24 CAPACITANCE AND DIELECTRICS 24.1 IDENTIFY: The capacitance depends on the geometry (area and plate separation) of the plates Q Q SET UP: For a parallel-plate capacitor, Vab = Ed , E = , and C = Vab ⑀0 A EXECUTE: (a) Vab = Ed = (4.00 × 106 V/m)(2.50 × 10−3 m) = 1.00 × 104 V (b) Solving for the area gives Q 80.0 × 10−9 C A= = = 2.26 × 10−3 m = 22.6 cm E⑀0 (4.00 × 106 V/m)[8.854 × 10−12 C2 /(N ⋅ m )] Q 80.0 × 10−9 C = = 8.00 × 10−12 F = 8.00 pF Vab 1.00 × 104 V EVALUATE: The capacitance is reasonable for laboratory capacitors, but the area is rather large ⑀ A Q IDENTIFY and SET UP: C = , C = and V = Ed d V (c) C = 24.2 (a) C = ⑀0 (b) V = A 0.00122 m = ⑀0 = 3.29 pF 0.00328 m d Q 4.35 × 10−8 C = = 13.2 kV C 3.29 × 10−12 F V 13.2 × 103 V = = 4.02 × 106 V/m 0.00328 m d EVALUATE: The electric field is uniform between the plates, at points that aren’t close to the edges IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Section 24.1 Q 0.148 × 10−6 C Q EXECUTE: (a) C = so Vab = = = 604 V Vab C 245 × 10−12 F (c) E = 24.3 (b) C = ⑀0 A d so A = (c) Vab = Ed so E = (d) E = Cd ⑀0 = (245 × 10−12 F)(0.328 × 10−3 m) 8.854 × 10−12 C2 /N ⋅ m = 9.08 × 10−3 m = 90.8 cm Vab 604 V = = 1.84 × 106 V/m −3 d 0.328 × 10 m σ so σ = E⑀ = (1.84 × 106 V/m)(8.854 × 10−12 C2 /N ⋅ m ) = 1.63 × 10−5 C/m ⑀0 EVALUATE: We could also calculate σ directly as Q/A σ = Q 0.148 × 10−6 C = = 1.63 × 10−5 C/m , A 9.08 × 10−3 m which checks © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-1 24-2 24.4 Chapter 24 IDENTIFY: C = ⑀ A = (3.0 × 10−2 m) is the area of each plate SET UP: (8.854 × 10−12 F/m)(3.0 × 10−2 m) = 1.59 × 10−12 F = 1.59 pF 5.0 × 10−3 m EVALUATE: C increases when A increases and C increases when d decreases Q ⑀ A IDENTIFY: C = C= d Vab EXECUTE: C = 24.5 A when there is air between the plates d SET UP: When the capacitor is connected to the battery, Vab = 12.0 V EXECUTE: (a) Q = CVab = (10.0 × 10−6 F)(12.0 V) = 1.20 × 10−4 C = 120 μ C (b) When d is doubled C is halved, so Q is halved Q = 60 μ C (c) If r is doubled, A increases by a factor of C increases by a factor of and Q increases by a factor of Q = 480 μ C 24.6 EVALUATE: When the plates are moved apart, less charge on the plates is required to produce the same potential difference With the separation of the plates constant, the electric field must remain constant to produce the same potential difference The electric field depends on the surface charge density, σ To produce the same σ , more charge is required when the area increases Q ⑀ A IDENTIFY: C = C= Vab d SET UP: When the capacitor is connected to the battery, enough charge flows onto the plates to make Vab = 12.0 V EXECUTE: (a) 12.0 V Q and Q is constant, so V doubles V = 24.0 V C (ii) When r is doubled, A increases by a factor of V decreases by a factor of and V = 3.0 V EVALUATE: The electric field between the plates is E = Q/⑀0 A Vab = Ed When d is doubled E is (b) (i) When d is doubled, C is halved Vab = 24.7 unchanged and V doubles When A is increased by a factor of 4, E decreases by a factor of so V decreases by a factor of ⑀ A IDENTIFY: C = Solve for d d SET UP: Estimate r = 1.0 cm A = π r EXECUTE: C = 24.8 ⑀0 A so d = ⑀0π r = ⑀0π (0.010 m)2 = 2.8 mm C d 1.00 × 10−12 F EVALUATE: The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it is a reasonable approximation to treat them as infinite sheets Q ⑀ A INCREASE: C = Vab = Ed C = d Vab SET UP: We want E = 1.00 × 104 N/C when V = 100 V EXECUTE: (a) d = A= r= Cd ⑀0 A π = Vab 1.00 × 102 V = = 1.00 × 10−2 m = 1.00 cm E 1.00 × 104 N/C (5.00 × 10−12 F)(1.00 × 10−2 m) 8.854 × 10−12 C /(N ⋅ m ) = 5.65 × 10−3 m A = π r so = 4.24 × 10−2 m = 4.24 cm (b) Q = CVab = (5.00 × 10−12 F)(1.00 × 102 V) = 5.00 × 10−10 C = 500 pC © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics EVALUATE: C = 24.9 ⑀0 A We could have a larger d, along with a larger A, and still achieve the required C d without exceeding the maximum allowed E IDENTIFY: The energy stored in a capacitor depends on its capacitance, which in turn depends on its geometry ⑀ A SET UP: C = Q/V for any capacitor, and C = for a parallel-plate capacitor d EXECUTE: (a) C = d= ⑀0 A C = Q 2.40 × 10−10 C ⑀ A = = 5.714 × 10−12 F Using C = gives V 42.0 V d (8.854 × 10−12 C /(N ⋅ m ))(6.80 × 10−4 m ) 5.714 × 10−12 F = 1.05 mm 5.714 × 10−12 F Q = 2.857 × 10−12 F V = , so V = 2(42.0 V) = 84.0 V d C EVALUATE: Doubling the plate separation halves the capacitance, so twice the potential difference is required to keep the same charge on the plates IDENTIFY: Capacitance depends on the geometry of the object 2π ⑀0 L (a) SET UP: The capacitance of a cylindrical capacitor is C = Solving for rb gives ln( rb /ra ) (b) d = 2.10 × 10−3 m C = 24.10 24-3 ⑀0 A = rb = e 2π ⑀0 L/C EXECUTE: Substituting in the numbers for the exponent gives 2π (8.85 × 10−12 C /N ⋅ m )(0.120 m) 3.67 × 10−11 F = 0.182 Now use this value to calculate rb : rb = e0.182 = (0.250 cm)e0.182 = 0.300 cm (b) SET UP: For any capacitor, C = Q/V and λ = Q/L Combining these equations and substituting the numbers gives λ = Q/L = CV/L EXECUTE: Numerically we get CV (3.67 × 10−11 F)(125 V) = = 3.82 × 10−8 C/m = 38.2 nC/m L 0.120 m 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 IDENTIFY: Apply the results of Example 24.4 C = Q/V λ= 24.11 SET UP: = 0.50 mm, rb = 5.00 mm EXECUTE: (a) C = L 2π ⑀0 (0.180 m)2π ⑀0 = = 4.35 × 10−12 F ln(rb /ra ) ln(5.00/0.50) (b) V = Q/C = (10.0 × 10−12 C)/(4.35 × 10−12 F) = 2.30 V C = 24.2 pF This value is similar to those in Example 24.4 The capacitance is determined L entirely by the dimensions of the cylinders IDENTIFY and SET UP: Use the expression for C/L derived in Example 24.4 Then use Eq (24.1) to calculate Q C 2π ⑀0 EXECUTE: (a) From Example 24.4, = L ln( rb /ra ) EVALUATE: 24.12 C 2π (8.854 × 