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Engenharia Civil ·
Física
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1. The minimum charge measurable is q_min = CV = (50 pF)(0.15 V) = 7.5 pC . 2. (a) The capacitance of the system is C = q/ΔV = 70 pC / 20 V = 3.5 pF . (b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes ΔV = q/C = 200 pC / 3.5 pF = 57 V . 3. 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 x 10^{-9} F)(12 V) = 30 x 10^{-9} C. 4. We verify the units relationship as follows: [κ0] = F/m = C/V·m = C / N·m/C = C^2 / N·m^2 . 5. (a) The capacitance of a parallel-plate capacitor is given by C = κ0A/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 = πR^2, where R is the radius of a plate. Thus, C = κ0πR^2/d = (8.85 x 10^{-12} F/m)(8.2 x 10^{-2} m)^2 / 1.3 x 10^{-4} m = 1.4 x 10^{-10} F = 140 pF . (b) The charge on the positive plate is given by q = CV , where V is the potential difference across the plates. Thus, q = (1.4 x 10^{-10} F)(12 V) = 1.7 x 10^{-9} C = 1.7 nC. 6. We use C = Aκ0/d. Thus C = (1.00 m^3)(8.85 x 10^{-12} C/V m) / 1.00 m = 8.85 x 10^{-12} F . 7. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4πεκ0R to find the capacitance. When the drops combine, the volume is doubled. It is then V = (4π/3)R^3. The new radius R' is given by 4π/3 (R')^3 = 2 * 4π/3 R^3 / 3 , so R' = 2^{1/3} R / 2^{1/3} = 2^{1/2} R . The new capacitance is C' = 4πεκ0 2^{1/2} R = 5.0κ0 πR . 8. (a) We use Eq. 26-17: C = 4πεἰo a^2 / (b-a) = (40.0 mm)(28.0 mm) / (8.99 x 10^9 N·m^2/C^2) ((40.0 mm - 28.0 mm) = 81.5 pF . (b) Let the areas required be A. Then C = κ0A/d, or A = C(d - a) / κ0 = (8.5 pF)(40.0 mm - 28.0 mm) / (8.85 x 10^{-12} N·m^2/C^2) = 191 cm^2 . 9. According to Eq. 29-17 the capacitance of a spherical capacitor is given by C = κ0 4π ab / (b - a) , where a and b are the radii of the spheres. If b and d are nearly the same then 4πab is nearly the surface area of either sphere. Replace 4πab with A and b - a with d to obtain C = κ0A / d . 10. The equivalent capacitance is C_eq = C_1 C_2/(C_1 + C_2) = (10.0 μF)(5.0 μF) / (10.0 μF + 5.0 μF) = 7.33 μF . 11. The equivalent capacitance is given by C_eq = 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, C_eq = NC, where C is the capacitance of one of them. Thus, N = q/(CV) and N = 1.0 C / (75 V)(1.0 x 10^{-6} F) = 9900 . 12. The charge that passes through meter A is q = C_V-V = 3CV = 3 x (25.0 pF)(420 V) = 0.315 C . 13. The equivalent capacitance is C_eq = (C_1 C_2)/(C_1 + C_2) = (10.0 μF)(5.0 μF)/(4.0 μF + 5.0 μF) = 3.16 μF . 14. (a) and (b) The original potential difference V_1 across C_1 is V_1 = C_1V/C_1 + C_2 = (3.16μF)(10 V) /(10.0 μF + 5.0 μF) = 21.1 V . Thus ΔV_1 = 100 V - 21.1 V = 79 V and Δq_1 = C_1ΔV_1 = (10.0 pF)(79 V) = 7.9 x 10^{-6} C. 15. Let x be the separation of the plates in the lower capacitor. Then the plate separation in the upper capacitor is c - b - x. The capacitance of the lower capacitor is C_L = κ0A/x and the capacitance of the upper capacitor is C_U = κ0A/(b - x), where A is the plate area. Since the two capacitors are in series, the equivalent capacitance is determined from 1 / C_eq = 1 / C_L + 1 / C_U . Thus, the equivalent capacitance is given by C_eq = κ0A/(a - b) - so C_eq is independent of x. 16. (a) The potential difference across C_3 is V_1 = V - V_2, thus, q = C_1V_1 = (10 μF)(10 V) = 1.0 x 10^{-4} C. (b) Let C_1 = 10 μF. We first consider the three-capacitor combination consisting of C_1 and its two closest neighbors, each of capacitance C'. The equivalent capacitance of this combination is C_eq = C_1C_2C_3/(C_1C_2 + C_2C_3 + C_3C_1) = 2/3 C' . Also, the voltage drop across this combination is the same as that across C_1 and those connected in series with it, the voltage difference across C_2 satisfies V_3 = V_2 / V_1/3. Thus q_2 = C_2(1/2) * (3V/2) = 20 x 10^{-9} V . 17. The charge initially on the charged capacitor is given by q = C_1v, where C_1 = 100 pF is the capacitance and V_0 = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor is wired in parallel to the first, the charge on the first capacitor is q_1 = C_1V_1, where v = 3 V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q = q - q1, whereas C_2 is the capacitance of the second capacitor. Substituting Clv0 for q1 and C1v1 for q, we obtain q = C_1(v0 - V). The potential difference across the second capacitor is also V_1, so the capacitance is C_2 = q_1 (v_0 - V) / V = (100 pF) = 3 pF . 