College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- A parallel-plate air-filled capacitor having area 38 cm² and plate spacing 1.4 mm is charged to a potential difference of 500 V. Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, (e) the energy density between the plates. (a) Number (b) Number (c) Number 24 (e) Number 12E-9 0.75 (d) Number 357142 0.056 Units Units Units Units Units pF μJ N/C or V/m J/m^3arrow_forwardA parallel-plate capacitor has circular plates of 10.8 cm radius and 1.23 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 125 V is applied? (a) Number i (b) Number i Units Unitsarrow_forwardTwo identical parallel-plate capacitors, each with capacitance 15.5 F, are charged to potential difference 46.0 V and then disconnected from the battery. They are then connected to each other in parallel with plates of like sign connected. Finally, the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the law of conservation of energy. O Positive work is done by the agent pulling the plates apart. Negative work is done by the agent pulling the plates apart. No work is done by pulling the agent pulling the plates apart.arrow_forward
- A slab of copper of thickness b = 1.68 mm is thrust into a parallel-plate capacitor of plate area A = 1.96 cm2 and plate separation d = 5.35 mm, as shown in the figure; the slab is exactly halfway between the plates. (a) What is the capacitance after the slab is introduced? (b) If a charge q = 2.68 µC is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? (d) Is the slab sucked in or must it be pushed in? Copper (a) Number 4.73e-13 Units (b) Number i 0.686 (c) Number i 5.36e-10 Units J (d) sucked inarrow_forwardTwo parallel-plate capacitors, 5.2 uF each, are connected in series to a 11 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the sum of the charges: the charge on the positive plate of one capacitor plus the charge on the positive plate of the other capacitor? Give your answers in Coulombs.arrow_forwardA parallel plate capacitor with a capacitance of 0.86 F with a has plate area A and are separated by a distance d. It is charged up to 11 V. The capacitor is now disconnected from the battery. If the area of the plates increases by a factor of 3 and the separation decreases by a factor of 10, find the new charge on the capacitor in Coulombs.arrow_forward
- An initially uncharged air-filled capacitor is connected to a 3.29 V charging source. As a result, the capacitor acquires 9.89 x 10- C of charge. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant k of this substance is 7.61. Find the voltage V across the capacitor and the charge Qf stored by it after the dielectric is inserted and the circuit has returned to a steady state. V = V Qt = Carrow_forwardA parallel-plate capacitor in air has a plate separation of 1.58 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC pC after (b) Determine the capacitance and potential difference after immersion. Cf = F AV = V (c) Determine the change in energy of the capacitor. nJ Need Help? Read Itarrow_forwardA paper-filled capacitor is charged to a potential difference of Vo = 2.1 V. The dielectric constant of paper is K = 3.7. The capacitor is then disconnected from the charging circuit and the paper filling is withdrawn, allowing air to fill the gap between the plates. Find the new potential difference V of the capacitor. Vị = V While the capacitor is disconnected from the charging circuit, an unknown substance is inserted between the plates. The plates then attain a potential difference that is 0.47 times the original potential difference Vo (when paper filled the capacitor). What is this unknown substance's dielectric constant K? Ku =arrow_forward
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