10−12 C2 /N ⋅ m ) = = 6.57 × 10−11 F/m = 66 pF/m L ln(3.5 mm/1.5 mm) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-4 Chapter 24 (b) C = (6.57 × 10−11 F/m)(2.8 m) = 1.84 × 10−10 F Q = CV = (1.84 × 10−10 F)(350 × 10−3 V) = 6.4 × 10−11 C = 64 pC 24.13 The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and −64 pC on the outer conductor EVALUATE: C depends only on the dimensions of the capacitor Q and V are proportional IDENTIFY: We can use the definition of capacitance to find the capacitance of the capacitor, and then relate the capacitance to geometry to find the inner radius (a) SET UP: By the definition of capacitance, C = Q/V EXECUTE: C = Q 3.30 × 10−9 C = = 1.50 × 10−11 F = 15.0 pF V 2.20 × 102 V (b) SET UP: The capacitance of a spherical capacitor is C = 4π ⑀0 rb rb − EXECUTE: Solve for and evaluate using C = 15.0 pF and rb = 4.00 cm, giving = 3.09 cm (c) SET UP: We can treat the inner sphere as a point charge located at its center and use Coulomb’s law, q E= 4π ⑀0 r EXECUTE: E = 24.14 (9.00 × 109 N ⋅ m /C2 )(3.30 × 10−9 C) = 3.12 × 104 N/C (0.0309 m) EVALUATE: Outside the capacitor, the electric field is zero because the charges on the spheres are equal in magnitude but opposite in sign IDENTIFY: Apply the results of Example 24.3 C = Q/V SET UP: = 15.0 cm Solve for rb 1⎛ r r ⎞ EXECUTE: (a) For two concentric spherical shells, the capacitance is C = ⎜ a b ⎟ kCrb − kCra = rb k ⎝ rb − ⎠ and rb = kCra k (116 × 10−12 F)(0.150 m) = = 0.175 m kC − k (116 × 10−12 F) − 0.150 m (b) V = 220 V and Q = CV = (116 × 10−12 F)(220 V) = 2.55 × 10−8 C EVALUATE: A parallel-plate capacitor with A = 4π rb = 0.33 m and d = rb − = 2.5 × 10−2 m has C= 24.15 ⑀0 A d = 117 pF, in excellent agreement with the value of C for the spherical capacitor IDENTIFY: For capacitors in series the voltage across the combination equals the sum of the voltages in the individual capacitors For capacitors in parallel the voltage across the combination is the same as the voltage across each individual capacitor SET UP and EXECUTE: (a) Connect the capacitors in series so their voltages will add (b) V = V1 + V2 + V3 + " = NV1, where N is the number of capacitors in the series combination, since the V 500 V = = 5000 V1 0.10 V EVALUATE: It requires many small cells to produce a large voltage surge IDENTIFY: The capacitors between b and c are in parallel This combination is in series with the 15 pF capacitor SET UP: Let C1 = 15 pF, C2 = 9.0 pF and C3 = 11 pF capacitors are identical N = 24.16 EXECUTE: (a) For capacitors in parallel, Ceq = C1 + C2 + " so C23 = C2 + C3 = 20 pF (b) C1 = 15 pF is in series with C23 = 20 pF For capacitors in series, 1 = + + " so Ceq C1 C2 1 CC (15 pF)(20 pF) and C123 = 23 = = + = 8.6 pF C123 C1 C23 C1 + C23 15 pF + 20 pF © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24.17 24-5 EVALUATE: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents In each equivalent network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the original circuit SET UP: Do parts (a) and (b) together The capacitor network is drawn in Figure 24.17a C1 = C2 = C3 = C4 = 4.00 μ F Vab = 28.0 V Figure 24.17a EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C1 and C2 are in series and are equivalent to C12 (Figure 24.17b) 1 = + C12 C1 C2 Figure 24.17b C1C2 (4.00 × 10−6 F)(4.00 × 10−6 F) = = 2.00 × 10−6 F C1 + C2 4.00 × 10−6 F + 4.00 × 10−6 F and C3 are in parallel and are equivalent to C123 (Figure 24.17c) C12 = C12 C123 = C12 + C3 C123 = 2.00 × 10−6 F + 4.00 × 10−6 F C123 = 6.00 × 10−6 F Figure 24.17c C123 and C4 are in series and are equivalent to C1234 (Figure 24.17d) 1 = + C1234 C123 C4 Figure 24.17d C123C4 (6.00 × 10−6 F)(4.00 × 10−6 F) = = 2.40 × 10−6 F C123 + C4 6.00 × 10−6 F + 4.00 × 10−6 F The circuit is equivalent to the circuit shown in Figure 24.17e C1234 = V1234 = V = 28.0 V Q1234 = C1234V = (2.40 × 10−6 F)(28.0 V) = 67.2 μ C Figure 24.17e © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-6 Chapter 24 Now build back up the original circuit, step by step C1234 represents C123 and C4 in series (Figure 24.17f) Q123 = Q4 = Q1234 = 67.2 μ C (charge same for capacitors in series) Figure 24.17f Then V123 = V4 = Q123 67.2 μ C = = 11.2 V C123 6.00 μ F Q4 67.2 μ C = = 16.8 V C4 4.00 μ F Note that V4 + V123 = 16.8 V + 11.2 V = 28.0 V, as it should Next consider the circuit as written in Figure 24.17g V3 = V12 = 28.0 V − V4 V3 = 11.2 V Q3 = C3V3 = (4.00 μ F)(11.2 V) Q3 = 44.8 μ C Q12 = C12V12 = (2.00 μ F)(11.2 V) Q12 = 22.4 μ C Figure 24.17g Finally, consider the original circuit, as shown in Figure 24.17h Q1 = Q2 = Q12 = 22.4 μ C (charge same for capacitors in series) Q 22.4 μ C V1 = = = 5.6 V C1 4.00 μ F V2 = Q2 22.4 μ C = = 5.6 V C2 4.00 μ F Figure 24.17h Note that V1 + V2 = 11.2 V, which equals V3 as it should Summary: Q1 = 22.4 μ C, V1 = 5.6 V Q2 = 22.4 μ C, V2 = 5.6 V Q3 = 44.8 μ C, V3 = 11.2 V Q4 = 67.2 μ C, V4 = 16.8 V (c) Vad = V3 = 11.2 V EVALUATE: V1 + V2 + V4 = V , or V3 + V4 = V Q1 = Q2 , Q1 + Q3 = Q4 and Q4 = Q1234 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24.18 IDENTIFY: The two capacitors are in series The equivalent capacitance is given by 24-7 1 = + Ceq C1 C2 SET UP: For capacitors in series the charges are the same and the potentials add to give the potential across the network 1 1 EXECUTE: (a) = + = + = 5.33 × 105 F−1 Ceq = 1.88 × 10−6 F Then Ceq C1 C2 (3.0 × 10−6 F) (5.0 × 10−6 F) Q = VCeq = (52.0 V)(1.88 × 10−6 F) = 9.75 × 10−5 C Each capacitor has charge 9.75 × 10−5 C (b) V1 = Q/C1 = 9.75 × 10−5 C/3.0 × 10−6 F = 32.5 V V2 = Q/C2 = 9.75 × 10−5 C/5.0 × 10−6 F = 19.5 V EVALUATE: V1 + V2 = 52.0 V, which is equal to the applied potential Vab The capacitor with the smaller 24.19 C has the larger V IDENTIFY: The two capacitors are in parallel so the