Fisica 2 - Cap 26 7 Edição

1. The minimum charge measurable is q_{min} = CV_{min} = (50 \text{pF})(0.15 \text{V}) = 7.5 \text{pC} . 2. (a) The capacitance of the system is C = \frac{q}{\Delta V} = \frac{70 \text{pC}}{20 \text{V}} = 3.5 \text{pF} . (b) The capacitance is independent of q; it is still 3.5 \text{pF} . (c) The potential difference becomes \Delta V = \frac{q}{C} = \frac{200 \text{pC}}{3.5 \text{pF}} = 57 \text{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 \times 10^{-6} \text{F})(20 \text{V}) = 3.0 \times 10^{-4} \text{C} . 4. We verify the units relationship as follows: \Bigg[\frac{\cdots}{P}\Bigg] = \frac{\frac{C^{2}}{N m}}{\frac{C}{m}} = \frac{C}{Vm} = \frac{F}{Nm(\cdots)/cm(m)} = \frac{C^{2}}{N \cdots} . 5. (a) The capacitance of a parallel-plate capacitor is given by C = \varepsilon_0 A/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 = \pi R^2, where R is the radius of a plate. Thus, C = \frac{\varepsilon_0 R^{2}}{d} = \Bigg(8.85 \times 10^{-12} \bigg) (8.82 \times 10^{-7})(m^{2}/m) \Bigg) = 1.4 \times 10^{-10} \text{F} = 14.0 \text{pF} . (b) The charge on the positive plate is given by q = CV, where V is the potential difference across the plates. Thus, q = (1.40 \times 10^{-10} \text{F})(20 \text{V}) = 1.7 \times 10^{-3} \text{C} = 1.7 \text{mC} . 6. We use t = A\varepsilon_0/d\pi. Thus, \frac{A\varepsilon_0}{C} = \Bigg(1.00\bigg) \Bigg(8.85 \times 10^{-9} \bigg) \bigg) = 1.0 \Big(t = \frac{8.85 \times 10^{-12} \bigg)}. Since d is much less than the size of an atom (~10^{-10} \bigg), this capacitor cannot be constructed. 7. Assuming conservation of volume, we find the radius of the combined spheres, then use C = 4\pi \varepsilon_0R to find the capacitance. When the drops combine, the volume is doubled. It is then V = 2(4\pi/3)R^{3}. The new radius R is given by \frac{4}{3} \pi R^3 = \frac{2}{3} \pi R^3 ,\not k = 1\equiv 1^{-\cdot 3}. The new capacitance is C' = 4\pi \Bigg(R^{2} - \frac{4\pi}{3R}. \not C^{2} = \frac{3.0 \times 10^{-8}}{2} \frac{4 \times 10^{-8}$ \left|C_s^2/r) is like for 1 cm diameter as \Bigg(\frac{1}{R_{2}}\Bigg is about 50\text{mF}). 8. We use Eq. 26-17: \cdots \cdots C = 4 \pi \varepsilon_0 \Bigg(\frac{d_{\cos 8}}{\cdots} \Bigg) \bigg) \text{cm}) \left(C_{eq}, \frac{\varepsilon_0}{}. \text{cm}^{3}. \Bigg(\pi^2 \left(\times 40 \text{Wa}) \right) 9. According to Eq. 29-17 the capacitance of a spheroidal capacitor is given by C = \frac{\frac{\cdots}{v}{V}}{2,v_{\cdots} = v{d}{f}_{ds} where a and b are the radii of the spheres. If a and b are nearly the same then \Delta r_{a\rightarrow a} is nearly the surface area of aether sphere. Replace \Delta r_{a\rightarrow b} with A and b = d \rightarrow d to obtain C = \frac{\varepsilon_0 A}{d}. 10. The equivalent capacitance is C_eq = \frac{\frac{C_1 \cdots}{C_{3}}_{\frac{2}{ ``|\right)}}C_{2}(2) \see, \frac{C_1}{ \mathbf{c} \textrm{like in C} \Bigg(\frac{\cdotsD.