18. (i) First, the equivalent capacitance of the two 4.0 μF capacitors connected in series is given by 4.0 μF / 2 = 2.0 μF . This combination is then connected in parallel with two other 2.0-μF capacitors (one on each side), resulting in an equivalent capacitance C = 3(2.0 μF) = 6.0 μF . This is now seen to be in series with another combination, which consists of the two 4.0-μF capacitors connected in parallel (which are themselves equivalent to C = 4(2.0 μF) = 6.0 μF ). 19. (a) After the switches are closed, the potential differences across the capacitors are the same and the two capacitors are in parallel. The potential difference from a to b is given by V_ab = Q/C_eq, where Q is the net charge on the combination and C_eq is the equivalent capacitance. The equivalent capacitance is C_eq = C_1 + C_2 = 4.0 x 10^{-6} F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q_1 = C_1V = (1.0 x 10^{-6} F)(10 V) = 1.0 x 10^{-5} C and the charge on capacitor 2 is q_2 = C_2V = (3.0 x 10^{-6} F)(10 V) = 3.0 x 10^{-5} C, so the net charge on the combination is 3.0 x 10^{-5} C - 1.0 x 10^{-5} C = 2.0 x 10^{-5} C. The potential difference is V_ab = 2.0 x 10^{-5} C / 4.0 x 10^{-6} F = 5.0 V. (b) The charge on capacitor 1 is now q_1 = C_1V_ab = (1.0 x 10^{-6} F)(5 V) = 5.0 x 10^{-6} C. (c) The charge on capacitor 2 is now q_2 = C_2V_ab = (3.0 x 10^{-6} F)(5 V) = 1.5 x 10^{-5} C. 20. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q_1 = q_3 = C_3C_1/(C_3 + C_1) = (1.0 μF)(3.0 μF)(12 V) / (1.0 μF + 3.0 μF) = 9.0 μC . Also, capacitors 2 and 4 are in series, so q_2 = q_4 = C_4C_2/(C_4 + C_2) = (2.0 μF)(4.0 μF)(12 V) / (2.0 μF + 4.0 μF) = 16 μC . (b) With switch 2 also closed, the potential difference V_1 across C_1 must equal the potential difference across C_3 and the sum of the potential differences across C_1 and C_2. V_1 = (C_3 + C_4)/(C_1 + C_3 + C_4) * V = (3.0 μF + 4.0 μF)[12 V] / (1.0 μF + 3.0 μF + 4.0 μF) = 8.4 V . Thus, q_1 = C_1V_1 = (1.0 μF)(8.4 V) = 8.4 μC, q_2 = C_2V_1 = (2.0 μF)(8.4 V) = 17 μC, q_3 = (C_3/C_1)V_1 = (3.0 μF)/(2.0 μF) * 12 μC, and q_4 = C_4V - (1.0 μF) * (2.0 μF) V = 14 μC. Express the total capacitance determined from 1 / C_1 = 1 / C_3 + 1 / C_4 . 21. Let V = 1.00 m^3. Using Eq. 26-23, the energy stored is U = 1/2 ε_0(E^2) V = 1/2 ε_0 (8.85 x 10^{-12} C/V·m)(150 V/m)^2(1.00 m^3) = 99.6 x 10^{-9} J . 23. The energy stored by a capacitor is given by U = 1/2 CV^2, where V is the potential difference across its plates. We convert the given value of the energy to Joules. Since a Joule is a watt-second, we multiply by (10^3 W/kW)(3600 s/h) to obtain 10 kW-h = 3.6 x 10^7 J. Thus, C = 2U / V^2 = (2)(3.6 x 10^7 J) / (1000 V)^2 = 72 F . 24. (a) The capacitance is \[ C = \frac {\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2})(40 \times 10^{-4} m^2)}{1.0 \times 10^{-1} m} = 3.5 \times 10^{-11} F = 35 pF. \] (b) \[ q = CV = (35 pF)(600 V) = 2.1 \times 10^{-8} C = 21 nC. \] U = \frac{1}{2} qV = \frac{1}{2} (35 pF)(21 nC)(6^{2}) = 6.3 \times 10^{-8} J = 6.3 pJ. (d) \[ E = V/d = 600V/1.0 \times 10^{-1} m = 6.0 \times 10^{3} V/m. \] (e) The energy density (energy per unit volume) is \[ \frac{U}{Ad} = \frac{6.3 \times 10^{-8} J}{(40 \times 10^{-4}m^{2})(1.0 \times 10^{-1} m)} = 1.6 J/m^{3}. \] 25. The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference \( V \) across the capacitors is the same and the total energy is \[ U = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (2.0 \times 10^{-6} F + 4.0 \times 10^{-6} F) (300 V)^2 = 0.27 J. \] 26. The total energy stored in the capacitor bank is \[ U = \frac{1}{2} C_{total}V^2 = \frac{1}{2} (2000)(0.06 \times 10^{-6} F)(XXXX V)^2 = 1.3 \times 10^{3} J. \] Thus, the cost is \[ \frac{1.3 \times 10^3 J}{3.6 \times 10^6 J/(\text{cent} / \text{kW h})} = 10 \text{ cents}. \] 27. (a) In the first case \[ U = q^{2}/2C \], and in the second case \[ U = q^{2}/2^{\prime} C = q^{2}/4C. \] So the energy is now \[ 4(3.0 J / 2) = 2.0 J. \] 28. (a) The potential difference across \( (b) \) is given by \[ V = \frac{C}{C + C_3} V_0 = \frac{(40.0 pF)(100 V)}{100.0 pF + 3.0 \mu F} = 21.1 V. \] Also, \[ V_2 = V - V_1 = V_3 = 100 V - 21.1 V = 79.9. Thus, \] \[ q_1 = C_1 V_1 = (10.0 pF)(21.1 V) = 2.11 \times 10^{-10} C\] \[ q_3 = C_3 V_3 = (50.0 pF)(79.9 V) = 3. 