voltage is the same on each, and equal to the applied voltage Vab SET UP: Do parts (a) and (b) together The network is sketched in Figure 24.19 EXECUTE: V1 = V2 = V V1 = 52.0 V V2 = 52.0 V Figure 24.19 C = Q/V so Q = CV Q1 = C1V1 = (3.00 μ F)(52.0 V) = 156 μ C Q2 = C2V2 = (5.00 μ F)(52.0 V) = 260 μ C 24.20 EVALUATE: To produce the same potential difference, the capacitor with the larger C has the larger Q IDENTIFY: For capacitors in parallel the voltages are the same and the charges add For capacitors in series, the charges are the same and the voltages add C = Q/V SET UP: C1 and C2 are in parallel and C3 is in series with the parallel combination of C1 and C2 EXECUTE: (a) C1 and C2 are in parallel and so have the same potential across them: Q2 40.0 × 10−6 C = = 13.33 V Therefore, Q1 = V1C1 = (13.33 V)(6.00 × 10−6 F) = 80.0 × 10−6 C C2 3.00 × 10−6 F Since C3 is in series with the parallel combination of C1 and C2 , its charge must be equal to their V1 = V2 = combined charge: Q3 = 40.0 × 10−6 C + 80.0 × 10−6 C = 120.0 × 10−6 C (b) The total capacitance is found from Ctot = 3.21 μ F Vab = 1 1 and = + = + Ctot C12 C3 9.00 × 10−6 F 5.00 × 10−6 F Qtot 120.0 × 10−6 C = = 37.4 V Ctot 3.21 × 10−6 F Q3 120.0 × 10−6 C = = 24.0 V Vab = V1 + V3 C3 5.00 × 10−6 F IDENTIFY: Three of the capacitors are in series, and this combination is in parallel with the other two capacitors SET UP: For capacitors in series the voltages add and the charges are the same; 1 = + + " For capacitors in parallel the voltages are the same and the charges add; Ceq C1 C2 EVALUATE: V3 = 24.21 Ceq = C1 + C2 + L C = Q V © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-8 Chapter 24 EXECUTE: (a) The equivalent capacitance of the 18.0 nF, 30.0 nF and 10.0 nF capacitors in series is 5.29 nF When these capacitors are replaced by their equivalent we get the network sketched in Figure 24.21 The equivalent capacitance of these three capacitors in parallel is 19.3 nF, and this is the equivalent capacitance of the original network Figure 24.21 (b) Qtot = CeqV = (19.3 nF)(25 V) = 482 nC (c) The potential across each capacitor in the parallel network of Figure 24.21 is 25 V Q6.5 = C6.5V6.5 = (6.5 nF)(25 V) = 162 nC 24.22 d) 25 V EVALUATE: As with most circuits, we must go through a series of steps to simplify it as we solve for the unknowns IDENTIFY: Simplify the network by replacing series and parallel combinations of capacitors by their equivalents 1 = + + " For SET UP: For capacitors in series the voltages add and the charges are the same; Ceq C1 C2 Q V EXECUTE: (a) The equivalent capacitance of the 5.0 μ F and 8.0 μ F capacitors in parallel is 13.0 μ 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 μ F (b) Qtot = CtotV = (3.47 μ F)(50.0 V) = 174 μ C capacitors in parallel the voltages are the same and the charges add; Ceq = C1 + C2 +" C = (c) Qtot 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 μ C Q EVALUATE: The voltages across each capacitor in Figure 24.22 are V10 = tot = 17.4 V, C10 V13 = Qtot Q = 13.4 V and V9 = tot = 19.3 V V10 + V13 + V9 = 17.4 V + 13.4 V + 19.3 V = 50.1 V The sum C13 C9 of the voltages equals the applied voltage, apart from a small difference due to rounding Figure 24.22 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24.23 24-9 IDENTIFY: Refer to Figure 24.10b in the textbook For capacitors in parallel, Ceq = C1 + C2 +" For capacitors in series, 1 = + + " Ceq C1 C2 SET UP: The 11 μ F, μ F and replacement capacitor are in parallel and this combination is in series with the 9.0 μ F capacitor EXECUTE: 24.24 ⎛ 1 1 ⎞ = =⎜ + ⎟ (15 + x ) μ F = 72 μ F and x = 57 μ F Ceq 8.0 μ F ⎝ (11 + 4.0 + x) μ F 9.0 μ F ⎠ EVALUATE: Increasing the capacitance of the one capacitor by a large amount makes a small increase in the equivalent capacitance of the network ⑀ A IDENTIFY: Apply C = Q/V C = The work done to double the separation equals the change in the d stored energy Q2 SET UP: U = CV = 2C EXECUTE: (a) V = Q/C = (2.55 μ C)/(920 × 10−12 F) = 2770 V (b) C = ⑀0 A says that since the charge is kept constant while the separation doubles, that means that the d capacitance halves and the voltage doubles to 5540 V Q (2.55 × 10−6 C) (c) U = = = 3.53 × 10−3 J If the separation is doubled while Q stays the same, the 2C 2(920 × 10−12 F) capacitance halves, and the energy stored doubles So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is 3.53 × 10−3 J 24.25 EVALUATE: The oppositely charged plates attract each other and positive work must be done by an external force to pull them farther apart IDENTIFY and SET UP: The energy density is given by Eq (24.11): u = 12 ⑀0 E Use V = Ed to solve for E EXECUTE: Calculate E: E = V 400 V = = 8.00 × 104 V/m d 5.00 × 10−3 m Then u = 12 ⑀0 E = 12 (8.854 × 10−12 C2 /N ⋅ m )(8.00 × 104 V/m)2 = 0.0283 J/m3 24.26 EVALUATE: E is smaller than the value in Example 24.8 by about a factor of so u is smaller by about a factor of 62 = 36 Q ⑀ A IDENTIFY: C = C = Vab = Ed The stored energy is 12 QV d Vab SET UP: d = 1.50 × 10−3 m μ C = 10−6 C EXECUTE: (a) C = (b) C = ⑀0 A d so A = 0.0180 × 10−6 C = 9.00 × 10−11 F = 90.0 pF 200 V Cd ⑀0 = (9.00 × 10−11 F)(1.50 × 10−3 m) 8.854 × 10−12 C2 /(N ⋅ m ) = 0.0152 m (c) V = Ed = (3.0 × 106 V/m)(1.50 × 10−3 m) = 4.5 × 103 V (d) Energy = 12 QV = 12 (0.0180 × 10−6 C)(200 V) = 1.80 × 10−6 J = 1.80 μ J EVALUATE: We could also calculate the stored energy as Q (0.0180 × 10−6 C) = = 1.80 J 2C 2(9.00 ì 1011 F) â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-10 24.27 Chapter 24 IDENTIFY and SET UP: Combine Eqs (24.9) and (24.2) to write the stored energy in terms of the separation between the plates Q2 xQ ⑀ A EXECUTE: (a) U = ; C = so U = x 2C 2⑀0 A (b) x → x + dx gives U = dU = ( x + dx)Q 2⑀0 A ( x + dx)Q xQ ⎛ Q ⎞ − =⎜ ⎟ dx 