A uniform electric field of 2000 NC points in the x directio

A uniform electric field of 2000 N/C points in the +x direction. (a) What is the change in potential energy of a +2.00 nC charged particle of mass 0.10 gram as it is moved from point a at x=-30.0 cm to point b at x=+50.0 cm? (b) If the same particle is released from rest at point a, what is its speed when it reaches point b? (c) If a negative charge instead of a positive charge were used in this problem, qualitatively how would your answers change?

Solution

According to the given problem,

Change in energy can be calculated as Q*d*E where d is the distance moved by a charge Q parallel to the direction of an electric field E.

a)When the charge mentioned in the question moves in the x direction the magnitude of the energy is 2.0*10^-9*2000*0.80 = 3.2 *10^-6J,

The direction of an electric field is the direction of the force on a positive charge in the field,

b)In this case the potential energy is converted in to K.E

1/2mv² = 3.2 *10^-6

v² = 0.064

v = 0.253 m/s

c) In this case there will be no change in P.E, the answers will not change qualitatively.