Answer:
a) It moved to a lower potential
b) ΔФ = - 86.28 Volt
Step-by-step explanation:
The energy of an charged particle and the electric potential are related by the potential al kinetic energies:

If we consider an electron:
m = 9.10938*10^-31 Kilogram
e = 1.60218*10^-19 Coulomb
And the potential diference may be calculed by:

Replacing all the values we get:
ΔФ = - 86.28 (Kilogram Meter^2)/(Coulomb Second^2) = -86.28 Volts