2⑀0 A 2⑀0 A ⎜⎝ 2⑀0 A ⎟⎠ (c) dW = F dx = dU , so F = Q2 2⑀0 A (d) EVALUATE: The capacitor plates and the field between the plates are shown in Figure 24.27a E= σ Q = ⑀0 ⑀0 A F = 12 QE , not QE Figure 24.27a The reason for the difference is that E is the field due to both plates If we consider the positive plate only and calculate its electric field using Gauss’s law (Figure 24.27b): G G v∫ E ⋅ dA = EA = E= Qencl ⑀0 σA ⑀0 σ 2⑀0 = Q 2⑀0 A Figure 24.27b The force this field exerts on the other plate, that has charge −Q, is F = 24.28 IDENTIFY: C = ⑀0 A The stored energy can be expressed either as Q2 2⑀0 A CV Q2 or as , whichever is more 2C d convenient for the calculation SET UP: Since d is halved, C doubles EXECUTE: (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and the stored energy, which was 8.38 J, decreases since U = Q /2C Therefore the new energy is 4.19 J (b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.8 J, using U = CV /2, when V is held constant throughout EVALUATE: When the capacitor is disconnected, the stored energy decreases because of the positive work done by the attractive force between the plates When the capacitor remains connected to the battery, Q = CV tells us that the charge on the plates increases The increased stored energy comes from the battery when it puts more charge onto the plates © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-18 Chapter 24 EXECUTE: C0 = A⑀0 (4.20 × 10−5 m )⑀0 = = 5.31 × 10−13 F The new value of C is d 7.00 × 10−4 m C = C0 + 0.25 pF = 7.81 × 10−13 F But C = A⑀0 A⑀ (4.20 × 10−5 m )⑀0 , so d ′ = = = 4.76 × 10−4 m d′ C 7.81 × 10−13 F Therefore the key must be depressed by a distance of 7.00 × 10−4 m − 4.76 × 10−4 m = 0.224 mm 24.53 EVALUATE: When the key is depressed, d decreases and C increases IDENTIFY: Some of the charge from the original capacitor flows onto the uncharged capacitor until the potential differences across the two capacitors are the same Q SET UP: C = Let C1 = 20.0 μ F and C2 = 10.0 μ F The energy stored in a capacitor is Vab Q2 C EXECUTE: (a) The initial charge on the 20.0 μ F capacitor is QV ab 2 = 12 CVab = 12 Q = C1(800 V) = (20.0 × 10−6 F)(800 V) = 0.0160 C (b) In the final circuit, charge Q is distributed between the two capacitors and Q1 + Q2 = Q The final circuit contains only the two capacitors, so the voltage across each is the same, V1 = V2 V = gives Q so V1 = V2 C Q1 Q2 C = Q1 = Q2 = 2Q2 Using this in Q1 + Q2 = 0.0160 C gives 3Q2 = 0.0160 C and C1 C2 C2 Q2 = 5.33 × 10−3 C Q = 2Q2 = 1.066 × 10−2 C V1 = Q1 1.066 × 10−2 C = = 533 V C1 20.0 × 10−6 F Q2 5.33 × 1023 C = = 533 V The potential differences across the capacitors are the same, as they C2 10.0 × 1026 F should be (c) Energy = 12 C1V + 12 C2V = 12 (C1 + C2 )V gives V2 = Energy = 12 (20.0 × 10−6 F + 10.0 × 10−6 F)(533 V)2 = 4.26 J (d) The 20.0 μ F capacitor initially has energy = 12 C1V = 12 (20.0 × 10−6 F)(800 V)2 = 6.40 J The decrease 24.54 in stored energy that occurs when the capacitors are connected is 6.40 J − 4.26 J = 14 J EVALUATE: The decrease in stored energy is because of conversion of electrical energy to other forms during the motion of the charge when it becomes distributed between the two capacitors Thermal energy is generated by the current in the wires and energy is emitted in electromagnetic waves IDENTIFY: Initially the capacitors are connected in parallel to the source and we can calculate the charges Q1 and Q2 on each After they are reconnected to each other the total charge is Q = Q2 − Q1 Q2 2C SET UP: After they are reconnected, the charges add and the voltages are the same, so Ceq = C1 + C2 , as U = 12 CV = for capacitors in parallel EXECUTE: Originally Q1 = C1V1 = (9.0 μ F) (36 V) = 3.24 × 10−4 C and Q2 = C2V2 = (4.0 μ F)(36 V) = 1.44 × 10−4 C Ceq = C1 + C2 = 13.0 μ F The original energy stored is U = 12 CeqV = 12 (13.0 × 10−6 F)(36 V)2 = 8.42 × 10−3 J Disconnect and flip the capacitors, so now the total charge is Q = Q2 − Q1 = 1.8 × 10−4 C and the equivalent capacitance is still the same, Ceq = 13.0 μ F © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics The new energy stored is U = 24-19 Q2 (1.8 × 10−4 C) = = 1.25 × 10−3 J The change in stored energy is 2Ceq 2(13.0 × 10−6 F) ΔU = 1.25 × 10−3 J − 8.42 × 10−3 J = −7.17 × 10−3 J 24.55 EVALUATE: When they are reconnected, charge flows and thermal energy is generated and energy is radiated as electromagnetic waves IDENTIFY: Simplify the network by replacing series and parallel combinations by their equivalent The stored energy in a capacitor is U = 12 CV SET UP: For capacitors in series the voltages add and the charges are the same; 1 = + + " For Ceq C1 C2 capacitors in parallel the voltages are the same and the charges add; Ceq = C1 + C2 + " C = Q V U = 12 CV 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 24.55a–c U tot = 12 CeqV = 12 (2.19 × 10−6 F)(12.0 V)2 = 1.58 × 10−4 J = 158 μ J (b) From Figure 24.55c, Qtot = CeqV = (2.19 × 10−6 F)(12.0 V) = 2.63 × 10−5 C From Figure 24.55b, Q4.8 = 2.63 × 10−5 C V4.8 = Q4.8 2.63 × 10−5 C = = 5.48 V C4.8 4.80 × 10−6 F U 4.8 = 12 CV = 12 (4.80 × 10−6 F)(5.48 V) = 7.21 × 10−5 J = 72.1 μ J This one capacitor stores nearly half the total stored energy Q2 EVALUATE: U = For capacitors in series the capacitor with the smallest C stores the greatest amount 2C of energy Figure 24.55 24.56 IDENTIFY: Apply the rules for combining capacitors in series and parallel For capacitors in series the voltages add and in parallel the voltages are the same SET UP: When a capacitor is a moderately good conductor it can be replaced by a wire and the potential across it is zero EXECUTE: (a) A network that has the desired properties is sketched in Figure 24.56a Ceq = C + C = C 2 The total capacitance is the same as