 100 F C C2 N ⋅m2 i = 8.85 × 10 −12 m (b) Since d is much less than the size of an atom (∼ 10–10 m), this capacitor cannot be constructed Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4πε0R to find the capacitance When the drops combine, the volume is doubled It is then V = 2(4π/3)R3 The new radius R' is given by 4π 4π ( R′ ) = R 3 Ÿ R ′ = 21 R The new capacitance is C ′ = 4πε R′ = 4πε 21 R = 5.04πε R With R = 2.00 mm, we obtain C = 5.04π ( 8.85 ×10−12 F m )( 2.00 ×10−3 m ) = 2.80 ×10−13 F (a) We use Eq 25-17: C = πε b gb g 40.0 mm 38.0 mm ab = = 84.5 pF b−a 8.99 × 109 NC⋅m2 40.0 mm − 38.0 mm d ib g (b) Let the area required be A Then C = ε0A/(b – a), or A= C (b − a ) ε0 = ( 84.5 pF )( 40.0 mm − 38.0 mm ) = 191cm (8.85 ×10 −12 C2 N ⋅m ) The equivalent capacitance is given by Ceq = q/V, where q is the total charge on all the capacitors and V is the potential difference across any one of them For N identical capacitors in parallel, Ceq = NC, where C is the capacitance of one of them Thus, NC = q / V and N= q 1.00C = = 9.09 ×103 VC (110V ) (1.00 ×10−6 F ) The equivalent capacitance is Ceq = C3 + b gb g 10.0 µF 5.00 µF C1C2 = 4.00µF + = 7.33 µF C1 + C2 10.0 µF + 5.00 µF The equivalent capacitance is Ceq = ( C1 + C2 ) C3 = (10.0 µ F + 5.00 µ F )( 4.00 µ F ) = 3.16 µ F C1 + C2 + C3 10.0 µ F + 5.00 µ F + 4.00 µ F 10 The charge that passes through meter A is b gb g q = CeqV = 3CV = 25.0 µF 4200 V = 0.315 C 70 The voltage across capacitor is V1 = q1 30 µC = = 3.0 V C1 10 µF Since V1 = V2, the total charge on capacitor is b gb g q2 = C2V2 = 20 µF V = 60 µC , which means a total of 90 µC of charge is on the pair of capacitors C1 and C2 This implies there is a total of 90 µC of charge also on the C3 and C4 pair Since C3 = C4, the charge divides equally between them, so q3 = q4 = 45 µC Thus, the voltage across capacitor is V3 = q3 45 µC = = 2.3 V C3 20 µF Therefore, |VA – VB| = V1 + V3 = 5.3 V 71 (a) The equivalent capacitance is Ceq = b gb g 6.00 µF 4.00µF C1C2 = = 2.40 µF C1 + C2 6.00 µF + 4.00 µF (b) q1 = CeqV = (2.40 µF)(200 V) = 4.80 × 10−4 C (c) V1 = q1/C1 = 4.80 ì 104 C/6.00 àF = 80.0 V (d) q2 = q1 = 4.80 × 10−4 C (e) V2 = V – V1 = 200 V – 80.0 V = 120 V 72 (a) Now Ceq = C1 + C2 = 6.00 µF + 4.00 µF = 10.0 µF (b) q1 = C1V = (6.00 µF)(200 V) = 1.20 × 10–3 C (c) V1=200 V (d) q2 = C2V = (4.00 àF)(200 V) = 8.00 ì 104 C (e) V2 = V1 = 200 V 73 We cannot expect simple energy conservation to hold since energy is presumably dissipated either as heat in the hookup wires or as radio waves while the charge oscillates in the course of the system “settling down” to its final state (of having 40 V across the parallel pair of capacitors C and 60 µF) We expect charge to be conserved Thus, if Q is the charge originally stored on C and q1, q2 are the charges on the parallel pair after “settling down,” then Q = q1 + q2 b g b g b gb g C 100 V = C 40 V + 60 µF 40 V which leads to the solution C = 40 µF 74 We first need to find an expression for the energy stored in a cylinder of radius R and length L, whose surface lies between the inner and outer cylinders of the capacitor (a < R < b) The energy density at any point is given by u = 21 ε E , where E is the magnitude of the electric field at that point If q is the charge on the surface of the inner cylinder, then the magnitude of the electric field at a point a distance r from the cylinder axis is given by E= q πε Lr (see Eq 25-12), and the energy density at that point is given by u= q2 ε 0E = 2 8π ε L r The energy in the cylinder is the volume integral U R = ∫ udv Now, dv = 2πrLdr , so UR = z R a q2 q2 rLdr = π 8π ε L2 r πε L z R a dr q2 R = ln r πε L a To find an expression for the total energy stored in the capacitor, we replace R with b: q2 b Ub = ln 4πε L a We want the ratio UR/Ub to be 1/2, so ln or, since R = ab b g d R b = ln a a i b g d i ln b / a = ln b / a , ln R / a = ln b / a This means R / a = b / a or 75 (a) Since the field is constant and the capacitors are in parallel (each with 600 V across them) with identical distances (d = 0.00300 m) between the plates, then the field in A is equal to the field in B: G V E = = 2.00 × 105 V m d G (b) | E |= 2.00 × 105 V m See the note in part (a) (c) For the air-filled capacitor, Eq 25-4 leads to σ= G q = ε E = 177 × 10−6 C m2 A (d) For the dielectric-filled capacitor, we use Eq 25-29: G σ = κε E = 4.60 × 10−6 C m2 (e) Although the discussion in the textbook (§25-8) is in terms of the charge being held fixed (while a dielectric is inserted), it is readily adapted to this situation (where comparison is made of two capacitors which have the same voltage and are identical except for the fact that one has a dielectric) The