College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- 4. An 8400 pF air-gap capacitor is charged by a power supply set to 9.1 V. It is removed from the power supply, and then a layer of mica (dielectric constant of 7) is inserted between the plates. a. What is the new capacitance of the plates? b. Consider just charge on the plates and the voltage across the plates. Which of them is the same before and after the mica is inserted? Why? c. Find the new value for the one that changed. d. Did the electric field increase, decrease, or stay the same when the mica was inserted? Explain how you know. e. Calculate the new potential energy stored by the capacitor. [5x10-8 J]arrow_forwardA parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between the plates is filled with a dielectric of constant κ = 2. What is the relationship between the stored energies of the capacitor, where PEi is the initial stored energy, and PEf is the stored energy after the dielectric is inserted? a. PEf = 0.25PEi b. PEf = 0.5PEi c. PEf = PEid. PEf = 2PEiarrow_forwardB8arrow_forward
- . A parallel-plate capacitor is made of two circular plates, each with a diameter of 2.50x10^-3 m. The plates of this capacitor are separated by a space of 1.40x10^-4 m. How much charge will be stored on each plate of this capacitor when it is connected across a potential difference of 0.12V? A. 3.7x10^-8 µC B. 4.2x10^-8 µC C. 5.6x10^-8 µC D. 6.1x10^-8 µCarrow_forward4. You need to design a capacitor capable of storing 4.0 x 10-7 C of charge. At your disposal, you have a 100 V power supply and two metal plates, each of area 0.385 m2. What is the limit of the separation of the plates? mmarrow_forwardA parallel-plate capacitor is made of two square plates 25 cm on a side and 1.5 mm apart. The capacitor is connected to a 70-V battery. Hint a. What is the energy stored in the capacitor? Energy stored in the capacitor is b. With the battery still connected, the plates are pulled apart to a separation of 3 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is HJ. µJ. c. This time, starting from situation in (a), with the batteries disconnected (but capacitors still charged), the plates are pulled apart to a separation of 3 mm. What is the energy stored in the capacitor now? Energy now stored in the capacitor is d. Comparing your results in (b) and (c) above, it makes sense that the energy stored in the capacitor increases in (c), because the work done in separating the plates is stored as the electrostatic potential energy. In (b), why does the energy decrease even though work is done in separating the plates? HJ. O A negative work is done in…arrow_forward
- The electric potential increases from 100 V to 800 V from the bottom plate to the top plate of a parallel-plate capacitor. a. What is the magnitude of the change in potential energy of a −3 × 10^−4 C charge that is moved from the bottom plate to the top plate? b. Does the potential energy increase or decrease in this process?arrow_forward100 There is a system with four capacitors connected, as shown in Figure P25.10. (a) Calculate the equivalent capacity between points a and b. (b) Find the charge amount of each capacitor when the potential difference AV between points a and bis 15.0 V. 11 (13). Find the equivalent capacitance a 15.0μF 3.00 μF 120.0 με 16: HH 6.00 με Figure P25.10arrow_forwardPart G Calculate the charge on capacitor C4. Express your answer in coulombs. ΑΣΦ ? Q4 = C %Darrow_forward
- A 20 µF capacitor is connected to a 60 V battery until it becomes fully charged. The capacitor is then disconnected from the battery and connected to a light bulb until 1/3rd of the charge is removed. Calculate the energy left in the capacitor.arrow_forwardb. Capacitor A and Capacitor B are two parallel plate capacitors. Capacitor B has twice the area and four times distance between plates of Capacitor A. (i) What is the ratio CB: CA, and (ii) What is the ratio of Energy stored UB:UA when the capacitors store the same charge on its plates?arrow_forwardIn (Figure 1), each capacitor has C= 5.00 μF and Vab = 35.0 V. Part A: Calculate the charge on capacitor C1. Q1=__C Part B: Calculate the potential difference across capacitor C1. V1=__V Part C: Calculate the charge on capacitor C2. Q2=__C Part D: Calculate the potential difference across capacitor C2. V2=__V Part E: Calculate the charge on capacitor C3. Q3=__C Part F: Calculate the potential difference across capacitor C3. V3=__V Part G: Calculate the charge on capacitor C4. Q4=__C Part H: Calculate the potential difference across capacitor C4. V4=__V Part I: Calculate the potential difference between points a and d. Vad=__Varrow_forward
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