each individual capacitor, and the voltage is divided equally between the two capacitors, so that V = 480 V (b) If one capacitor is a moderately good conductor, then it can be treated as a “short” and thus removed from the circuit, and one capacitor will have greater than 600 V across it EVALUATE: An alternative solution is two in parallel in series with two in parallel, as sketched in Figure 24.56b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-20 Chapter 24 Figure 24.56 24.57 (a) IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents SET UP: The network is sketched in Figure 24.57a C1 = C5 = 8.4 μ F C2 = C3 = C4 = 4.2 μ F Figure 24.57a EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C3 and C4 are in series and can be replaced by C34 (Figure 24.57b): 1 = + C34 C3 C4 C + C4 = C34 C3C4 Figure 24.57b C34 = C3C4 (4.2 μ F)(4.2 μ F) = = 2.1 μ F C3 + C4 4.2 μ F + 4.2 μ F C2 and C34 are in parallel and can be replaced by their equivalent (Figure 24.57c): C234 = C2 + C34 C234 = 4.2 μ F + 2.1 μ F C234 = 6.3 μ F Figure 24.57c C1, C5 and C234 are in series and can be replaced by Ceq (Figure 24.57d): © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24-21 1 1 = + + Ceq C1 C5 C234 = + Ceq 8.4 μ F 6.3 μ F Ceq = 2.5 μ F Figure 24.57d EVALUATE: For capacitors in series the equivalent capacitor is smaller than any of those in series For capacitors in parallel the equivalent capacitance is larger than any of those in parallel (b) IDENTIFY and SET UP: In each equivalent network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the original circuit EXECUTE: The equivalent circuit is drawn in Figure 24.57e Qeq = CeqV Qeq = (2.5 μ F)(220 V) = 550 μ C Figure 24.57e Q1 = Q5 = Q234 = 550 μ C (capacitors in series have same charge) V1 = Q1 550 μ C = = 65 V C1 8.4 μ F V5 = Q5 550 μ C = = 65 V C5 8.4 μ F V234 = Q234 550 μ C = = 87 V C234 6.3 μ F Now draw the network as in Figure 24.57f V2 = V34 = V234 = 87 V capacitors in parallel have the same potential Figure 24.57f Q2 = C2V2 = (4.2 μ F)(87 V) = 370 μ C Q34 = C34V34 = (2.1 μ F)(87 V) = 180 μ C Finally, consider the original circuit (Figure 24.57g) Q3 = Q4 = Q34 = 180 μ C capacitors in series have the same charge Figure 24.57g © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-22 Chapter 24 V3 = Q3 180 μ C = = 43 V C3 4.2 μ F V4 = Q4 180 μ C = = 43 V C4 4.2 μ F Summary: Q1 = 550 μ C, V1 = 65 V Q2 = 370 μ C, V2 = 87 V Q3 = 180 μ C, V3 = 43 V Q4 = 180 μ C, V4 = 43 V Q5 = 550 μ C, V5 = 65 V EVALUATE: V3 + V4 = V2 and V1 + V2 + V5 = 220 V (apart from some small rounding error) Q1 = Q2 + Q3 and Q5 = Q2 + Q4 24.58 IDENTIFY: We can make series and parallel combinations 1 SET UP: For capacitors in series, = + + ", so for N equivalent capacitors in series, Ceq = C/N Ceq C1 C2 For capacitors in parallel, Ceq = C1 + C2 + ", so for N equivalent capacitors in parallel, Ceq = NC EXECUTE: There are many ways to achieve the required equivalent capacitance In each case one simple solution is shown in Figure 24.58 Figure 24.58 24.59 EVALUATE: By combining capacitors in series and parallel combinations, we can produce a wide variety of equivalent capacitances IDENTIFY: Capacitors in series carry the same charge, while capacitors in parallel have the same potential difference across them SET UP: Vab = 150 V, Q1 = 150 μ C, Q3 = 450 μ C, and V = Q/C EXECUTE: C1 = 3.00 μ F so V1 = V3 = 100 V C3 = Q1 150 μ C = = 50.0 V and V1 = V2 = 50.0 V V1 + V3 = Vab so C1 3.00 μ F Q3 450 μ C = = 4.50 μ F Q1 + Q2 = Q3 so Q2 = Q3 − Q1 = 450 μ C − 150 μ C = 300 μ C V3 100 V Q2 300 μ C = = 6.00 μ F V2 50.0 V EVALUATE: Capacitors in parallel only carry the same charge if they have the same capacitance and C2 = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24.60 24-23 IDENTIFY: Apply the rules for combining capacitors in series and in parallel SET UP: With the switch open, each pair of 3.00 μ F and 6.00 μ F 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 μ F and 6.00 μ F capacitors are in parallel with each other and the two pairs are in series ⎛⎛ 1 ⎞ + EXECUTE: (a) With the switch open Ceq = ⎜ ⎜ ⎟ ⎜ ⎝ μF μF ⎠ ⎝ −1 ⎛ 1 ⎞ +⎜ + ⎟ ⎝ μF μF ⎠ −1 ⎞ ⎟ = 4.00 μ F ⎟ ⎠ Qtotal = CeqV = (4.00 μ F)(210 V) = 8.40 × 10−4 C By symmetry, each capacitor carries 4.20 × 10−4 C The voltages are then calculated via V = Q/C This gives Vad = Q/C3 = 140 V and Vac = Q/C6 = 70 V Vcd = Vad − Vac = 70 V (b) When the switch is closed, the points c and d must be at the same potential, so the equivalent ⎛ ⎞ 1 + capacitance is Ceq = ⎜ ⎟ ⎝ (3.00 + 6.00) μ F (3.00 + 6.00) μ F ⎠ −1 = 4.5 μ F Qtotal = CeqV = (4.50 μ F)(210 V) = 9.5 × 10−4 C, and each 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 C1 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, Q1 = Q2 and Q2 − Q1 = After the switch is closed, Q2 − Q1 = 315 μ C, so 315 μ C of charge flowed through the switch 24.61 EVALUATE: When the switch is closed the charge must redistribute to make points c and d be at the same potential (a) IDENTIFY: Replace the three capacitors in series by their equivalent The charge on the equivalent capacitor equals the charge on each of the original capacitors SET UP: The three capacitors can be replaced by their equivalent as shown in Figure 24.61a Figure 24.61a EXECUTE: C3 = C1/2 so 1 1 = + + = and Ceq = 8.4 μ F/4 = 2.1 μ F Ceq C1 C2 C3 8.4 μ F Q = CeqV = (2.1 μ F)(36 V) = 76 μ C The three capacitors are in series so they each have the same charge: Q1 = Q2 = Q3 = 76 μ C EVALUATE: The equivalent capacitance for capacitors in series is smaller than each of the original capacitors (b) IDENTIFY and SET UP: Use U = 12 QV We know each Q and we know that V1 + V2 + V3 = 36 V EXECUTE: U = 12 Q1V1 + 12 Q2V2 + 12 Q3V3 But Q1 = Q2 = Q3 = Q so U = 12 Q(V1 + V2 + V3 ) But also V1 + V2 + V3 = V = 36 