Two spherical, hollow conductors are concentrically nested as shown in the cross-sectional diagram below and electrically isolated from each other. A net charge of -3 nC is divided between the conductors, with a total of -12 nC on the inner conductor and +9 nC on the outer one. The charges are, of course, free to move between the surfaces within each shell but cannot move from one shell to the other. The inner conductor has an inner radius of a=2 cm, outer radius of b=3 cm. The outer conductor has an inner radius of c=6 cm, an outer radius of d=8 cm. a bv (a) In equilibrium, all of the excess charge is found on the surfaces. Why is there no charge between a and b, nor between c and d? (b) Determine the net charge on each surface (ie how much charge is there at r = a, at r = b, at r = c and at r = d). (c) Is the charge on the outer surface equal to the net charge on the outer container, the net charge of the system, or some other amount? Explain briefly. (d) Find the surface charge density at the outer surface (r = d.) (e) Find the electric field E(r) as a function of r from r=0 to r=10 cm. (Note that the functional expression may be different in different regions.) Plot the field as a function of r, using positive values if the field is outward and negative if inward. (f) Check that the charges you found in part b for the charge at r=a, b and c and the field you plotted in part e for the field between r=b and r=c are consistent with Gauss's law. Check also that the field just outside the outermost surface (at r just past 0.08 m) and the total charge enclosed within that radius are consistent with Gauss's law.

Answered: Two spherical, hollow conductors are concentrically nested as shown in the cross-sectional diagram below and electrically isolated from each other. A net charge…

College Physics

11th Edition

ISBN: 9781305952300

Author: Raymond A. Serway, Chris Vuille

Publisher: Cengage Learning

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Question

Two spherical, hollow conductors are concentrically nested as shown in the cross-sectional diagram
below and electrically isolated from each other. A net charge of -3 nC is divided between the
conductors, with a total of -12 nC on the inner conductor and +9 nC on the outer one. The charges
are, of course, free to move between the surfaces within each shell but cannot move from one shell to
the other. The inner conductor has an inner radius of a=2 cm, outer radius of b=3 cm. The outer
conductor has an inner radius of c=6 cm, an outer radius of d=8 cm.
a
(a) In equilibrium, all of the excess charge is found on the surfaces. Why is there no charge between
a and b, nor between c and d?
(b) Determine the net charge on each surface (ie how much charge is there at r = a, at r = b, at r
= c and at r= = d).
(c) Is the charge on the outer surface equal to the net charge on the outer container, the net charge
of the system, or some other amount? Explain briefly.
(d) Find the surface charge density at the outer surface (r = d.)
(e) Find the electric field E(r) as a function of r from r=0 to r=10 cm. (Note that the functional
expression may be different in different regions.) Plot the field as a function of r, using positive
values if the field is outward and negative if inward.
(f) Check that the charges you found in part b for the charge at r=a, b and c and the field you
plotted in part e for the field between r=b and r=c are consistent with Gauss's law. Check also
that the field just outside the outermost surface (at r just past 0.08 m) and the total charge
enclosed within that radius are consistent with Gauss's law.

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Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately specifying the sub parts needed to be answered.

Given : a=2 cm, b=3cm, c=6 cm, d=8 cm.

a) The electric field inside a conducting sphere is zero and the potential is constant. The volume between a and b and c and d is a conductor. Thus the charge is zero.

Because these surfaces are conductors.

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