1) Potential difference: 1 V
2) ![V_b-V_a = -1 V](https://tex.z-dn.net/?f=V_b-V_a%20%3D%20-1%20V)
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
![\Delta U=q\Delta V](https://tex.z-dn.net/?f=%5CDelta%20U%3Dq%5CDelta%20V)
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
![\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cfrac%7B%5CDelta%20U%7D%7Bq%7D%3D%5Cfrac%7B4.80%5Ccdot%2010%5E%7B-19%7D%7D%7B4.80%5Ccdot%2010%5E%7B-19%7D%7D%3D1%20V)
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:
![V_b-V_a = - 1V](https://tex.z-dn.net/?f=V_b-V_a%20%3D%20-%201V)