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
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Chapter16: Electrical Energy And Capacitance
Section: Chapter Questions
Problem 5CQ: A parallel-plate capacitor with capacitance C0 stores charge of magnitude Q0 on plates of area A0...
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Both of them, please.
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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