}{\cdots} \textrm{in }\Bigg)``\cdots \begin{pm}\ " break}\equiv, \Bigg) 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, NC = q/V and N = \frac{q}{V \times C} = \frac{1.00 \text{C}}{(10.0 \text{V})(1.00 \times 10^{-4} \text{F})} = 9500 . 12. The charge that passes through meter A is q = C_{A}V = 3CV = 3(25.0 \text{pF})(220 \text{V}) = 0.215 \text{C} . 13. The equivalent capacitance is C_eq = \frac{(C_1+C_2)C_3}{C_1+C_2+C_3} = \frac{(10.0 \mu \text{F} + 5.00 \mu \text{F})(4.00 \mu \text{F})}{10.0 \mu \text{F} + 5.00 \mu \text{F} + 4.00 \mu \text{F}} = 3.16 \mu \text{F} . 14. (a) and (b) The original potential difference V_i across C_1 is V_i = \frac{C_{1}V}{C_{1}+C_{2}} = \frac{(3.16 \mu \text{F})(10 \text{V})}{10.0 \mu \text{F} + 3.00 \mu \text{F}} = 2.11 \text{V}. Thus, \Delta V_1 = 100 \text{V} - 2.11\text{V} = 79 \text{V} and \Delta q_1 = C_1 \Delta V_1 = (10.0 \mu \text{F})(79 \text{V}) = 7.9 \times 10^{-4} \text{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 = \varepsilon_0 A/x and the capacitance of the upper capacitor is C_u = \varepsilon_0 A/(c - b - x), where A is the plate area. Since the two capacitors are in series, the equivalent capacitance is determined from \frac{1}{C_eq} = \frac{1}{C_l} + \frac{1}{C_u} = \frac{x}{\varepsilon_0 A} + \frac{c - b - x}{\varepsilon_0 A} = \frac{a - b}{\varepsilon_0 A} - \frac{a - b}{\varepsilon_0 A}. Thus, the equivalent capacitance is given by C_eq = \varepsilon_0 A/(a - b) and is independent of x. 16. (a) The potential difference across C_1 is V_i = 10 \text{V}. Thus, q_1 = C_1 V_i = (10 \text{pF})(10 \text{V}) = 1.00 \times 10^{-4} \text{C}. (b) Let C = 10 \mu \text{F}. 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 + C \Bigg(\frac{C}{C+C} \Bigg) = 1.5C . Also, the voltage drop across this combination is V = \frac{V_1}{C+C} = \frac{V_1}{C+1} . Since this voltage difference is divided equally between C_2 and the one connected in series with it, the voltage difference across C_2 satisfies V_2 = V/2 = V_1/3. Thus q_2 = C_2 V_2 = \Bigg( 10 \mu \text{F} \Bigg) \Bigg( \frac{1.00 \text{V}}{3} \Bigg) = 2.0 \times 10^{-4} \text{C} . 17. The charge initially on the charged capacitor is given by q = C_{1}V_0, where C_{1} = 100 \text{pF} is the capacitance and V_0 = 50 \text{V} is the initial potential difference. After the battery is disconnected and the second capacitor is placed parallel to the first, the charge on the first capacitor is q_1 = C_{1}(V_0 - V) and the second charge is q = q_1. Since charge is conserved in the process, the charge on the second capacitor is q = q_1 - q_n, where C_2 is the capacitance of the second capacitor. Substituting C_{1}V_0 for q and C_{1}V for q_1, we obtain q = C_{1}(V_0 - V). The potential difference across the second capacitor at a half V_0, so the capacitance is C_2 = \frac{V_{0} - V_{1}}{V_{2}} C_1 = \frac{50 \text{V} - 38 \text{V}}{100 \text{pF}} = 3 \text{pF} . 