1. 1. Gauss' law states that . Thus, for our radial electric field in spherical coordinates,

      

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

      

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

      

      Here the surface integral for flux goes over the spherical surface at radius R, with normal . is then . Furthermore is constant over our surface of fixed r=R, and may be factored out of the integral. Thus Gauss' law gives

      

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

      

      in agreement with the direct integration above.

2. 
   

   

   The charge enclosed by the Gaussian surface is just its z-height times the uniform charge line density :

   

   The surface integral for flux goes over the complete Gaussian surface: the 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 .). Equating the 2 expressions for enclosed charge,

   

   or

   

3. Here our Gaussian surface is a sphere of radius R, with normal . By symmetry, since depends only on r, the electric field is radial and depends only on r. Thus is , with constant over the Gaussian surface of fixed r, so that it may be factored out of the flux integral. Thus Gauss' law again gives

   

   We may also directly integrate the charge enclosed

   

   Equating the 2 expressions for enclosed charge,

   

   or

   

4. 
   

   

   Note that since points in the y-direction, only the top and bottom surfaces, with normals , can make any contribution to the flux integral. On the bottom surface, at y=0, so there is no contribution. So the total flux integral for the Gaussian surface shown is

   

   since is independent of x and z and can be factored out of the area integral.

   The charge enclosed by the surface is

   

   If Y;SPMlt;d, our Gaussian volume lies completely in the slab where is constant, so

   

   If Y;SPMgt;d, is not constant within our Gaussian volume, but vanishes for y;SPMgt;d. The enclosed charge is thus

   

   since is constant where it is nonzero. Using Gauss' law, then

   

   or

   

5. 1. 

      so this is an impossible electrostatic field.

   2. 

      so this is a viable electrostatic field.

      

      This path is chosen because along the first leg, vanishes when y=0, and along the second leg, vanishes when y=0. There is thus a nonzero only on the third leg, where and is nonzero. Thus

      

      which is easily checked to reproduce the correct .

6. We're asked to find the potential for the infinitely long straight wire of uniform charge line density , whose electric field we found in problem 2 to be

   

   The potential is just

   

   since the electric field points in the direction. Here we've chosen our reference point not at infinity, but at some finite radius a, since we will find that the potential diverges at infinity. This gives

   

   Note that this potential reproduces the correct electric field. It approaches minus infinity infinitely far from the wire, and plus infinity as . This means, for example, that all negative charges are bound by the wire, no matter how energetic they are.

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