Answer, Key – Homework 10 – David McIntyre This print-out should have 22 questions, check that it is complete.Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapter 26 problems. 001 (part 1 of 1) 0 points A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential a cross it? 1. The charge increases and the electric potential decreases. 2. The charge decreases and the electric potential increases. 3. The charge and the electric potential increase. 4. The charge decreases and the electric potential remains fixed. 5. The charge and the electric potential decrease. 6. The charge remains fixed and the electric potential increases. correct 7. The charge and the electric potential remain fixed. 8. The charge increases and the electric potential remains fixed. 9. The charge remains fixed and the electric potential decreases. Explanation: Charge is conserved, so it must remain constant since it is stuck on the plates. With the battery disconnected, Q is fixed. C= A d 1 Q is smaller. Thus the new potential V = C larger. 002 (part 1 of 2) 5 points Given: A capacitor network is shown below. 14 µF 20 µF a 7 µF b 116 V Find the equivalent capacitance Cab between points a and b for the group of capacitors. Correct answer: 15.2353 µF. Explanation: Given : C1 C2 C3 EB C1 a C3 = 14 µF , = 20 µF , = 7 µF , and = 116 V . C2 b EB The capacitors C1 and C2 are in series, so 1 1 + C1 C2 C1 C2 = . C1 + C 2 Cs is parallel with C3 , so Cs = −1 Cab = Cs + C3 C1 C2 = + C3 C1 + C 2 (14 µF) (20 µF) = + 7 µF 14 µF + 20 µF = 8.23529 µF + 7 µF = 15.2353 µF . A larger d makes the fraction smaller, so C is c C2 C3 EB d and C1 a 005 (part 2 of 2) 5 points What is the charge on the 35 µF uppercentered capacitor? Correct answer: 2241. respectively. C4 006 (part 1 of 2) 0 points .1068 µF = 128. C4 . 004 (part 1 of 2) 5 points Four capacitors are connected as shown in the figure. = 97 V .107 µF . = 68 µF . Explanation: The voltages across C2 and C3 . i. the same potential EB = V is across both.. Cab = C1 + C4 + C23 = 21 µF + 84 µF + 23. = 84 µF . Find the capacitance between points a and b. and we have b Q23 = Q3 = Q2 = Vab C23 = (97 V) (23.1068 µF) = 2241.Answer. and C23 . so q3 = C 3 V = (7 µF) (116 V) = 812 µC . a C2 C1 C4 C3 b Q The definition of capacitance is C ≡ . Explanation: Given : C1 C2 C3 C4 E = 21 µF .1068 µF . = 35 µF .107 µF. Correct answer: 128. V The series connection of C2 and C3 gives the equivalent capacitance C23 = 1 1 1 + C2 C3 C2 C3 = C2 + C 3 (35 µF) (68 µF) = 35 µF + 68 µF = 23. c F 1µ 2 68 µF a 35 µF b 2 A good rule of thumb is to eliminate junctions connected by zero capacitance. Key – Homework 10 – David McIntyre 003 (part 2 of 2) 5 points What charge is stored on the 7 µF capacitor on the lower portin of the parallel circuit? Correct answer: 812 µC.e. (the voltage between a and b) are Vab = V23 = 97 V . 97 V d 8 F 4µ The total capacitance Cab between a and b can be obtained by calculating the capacitance in the parallel combination of the capacitors C1 .36 µC. Explanation: Since Cs and C3 are parallel.36 µC . 4.78 µF) (5.8 µF Each capacitor has voltage V.78 µF + 5.3788 µF .215371 J . Explanation: When in series the equivalent capacitance is Cseries = C1 C2 C1 + C 2 (21 µF) (4. .1 V .6 µF) + 13.78 µF a 17. Explanation: Since C1 and C2 are in series they carry the same charge C1 V1 = C 2 V2 .1 µF) (131 V)2 2 = 0. and they are together are parallel with C3 . 