18. (a) First, the equivalent capacitance of the two 4.0 \mu \text{F} capacitors connected in series is given by \frac{1}{4.0}\text{pF} = 2.0\text{pF}. This combination is then connected in parallel with two other 2.0\mu \text{F} capacitors (one on each side), resulting in an equivalent capacitance C_{eq} = (3/2)(20.0\text{pF}) = 6.0\text{pF}. This is now seen to be in series with another combination, which consists of the two 4.0\mu \text{F} capacitors connected in parallel (which are themselves equal to C_{eq} = (2\times30\text{pF}) = 6.0\text{pF}). Thus, the equivalent capacitance of the circuit is C_eq = \frac{CC'}{C + C'} = \frac{(6.0 \mu \text{F})(20 \text{V})}{1.0 \text{pF} + 3.0 \text{pF}} . (b) Let V = 20 \text{V} be the potential difference supplied by the battery. Then q = C_q V = (3.0\text{pF})(20 \text{V}) = 6.0 \times 10^{-5} \text{C}. (c) The potential difference across C_1 is given by q_1 = C_1 V_1 where q_1 = \frac{C_{1}V}{C_{1} + C_{2}} = \frac{(6.6 \mu \text{F})(20 \text{V})}{6.0 \mu \text{F} + 6.0 \mu \text{F}} = 6.0 \times 10^{-5} \text{C} . and the charge carried by C_1 is q_1 = C_1 V_1 = (3.0\text{pF})(20\.V) = 3.00 \times 10^{-5} \text{C} . (d) The potential difference across \text{C_3} is given by V_3 = V_1 - V_2 = 25 \text{V} - 10 \text{V}. Consequently, the charge carried by C_3 is q_3 = C_3V_3 = (6.0\text{pF})(10 \text{V}) = 20 \times 10^{-5} \text{C} . (e) Since this voltage difference V_3 is divided equally between C_3 and the other 4.0\mu \text{F} capacitor connected in series with it, the voltage difference across C_3 is given by V_3 = V_2/2 = 10 \text{V}/2 = 5.0 \text{V}. Thus, q_3 = C_3V_3 = (4.0\text{pF})(10 \text{V}) = 2.0 \times 10^{-5} \text{C} . 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_3 + C_4 + 6.0 \times 10^{-4} \text{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_{1} V = \Big(1.0 \times 10^{-6} \text{F}\Big)(100 \text{V}) = 1.0 \times 10^{-4} \text{C} and the charge on capacitor 2 is q_2 = C_{2} V = \Big(3.0 \times 10^{-6} \text{F}\Big)(100 \text{V}) = 3.0 \times 10^{-4} \text{C} , so the net charge on the combination is 3.0 \times 10^{-4} \text{C} - 1.0 \times 10^{-4} \text{C} = 2.0 \times 10^{-4} \text{C}. The potential difference is V_{ab} = \frac{2.0 \times 10^{-4} \text{C}}{4.0 \times 10^{-6} \text{F}} = 50 \text{V} . (b) The charge on capacitor 1 is now q_1 = C_{1} V_{ab} = \Big(1.0 \times 10^{-6} \text{F}\Big)(50 \text{V}) = 5.0 \times 10^{-6} \text{C} . (c) The charge on capacitor 2 is now q_2 = C_{2} V_{ab} = \Big(3.0 \times 10^{-6} \text{F}\Big)(50 \text{V}) = 1.5 \times 10^{-4} \text{C} . 20. