The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 4.80×10−19 JJ . In what direction and through what potential difference Vb−VaVb−Va does the particle move?

1) Potential difference: 1 V

2)

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

where

q is the charge's magnitude

is the potential difference between the initial and final position

In this problem, we have:

is the magnitude of the charge

is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray