Question

A proton with an initial speed of 800000 m/s is brought to rest by an electric field.

Answer 1

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:

$K = \Delta U\\\frac{1}{2}mv^2 = q\Delta V$

where

$m=1.67\cdot 10^{-27}kg$ is the mass of the proton

$v=8\cdot 10^5 m/s$ is the proton's initial speed

$q=1.6\cdot 10^{-19}C$ is the proton's charge

$\Delta V$ is the potential difference

Solving for $\Delta V$, we find

$\Delta V=\frac{mv^2}{2q}=\frac{(1.67\cdot 10^{-27}kg)(8\cdot 10^5 m/s)^2}{2(1.6\cdot 10^{-19}C)}=3340 V$

C) 3340 eV

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

$\Delta U=q\Delta V=(1 e)(3340 V)=3340 eV$

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