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q_1 = \frac{C_2 C_3}{C_1 + C_3} V = \frac{(10\text{pF})(30 \text{pF})(12 \text{V})}{1.0 \text{pF} + 3.0 \text{pF}} = 9.0 \mu \text{C} . Also, capacitors 2 and 4 are in series: q_2 = q_4 = \frac{C_2 C_4}{C_2 + C_4} V_1 = \frac{(20 \text{pF})(4.0 \text{pF})(12 \text{V})}{2.0 \text{pF} + 4.0 \text{pF}} = 6.0 \mu \text{C} . (b) With switch 2 also closed, the potential difference V_1 across C_1 must equal the potential difference across C_3 and C_4; therefore, V_i = \frac{C_1}{C_1 + C_3 + C_4} V_{i} = \frac{(3.0 \text{pF})(12 \text{V})}{1.0 \text{pF} + 2.0 \text{pF} + 3.0 \text{pF}} = 8.4 \text{V} . Thus, q_1 = C_{1} V_1 = \Big(1.0 \mu \text{F}\Big)(8.4 \text{V}) = 8.4 \mu \text{C} . q_2 = C_{2} V_2 = \Big(20\text{pF}\Big)(8.4\text{V}) = 16.7 \text{C} = 17 \mu \text{C} , \qquad q = (C - C_1) V = \Big(3.0\text{pF})(12.0 - 8.4 \text{V}) = 1\mu \text{C}, and q = C_{2} V = (25.0\text{pF})(12.0 - 8.4 \text{V}) = 4.8 \mu \text{C} . 21. The charge on capacitors 2 and 3 are the same because these capacitors may be replaced by an equivalent capacitance determined from \frac{1}{C_{eq}} = \frac{1}{C_2} + \frac{1}{\frac{C_2 C_4}{C_3 + C_4}} + \frac{C_4 C_5}{C_3} . Thus, C_{eq} = \frac{C_2 C_3}{(C_3 + C_4)}. 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 given by q/C_{eq}. The potential difference across capacitor 1 is q_1 = q_1/C_1, where q_1 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 q_1/C_1 = q_p/(C_3 + C_4). Now some of the charge originally on capacitor 1 flows to the combination of 2 and 3. If q_o is the original charge, conservation of charge yields q_1 + q_p = q = C_1 V_0, where V_0 is the original potential difference across capacitor 1. Solving the two equations \frac{q_1}{C_1} = \frac{q_p}{C_2} + \frac{q_o}{C_1} = \frac{q_o}{2} = C_{1} V_0 is equivalent to replacing the two equations q_1 = C_1 V_o - q_s and \Bigg(\frac{q}{C_0}\Bigg) = \Bigg(\frac{q_r}{C_2} + \frac{q_4}{C_1}\Bigg) \Bigg(\frac{C_2 C_3 C_4}{(C_3 + C_4)(C_5 + C_6) + C_5 C_6}\Bigg) = \frac{C_2 C_3 C_4}{C_{4, 6 + C_{5, 8}}} \left(C_3 C_4 C_5 C_4 C_5 C_6 \right). The charges on capacitors 2 and 3 are q_2 = q_3 = C_3 V_o - q_{s, i} = C_1 V_o - q_{q, i} = \frac{C_2(1)C_4(V_o)}{C_o(C_o+C_1)} = \left(C_{3}(1)\left(C_4 C_4 C_5 C_6 \right) \left(C_5 + C_4 C_6 \right) + C_o C_5\right) q_2 \Bigg(\frac{C(C_c) C}{C_c C_{3, a} C_{4, b} C_{5, c} C_{4, d}\right)}. 22. Let V = 1.00 \times 10^{-3} \text{V}. Using Eq. 26-23, the energy stored is U = \frac{w^{2}}{2\omega_{r}^{2}} = \frac{\varepsilon_{0} \varepsilon_{r}}{2 \times 10^{-3}} \frac{\text{N/m}^{2}}{ ext{N/m}^{2}} \left(150 \text{kWh}/(m^{3})^{1}(10\text{m}^{3}) \right) = 9.96 \times 10^{3} \text{J} . 23. The energy stored by a capacitor is given by U = \frac{1}{2}CV^{2}, where V is the potential difference across its plates. We convert the given value of the energy to kilojoules. Since a Joule is a watt-second, we multiply by \Bigg(1 \times 10^{-4}\text{kWh}/W) \Bigg((1/ \omega_{L} ) (\text{kWh})/2000 \text{J}) \bigg(10\text{kW} = 3.6 \times 10^{7}\text{J} \bigg(2.0 \times 10^{3}). Thus, \Bigg(\frac{30}{2} \Bigg(3.6 \times 10^{-8} \Bigg) \frac{30}{100} \Bigg(10^{2}(J)} = 72 F. 