Aproton with an initial speed of 8 m/s is brought to rest by an electric field.

part a- did the proton move into a region of higher potential or lower potential?

part b - what was the potential difference that stopped the proton?
δu =

part c - what was the initial kinetic energy of the proton, in electron volts?
ki =

A) Higher potential

In order to stop the proton, it must be that the proton is moving against the direction of the electric field (=towards a positive charge, since it feels a repulsive force). This also means that the proton is moving from a region at lower electric potential to a region at higher electric potential (in fact, the electric potential is maximum near to the charge source of the field and it decreases as we move far away from it).

B) 3340 V

When the proton stops, the initial kinetic energy of the proton has been all converted into electric potential energy due to the field:

where

is the mass of the proton

is the proton's initial speed

is the proton's charge

is the potential difference

Solving for , we find

C) 3340 eV

Since the proton has charge of q = +1 e, the electric potential energy gained by the proton is

And due to the conservation of energy, the initial kinetic energy of the proton must be equal to this value.

neither accurate nor precise.