Fisica 2 - Cap 26 7 Edição

1. The minimum charge measurable is qmin = CVmin = (50 pF)(0.15 V) = 7.5 pC. 2. (a) The capacitance of the system is C = q/V = 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^-6 F)(12 V) = 30 x 10^-6 C. 4. We verify the units relationship as follows: [C] = F/V = Nm/(Vm) = C/V = C^2/Nm^2. 5. (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 = πR^2, where R is the radius of a plate. Thus, C = ε₀πR^2/d = (8.85 x 10^-12 F/m)(8.2 x 10^-2 m)^2/1.3 x 10^-3 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)(20 V) = 17 x 10^-9 C = 17 nC. 6. We use s = At₀/c₀. Thus At₀/c₀ = (1.00 nm)/(s) = (8.85 x 10^-12 C^2/Nm^2)/(μm) = 8.85 x 10^-12 m. Since d is much less than the size of an atom (~ 10^-10 m), this capacitor cannot be constructed. 7. 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 = (4π/3)R^3. The new radius R’ is given by 4π/3 (R’^3) = 2*4π/3*R^3, so R’^2 = 2*1/2*R^2, The new capacitance is C’ = 4πε₀R’ = 4πε₀2^1/2*R = 5.0Rπε₀. 8. (a) We use Eq. 26-17: C = 4πε₀b*a/(b-a) = (40.0 mm)(28.0 mm)/((8.99 x 10^9 Nm^2/C^2 )(40.0 mm - 38.0 mm)) = 51.5 pF. (b) Let the area required be A. Then C = ε₀A/(b - a), or A = C(b-a)/ε₀ = 5.5 pF/((10.0 mm - 38.0 mm) * (8.85 x 10^-12 Nm^2/C^2)) = 191 cm². 9. According to Eq. 26-17 the capacitance of a spherical capacitor is given by C = 4πε₀ab/b - a, where a and b are the radii of the spheres. If a and b are nearly the same then 4πε₀ab is nearly the surface area of either sphere. Replace 4πε₀ with A and b - a with d to obtain C = ε₀A/d. 10. The equivalent capacitance is Ceq = C1 + C2*C3/C1 + C3 = 4.00 pF + (10.0 pF)(5.00 pF)/10.0 pF + 5.00 pF = 7.33 pF. 11. 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/C = 1.00 C/(10.0 V)(1.0 x 10^-6 F) = 9090. 12. The charge that passes through meter A is q = CΔV = 3CV = 3(25.0 pF)(420 V) = 0.315 C. 13. The equivalent capacitance is Ceq = C1 + C2*C3/C1 + C3 = (10.0 pF + 5.00 pF)(4.00 pF)/10.0 pF + 5.00 pF = 3.16 pF. 14. (a) and (b) The original potential difference V1 across C1 is V1 = C2/C1 + C3*V = (3.16 pF)(10 V)/(10.0 pF + 3.00 pF) = 21.1 V. Thus ΔV1 = 100 V - 21.1 V = 79 V and Δq1 = C1ΔV1 = (10.0 pF)(79 V) = 79 x 10^-9 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 C2 = ε₀A/x and the capacitance of the upper capacitor is C3 = πA/(c - b - x), where A is the plate area. Since the two capacitors are in series, the equivalent capacitance is determined from 1/Ceq = 1/C2 + 1/C3 = x/ε₀A + (a - b - x)/ε₀A. Thus, the equivalent capacitance is given by Ceq = ε₀A/(a - b) and is independent of x. 16. (a) The potential difference across C1 is V1 = eV. Thus, q = C1V1 = (10.0 pF)(10 V) = 1.0 x 10^-6 C. (b) Let C = 10 μF. If we first consider the three-capacitor combination consisting of C1 and its two closest neighbors, each of capacitance C/, the equivalent capacitance of this combination is Ceq = C1 + C/C1 = 1.5 C. Also, the voltage