009 (part 2 of 7) 2 points What is the voltage across the 5. Since U = Ceq V 2 . Key – Homework 10 – David McIntyre Two capacitors of 21 µF and 4.Answer.6 µF b 3 13. = 13.6 µF upper right-hand capacitor? Correct answer: 7. C1 a EB and 007 (part 2 of 2) 0 points What potential difference would be required across the same two capacitors connected in series in order for the combination to store the same energy as in the first part? Correct answer: 354. So Cab = C1 C2 + C3 C1 + C 2 (4. 008 (part 1 of 7) 2 points Given: A capacitor network is shown in the following figure. so U= 1 (C1 + C2 ) V 2 2 1 = (21 µF + 4. Explanation: U= Cparallel CV2 2 = C1 + C2 .78 µF .1 µF are connected parallel and charged with a 131 V power supply.6 µF .1 µF = 3. Explanation: Given: C1 C2 C3 EB = 4.8 µF = 4.359 V.1 V 5.43028 µF 1F = 354.87457 V. = 17.8 µF .215371 J. = 5. Correct answer: 0. Calculate the total energy stored in the two capacitors.3788 µF. we have 2 2U Ceq V = 2 (0. C2 b C3 C1 and C2 are in series with each other.1 µF) = 21 µF + 4. What is effective capacitance Cab of the entire capacitor network? Correct answer: 16.43028 µF .6 µF = 16.359 V .215371 J) 106 µF = · 3. 98 µc Q1 = Q 2 = C12 V = (3.54315 V.78 µF + (4. 011 (part 4 of 7) 1 points If the battery is disconnected and then the dielectric is removed. = 14.1 V)(5.78 µF) = 4.8416 µc .18 is inserted in the 5.6 µF) = (13.8416 µc = 5. 012 (part 5 of 7) 1 points What is now the voltage drop across the 5.6 µF top right-hand capacitor? Correct answer: 8.78 µF top left-hand capacitor? Correct answer: 14.98 µc = 13.6 µF = 8.1 V)(4.78 µF + 5. and the voltages on C3 and on C12 (where C12 is the equivalent capacitance of C1 and C2 in series) are equal to each other Q1 + Q 3 = Q 1 + Q 3 Q1 Q = 3.57881 µF = 47.2003 V . 4 When we remove the dielectric. Key – Homework 10 – David McIntyre and their voltages add up to V . When the dielectric is inserted. the capacitance formerly C2 becomes C2 = κ C 2 .8772 µc + 235. C12 C3 Therefore Q1 = Q1 + Q 3 C3 1+ C12 67. what is the charge on 4.96943 µF)(17. Explanation: Immediately before the battery was disconnected the charges on the capacitors had been Q3 = C 3 V V2 = 013 (part 6 of 7) 1 points What was the energy stored in the system .8 µF)(17.8772 µc . 010 (part 3 of 7) 2 points If a dielectric of constant 4.87457 V .2003 V. what is the electric potential across the 4.Answer. Explanation: Given : κ = 4. Explanation: Q2 C2 Q = 1 C2 47.18 .6 µF top right-hand capacitor (when the battery is connected).54315 V .8 µF 1+ 2.8416 µc.18)(5.6 µF = 7.78 µF top left-hand capacitor? Correct answer: 47.1 V) = 235.1 V) = 67.6 µF) = 4. and the new voltage across C1 is V1 = V C2 C1 + C 2 κ V C2 = C1 + κ C 2 (4.18)(17. voltage of the battery V1 + V 2 = V C2 V2 + V2 = V C1 C2 V2 + C 1 V2 = V C 1 V C1 V2 = C1 + C 2 (17. the sum of the charges stays the same. 00259798 J . correct 2.00259798 J. The electric field between the capacitor plates increases. Therefore.8 µF = 17. 5 4. it becomes U= 1 κC V 2 2 . correct 3.56 µJ = 0.00281856 J . from Q V = .1 V) = 303. Therefore. Explanation: Since the battery is disconnected. 014 (part 7 of 7) 1 points What is the energy stored in the system after the dielectric was removed? Correct answer: 0.96943 µF + 13. 