105 \times 10^{-4} C\] \[ q_{e} = q_h + q_2 = 2.11 \times 10^{-4c} + 1.05 \times 10^{-4} C = 3.16 \times 10^{-5} C. \] (b) The potential differences were found in the course of solving for the charges in part \((a)\). (c) The stored energies are as follows: \[ U_1 = \frac{1}{2} \frac{9}{10.0 pF)(21.1 V)^2 = 2.22 \times 10^{-3} J,\] \[ U_2 = \frac{1}{2} \frac{3.0 \mu F)(21.1 V)^2 = 1.11 \times 10^{-3} J,\] \[ U_3 = \frac{1}{2} \frac{9}{4.00 pF)(78.9 V)^2 = 1.25 \times 10^{-2} J. \] 29. (a) Let \( q \) be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by \( C = \varepsilon_0 A / d \), the charge is \( q = CV = \varepsilon_0 A V / d \). After the plates are pulled apart, their separation is \( 2d \) and the potential difference is \( V' \). Then \( q = \varepsilon_0 A V'/ 2d \) and \( V' = 2d \varepsilon_0 A q = q d A/ \varepsilon_0 = 2 V. \) (b) The initial energy stored in the capacitor is \( U_1 = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d} \) and the final energy stored is \( U_f = \frac{\varepsilon_0 A V^2}{4d} \) and \( U_i - U_f = \varepsilon_0 A V^2 / 2d \). (c) The work done to pull the plates apart is the difference in the energy: \( W = U_i - U_f = \varepsilon_0 A V^2 / 2d. \) 30. (a) The charge in the Figure is \[ q_n = C_n V = (4.00 \mu F)(100 V) = 4.00 \times 10^{-4} mQ ,\] \[ q_n = \frac{C_n V^2}{C_1 + C_2} = \frac{(10.0 pF)(3.00 \mu F)(100 V)}{100 \mu F + 3.00 \mu F} = 3.33 \times 10^{-6} C . \] \[ V_2 = (10 \mu C) /( 10.00 pF) = 333. V, V_3 = V - V_1 = 10 W - 333 V = 67. 7. \], \( V_3 = V - 10W. \) (c) We use \( \, U_1 = 1, 2, 3. \) The answers are \( U_1 = 5.6mJ, U_1 = 11 mJ, \text{ and } U_1 = 20 mJ. \) 31. 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 \(u = \frac{1}{2} \varepsilon_{0} E^{2}\), 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 \[ E = \frac{q}{2 \pi \varepsilon_{0} L r} \] (see Eq. 26-12), and the energy density at that point is given by \[ u = \frac{1}{2} \frac{q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} L^{2} r^{2}} \] The energy in the cylinder is the volume integral \[ U_{L} = \int u \, dV . \] Now, \( dV = 2 \pi r L \, dr \), so \[ U = \int \frac{q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} L^{2}} 2 \pi r L \, dr = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \int_{a}^{b} \frac{dr}{r} = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} . \] To find an expression for the total energy stored in the capacitor, we replace \( R \) with \( b \): \[ U_{b} = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} . \] We want the ratio \( U_{b} / U_{a} \) to be \(1/2\), so \[ \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} = \frac{1}{2} \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} = \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} \] or, since \( \frac{1}{2} \ln(\frac{b}{a}) = \ln(\sqrt{\frac{b}{a}}), \ln(R/a) = \ln(\sqrt{a b}) \). This means \( R/a = \sqrt{a b} \) or \( R = \sqrt{a b}. \] 32. We use \( E = q/4 \pi \varepsilon L = V / R . \) Thus \[ q = \frac{1}{2 \pi} \frac{(V)}{(R)} = ( \frac{8.85 \times 10^{-12} \frac{C^2}{N \cdot m^{2}}) (8000 \, V \,) \] / (0.050 m) = 0.011 J/m^{3}. 33. The charge is held constant while the plates are being separated, so we write the expression for the stored energy as \( U = q^2/2C. \), where q is the charge and C is the capacitance. The capacitance of a parallel-plate capacitor is given by \( C = \varepsilon_{0} A / x \), where A is the plate area and x is the plate separation, so \[ U = \frac{q^{2}}{2 \frac{\varepsilon_{0} A}{x}} = \frac{q^{2} x}{2 \varepsilon_{0} A}. \] If the plate separation increases by dx, the energy increases by dU := \( (q^{2}/2\varepsilon_{0} A) \) dx. Suppose the agent pulling the plate apart exerts force F. Then the agent does work Fdx and if the plates begin and end at rest, this must equal the increase in stored energy. Thus, \[ Fdx = (q^{2}/2\varepsilon_{0} A) dx \] and \[ F = \frac{q^{2}}{2\varepsilon_{0} A}. \] The net force on a plate is zero, so this must also be the magnitude of the force one plate exerts on the other. The force can also be computed as the product of the charge q on one plate and the electric field \( E_1 \) due to the charge on the other plate. Recall that the field produced by a uniform plane surface of charge is \( E_1 = q/2\varepsilon_{0} A. \). Thus, \[ F = q/2\varepsilon_{0} A. \] 34. If the original capacitance is given by \( C = \varepsilon_{0} A/2d, \) then the new capacitance is \( C' = \varepsilon A/2d. \) Thus \[ C'/C = \varepsilon/\varepsilon_{0} = 2(2.6 pF/1.3 pF) = 4.0. \] 35. The capacitance with the dielectric in place is given by \( C = \kappa C_0 \), where \( C_0 \) is the capacitance before the dielectric is inserted. The energy stored is given by \( U = q^{2}/C = \frac{q^{2} \varepsilon_{0} A/L} \) so \( \kappa = \frac{2U}{C} \) at 4.7. \) According to Table 26-1, you should use Pyrex. 