1. For potential energy ,

   (a) At the potential V increases (spatially) most rapidly in the direction of the gradient, calculated as in Cartesian coordinates,

   

   This means that potential energy increases most rapidly in the direction ( counterclockwise from the x-axis). It's rate of increase, spatially, has magnitude e (in whatever units are presumably given).

   The force is with magnitude e and direction counterclockwise from the negative x-axis.

   (b) To find the rate of change of V with distance in a particular direction, we need the directional derivative

   

   Here at the point . (Note that we normalized the direction vector.) Thus

   

   The force component in the direction is just

   

   that is, it is minus the directional derivative of the potential energy in the direction.

   (c)

   

   The electric field thus points along the positive x-axis with magnitude 1, in units presumed given.

   (d) At x = -1,

   

   with magnitude at any y.

2. For we have

   

   with and independent of y.

   1. is conservative if and only if . Recalling the definition

      

      

      Thus is the zero vector and is conservative.

   2. Taking the origin as our reference position (with potential 0),

      

      for any path C from the origin to .

      

      The line integral for the potential then gives: for leg 1, in the x-direction, so ; similarly, for leg 2, so ; and for leg 3, so . Noting that so that no work is done on the second leg,

      

   3. Double-checking our result,

      

      as given.

3. 1. Solution appears in Griffiths, Example 1.5 (p. 19).
   2. 
      

      

      This gives the loop integral

      

      1. For , this loop integral gives

         

      2. For , this loop integral gives

         

4. 1. Solution appears in Griffiths, Example 1.6 (p. 25).
   2. We have . Using the determinant notation,

      

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