015 (part 1 of 1) 0 points A sheet of mica is inserted between the plates of an isolated charged parallel-plate capacitor. the capacitance is increased by introducing a dielectric. the energy stored is Q 2 U = 2 Cab = 2818. The energy of the capacitor does not change. the charge on the capacitor plates remains the same. The potential difference across the capacitor decreases.7694 µF)(17.00281856 J. Explanation: The total capacitance of the system before the dielectric was removed had been Cab = C12 + C3 = 3. decrease. increase. The capacitance decreases. While the battery is still connected. The stored energy will 1.98 µJ 3. The charge on the capacitor plates decreases.Answer. V V = 0.7694 µF so the energy stored in the system was U = 1 C V2 2 ab = 2597. Which of the following statements is true? 1. 5. On the other hand. the potential difference across the C capacitor is decreased. a glass slab is inserted so as to just fill the space between the capacitor plates.857 µc . 016 (part 1 of 1) 0 points A parallel plate capacitor is attached to a battery which maintains a constant potential difference of V between the plates. the stored energy is U0 = 1 CV2 2 After inserting the glass. 2. remain the same. Explanation: The energy stored in the capacitor is given by Q2 κC 2 U= = V 2 κC 2 Without the glass. the total charge is conserved: Q =Q = Cab V = (17. Explanation: Since the capacitor is isolated. Key – Homework 10 – David McIntyre before the dielectric was removed? Correct answer: 0. 25 mm .53) is inserted to completely fill the space between the plates. The battery is then disconnected.42546 × 10−10 C.52456 × 10−10 F. 021 (part 5 of 6) 1 points What is the potential difference across the plates before the dielectric is inserted? Correct answer: 5. 017 (part 1 of 6) 2 points A parallel-plate capacitor has a plate area of 105 cm2 and a plate separation of 2. The charge on the plates before the dielectric is inserted is given by Q1 = C 1 V = (4. Explanation: Given : V = 5.688159 V .87 V) = 2. 022 (part 6 of 6) 1 points What is the potential difference across the plates after the dielectric is inserted? Correct answer: 0. so it is given by Q2 = Q1 = 2.52456 × 10−10 F .53 .Answer.85419 × 10−12 C2 /N m2 (105 cm2 ) = (2.87 V.85419 × 10−12 C2 /N m2 . A potential difference of 5. Explanation: Given : κ = 8.13195 × 10−11 F) (5. 018 (part 2 of 6) 2 points What is the capacitance after the dielectric is inserted? Correct answer: 3.688159 V.13195 × 10−11 F. and d = 2.25 mm) = 4. 0 6 What is the charge on the plates before the dielectric is inserted? Correct answer: 2.42546 × 10−10 C. A = 105 cm2 .25 mm.87 V is applied across the plates with only air between the plates. Explanation: The potential difference across the plates after the dielectric is inserted is given by V2 = V 5.42546 × 10−10 C . 019 (part 3 of 6) 2 points . the stored energy will increase.53 The capacitance before the dielectric is inserted is C1 = 0A d 8. What is the capacitance before the dielectric is inserted? Correct answer: 4. Key – Homework 10 – David McIntyre Since κ > 1. Explanation: The charge on the plates doesn’t change after the dielectric is inserted. and a piece of glass (κ = 8. κ 8. Explanation: The potential difference across the plates before the dielectric is inserted is given by V1 = V = 5.42546 × 10−10 C . The capacitance after the dielectric is inserted is C2 = κ 0A d (8.25 mm) = 3.13195 × 10−11 F . 020 (part 4 of 6) 2 points What is the charge on the plates after the dielectric is inserted? Correct answer: 2.87 V .53) ( 0 ) (105 cm2 ) = (2.87 V = = 0.87 V . Explanation: Given : = 8.