37. The capacitance of a cylindrical capacitor is given by \[ C = \frac{2 \pi \varepsilon_{0} L}{\ln(b/a)} , \] where \( C_0 \) is the capacitance without the dielectric, \( \varepsilon \) is the dielectric constant, \( L \) is the length, \( a \) is the inner radius, and \( b \) is the outer radius. The capacitance per unit length of the cable is \[ \frac{C}{L} = \frac{2 \pi \varepsilon_{0}}{\ln(b/a)} = \frac{2\pi(8.85 \times 10^{-12} F/m)}{\ln(0.9 mm/0.1 mm)} = 8.1 \times 10^{-11} F/m = 81 pF/m . \] 38. (a) We use Eq. 26-14: \[ C = \frac{2\pi \varepsilon_{0} L}{\ln(b/a)} (4.7/10.18 m) = \frac{(2.99 \times 10^{-19} )} / \ln(8.3 cm/3.6 cm) = 0.73 nF . \] 39. The breakdown potential is (14kV/mm)(28 cm − 3.6 cm) = 28kV. 40. The capacitance is given by \( C = nC_{0} = nC_{0} A/xd, \) where \( C_0 \) is the capacitance without the dielectric, \( \kappa \) is the dielectric constant, \( A \) is the plate area, and \( x \) is the plate separation. The electric field between the plates is given by \( E = V/d, \) where \( V \) is the potential difference between the plates. Thus, \( d = A \eta/ \varepsilon_0 \). Thus, \[ A = /\text{RE}. \] 41. We assume there is charge +q on one plate and charge −q on the other. The electric field in the lower half of the region between the plates is \( E = \frac{\sigma}{\varepsilon_{0}} \frac{d}{4} \) at 9 \(/2. \) where \( A \) is the plate area. The electric field is then in each direction of each dielectric. Since the field is uniform in each region, the potential difference between the plates is \[ V = \frac{E}_{d}{d} \] \( = 2(1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{\varepsilon}_{0}{\varepsilon}_{0} \varepsilon_{1}}}) + ...] = \xi_{d}( \frac{1}{1 + \frac{1}{\varepsilon}_{0} \varepsilon_{0} = \frac{\kappa \varepsilon_{0}E/d}{d} \frac{1}{1} \frac{dx}{d} \]. \] \[ C \equiv \frac{E_{q2}^{e}}{1} \frac{1}{1 + \frac{ \kappa_{1} + \kappa_{2} }{r} } \] 42. Let \( C_n = \varepsilon_{0} A/(2kx) 2x md = \varepsilon_{0} x \varepsilon \varepsilon_{0} \varepsilon_{0}/ dx \) and \( C_m \), \( C_m \), \( C_n, \) and \( C_{3}^{.} \). Note that \( C_{1}\) and \( C_{3}^{.} \, \) where \( C_x ) = 2(C_n + C_{2}) \, \) \( C_3 \) are effectively connected in series, while \( C_2 \) is effectively connected in parallel with the \( C_{2}, C_{3}, \) containment. Thus, \[ C = \frac{C_{1} C_{2} / 2} {C_{2} - \kappa \varepsilon_{0} A/(dx)k \varepsilon_{0} \varepsilon_{0} A. \] 43. (a) The electric field is the region between the plates is given by \( E = V/d, \) where \( V \) is the potential difference between the plates and \( d \) is the plate separation. The capacitance is given by \( C = \varepsilon_{0} A/c , \) \( where \( A \) is the plate area and \( x \) is the dielectric constant, so \( x = \varepsilon_{0} A/c \) and \[ C = \frac{2x}{\varepsilon \varepsilon_{0} A \varepsilon_{0}/A/ST x/)2, \]. \] C = \frac{V}{RD}.\] (b) The free charge on the plates is \( q = C_0 V = (100 \times 10^{-12}F) (5V) = 5.0 \times 10^{-9} C. \) (c) The electric field is produced by both the free and induced charge. Since the field of a large uniform layer of charge is \( q/2\varepsilon_{0} A, \), the field between the plates is \[ q - q\] = q \varepsilon_{0} = (100 \times 10^{-12}F ) (100 \times 10^{-12}m) q \varepsilon_{0} / \varepsilon_{0}. \] 5.0 \times 10^{-9} C - (8.85 \times 10^{-12} F)(100 \times 10^{-9})^{-1} \varepsilon_{0}. \] 4.1 \times 10^{-9} C = 41 nC. \] where the first term is due to the positive free charge on one plate, the second is due to the negative free charge on the other plate, the third is due to the positive induced charge on one dielectric surface, and the fourth is due to the negative induced charge on the other dielectric surface. Note that the field due to the induced charge is opposite the field due to the free charge, so they tend to cancel. The induced charge is therefore q - = 5.0 \varepsilon_{0} A, 5.0 \times 10^{-9} C - (8.85 \times 10^{-12}F F / m)(100 \times 10^{-9} m)(1.0 \times 10^{-12} V / m) = 4.1 \times 10^{-9} C = 41 nC .
