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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K = 3.60. The I can't found part c resultant electric field in the dielectric is 1.20 × 106 volts per meter. Part A Compute the magnitude of the charge per unit area o on the conducting plate. Express your answer in coulombs per square meter to three significant figures. • View Available Hint(s) o = 3.82x10-5 C/m² Submit Previous Answers Correct Part B Compute the magnitude of the charge per unit area o1 on the surfaces of the dielectric. Express your answer using three significant figures. • View Available Hint(s) 01 = 2.76x10-5 C/m² Submit Previous Answers Correct Note that the charges on the dielectric will be polarized to counteract the charges (and electric field) created by the capacitor. For example, near the positive surface of the capacitor the dielectric will have a negative charge. However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. Part C Find the total electric-field energy U stored in the capacitor.

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K = 3.60. The I can't found part c resultant electric field in the dielectric is 1.20 × 106 volts per meter. Part A Compute the magnitude of the charge per unit area o on the conducting plate. Express your answer in coulombs per square meter to three significant figures. • View Available Hint(s) o = 3.82x10-5 C/m² Submit Previous Answers Correct Part B Compute the magnitude of the charge per unit area o1 on the surfaces of the dielectric. Express your answer using three significant figures. • View Available Hint(s) 01 = 2.76x10-5 C/m² Submit Previous Answers Correct Note that the charges on the dielectric will be polarized to counteract the charges (and electric field) created by the capacitor. For example, near the positive surface of the capacitor the dielectric will have a negative charge. However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. Part C Find the total electric-field energy U stored in the capacitor.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K = 3.60. The I can't found part c resultant electric field in the dielectric is 1.20 x 10° volts per meter. Part A Compute the magnitude of the charge per unit area o on the conducting plate. Express your answer in coulombs per square meter to three significant figures. • View Available Hint(s) o = 3.82x10-5 C/m² Submit Previous Answers Correct Part B Compute the magnitude of the charge per unit area o1 on the surfaces of the dielectric. Express your answer using three significant figures. • View Available Hint(s) ơ1 = 2.76x10-5 C/m² Submit Previous Answers Correct Note that the charges on the dielectric will be polarized to counteract the charges (and electric field) created by the capacitor. For example, near the positive surface of the capacitor the dielectric will have a negative charge. However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. Part C Find the total electric-field energy U stored in the capacitor. Express your answer in joules to three significant figures. » View Available Hint(s) ? U = 10.322 J Submit Previous Answers
Transcribed Image Text:Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of K = 3.60. The I can't found part c resultant electric field in the dielectric is 1.20 x 10° volts per meter. Part A Compute the magnitude of the charge per unit area o on the conducting plate. Express your answer in coulombs per square meter to three significant figures. • View Available Hint(s) o = 3.82x10-5 C/m² Submit Previous Answers Correct Part B Compute the magnitude of the charge per unit area o1 on the surfaces of the dielectric. Express your answer using three significant figures. • View Available Hint(s) ơ1 = 2.76x10-5 C/m² Submit Previous Answers Correct Note that the charges on the dielectric will be polarized to counteract the charges (and electric field) created by the capacitor. For example, near the positive surface of the capacitor the dielectric will have a negative charge. However, this does not mean that the charge on the capacitor plates changes, only that the dielectric has an induced charge on each of its surfaces that will oppose the effects of the charges on the plates. Part C Find the total electric-field energy U stored in the capacitor. Express your answer in joules to three significant figures. » View Available Hint(s) ? U = 10.322 J Submit Previous Answers
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