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/K = q/(κε0A). Thus, q Vg = E1 (d - b) + E2b = ------- (d - b + - b ) ε0A K and the capacitance is q κκε0A C = ------- = --------------- Vg q κκε0A (8.85 x 10^-12 F/m) (115 x 10^-4m)(2/261) = --------------- = ------------------------------------------- (d - b) + b [2.61 (0.012 m - 0.00780 m) + 1 (0.00780 m)] = 13.4 pF (b) q = CV = (13.4 x 10^-12 F)(5.3 V) = 1.13 nC. (c) The magnitude of the electric field in the gap is q/ε0A 1.13 x 10^-9C E1 = --------------- = ----------------------- = 1.13 x 10^4N/C. (8.85 x 10^-12 F/m)(115 x 10^-4 m) (d) Using Eq. 26-32, we obtain E1 1.13 x 10^4 N/C E2 = ----------- = ------------------ = 4.32 x 10^3 N/C. 261 45. (a) According to Eq. 29-17 the capacitance of an air-filled spherical capacitor is given by ab C0 = 4πε0 ----------------- , a - b 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 κab C = 4πε0 ----------------- . a - b (b) The charge on the positive plate is qq' - q q' = CV = ------------- ΔV = ------------------ κo (a - b) (c) Let the charge on the inner conductor to be -q1. Immediately adjacent to it is the induced charge -q' and the net charge is less by a factor 1/κ than the field when no dielectric is present, then q' = q/κ - q. Thus, q' = --1-- (q/q' - 1/κq) = --1-- q(1 - 1/κ) - q. 46. (a) We apply Gauss's law with dielectric: q/ε0 = κκEA, and solve for κ: q κ = --------------- 9.0 x 10^-7 C ------------------------------------------------ (8.85 x 10^-12 F/m)(4.0 x 10^-4 m)(180 x 10^-4 m) = 7.2 . (b) The charge induced is (1 1) q' = q( -- - -- ) = ( 8.9 x 10^-7 C) (-- - -------- ) = 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 them to find an expression for the potential difference V between the plates. The capacitance is q E = ----------------- . The electric field in the dielectric is E = q/(κε0A), where κ is the dielectric constant and A is the plate area. Outside the dielectric (but still between the capacitor plates) the field is E = q/(ε0A). The field is uniform in each region so the potential difference across the plates is V = E1b + E2(d - b). q q(d-b) -------------- = ----------- + ------------------ . ε0A ε0A κ ο (ε0A κ (d + b - b) ε0A The capacitance is κε0A (d - b) + b/κ C = κε0A/d-b = -------------- + ---------------, κε0A - b/ (d - b) 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 28-5, a = 2.61, A = 115cm^2 = 115 x 10^-4m^2, d = 1.24cm = 1.24 x 10^-2 m, and b = 0.78 cm = 0.78 x 10^-2 m, κε0A( 8.85 x 10^-12 F/m)(115 x 10^-4 m2/)( ---------------------- κ) (2/261 - b) C = ------------------------- 2.61.(2) 0.87.0.32,) (5.(8.85 x 10^-12 F/m)(0.078)(2(2 -) (7.82)... 12 .7 p>1 1.34 x 10^-11 m/3 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 b( 20K x 10^-11 F) ( U' = CV^2/2 = --------7)( 7.0 x 10^5V ) = 4.9 x 10^-3 J. 9 6.9 x 10^-12 (b) Our result from part (a) is much less than the required 120 mJ, so such a spark should not have set off an explosion.