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1. The minimum charge measurable is q_min = CV = (50 pF)(0.15 V) = 7.5 pC . 2. (a) The capacitance of the system is C = q/ΔV = 70 pC / 20 V = 3.5 pF . (b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes ΔV = q/C = 200 pC / 3.5 pF = 57 V . 3. 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 x 10^{-9} F)(12 V) = 30 x 10^{-9} C. 4. We verify the units relationship as follows: [κ0] = F/m = C/V·m = C / N·m/C = C^2 / N·m^2 . 5. (a) The capacitance of a parallel-plate capacitor is given by C = κ0A/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 = πR^2, where R is the radius of a plate. Thus, C = κ0πR^2/d = (8.85 x 10^{-12} F/m)(8.2 x 10^{-2} m)^2 / 1.3 x 10^{-4} m = 1.4 x 10^{-10} F = 140 pF . (b) The charge on the positive plate is given by q = CV , where V is the potential difference across the plates. Thus, q = (1.4 x 10^{-10} F)(12 V) = 1.7 x 10^{-9} C = 1.7 nC. 6. We use C = Aκ0/d. Thus C = (1.00 m^3)(8.85 x 10^{-12} C/V m) / 1.00 m = 8.85 x 10^{-12} F . 7. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4πεκ0R to find the capacitance. When the drops combine, the volume is doubled. It is then V = (4π/3)R^3. The new radius R' is given by 4π/3 (R')^3 = 2 * 4π/3 R^3 / 3 , so R' = 2^{1/3} R / 2^{1/3} = 2^{1/2} R . The new capacitance is C' = 4πεκ0 2^{1/2} R = 5.0κ0 πR . 8. (a) We use Eq. 26-17: C = 4πεἰo a^2 / (b-a) = (40.0 mm)(28.0 mm) / (8.99 x 10^9 N·m^2/C^2) ((40.0 mm - 28.0 mm) = 81.5 pF . (b) Let the areas required be A. Then C = κ0A/d, or A = C(d - a) / κ0 = (8.5 pF)(40.0 mm - 28.0 mm) / (8.85 x 10^{-12} N·m^2/C^2) = 191 cm^2 . 9. According to Eq. 29-17 the capacitance of a spherical capacitor is given by C = κ0 4π ab / (b - a) , where a and b are the radii of the spheres. If b and d are nearly the same then 4πab is nearly the surface area of either sphere. Replace 4πab with A and b - a with d to obtain C = κ0A / d . 10. The equivalent capacitance is C_eq = C_1 C_2/(C_1 + C_2) = (10.0 μF)(5.0 μF) / (10.0 μF + 5.0 μF) = 7.33 μF . 11. The equivalent capacitance is given by C_eq = 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, C_eq = NC, where C is the capacitance of one of them. Thus, N = q/(CV) and N = 1.0 C / (75 V)(1.0 x 10^{-6} F) = 9900 . 12. The charge that passes through meter A is q = C_V-V = 3CV = 3 x (25.0 pF)(420 V) = 0.315 C . 13. The equivalent capacitance is C_eq = (C_1 C_2)/(C_1 + C_2) = (10.0 μF)(5.0 μF)/(4.0 μF + 5.0 μF) = 3.16 μF . 14. (a) and (b) The original potential difference V_1 across C_1 is V_1 = C_1V/C_1 + C_2 = (3.16μF)(10 V) /(10.0 μF + 5.0 μF) = 21.1 V . Thus ΔV_1 = 100 V - 21.1 V = 79 V and Δq_1 = C_1ΔV_1 = (10.0 pF)(79 V) = 7.9 x 10^{-6} C. 15. Let x be the separation of the plates in the lower capacitor. Then the plate separation in the upper capacitor is c - b - x. The capacitance of the lower capacitor is C_L = κ0A/x and the capacitance of the upper capacitor is C_U = κ0A/(b - x), where A is the plate area. Since the two capacitors are in series, the equivalent capacitance is determined from 1 / C_eq = 1 / C_L + 1 / C_U . Thus, the equivalent capacitance is given by C_eq = κ0A/(a - b) - so C_eq is independent of x. 16. (a) The potential difference across C_3 is V_1 = V - V_2, thus, q = C_1V_1 = (10 μF)(10 V) = 1.0 x 10^{-4} C. (b) Let C_1 = 10 μF. We first consider the three-capacitor combination consisting of C_1 and its two closest neighbors, each of capacitance C'. The equivalent capacitance of this combination is C_eq = C_1C_2C_3/(C_1C_2 + C_2C_3 + C_3C_1) = 2/3 C' . Also, the voltage drop across this combination is the same as that across C_1 and those connected in series with it, the voltage difference across C_2 satisfies V_3 = V_2 / V_1/3. Thus q_2 = C_2(1/2) * (3V/2) = 20 x 10^{-9} V . 17. The charge initially on the charged capacitor is given by q = C_1v, where C_1 = 100 pF is the capacitance and V_0 = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor is wired in parallel to the first, the charge on the first capacitor is q_1 = C_1V_1, where v = 3 V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q = q - q1, whereas C_2 is the capacitance of the second capacitor. Substituting Clv0 for q1 and C1v1 for q, we obtain q = C_1(v0 - V). The potential difference across the second capacitor is also V_1, so the capacitance is C_2 = q_1 (v_0 - V) / V = (100 pF) = 3 pF . 