V, so U = 12 QV = 12 (76 μ C)(36 V) = 1.4 × 10−3 J EVALUATE: We could also use U = Q /2C and calculate U for each capacitor (c) IDENTIFY: The charges on the plates redistribute to make the potentials across each capacitor the same © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-24 Chapter 24 SET UP: The capacitors before and after they are connected are sketched in Figure 24.61b Figure 24.61b EXECUTE: The total positive charge that is available to be distributed on the upper plates of the three capacitors is Q0 = Q01 + Q02 + Q03 = 3(76 μ C) = 228 μ C Thus Q1 + Q2 + Q3 = 228 μ C After the circuit is completed the charge distributes to make V1 = V2 = V3 V = Q/C and V1 = V2 so Q1/C1 = Q2 /C2 and then C1 = C2 says Q1 = Q2 V1 = V3 says Q1/C1 = Q3/C3 and Q1 = Q3 (C1/C3 ) = Q3 (8.4 μ F/ 4.2 μ F) = 2Q3 Using Q2 = Q1 and Q1 = 2Q3 in the above equation gives 2Q3 + 2Q3 + Q3 = 228 μ C 5Q3 = 228 μ C and Q3 = 45.6 μ C, Q1 = Q2 = 91.2 μ C Then V1 = 91.2 μ C Q 45.6 μ C Q1 91.2 μ C Q = 11 V = = 11 V, V2 = = = 11 V, and V3 = = C3 4.2 μ F 8.4 μ F C1 8.4 μ F C2 The voltage across each capacitor in the parallel combination is 11 V (d) U = 12 Q1V1 + 12 Q2V2 + 12 Q3V3 But V1 = V2 = V3 so U = 12 V1 (Q1 + Q2 + Q3 ) = 12 (11 V)(228 μ C) = 1.3 × 10−3 J 24.62 EVALUATE: This is less than the original energy of 1.4 × 10−3 J The stored energy has decreased, as in Example 24.7 ⑀ A Q IDENTIFY: C = C = V = Ed U = 12 QV d V SET UP: d = 3.0 × 103 m A = π r , with r = 1.0 × 103 m EXECUTE: (a) C = ⑀0 A d = (8.854 × 10−12 C2 /N ⋅ m )π (1.0 × 103 m)2 (b) V = Q 20 C = = 2.2 × 109 V C 9.3 × 10−9 F (c) E = V 2.2 × 109 V = = 7.3 × 105 V/m d 3.0 × 103 m 3.0 × 103 m = 9.3 × 10−9 F (d) U = 12 QV = 12 (20 C)(2.2 × 109 V) = 2.2 × 1010 J 24.63 EVALUATE: Thunderclouds involve very large potential differences and large amounts of stored energy IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents In each equivalent network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the original circuit (a) SET UP: The network is sketched in Figure 24.63a C1 = 6.9 μ F C2 = 4.6 μ F Figure 24.63a EXECUTE: Simplify the network by replacing the capacitor combinations by their equivalents Make the replacement shown in Figure 24.63b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24-25 = Ceq C1 Ceq = C1 6.9 μ F = = μ F 3 Figure 24.63b Next make the replacement shown in Figure 24.63c Ceq = 2.3 μ F + C2 Ceq = 2.3 μ F + 4.6 μ F = 6.9 μ F Figure 24.63c Make the replacement shown in Figure 24.63d = + = Ceq C1 6.9 μ F 6.9 μ F Ceq = 2.3 μ F Figure 24.63d Make the replacement shown in Figure 24.63e Ceq = C2 + 2.3 μ F = 4.6 μ F + 2.3 μ F Ceq = 6.9 μ F Figure 24.63e Make the replacement shown in Figure 24.63f = + = Ceq C1 6.9 μ F 6.9 μ F Ceq = 2.3 μ F Figure 24.63f (b) Consider the network as drawn in Figure 24.63g From part (a) 2.3 μ F is the equivalent capacitance of the rest of the network Figure 24.63g © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-26 Chapter 24 The equivalent network is shown in Figure 24.63h The capacitors are in series, so all three capacitors have the same Q Figure 24.63h But here all three have the same C, so by V = Q/C all three must have the same V The three voltages must add to 420 V, so each capacitor has V = 140 V The 6.9 μ F to the right is the equivalent of C2 and the 2.3 μ F capacitor in parallel, so V2 = 140 V (Capacitors in parallel have the same potential difference.) Hence Q1 = C1V1 = (6.9 μ F)(140 V) = 9.7 × 10−4 C and Q2 = C2V2 = (4.6 μ F)(140 V) = 6.4 × 10−4 C (c) From the potentials deduced in part (b) we have the situation shown in Figure 24.63i From part (a) 6.9 μ F is the equivalent capacitance of the rest of the network Figure 24.63i The three right-most capacitors are in series and therefore have the same charge But their capacitances are also equal, so by V = Q/C they each have the same potential difference Their potentials must sum to140 V, so the potential across each is 47 V and Vcd = 47 V EVALUATE: In each capacitor network the rules for combining V for capacitors in series and parallel are obeyed Note that Vcd < V , in fact V − 2(140 V) − 2(47 V) = Vcd 24.64 IDENTIFY: Find the total charge on the capacitor network when it is connected to the battery This is the amount of charge that flows through the signal device when the switch is closed Circuit(a): SET UP: For capacitors in parallel, Ceq = C1 + C2 + C3 + " EXECUTE: Cequiv = C1 + C2 + C3 = 60.0 μ F Q = CV = (60.0 μ F)(120 V) = 7200 μ C EVALUATE: More charge is stored by the three capacitors in parallel than would be stored in each capacitor used alone Circuit(b): −1 ⎛ 1 ⎞ SET UP: Cequiv = ⎜ + + ⎟ ⎝ C1 C2 C3 ⎠ EXECUTE: Cequiv = 5.45 μ F Q = (5.45 μ F)(120V) = 654 μ C 24.65 EVALUATE: Less charge is stored by the three capacitors in series than would be stored in each capacitor used alone (a) IDENTIFY and SET UP: Q is constant C = KC0 ; use Eq (24.1) to relate the dielectric constant K to the ratio of the voltages without and with the dielectric EXECUTE: With the dielectric: V = Q/C = Q/(KC0 ) without the dielectric: V0 = Q/C0 V0 /V = K , so K = (45.0 V)/(11.5 V) = 3.91 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24-27 EVALUATE: Our analysis agrees with Eq (24.13) (b) IDENTIFY: The capacitor can be treated as equivalent to two capacitors C1 and C2 in parallel, one with area 2A/3 and air between the plates and one with area A/3 and dielectric between the plates SET UP: The equivalent network is shown in Figure 24.65 Figure 24.65 EXECUTE: Let C0 = ⑀0 A/d be the capacitance with only air between the plates C1 = KC0 /3, C2 = 2C0 /3; Ceq = C1 + C2 = (C0 /3)( K + 2) V= 24.66 Q Q⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ = V0 ⎜ ⎟ = (45.0 V) ⎜ ⎟ = 22.8 V Ceq C0 ⎝ K + ⎠ ⎝ K +2⎠ ⎝ 5.91 ⎠ EVALUATE: The