CATAGORY: Electricity & Magnetism
QUESTION: An electron is placed ar rest in a uniform electric field,
and when it is released it begins to move to the right.
Which direction does the field point in? How does the electron's
electric potential energy change as it moves? How does the
potential that it feels change as it moves?

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The force on a charged particle is given by

      ->    ->
(1)   F  = qE
                                      ->                          ->
Where q is the charge of the particle E is the electric field and F is
the force.

(2) Since The force is to the right and q = -e (e =1.602 x 10^-19C) is
less than zero then the electric field points in the opposite
direction or to the left. 

Since the field is uniform the change in the electric potential energy
is given by negative the work done in moving the electron from one point
to another. So the idea here is that energy is being taken out of the
field and being put into work done in moving the charge so what one gains
the other looses. Thus the change in electrical potential energy is

                                ->  ->
(2)    U = -work on electron = -F * d 
    
      ->
where d is the displacement of the electron.  The displacement is in
the same direction as the force so

    ->  ->
(3) F * d = |F||d|
                              ->    ->
Where |F| is the magnitude of F and d is the magnitude of |d|.
Equation (2) then becomes

(4) U = -|F||d| = -|-eE||d| = -|eE||d| = -|e||E||d| = -e|E||d| < 0

So the change in electrical potential energy is negative. 
the potential is defined as

                   U
(5) V =   lim     ---
         q --> 0   q

so in this case we have

          -e|E||d|
(6) V =  --------  = |E||d| > 0
           -e
           
So the change of electric potential is positive.