18. (i) First, the equivalent capacitance of the two 4.0 μF capacitors connected in series is given by 4.0 μF / 2 = 2.0 μF . This combination is then connected in parallel with two other 2.0-μF capacitors (one on each side), resulting in an equivalent capacitance C = 3(2.0 μF) = 6.0 μF . This is now seen to be in series with another combination, which consists of the two 4.0-μF capacitors connected in parallel (which are themselves equivalent to C = 4(2.0 μF) = 6.0 μF ). 19. (a) After the switches are closed, the potential differences across the capacitors are the same and the two capacitors are in parallel. The potential difference from a to b is given by V_ab = Q/C_eq, where Q is the net charge on the combination and C_eq is the equivalent capacitance. The equivalent capacitance is C_eq = C_1 + C_2 = 4.0 x 10^{-6} F. The total charge on the combination is the net charge on either pair of connected plates. The charge on capacitor 1 is q_1 = C_1V = (1.0 x 10^{-6} F)(10 V) = 1.0 x 10^{-5} C and the charge on capacitor 2 is q_2 = C_2V = (3.0 x 10^{-6} F)(10 V) = 3.0 x 10^{-5} C, so the net charge on the combination is 3.0 x 10^{-5} C - 1.0 x 10^{-5} C = 2.0 x 10^{-5} C. The potential difference is V_ab = 2.0 x 10^{-5} C / 4.0 x 10^{-6} F = 5.0 V. (b) The charge on capacitor 1 is now q_1 = C_1V_ab = (1.0 x 10^{-6} F)(5 V) = 5.0 x 10^{-6} C. (c) The charge on capacitor 2 is now q_2 = C_2V_ab = (3.0 x 10^{-6} F)(5 V) = 1.5 x 10^{-5} C. 20. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q_1 = q_3 = C_3C_1/(C_3 + C_1) = (1.0 μF)(3.0 μF)(12 V) / (1.0 μF + 3.0 μF) = 9.0 μC . Also, capacitors 2 and 4 are in series, so q_2 = q_4 = C_4C_2/(C_4 + C_2) = (2.0 μF)(4.0 μF)(12 V) / (2.0 μF + 4.0 μF) = 16 μC . (b) With switch 2 also closed, the potential difference V_1 across C_1 must equal the potential difference across C_3 and the sum of the potential differences across C_1 and C_2. V_1 = (C_3 + C_4)/(C_1 + C_3 + C_4) * V = (3.0 μF + 4.0 μF)[12 V] / (1.0 μF + 3.0 μF + 4.0 μF) = 8.4 V . Thus, q_1 = C_1V_1 = (1.0 μF)(8.4 V) = 8.4 μC, q_2 = C_2V_1 = (2.0 μF)(8.4 V) = 17 μC, q_3 = (C_3/C_1)V_1 = (3.0 μF)/(2.0 μF) * 12 μC, and q_4 = C_4V - (1.0 μF) * (2.0 μF) V = 14 μC. Express the total capacitance determined from 1 / C_1 = 1 / C_3 + 1 / C_4 . 21. Let V = 1.00 m^3. Using Eq. 26-23, the energy stored is U = 1/2 ε_0(E^2) V = 1/2 ε_0 (8.85 x 10^{-12} C/V·m)(150 V/m)^2(1.00 m^3) = 99.6 x 10^{-9} J . 23. The energy stored by a capacitor is given by U = 1/2 CV^2, where V is the potential difference across its plates. We convert the given value of the energy to Joules. Since a Joule is a watt-second, we multiply by (10^3 W/kW)(3600 s/h) to obtain 10 kW-h = 3.6 x 10^7 J. Thus, C = 2U / V^2 = (2)(3.6 x 10^7 J) / (1000 V)^2 = 72 F . 24. (a) The capacitance is \[ C = \frac {\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2})(40 \times 10^{-4} m^2)}{1.0 \times 10^{-1} m} = 3.5 \times 10^{-11} F = 35 pF. \] (b) \[ q = CV = (35 pF)(600 V) = 2.1 \times 10^{-8} C = 21 nC. \] U = \frac{1}{2} qV = \frac{1}{2} (35 pF)(21 nC)(6^{2}) = 6.3 \times 10^{-8} J = 6.3 pJ. (d) \[ E = V/d = 600V/1.0 \times 10^{-1} m = 6.0 \times 10^{3} V/m. \] (e) The energy density (energy per unit volume) is \[ \frac{U}{Ad} = \frac{6.3 \times 10^{-8} J}{(40 \times 10^{-4}m^{2})(1.0 \times 10^{-1} m)} = 1.6 J/m^{3}. \] 25. The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference \( V \) across the capacitors is the same and the total energy is \[ U = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (2.0 \times 10^{-6} F + 4.0 \times 10^{-6} F) (300 V)^2 = 0.27 J. \] 26. The total energy stored in the capacitor bank is \[ U = \frac{1}{2} C_{total}V^2 = \frac{1}{2} (2000)(0.06 \times 10^{-6} F)(XXXX V)^2 = 1.3 \times 10^{3} J. \] Thus, the cost is \[ \frac{1.3 \times 10^3 J}{3.6 \times 10^6 J/(\text{cent} / \text{kW h})} = 10 \text{ cents}. \] 27. (a) In the first case \[ U = q^{2}/2C \], and in the second case \[ U = q^{2}/2^{\prime} C = q^{2}/4C. \] So the energy is now \[ 4(3.0 J / 2) = 2.0 J. \] 28. (a) The potential difference across \( (b) \) is given by \[ V = \frac{C}{C + C_3} V_0 = \frac{(40.0 pF)(100 V)}{100.0 pF + 3.0 \mu F} = 21.1 V. \] Also, \[ V_2 = V - V_1 = V_3 = 100 V - 21.1 V = 79.9. Thus, \] \[ q_1 = C_1 V_1 = (10.0 pF)(21.1 V) = 2.11 \times 10^{-10} C\] \[ q_3 = C_3 V_3 = (50.0 pF)(79.9 V) = 3. 