voltage is reduced by the dielectric The voltage reduction is less when the dielectric doesn’t completely fill the volume between the plates IDENTIFY: This situation is analogous to having two capacitors C1 in series, each with separation (d − a ) SET UP: For capacitors in series, 1 = + Ceq C1 C2 −1 ⎛ 1 ⎞ ⑀0 A ⑀ A EXECUTE: (a) C = ⎜ + ⎟ = 12 C1 = 12 = C C ( d a )/2 d − −a 1⎠ ⎝ ⑀ A ⑀ A d d = C0 (b) C = = d −a d d −a d −a (c) As a → 0, C → C0 The metal slab has no effect if it is very thin And as a → d , C → ∞ V = Q/C V = Ey is the potential difference between two points separated by a distance y parallel to a uniform 24.67 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 IDENTIFY: The conductor can be at some potential V, where V = far from the conductor This potential depends on the charge Q on the conductor so we can define C = Q/V where C will not depend on V or Q SET UP: Use the expression for the potential at the surface of the sphere in the analysis in part (a) EXECUTE: (a) For any point on a solid conducting sphere V = Q/4π ⑀0 R if V = at r → ∞ C= Q = V Q Q = 4π ⑀0 R 4π ⑀0 R (b) C = 4π ⑀ R = 4π (8.854 × 10−12 F/m)(6.38 × 106 m) = 7.10 × 10−4 F = 710 μ F 24.68 EVALUATE: The capacitance of the earth is about ten times larger than typical electronic circuit capacitances in the range of 10 pF to 100 pF Nevertheless, the capacitance of the earth is quite small, in view of its large size IDENTIFY: Capacitors in series carry the same charge, but capacitors in parallel have the same potential difference across them SET UP: Vab = 48.0 V C = Q/V and U = CV For capacitors in parallel, C = C1 + C2 , and for capacitors in series, 1/C = 1/C1 + 1/C2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-28 Chapter 24 2U 2(2.90 × 10−3 J) EXECUTE: Using U = CV gives C = = = 2.517 × 10−6 F, which is the equivalent (48.0 V) V capacitance of the network The equivalent capacitance for C1 and C2 in series is C12 = (4.00 μ F) = 2.00 μ F If C123 is the equivalent capacitance for C12 and C3 in parallel, then 1 + = Solving for C123 gives C123 C4 C 1 1 = − = − = 2.722 × 105 F−1, so C123 = 3.673 × 10−6 F C123 C C4 2.517 × 10−6 F 8.00 × 10−6 F C12 + C3 = C123 C3 = C123 − C12 = 3.673 μ F − 2.00 μ F = 1.67 μ F 24.69 EVALUATE: As with most circuits, it is necessary to solve them in a series of steps rather than using a single step IDENTIFY: We model the earth as a spherical capacitor rr SET UP: The capacitance of the earth is C = 4π ⑀ a b and, the charge on it is Q = CV , and its stored rb − energy is U = 12 CV EXECUTE: (a) C = (6.38 × 106 m)(6.45 × 106 m) 2 6 9.00 × 10 N ⋅ m /C 6.45 × 10 m − 6.38 × 10 m = 6.5 × 10−2 F (b) Q = CV = (6.54 × 10−2 F)(350,000 V) = 2.3 × 104 C (c) U = 12 CV = 12 (6.54 × 10−2 F)(350,000 V)2 = 4.0 × 109 J 24.70 EVALUATE: While the capacitance of the earth is larger than ordinary laboratory capacitors, capacitors much larger than this, such as F, are readily available Q2 IDENTIFY: The electric field energy density is u = 12 ⑀0 E U = 2C SET UP: For this charge distribution, E = for r < , E = Example 24.4 shows that λ 2π ⑀0r for < r < rb and E = for r > rb U 2π ⑀ for a cylindrical capacitor = L ln(rb /ra ) ⎛ λ ⎞ λ2 EXECUTE: (a) u = 12 ⑀0 E = 12 ⑀0 ⎜ ⎟ = 2 8π ⑀0r ⎝ 2π ⑀0r ⎠ (b) U = ∫ udV = 2π L ∫ urdr = (c) Using Eq (24.9), U = Lλ rb dr U λ2 and = ln(rb /ra ) 4π ⑀0 ∫ra r L 4π ⑀0 Q2 Q2 λ 2L = ln(rb /ra ) = ln(rb /ra ) This agrees with the result of part (b) 2C 4π ⑀0 L 4π ⑀0 EVALUATE: We could have used the results of part (b) and U = 24.71 Q2 to calculate C/L and would obtain 2C the same result as in Example 24.4 IDENTIFY: The increase in temperature of the wire allows us to find out how much heat it gained, which is the energy initially stored in the capacitor We can use this energy to find the capacitance of the capacitor SET UP: The heat in the wire is Q = mcΔT and the energy stored in the capacitor is U = CV , which is equal to the heat Q © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24-29 EXECUTE: The heat that goes into the wire is 24.72 Q = mcΔT = (12.0 × 10−3 kg)(910 J/(kg ⋅ K))(11.2 K) = 122.3 J For the capacitor, U = CV 2U 2(122.3 J) −5 C= = = 4.83 × 10 F = 48.3 μ F V (2.25 × 103 V) EVALUATE: A capacitance of 48.3 μ F is quite reasonable for ordinary laboratory capacitors IDENTIFY: The capacitor is equivalent to two capacitors in parallel, as shown in Figure 24.72 Figure 24.72 SET UP: Each of these two capacitors have plates that are 12.0 cm by 6.0 cm For a parallel-plate A capacitor with dielectric filling the volume between the plates, C = K ⑀0 For two capacitors in parallel, d C = C1 + C2 The energy stored in a capacitor is U = 12 CV EXECUTE: (a) C = C1 + C2 C2 = ⑀ A (8.854 × 10−12 F/m)(0.120 m)(0.060 m) = = 1.42 × 10−11 F d 4.50 × 10−3 m C1 = KC2 = (3.40)(1.42 × 10−11 F) = 4.83 × 10−11 F C = C1 + C2 = 6.25 × 10−11 F = 62.5 pF (b) U = 12 CV = 12 (6.25 × 10−11 F)(18.0 V) = 1.01 × 10−8 J (c) Now C1 = C2 and C = 2(1.42 × 10−11 F) = 2.84 × 10−11 F U = 12 CV = 12 (2.84 × 10−11 F)(18.0 V) = 4.60 × 10−9 J 24.73 EVALUATE: The plexiglass increases the capacitance and that increases the energy stored for the same voltage across the capacitor IDENTIFY: The two slabs of dielectric are in series with each other 1 CC + = , which gives C = SET UP: The capacitor is equivalent to C1 and C2 in series, so C1 C2 C C1 + C2 EXECUTE: With d = 1.90 mm, C1 = K1⑀0 A K⑀ A and C2 = d d ⎛ K K ⎞ ⑀ A (8.854 × 10−12 C2 /(N ⋅ m )(0.0800 m) ⎛ (4.7)(2.6) ⎞ −11 C =⎜ ⎟ = ⎜ ⎟ = 4.992 × 10 F 1.90 × 10−3 m ⎝ 4.7 + 2.6 ⎠ ⎝ K1 + K ⎠ d 1 U = CV = (4.992 × 10−11 F)(86.0 V)2 = 1.85 × 10−7 J 2 EVALUATE: The dielectrics increase the capacitance, allowing the capacitor to store more energy than if it were air-filled © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-30 24.74 Chapter 