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 r of the sphere. Instead, youl have to do an integral. Part A Consider a thin spherical shell centered on the proton, with radius ri and infinitesimal thickness dri. 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 ri? Express your answer in terms of the variables ri, dri, and any appropriate constants.
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 r of the sphere. Instead, youl have to do an integral. Part A Consider a thin spherical shell centered on the proton, with radius ri and infinitesimal thickness dri. 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 ri? Express your answer in terms of the variables ri, dri, and any appropriate constants.
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Transcribed Image Text:A hydrogen atom is made up of a proton of charge
+Q = -1.60 × 10 19 Cand an electron of charge
-Q = -1.60 x 10 19 C. The proton may be regarded as a
point charge at r = 0, the center of the atom. The motion of
the electron causes its charge to be "smeared out" into a
spherical distribution around the proton (Figure 1), so that the
electron is equivalent to a charge per unit volume of
p(r) = -(Q/Tao )e 2r/40, where ao = 5.29 x 10 11 m
is called the Bohr radius. (a) Find the total amount of the
hydrogen atom's charge that is enclosed within a sphere with
radius r centered on the proton. (b) Find the electric field
(magnitude and direction) caused by the charge of the
hydrogen atom as a function ofr.c) Make a graph as a
function of r of the ratio of the electric-field magnitude E to
the magnitude of the field due to the proton alone.
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 dr1. 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 shellI. What is dV in terms of r1?
Express your answer in terms of the variables r1, dri, and any appropriate constants.
Figure
1 of 1
Vo AEO
?
Proton:
dV =
point charge +Q
Electron:
charge -Q "smeared out"
in a spherical distribution
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