105 \times 10^{-4} C\] \[ q_{e} = q_h + q_2 = 2.11 \times 10^{-4c} + 1.05 \times 10^{-4} C = 3.16 \times 10^{-5} C. \] (b) The potential differences were found in the course of solving for the charges in part \((a)\). (c) The stored energies are as follows: \[ U_1 = \frac{1}{2} \frac{9}{10.0 pF)(21.1 V)^2 = 2.22 \times 10^{-3} J,\] \[ U_2 = \frac{1}{2} \frac{3.0 \mu F)(21.1 V)^2 = 1.11 \times 10^{-3} J,\] \[ U_3 = \frac{1}{2} \frac{9}{4.00 pF)(78.9 V)^2 = 1.25 \times 10^{-2} J. \] 29. (a) Let \( q \) be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor is given by \( C = \varepsilon_0 A / d \), the charge is \( q = CV = \varepsilon_0 A V / d \). After the plates are pulled apart, their separation is \( 2d \) and the potential difference is \( V' \). Then \( q = \varepsilon_0 A V'/ 2d \) and \( V' = 2d \varepsilon_0 A q = q d A/ \varepsilon_0 = 2 V. \) (b) The initial energy stored in the capacitor is \( U_1 = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d} \) and the final energy stored is \( U_f = \frac{\varepsilon_0 A V^2}{4d} \) and \( U_i - U_f = \varepsilon_0 A V^2 / 2d \). (c) The work done to pull the plates apart is the difference in the energy: \( W = U_i - U_f = \varepsilon_0 A V^2 / 2d. \) 30. (a) The charge in the Figure is \[ q_n = C_n V = (4.00 \mu F)(100 V) = 4.00 \times 10^{-4} mQ ,\] \[ q_n = \frac{C_n V^2}{C_1 + C_2} = \frac{(10.0 pF)(3.00 \mu F)(100 V)}{100 \mu F + 3.00 \mu F} = 3.33 \times 10^{-6} C . \] \[ V_2 = (10 \mu C) /( 10.00 pF) = 333. V, V_3 = V - V_1 = 10 W - 333 V = 67. 7. \], \( V_3 = V - 10W. \) (c) We use \( \, U_1 = 1, 2, 3. \) The answers are \( U_1 = 5.6mJ, U_1 = 11 mJ, \text{ and } U_1 = 20 mJ. \) 31. 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 \(u = \frac{1}{2} \varepsilon_{0} E^{2}\), 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 \[ E = \frac{q}{2 \pi \varepsilon_{0} L r} \] (see Eq. 26-12), and the energy density at that point is given by \[ u = \frac{1}{2} \frac{q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} L^{2} r^{2}} \] The energy in the cylinder is the volume integral \[ U_{L} = \int u \, dV . \] Now, \( dV = 2 \pi r L \, dr \), so \[ U = \int \frac{q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} L^{2}} 2 \pi r L \, dr = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \int_{a}^{b} \frac{dr}{r} = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} . \] To find an expression for the total energy stored in the capacitor, we replace \( R \) with \( b \): \[ U_{b} = \frac{q^{2}}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} . \] We want the ratio \( U_{b} / U_{a} \) to be \(1/2\), so \[ \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} = \frac{1}{2} \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} = \frac{1}{4 \pi \varepsilon_{0} L} \ln \frac{b}{a} \] or, since \( \frac{1}{2} \ln(\frac{b}{a}) = \ln(\sqrt{\frac{b}{a}}), \ln(R/a) = \ln(\sqrt{a b}) \). This means \( R/a = \sqrt{a b} \) or \( R = \sqrt{a b}. \] 32. We use \( E = q/4 \pi \varepsilon L = V / R . \) Thus \[ q = \frac{1}{2 \pi} \frac{(V)}{(R)} = ( \frac{8.85 \times 10^{-12} \frac{C^2}{N \cdot m^{2}}) (8000 \, V \,) \] / (0.050 m) = 0.011 J/m^{3}. 33. The charge is held constant while the plates are being separated, so we write the expression for the stored energy as \( U = q^2/2C. \), where q is the charge and C is the capacitance. The capacitance of a parallel-plate capacitor is given by \( C = \varepsilon_{0} A / x \), where A is the plate area and x is the plate separation, so \[ U = \frac{q^{2}}{2 \frac{\varepsilon_{0} A}{x}} = \frac{q^{2} x}{2 \varepsilon_{0} A}. \] If the plate separation increases by dx, the energy increases by dU := \( (q^{2}/2\varepsilon_{0} A) \) dx. Suppose the agent pulling the plate apart exerts force F. Then the agent does work Fdx and if the plates begin and end at rest, this must equal the increase in stored energy. Thus, \[ Fdx = (q^{2}/2\varepsilon_{0} A) dx \] and \[ F = \frac{q^{2}}{2\varepsilon_{0} A}. \] The net force on a plate is zero, so this must also be the magnitude of the force one plate exerts on the other. The force can also be computed as the product of the charge q on one plate and the electric field \( E_1 \) due to the charge on the other plate. Recall that the field produced by a uniform plane surface of charge is \( E_1 = q/2\varepsilon_{0} A. \). Thus, \[ F = q/2\varepsilon_{0} A. \] 34. If the original capacitance is given by \( C = \varepsilon_{0} A/2d, \) then the new capacitance is \( C' = \varepsilon A/2d. \) Thus \[ C'/C = \varepsilon/\varepsilon_{0} = 2(2.6 pF/1.3 pF) = 4.0. \] 35. The capacitance with the dielectric in place is given by \( C = \kappa C_0 \), where \( C_0 \) is the capacitance before the dielectric is inserted. The energy stored is given by \( U = q^{2}/C = \frac{q^{2} \varepsilon_{0} A/L} \) so \( \kappa = \frac{2U}{C} \) at 4.7. \) According to Table 26-1, you should use Pyrex. 