Fisica 2 - Cap 26 7 Edição

1. The minimum charge measurable is qmin = CVmin = (50 pF)(0.15V) = 7.5 pC. 2. (a) The capacitance of the system is C = qV = 70 pC20V = 3.5 pF, (b) The capacitance is independent of q; it is still 3.5 pF, (c) The potential difference becomes ΔV = qC = 200 pC3.5 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 10-12 F)(120 V) = 3.0 x 10-9 C. 4. We verify the units relationship as follows: [q] = F V = C V = C (N m)(C) = N mC2 N mC2. 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 = πR2, where R is the radius of a plate. Thus, C = ε0πR2d = (8.85 x 10-12 F/m)(8.8 x 10-3 m)21.3 x 10-3 m = 1.4 x 10-12 F = 1.4 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 10-12 F)(120 V) = 1.7 x 10-10 C = 0.17 nC. 6. We use C = Ae0/d. Thus d = Ae0 C = (1.0 m2) (8.85 x 10-12 CN2)(1.00 F) = 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 = 2(4/3)πR3. The new radius R' is given by 4π3 (R')3 = 2 4π3R3, so (R')3 = 2(R)3, The new capacitance is C' = 4πε0R' = 4πε02 1/3R = 5.0 cm/m. 8. (a) We use Eq. 26-17: C = 4πε0 ab b - a = (40.0 mm)(28.0 mm)(8.99 x 109 N m2/C2)(40.0 mm - 28.0 mm) = 51.5 pF. (b) Let the areas required be A. Then C = ε0A/(c - d), or A = C(c - d)ε0 = (51.5 pF)(0.6 m - 0.2 m)(8.85 x 10-12 Nm2/C2) = 191 cm2. 9. According to Eq. 26-17 the capacitance of a spherized 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 4πab is nearly the surface area of either sphere. Replace 4πab with A and b - a with d to obtain C = ε0Ad . 10. The equivalent capacitance is Ceq = C1 + C2 C1 + C2 = 4.00 pF (10.0 pF)(5.00 pF)10.0 pF + 5.00 pF = 7.33 pF. 44. (a) The electric field E1 in the free space between the two plates is E1 = q / ε0A while that inside the slab is E2 = E1 / κ = q / κε0A. Thus, V0 = E1 (d - b) + E2 b = ( q / ε0A ) ( d - b + b / κ ) and the capacitance is C = q / V0 = qε0A / q ( d - b + b / κ ) = (8.85 x 10^-12 m^2) (115 x 10^-4m) / ( (26.0μm) (0.012m) + 0.00750m) + (0.00700m) ) = 1.34 μF (b) q = CV = (13.4 x 10^-12 F)(3.5V) = 1.15nC. (c) The magnitude of the electric field in the gap is E1 = q / ε0Ad = 1.15 x 10^-9 C / (8.85 x 10^-12 m^2) (115 x 10^-4 m^2) = 1.13 x 10^3 N/C, (d) Using Eq. 26-32, we obtain E2 = E1 / κ = 1.13 x 10^3 N/C / 2.61 = 4.32 x 10^3 N/C. 45. (a) According to Eq. 29-17 the capacitance C0 of an air-filled spherical capacitor is given by C0 = 4πε0 ab / b - a When the dielectric is inserted between the plates the capacitance is greater by a factor of the dielectric constant κ. Consequently, the new capacitance is C = 4πε0κ ab / b - a (b) The charge on the positive plate is q = CV = εκ0 ab / b - a V. (c) Let the charge on the inner conductor to be -q0. Immediately adjacent to it is the induced charge + q', where q' is found to be less by a factor 1/κ than the field when no dielectric is present, then q' = q' - q/κ. Thus, q - q' = -q + q/k 46. (a) We apply Gauss's law with dielectric: q / κ = κε0A, and solve for q: q = κε0A q / κ = (8.9 x 10^-7 C) / (1.4 x 10^-4 m) (100 x 10^-4 m^2) = 7.2 a, (b) The charge induced is q' = q( 1 - 1/κ ) = (8.9 x 10^-7 C) (1 - 1/2.8) = 7.7 x 10^-7 C. 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 = q / V . The electric field in the dielectric is E1 = q / κAε0, where κ is the dielectric constant and A is the plate size. Outside the dielectric (but still between the capacitor plates) the field is E2 = q / Aε0 . The field is uniform in each region so the potential difference across the plates is V = E1 (b) + E2 (d-b) = qbd / κAε0 + ( qd-b ) / Aε0 q / κAε0 + ( kq(d-b) ) / κ Aκ0ε0 κA - kAε0 ) The capacitance is C = κε0A / (d - b/κ + b/κ) . . The result does not depend on where the dielectric is located between the plates; it might be touching one plate or it might have a vacuum gap on each side. . . For the capacitor of Sample Problem 28-8, κ = 2.61, A = 115cm^2 = 115 x 10^-4m^2, d = 1.24cm = 1.24 x 10^-3 m, and b = 0.78 cm = 0.78 x 10^-3m, C =καε0Δ / ( κ ( 115 x 10^-4 / m ) ( 115 x 10^-4 m^2 ) / ( (26.0μm) (0.012m) + 0.00750m ) 1.34 x 10^-11 m^2 ) = 1.34 x 10^-11 x 10^-3 F = 1.34 pF, in agreement with the result found in the sample problem. If b = 0 and κ = 1, then the expression derived above yields C = κε0A / d, the correct expression for a parallel-plate capacitor with no dielectric. If b = d, then the derived expression yields C = κε0A / d, the correct expression for a parallel-plate capacitor completely filled with a dielectric. 48. (a) Eq. 26-22 yields Uc = 1/2 (κε0)^2 V ^2 = (200V x 10^-12 F ) / (7.0 x 10^9 V^2) = 4.9 x 10^-9 J. (b) Our result from part (a) is much less than the required 120mJ, so such a spark should not have set off an explosion.