1. In Example 2.7, Griffiths' finds that a uniformly charged spherical shell, with charge q and radius R, has potential

   

   This yields an electric field

   

   Here ``above'' means , and ``below'' means , with the normal to our charged spherical shell surface being the radial direction . Thus

   

   where both fields have been evaluated just above and below r= R. Since the charge density on the spherical shell is just , we have indeed shown

   

2. 1. 
      

      

      This means the metal sphere carries its entire charge q at radius R, with uniform charge density

      

      Within the bulk of the shell (a;SPMlt;r;SPMlt;b) the electric field vanishes, so by Gauss' law the total charge enclosed in must be zero. Thus the total charge on the inner (r=a) surface of the conducting shell must be -q, with uniform charge density

      

      Finally, the spherical shell is uncharged, so compensating charge +q must be uniformly distributed on the outer surface of the spherical shell:

      

   2. 

      Thus

      

   3. If instead the conductor were grounded, all the charge on the outer shell would escape down the grounding wire. This is the only way the outer shell can have the same potential as at infinity (with no gradient, or electric field in between them -- this is possible only if there is no net charge enclosed for the sphere with r=b). Because within the spherical shell, we still must have screening charge -q on the inner shell. Thus answers (a) and (b) change to

      

      and

      

      since now the only nonzero electric field is between the sphere and inner shell surface.

3. See worked Griffiths example 2.11.
4. See worked Griffiths example 3.3.
5. See worked Griffiths example 3.4.
6. See worked Griffiths example 3.5.

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