37. The capacitance of a cylindrical capacitor is given by \[ C = \frac{2 \pi \varepsilon_{0} L}{\ln(b/a)} , \] where \( C_0 \) is the capacitance without the dielectric, \( \varepsilon \) is the dielectric constant, \( L \) is the length, \( a \) is the inner radius, and \( b \) is the outer radius. The capacitance per unit length of the cable is \[ \frac{C}{L} = \frac{2 \pi \varepsilon_{0}}{\ln(b/a)} = \frac{2\pi(8.85 \times 10^{-12} F/m)}{\ln(0.9 mm/0.1 mm)} = 8.1 \times 10^{-11} F/m = 81 pF/m . \] 38. (a) We use Eq. 26-14: \[ C = \frac{2\pi \varepsilon_{0} L}{\ln(b/a)} (4.7/10.18 m) = \frac{(2.99 \times 10^{-19} )} / \ln(8.3 cm/3.6 cm) = 0.73 nF . \] 39. The breakdown potential is (14kV/mm)(28 cm − 3.6 cm) = 28kV. 40. The capacitance is given by \( C = nC_{0} = nC_{0} A/xd, \) where \( C_0 \) is the capacitance without the dielectric, \( \kappa \) is the dielectric constant, \( A \) is the plate area, and \( x \) is the plate separation. The electric field between the plates is given by \( E = V/d, \) where \( V \) is the potential difference between the plates. Thus, \( d = A \eta/ \varepsilon_0 \). Thus, \[ A = /\text{RE}. \] 41. We assume there is charge +q on one plate and charge −q on the other. The electric field in the lower half of the region between the plates is \( E = \frac{\sigma}{\varepsilon_{0}} \frac{d}{4} \) at 9 \(/2. \) where \( A \) is the plate area. The electric field is then in each direction of each dielectric. Since the field is uniform in each region, the potential difference between the plates is \[ V = \frac{E}_{d}{d} \] \( = 2(1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{\varepsilon}_{0}{\varepsilon}_{0} \varepsilon_{1}}}) + ...] = \xi_{d}( \frac{1}{1 + \frac{1}{\varepsilon}_{0} \varepsilon_{0} = \frac{\kappa \varepsilon_{0}E/d}{d} \frac{1}{1} \frac{dx}{d} \]. \] \[ C \equiv \frac{E_{q2}^{e}}{1} \frac{1}{1 + \frac{ \kappa_{1} + \kappa_{2} }{r} } \] 42. Let \( C_n = \varepsilon_{0} A/(2kx) 2x md = \varepsilon_{0} x \varepsilon \varepsilon_{0} \varepsilon_{0}/ dx \) and \( C_m \), \( C_m \), \( C_n, \) and \( C_{3}^{.} \). Note that \( C_{1}\) and \( C_{3}^{.} \, \) where \( C_x ) = 2(C_n + C_{2}) \, \) \( C_3 \) are effectively connected in series, while \( C_2 \) is effectively connected in parallel with the \( C_{2}, C_{3}, \) containment. Thus, \[ C = \frac{C_{1} C_{2} / 2} {C_{2} - \kappa \varepsilon_{0} A/(dx)k \varepsilon_{0} \varepsilon_{0} A. \] 43. (a) The electric field is the region between the plates is given by \( E = V/d, \) where \( V \) is the potential difference between the plates and \( d \) is the plate separation. The capacitance is given by \( C = \varepsilon_{0} A/c , \) \( where \( A \) is the plate area and \( x \) is the dielectric constant, so \( x = \varepsilon_{0} A/c \) and \[ C = \frac{2x}{\varepsilon \varepsilon_{0} A \varepsilon_{0}/A/ST x/)2, \]. \] C = \frac{V}{RD}.\] (b) The free charge on the plates is \( q = C_0 V = (100 \times 10^{-12}F) (5V) = 5.0 \times 10^{-9} C. \) (c) The electric field is produced by both the free and induced charge. Since the field of a large uniform layer of charge is \( q/2\varepsilon_{0} A, \), the field between the plates is \[ q - q\] = q \varepsilon_{0} = (100 \times 10^{-12}F ) (100 \times 10^{-12}m) q \varepsilon_{0} / \varepsilon_{0}. \] 5.0 \times 10^{-9} C - (8.85 \times 10^{-12} F)(100 \times 10^{-9})^{-1} \varepsilon_{0}. \] 4.1 \times 10^{-9} C = 41 nC. \] where the first term is due to the positive free charge on one plate, the second is due to the negative free charge on the other plate, the third is due to the positive induced charge on one dielectric surface, and the fourth is due to the negative induced charge on the other dielectric surface. Note that the field due to the induced charge is opposite the field due to the free charge, so they tend to cancel. The induced charge is therefore q - = 5.0 \varepsilon_{0} A, 5.0 \times 10^{-9} C - (8.85 \times 10^{-12}F F / m)(100 \times 10^{-9} m)(1.0 \times 10^{-12} V / m) = 4.1 \times 10^{-9} C = 41 nC .