fact that capacitor B has a relatively large charge but only produces the field that A produces (with its smaller charge) is in line with the point being made (in the text) with Eq 25-34 and in the material that follows Adapting Eq 25-35 to this problem, we see that the difference in charge densities between parts (c) and (d) is due, in part, to the (negative) layer of charge at the top surface of the dielectric; consequently, c h c h σ ′ = 177 × 10−6 − 4.60 × 10−6 = −2.83 × 10−6 C m2 76 (a) The equivalent capacitance is Ceq = C1C2/(C1 + C2) Thus the charge q on each capacitor is q = q1 = q2 = CeqV = C1C2V (2.00 F)(8.00à F)(300V) = = 4.80 ì104 C C1 + C2 2.00 µ F + 8.00 µ F (b) The potential difference is V1 = q/C1 = 4.80 × 10–4 C/2.0 µF = 240 V (c) As noted in part (a), q2 = q1 = 4.80 ×10−4 C (d) V2 = V – V1 = 300 V – 240 V = 60.0 V Now we have q'1/C1 = q'2/C2 = V' (V' being the new potential difference across each capacitor) and q'1 + q'2 = 2q We solve for q'1, q'2 and V: (e) q '1 = (f) V1 = 2C1q 2(2.00à F)(4.80 ì 104C ) = = 1.92 ì 104 C C1 + C2 2.00à F + 8.00 F q1 1.92 ì104 C = = 96.0 V C1 2.00 µ F (g) q '2 = 2q − q1 = 7.68 × 10−4 C (h) V2′ = V1′ = 96.0 V (i) In this circumstance, the capacitors will simply discharge themselves, leaving q1 =0, (j) V1=0, (k) q2 = 0, (l) and V2 = V1 = 77 We use U = 21 CV As V is increased by ∆V, the energy stored in the capacitor increases correspondingly from U to U + ∆U: U + ∆U = 21 C (V + ∆V ) Thus, (1 + ∆V/V)2 = + ∆U/U, or ∆V ∆U = 1+ − = + 10% − = 4.9% V U 78 (a) The voltage across C1 is 12 V, so the charge is q1 = C1V1 = 24 µC (b) We reduce the circuit, starting with C4 and C3 (in parallel) which are equivalent to µ F This is then in series with C2, resulting in an equivalence equal to 43 µ F which would have 12 V across it The charge on this 43 µ F capacitor (and therefore on C2) is ( 43 µ F)(12 V) = 16 µ C Consequently, the voltage across C2 is V2 = q2 16 µ C = = V C2 µF This leaves 12 – = V across C4 (similarly for C3) 79 We reduce the circuit, starting with C1 and C2 (in series) which are equivalent to µF This is then parallel to C3 and results in a total of µF, which is now in series with C4 and can be further reduced However, the final step in the reduction is not necessary, as we observe that the µF equivalence from the top capacitors has the same capacitance as C4 and therefore the same voltage; since they are in series, that voltage is then 12/2 = 6.0 V 80 We use C = ε0κA/d ∝ κ/d To maximize C we need to choose the material with the greatest value of κ/d It follows that the mica sheet should be chosen 81 We may think of this as two capacitors in series C1 and C2, the former with the κ1 = 3.00 material and the latter with the κ2 = 4.00 material Upon using Eq 25-9, Eq 25-27 and then reducing C1 and C2 to an equivalent capacitance (connected directly to the battery) with Eq 25-20, we obtain § κ1 ã A Ceq = ă = 1.52 ì 1010 F â + d Therefore, q = CeqV = 1.06 × 10−9 C 82 (a) The length d is effectively shortened by b so C' = ε0A/(d – b) = 0.708 pF (b) The energy before, divided by the energy after inserting the slab is U q / 2C C ′ ε A /(d − b) d 5.00 = = = = = = 1.67 U ′ q / 2C ′ C ε0 A / d d − b 5.00 − 2.00 (c) The work done is W = ∆U = U ′ − U = q2 § 1 · q2 q 2b − = − − = − = 5.44 J ( d b d ) ă â C C A A (d) Since W < the slab is sucked in 83 (a) C' = ε0A/(d – b) = 0.708 pF, the same as part (a) in problem 82 (b) Now, ε0 A / d U C d − b 5.00 − 2.00 CV =1 = = = = = 0.600 U ′ C ′V C ′ ε A /(d − b) d 5.00 (c) The work done is W = ∆U = U '− U = ε A§ ε AbV 1· − ¸V = = 1.02 ×10−9 J (C ′ − C )V = ă 2 â d b d 2d ( d − b) (d) In Problem 82 where the capacitor is disconnected from the battery and the slab is sucked in, F is certainly given by −dU/dx However, that relation does not hold when the battery is left attached because the force on the slab is not conservative The charge distribution in the slab causes the slab to be sucked into the gap by the charge distribution on the plates This action causes an increase in the potential energy stored by the battery in the capacitor 84 We not employ energy conservation since, in reaching equilibrium, some energy is dissipated either as heat or radio waves Charge is conserved; therefore, if Q = 48 µC, and q1 and q3 are the charges on C1 and C3 after the switch is thrown to the right (and equilibrium is reached), then Q = q1 + q3 We note that V1 and = V3 because of the parallel arrangement, and V1 = 21 V1 and since they are identical capacitors This leads to 2V1 = V3 q1 q3 = C1 C3 2q1 = q3 where the last step follows from multiplying both sides by 2.00 µF Therefore, Q = q1 + (2q1 ) which yields q1 = 16.0 µC and q3 = 32.0 µC