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9. Find the capacitance of a spherical capacitor. Use Gauss Law to find the electric field between the inner sphere of radius "a" and the outer sphere of radius "b". Then use Faraday's Law to find the potential difference between the spheres by integrating the electric field along a line from "a" to "b". finally use the definition of capacitance C=Q/V to find the capacitance.

9. Find the capacitance of a spherical capacitor. Use Gauss Law to find the electric field between the inner sphere of radius "a" and the outer sphere of radius "b". Then use Faraday's Law to find the potential difference between the spheres by integrating the electric field along a line from "a" to "b". finally use the definition of capacitance C=Q/V to find the capacitance.

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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9. Find the capacitance of a spherical capacitor. Use Gauss Law to find the electric field between the inner sphere of radius "a" and the outer sphere of radius "b". Then use Faraday's Law to find the potential difference between the spheres by integrating the electric field along a line from "a" to "b". finally use the definition of capacitance C=Q/V to find the capacitance.
Transcribed Image Text:9. Find the capacitance of a spherical capacitor. Use Gauss Law to find the electric field between the inner sphere of radius "a" and the outer sphere of radius "b". Then use Faraday's Law to find the potential difference between the spheres by integrating the electric field along a line from "a" to "b". finally use the definition of capacitance C=Q/V to find the capacitance.
7. A parallel plate capacitor is constructed of two sheets of aluminum , 20.0 cm by 80.0 cm, separated by a layer of air, dielectgric constant K = 1, which is 0.025 mm thick. (a) What is its capacitance? (b) If it is connected to a 40.0 volt battery, what will be the charge on the capacitor? (c) What is the energy stored on the capacitor? (d) What is the electric field insie the capacitor? (e) a dielectric material with a dielectric constant K = 4.50 replaces the air between the plates, what is the new capacitance?
Transcribed Image Text:7. A parallel plate capacitor is constructed of two sheets of aluminum , 20.0 cm by 80.0 cm, separated by a layer of air, dielectgric constant K = 1, which is 0.025 mm thick. (a) What is its capacitance? (b) If it is connected to a 40.0 volt battery, what will be the charge on the capacitor? (c) What is the energy stored on the capacitor? (d) What is the electric field insie the capacitor? (e) a dielectric material with a dielectric constant K = 4.50 replaces the air between the plates, what is the new capacitance?
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