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Part A You find that if you place charges of + 1.10 C on two separated metal objects, the potential difference between them is 11.6 V . What is their capacitance? VO ? undo CE Submit Request Answer Part B A capacitor has a capacitance of 7.24 /F . What amount of excess charge must be placed on each of its plates to make the potential difference between the plates equal to 25.6 V ? AEd ? Q = Submit Request AnswerPart A The plates of a parallel-plate capacitor are 3.00 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuum. The electric field between the plates has a magnitude of 5.00x10 V/m . What is the potential difference between the plates? ? V = V Submit Request Answer Part B What is the area of each plate? ? S = m2 Submit Request Answer V Part C What is the capacitance? 0 8 ? C = F Submit Request AnswerReview Part A A 11.0 AF parallel-plate capacitor with circular plates is connected to a 11.0 V battery. What is the charge on each plate? AZQ 91= UC Submit Request Answer Part B How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? ? 92 Submit Request Answer V Part C How much charge would be on the plates if the capacitor were connected to the 11.0 V battery after the radius of each plate was doubled without changing their separation? ? 93 uC Submit Request AnswerPart A A 10.0 uF parallel-plate capacitor is connected to a 20.0 V battery. After the capacitor is fully charged, the battery is disconnected without loss of any of the charge on the plates. A voltmeter is connected across the two plates without discharging them. What does it read? ? V Submit Request Answer Part B What would the voltmeter read if the plate separation were doubled. AEd V = V Submit Request Answer Part C What would the voltmeter read if the radius of each plate was doubled, but the separation between the plates was unchanged? AEQ ? V : Submit Request AnswerPart A In the Figure (Figure 1), C1 = 2.80 MF and Vab = 150 V . The charge on capacitor C1 is 150 uC . Calculate the voltage across C2. AZO ? V2 : V Submit Request Answer Part B Calculate the voltage across C3. AEd V3 V Submit Request Answer Provide Feedback Figure C3Learning Goal: To understand how to calculate capacitance, voltage, and charge for a combination of capacitors connected in eries Part B Consider the combination of capacitors shown in the figure.(Figure 1) Three capacitors are connected to each other in series, and then to the battery. The values of the capacitances are C, 2C, and 3C, and the applied voltage is A V. Initially, all of the capacitors are completely discharged; after the battery is connected, the charge Part C on plate 1 is Q. Find the voltage Vi across the first capacitor. Express your answer in terms of AV. View Available Hint(s) AEd AVI = 6VI1 Submit Previous Answers X Incorrect; Try Again; 5 attempts remaining Part D Find the charge Q on the first capacitor. Express your answer in terms of C and AV1. IVO Q = Submit Request Answer Part E Using the value of Q just calculated, find the equivalent capacitance Ceq for this combination of capacitors in series. Express your answer in terms of C. View Available Hint(s) Figure IVO AEd 30 Ceq = 3 4 Submit Provide FeedbackPart A Consider the combination of capacitors shown in the diagram, where C1 = 3.00 UF , C2 = 11.0 MF , C3 = 3.00 MF , and C4 = 5.00 uF . (Figure 1) Find the equivalent capacitance CA of the network of capacitors. Express your answer in microfarads. View Available Hint(s) ? CA Submit Part B Provide Feedback Figure Part A Consider the combination of capacitors shown in the diagram, where C1 = 3.00 UF , C2 = 11.0 MF , C3 = 3.00 MF , and C4 = 5.00 uF . (Figure 1) Part B Two capacitors of capacitance Cs = 6.00 uF and C6 = 3.00 MF are added to the network, as shown in the diagram.(Figure 2) Find the equivalent capacitance CB of the new network of capacitors. Express your answer in microfarads. View Available Hint(s) AEd 0 EX ? CB= Submit Provide Feedback Next > Figure Part A For the capacitor network shown in the figure below(Figure 1), the potential difference across ab is 14.0 V . Find the total energy stored in this network. ? U = MJ Submit Request Answer Part B Find the energy stored in the 4.80 uF capacitor. AEQ 0 8 ] ? U4.80 MF = MJ Submit Request Answer Provide Feedback Figure 6.20 MF 11.8 ALF 8.60 uLF HK 4.80 MF 3.50 MF