IDENTIFY and SET UP The charge distribution in this problem is spherically symmetric, so you can solve it with Gauss's law. The charge within a sphere of radius r includes the proton charge +Q plus the portion of the electron charge distribution that lies within the sphere. The electron charge distribution is not uniform, so the charge enclosed within a sphere of radius r is not simply the charge density multiplied by the volume Tr* of the sphere. Instead, you'll have to do an integral. Part A Consider a thin spherical shell centered on the proton, with radius r and infinitesimal thickness dr. Since the shell is so thin, every point within the shell is at essentially the same radius from the proton. Hence the amount of electron charge within this shell is equal to the electron charge density p(ri) at this radius multiplied by the volume dV of the shell. What is dV in terms of r1? Express your answer in terms of the variables r1, dri, and any appropriate constants. • View Available Hint(s) Hνα ΑΣφ Submit Previous Answers

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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IDENTIFY and SET UP
The charge distribution in this problem is spherically symmetric, so you can solve it with Gauss's law.
The charge within a sphere of radius r includes the proton charge +Q plus the portion of the electron charge distribution that lies within the sphere. The electron charge
distribution is not uniform, so the charge enclosed within a sphere of radius r is not simply the charge density multiplied by the volume Tr* of the sphere. Instead, you'll
have to do an integral.
Part A
Consider a thin spherical shell centered on the proton, with radius r and infinitesimal thickness dr. Since the shell is so thin, every point within the shell is at
essentially the same radius from the proton. Hence the amount of electron charge within this shell is equal to the electron charge density p(ri) at this radius multiplied
by the volume dV of the shell. What is dV in terms of r1?
Express your answer in terms of the variables r1, dri, and any appropriate constants.
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Hνα ΑΣφ
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Transcribed Image Text:IDENTIFY and SET UP The charge distribution in this problem is spherically symmetric, so you can solve it with Gauss's law. The charge within a sphere of radius r includes the proton charge +Q plus the portion of the electron charge distribution that lies within the sphere. The electron charge distribution is not uniform, so the charge enclosed within a sphere of radius r is not simply the charge density multiplied by the volume Tr* of the sphere. Instead, you'll have to do an integral. Part A Consider a thin spherical shell centered on the proton, with radius r and infinitesimal thickness dr. Since the shell is so thin, every point within the shell is at essentially the same radius from the proton. Hence the amount of electron charge within this shell is equal to the electron charge density p(ri) at this radius multiplied by the volume dV of the shell. What is dV in terms of r1? Express your answer in terms of the variables r1, dri, and any appropriate constants. • View Available Hint(s) Hνα ΑΣφ Submit Previous Answers
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