24 IDENTIFY: 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 ⑀ A SET UP: Define K eff by Ceq = eff , where A = wL For two capacitors in parallel, Ceq = C1 + C2 d ⑀ w( L − h) K ⑀0 wh ⑀0 wL ⎛ Kh h ⎞ EXECUTE: (a) The capacitors are in parallel, so C = + = − ⎟ This ⎜1 + d d d ⎝ L L⎠ ⎛ Kh h ⎞ gives K eff = ⎜1 + − ⎟ L L⎠ ⎝ (b) For gasoline, with K = 1.95: 1 L⎞ L⎞ ⎛ ⎛ full: K eff ⎜ h = ⎟ = 1.24; full: K eff ⎜ h = ⎟ = 1.48; 4 2⎠ ⎝ ⎠ ⎝ 3L ⎞ ⎛ full: K eff ⎜ h = ⎟ = 71 ⎠ ⎝ (c) For methanol, with K = 33: L⎞ L⎞ 1 ⎛ ⎛ full: K eff ⎜ h = ⎟ = 9; full: K eff ⎜ h = ⎟ = 17; 4⎠ 2⎠ ⎝ ⎝ 3L ⎞ ⎛ full: K eff ⎜ h = ⎟ = 25 4 ⎠ ⎝ (d) This kind of fuel tank sensor will work best for methanol since it has the greater range of K eff values EVALUATE: When h = 0, K eff = When h = L, K eff = K 24.75 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, C = d EXECUTE: (a) The charge distribution on the plates is shown in Figure 24.75 ⎛ ⑀ A ⎞ 2(4.2)⑀0 (0.120 m) = 2.38 × 10−9 F (b) C = ⎜ ⎟ = −4 d 4.5 × 10 m ⎝ ⎠ EVALUATE: If two of the plates are separated by both sheets of paper to form a capacitor, ⑀ A 2.38 × 10−9 F C= = , smaller by a factor of compared to the capacitor in the problem 2d Figure 24.75 24.76 IDENTIFY: The force on one plate is due to the electric field of the other plate The electrostatic force must be balanced by the forces from the springs SET UP: The electric field due to one plate is E = σ 2⑀0 The force exerted by a spring compressed a distance z0 − z from equilibrium is k ( z0 − z ) EXECUTE: (a) The force between the two parallel plates qσ q2 (CV ) ⑀02 A2 V ⑀ AV = = = = is F = qE = 2⑀0 2⑀0 A 2⑀0 A z 2⑀0 A 2z © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Capacitance and Dielectrics 24-31 (b) When V = 0, the separation is just z0 When V ≠ 0, the total force from the four springs must equal the electrostatic force calculated in part (a) F4 springs = 4k ( z0 − z ) = ⑀0 AV 2z and z − z z0 + ⑀0 AV 4k = (c) For A = 0.300 m , z0 = 1.2 × 10−3 m, k = 25 N/m and V = 120 V, so z − (2.4 × 10−3 m) z + 3.82 × 10−10 m3 = The physical solutions to this equation are z = 0.537 mm and 1.014 mm 24.77 EVALUATE: (d) Stable equilibrium occurs if a slight displacement from equilibrium yields a force back toward the equilibrium point If one evaluates the forces at small displacements from the equilibrium positions above, the 1.014 mm separation is seen to be stable, but not the 0.537 mm separation 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 K ⑀0 A SET UP: C = for a parallel-plate capacitor with plate area A d ⑀ ⑀ L EXECUTE: (a) C = (( L − x) L + xKL) = ( L + ( K − 1) x) D D ⑀0 L ⑀ L (b) dU = (dC )V , where C = C0 + (− dx + dxK ), with C0 = ( L + ( K − 1) x) This gives D D ( K − 1)⑀0V L ⎛ ⑀ Ldx ⎞ dU = 12 ⎜ ( K − 1) ⎟V = dx 2D ⎝ D ⎠ (c) If the charge is kept constant on the plates, then Q = ⑀0 LV ( L + ( K − 1) x) and D ⎛C ⎞ ⎞ ⑀ L ( K − 1)⑀0V L C V2 ⎛ dx U = 12 CV = 12 C0V ⎜ ⎟ U ≈ ⎜1 − ( K − 1)dx ⎟ and ΔU = U − U = − 2D ⎝ DC0 ⎝ C0 ⎠ ⎠ ( K −1)⑀0V L dx, the force is in the opposite direction to the motion dx, meaning (d) Since dU = − Fdx = − 2D that the slab feels a force pushing it out EVALUATE: (e) 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 the plates Comparing the results for dU in part (c) to dU = − Fdx gives ( K − 1)⑀0V L 2D IDENTIFY: C = Q/V Apply Gauss’s law and the relation between potential difference and electric field rb G rb G G G SET UP: Each conductor is an equipotential surface Va − Vb = ∫ E U ⋅ d r = ∫ EL ⋅ d r , so EU = EL , F= 24.78 ra where these are the fields between the upper and lower hemispheres The electric field is the same in the air space as in the dielectric ⎛ rr ⎞ EXECUTE: (a) For a normal spherical capacitor with air between the plates, C0 = 4π ⑀0 ⎜ a b ⎟ The ⎝ rb − ⎠ capacitor in this problem is equivalent to two parallel capacitors, CL and CU , each with half the plate area of the normal capacitor CL = ⎛ rr ⎞ ⎛ rr ⎞ KC0 C = 2π K ⑀0 ⎜ a b ⎟ and CU = = 2π ⑀0 ⎜ a b ⎟ r r 2 − ⎝ b a⎠ ⎝ rb − ⎠ ⎛ rr ⎞ C = CU + CL = 2π ⑀0 (1 + K ) ⎜ a b ⎟ ⎝ rb − ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 24-32 Chapter 24 (b) Using a hemispherical Gaussian surface for each respective half, EL and EU QL 4π r QL , so EL = = , K ⑀0 2π K ⑀0 r 4π r QU QU , so EU = = But QL = VCL and QU = VCU Also, QL + QU = Q Therefore, ⑀0 2π ⑀0r QL = KQ Q Q KQ VC0 K , QL = This gives EL = and = KQU and QU = = 1+ K 1+ K + K 2π K ⑀0r + K 4π ⑀0r 2 EU = Q Q = We find that EU = EL + K 2π K ⑀ r + K 4π K ⑀0 r (c) The free charge density on upper and lower hemispheres are: (σ f,ra ) U = (σ f,rb ) U = QU 2π rb2 = Q 2π rb2 (1 + K ) ; (σ f,ra ) L = QL 2π ra2 = KQ 2π ra2 (1 + K ) QU 2π ra2 and (σ f,rb )L = = QL Q 2π ra2 (1 + K ) 2π rb2 = and KQ 2π rb2 (1 + K ) ⎛ ( K − 1) ⎞ Q ⎛ K ⎞ ⎛ K − ⎞ Q (d) σ i,ra = σ f,ra (1 − 1/K ) = ⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎝ K ⎠ 2π ra2 ⎝ K + ⎠ ⎝ K + ⎠ 2π ra2 ⎛ ( K − 1) ⎞ Q ⎛ K ⎞ ⎛ K − ⎞ Q ⎟ ⎜ ⎟=⎜ ⎟ ⎝ K ⎠ 2π rb2 ⎝ K + ⎠ ⎝ K + ⎠ 2π rb2 (e) There is zero bound charge on the flat surface of the dielectric-air 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 σ i,rb = σ f,rb (1 − 1/K ) = ⎜ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... in Figure 24.22 are V10 = tot = 17.4 V, C10 V13 = Qtot Q = 13. 4 V and V9 = tot = 19.3 V V10 + V13 + V9 = 17.4 V + 13. 4 V + 19.3 V = 50.1 V The sum C13 C9 of the voltages equals the applied voltage,... )⑀0 = = 5.31 × 10 13 F The new value of C is d 7.00 × 10−4 m C = C0 + 0.25 pF = 7.81 × 10 13 F But C = A⑀0 A⑀ (4.20 × 10−5 m )⑀0 , so d ′ = = = 4.76 × 10−4 m d′ C 7.81 × 10 13 F Therefore the... = 12 (12.5 × 10−6 F)(24.0 V) = 3.60 mJ after: U = 12 CV = 12 (46.9 × 10−6 F)(24.0 V) = 13. 5 mJ (b) ΔU = 13. 5 mJ − 3.6 mJ = 9.9 mJ The energy increased EVALUATE: The power supply must put additional