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    Use Gauss's Law to find the electic field, electric potential

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    A solid insulating sphere has a radius A and net charge of +Q. The insulating sphere has a uniform charge density. a conducting sphere of inner radius B and outer radius C surrounds it.

    I.
    Use Gauss's Law to find the Electic Field for R<A
    II.
    Use Gauss's Law to find the Electic Field for A<R<B
    III.
    Use Gauss's Law to find the Electic Field for B<R<C
    IV.
    Use Gauss's Law to find the Electic Field for R>C
    V.
    What is the electric potential for R < A
    VI.
    What is the electric potential for R = A

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    Solution Preview

    1) For R<A
    Choose a sphere of radius R<A as your Gaussian surface. Let p be the charge density of the insulator, and let V' be the volume of your Gaussian sphere. Then the charge enclosed by the Gaussian surface is:

    Q_in = pV'
    where V' = (4/3)*pi*R^3

    Since the magnitude of the electric field is constant everywhere on the spherical Gaussian surface and is normal to the surface at each point, gauss' law gives:

    Electric flux through surface = E*SurfaceArea
    = E*4*pi*R^2
    = ...

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