1. 1. Gauss' law states that . Thus

      

      using the expression for the divergence in cylindrical coordinates.

   2. First, we may simply integrate up the charge in the cylinder:

      

      Second, we may use Gauss's law for the cylinder:

      

      Here the surface integral for flux goes over the surface enclosing the cylinder: that is, the surface including a cylindrical shell at s = S (with normal ), a top circle at z=h (with normal ), and a bottom circle at z = -h (with normal ). Since points radially outward (the direction), the only part of the surface with nonzero is the cylindrical shell, where and the surface's normal vector align. Note that this surface is at constant s, so is constant over the surface, and may be factored out of the integral. Thus Gauss' law gives

      

      using the surface area for the cylinder (easily integrated, noting that .) Plugging in the given gives

      

      in agreement with the direct integration above.

2. See worked Griffiths example 2.3.
3. See worked Griffiths example 2.4.
4. See worked Griffiths example 2.5.
5. The uniformly charged sphere has constant charge density . points in the radial direction, and depends only on r, by symmetry. We use Gauss' law with our Gaussian surface being the spherical shell at radius r. This surface has normal , so

   

   using the fact that depends only on r and so is constant over the surface and can be factored out of the integral, and that the surface area of a sphere is . We then set this equal to the charge enclosed divided by . For , is constant within the sphere, so

   

   where we've used the fact that the volume of the sphere of radius r is . This gives

   

   For , vanishes once r exceeds R so the volume integrand extends only to r=R, giving the volume of the charged sphere (alternately, precisely the full charge Q is enclosed). Thus

   

   and

   

   Note this is the expected result for the electric field outside a charge, and links continuously to the electric field value inside the charged sphere.

6. We calculate the potential for the configuration in the previous problem, as

   

   Since the electric field is radial, only radial elements of the path length contribute, giving

   

   Here we just use our values for above. For ,

   

   as expected outside a charge. For , this gives

   

   which again links continuously to the value of V outside the sphere. We check that the gradient of V gives :

   

   exactly in agreement with problem 5. I leave the plotting of V to the reader.

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