drop across this combination is V = C1 + C1 = CS/V. Since the voltage difference is divided equally between C2 and the one connected in series with it, the voltage difference across C2 satisfies V2 = V/2 = V1/3. Thus q2 = C2V2 = (10 pF)(10 V) = 2.0 x 10^-9 C. 17. The charge initially on the charged capacitor is given by q₀ = C1V₀, where C1 = 100 pF is the capacitance and V₀ = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor is joined parallel to the first, the charge on the first capacitor is q₁ = q₀V₂/V₀, where q = 3 V is the new potential difference. Since charge is conserved in the process, the charge on the second capacitor is q₂ = q₀ - q₁, where C₂ is the capacitance of the second capacitor. Substituting C1V₀ for q₀ and C1V for q₁, we obtain q = C₂(Vo - V/). The potential difference across the second capacitor is V/0, so the capacitance is C₂ = V₂ - V₀/C1V₀ = 20 V/ - 3V/100 pF = 3 pF. 18. (a) First, the equivalent capacitance of the two 4.0 μF capacitors connected in series is given by 1/4.0F/2 = 3.0 μF. This combination is then connected in parallel with two other 2.0 μF capacitors (one on each side), resulting in an equivalent capacitance C₂ = 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 C₂ = 23(0.3 pF) = 6.0 μF). Thus, the equivalent capacitance of the circuit is Ceq = C*/(C + C*) = (6.0 μF)(20 μF)/6.0 μF + 20 μF = 30 pF. (b) Let V = 20 V be the potential difference supplied by the battery. Then q = C₂*Vo = (3.0 μF)(20 V) = 30 x 10^-6 C. (c) The potential difference across C3 is given by q3 = CV = (6.0 μF)(20 V) = 3(20pF)(4.0 pF)/6.0 pF. and the charge carried by C1 is q₁ = q₁ = C3(V₀ - V) = 3 (3.8 pF)(10 V) = 3.0 x 10^-6 C. (d) The potential difference across C2 is given by V₂ = Vo - V₁ = 2(V - V)/V = 0.0 V. Consequently, the charge carried by C2 is q₂ = C2V₂ = (20 μF)(10 V) = 2.0 x 10^-5 C. (e) Since the voltage difference V₃ is divided equally between C2 and the two other 4.0 μF capacitors connected in series with it, the voltage difference across C3 is given by V₃ = V/2 = 10 V/2 = V/2 = 5.0 V. Thus, q3 = C/L*V₂ = (4.0 μF/4.0 μF)(10 V)/(2.0 x 10^-3 C - 10^-2 C - 12), Thus, q3 = (f) 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 Vab = q/Cab, where Cq is the net charge on the combination and Cq is the equivalent capacitance. The equivalent capacitance is Ceq = C1 + C2 + 6.0 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 q₁ = CV₁ = (1.0 x 10^-6 F)(10 V) = 1.0 x 10^-4 C and the charge on capacitor 2 is q₂ = CV² = (3.0 x 10^-6 F)(10 V) = 3.0 x 10^-6 C, so the net charge on the combination is 3.0 x 10^-4 C - 1.0 x 10^-4 C = 2.0 x 10^-4 C. The potential difference is Vab = 2.0 x 10^-4 C/4.0 x 10^-5 C = 50 V. (b) The charge on capacitor 1 is now q₁ = C₁Vab = (1.0 x 10^-6 F)(50 V) = 5.0 x 10^-6 C. (c) The charge on capacitor 2 is now q₂ = C₂Vab = (3.0 x 10^-6 F)(50 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: q1 = q3 = C/(C1 + C3)(C3)(12 V)/1.0 μF + 3.0 μF = 9.0 μC. Also, capacitors 2 and 4 are in series: q2 = q4 = C/(C2 + C4V) = 20pF)(4.0 μF)(12 V)/2.0 μF + 4.0 μF = 1.8 μC. (b) With switch 2 also closed, the potential difference V1 across C1 must equal the potential difference across C3 and so be V1 = C/(C1 + C + C3V) = (3.0 pF)(4.0 pF)(12 V)/(1.0 μF + 2.0 pF)+3.0 μF = 8.4 V. Thus, q1 = C1V1 = (1.0 μF)(8.4 V) = 8.4 μC, q3 = q, and q = (C/(C3V) = (20 pF)(4.0 pF)(17 V)/(1.0 pF)(10 V/(2.8 V - 8.4 V)) = 11.0 μC, and q1 = V + V1) = (V + 12) + 8.4 V/(17 dV). 21. The charges on capacitors 2 and 3 are the same and have opposite signs may be replaced by an equivalent capacitance determined from 1/2 = 1/C1 + 1/C2 + C3(level + Q)(level C4 level)/level C1 + C1/(C2,(1) = 2pF) + C3C1 +. Thus, Ceq, = C/C³/(C3 + C⁴). The charge on the equivalent capacitor is the same as the charge on either of the two capacitors in the combination and the potential difference across the equivalent capacitor is 1 across by q/C². The potential difference across capacitor 1 is q1/C1, where q1 is the charge on this capacitor. The potential difference across the combination of capacitors 2 and 3 must be the same as the potential difference across capacitor 1, so q1/C1 = q²/q.. Now some of the charge originally on capacitor 1 flows to the combination of 2 and 3, ¶ is the original charge, conservation of charge yields q1 + q1 = C1V/o. Where Y, is the optional potential difference across capacitor 1. Solving the two equations q₃ C3 + S1; S1 + q1 = q₁. for q1 and q2, we find q1 = C3V/o - q1 and q1 = C1V/o - q1 q1/C1 + C1 q/C3 + C1V/o/+ The charges on capacitors 2 and 3 are q2/C1V/o. = C1V/o = C1V/o = C1V/o/C3 + C3C3/level^3 22. Let V = 1.00 m^3. Using Eq. 26-23, the energy stored is U = w²/2ε = 1/+2 + S²/+ ε = (8.85 x 10^-12 F)(150 V/m)?(1.00 m²) = 9.96 x 10^-5 J. 23. The energy stored by a capacitor is given by U = 1/2C², 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^(8 W)/kW)(206W/h) to obtain 10 kW-h = 3.6 x 10^7 J. Thus, C = U/½, V² (15.0 mF) (15.0) (100 V)= U/V (20.0)(100) (3.0) (100 V) = 72 F. 44. (a) The electric field E_i in the free space between the two plates is E_i = q/ε_0 A while that inside the slab is E_2 = E_1/κ = q/κε_0 A. Thus, V_b = E_1(a - b) + E_2b = q (ε_0 A) (a - b) + (κε_0 A) b \ and the capacitance is \ C = q/V_b = κε_0 A (ε_0 A)(a - b) + κb \ = (8.85 x 10^-12 m^2) (8.85 x 10^-12 m^2) \ [2.61 (0.00211 m - 0.000750 m) + (0.000750 m)] \ \ = 13.4 pF . \ (b) q = CV = (13.4 x 10^-12 F)(85.3 V) = 1.15 nC . \ (c) The magnitude of the electric field in the gap is \ E_1 = \ q/(ε_0 A) = (8.85 x 10^-12 m) \ [2.61 (0.00211 - 0.000750) m] \ = 1.15 x 10^-9 C = 1.13 x 10^5 N/C \ (d) Using Eq. 26-32, we obtain \ E_2 = E_1/κ = 2.61 \ \ \ n \ 252 \ \ E_2 = \ n 2.61 \ \ n \ 2 \ \ n / κ = \ n 2.61 \ \ q = q = q/(εκ_0 A)\ q/(εκ_0 A) = (εκ)q/(εκ_0 A) \ \ \ q = q = q/κε_0 A \ \ (εκ) \ n \ 2 \ \ n \ 45. (a) According to Eq. 29-17 the capacitance of an air-filled spherical capacitor is given by \ C = 4πκε_0 ab \ (b - a) \ \ (εκ) \ q = \ q \ in the free space. Thus,\ \ (εκ) = \ q_(1 - \ κ) - 2.61 \ \ n \ 2.61 \ q = \ in \ the \ free \ space. \ b - a \ \ \ (εκ) \ q = \ q = \ q \ (εκ_0 A) \ \ q \ n \ (εκ) \ b - a \ \ \ in \ the \ plate. \ a = \ 78 \ ks - 1 \ \ (εκ) \ n \ 2 \ \ n \ (εκ) \ q = \ q = \ q \ (εκ_0 A) \ \ q \ in the free space. \ a = a \ q \ (εκ_0 A) \ \ b - a \ \ \ (εκ) \ a = a \ q \ (εκ_0 A) \ (εκ) \ a = a \ 315 \ (εκ) = 1 ms \ \ \ (εκ) \ q - \ a κ \ \ b - a \ \ \ (εκ) \ a \ κ \ \ b - a \ \ \ (εκ) \ q = q = q \ \ (εκ_0 A) \ \ a = a \ q \ \ (εκ_0 A) \ (εκ) = 1 ms \ \ \ (εκ) \ q = q = \ b - a \ प्रदर्शन \ \ इंस्ट्रक्शन \ \ " & \ विडियो \ फिंगर \ \ n \ एफ \ उफ़ \ में \ q \ \ = \ का \ q \ \ n \ (εκ) \ n \ q = \ q = q \ q = \ q \ q \ \ E_0 \ (εκ_0 A) \ \ n \ q \ \ (εκ) = (q - q) \ (εκ)\ \ \ 3 \ n \ (εκ_0 A) . \ \ (εκ_0 A) \ n \ q \ \ (εκ_0 A) \ \ n \ q \ \ (εκ_0 A) \ \ n \ q \ \ (εκ_0 A) \ \ n \ q \ \ (εκ_0 A) .\ 45. (a) According to Eq. 29-17 the capacitance of an air-filled spherical capacitor is given by \ C= 4πκε_0 ab \ (b - a) \ \ Thus, \ q \ equals \ \ Incapacity of parallel-plate capacitor is \ \ therefore, \ a =0. \ 46. (a) We apply Gauss’s law with dielectric: q/ε_0 = κΕA, and solve for κ: \ q/ε_0 = 1/(8.85 x 10^-12 )(2.5 x 10^-4 \ (4.1 x 10^-4)/(11))(10)(8.4 x 10^-8 F/m) \ \ (b) The charge induced is \ q1 = κ_0 q1 \ (1 - κ) q1 = (1 - ε) q_static \ (1/κ - 1) . q_static = κ_static \ 47. Assuming the charge on one plate is +q and the charge on the other plate is -q, we find an expression for the electric field in each region, in terms of q, then use the result to find an expression for the potential difference V between the plates. The capacitance is \ C= κε_0 Α/d \ \ The electric field in the dielectric is E′= q/κε_0 Α, where κ is the dielectric constant and A is the plate area. Outside the dielectric (but still between the capacitor plates) the field E in free space is therefore uniform in each region so the potential difference across the plates is \ V = E_unblocked +(E’ - E_bound). (κΕ/d - qd/κΕ (11)) \ \ (κ*/d - (8.9 x 10^-4)\ \ The capacitance is \ q = κε_0 Α/d+ɸ_ic \ q/κε_0 Α =(εκ_static) \ (Conversion) \ \ n \ this \ outside \ k \ area \ thus \ ++) \ Expression for the potential \ V = κε_0_A/κ + q/lə \ \(un) \ Conversion \ \ n \ q/d \ d \ outside \ q/q-capacitor \ \ = \ outside \ के \ \ Thus \ unexpressed \ therefore \ outside \ Unmeaning \ for \ \ _sc \ \ (once) \ k \ outside \ 48. (Procedure) . \ Once \ outside \ once \ central \ outside \ Thus \ \ once \ unrepresented \ increase \ Incremental \ requisite \ \ widespread \ Equivalent \ therefore \ therefore \_OS \ \ Thus \ _OS \ broadway \ Static \ Thus \ or \ indifference \ _OS \ once \ process \ Procedure \ procedure \ indicator \\