Answer:
a) It moved to a lower potential
b) ΔФ = - 86.28 Volt
Explanation:
The energy of an charged particle and the electric potential are related by the potential al kinetic energies:
[tex]\frac{\text{mv}^2}{2} = e\phi[/tex]
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:
[tex]\Delta \phi =\frac{m \left(v_f^2-v_i^2\right)}{2 e}[/tex]
Replacing all the values we get:
ΔФ = - 86.28 (Kilogram Meter^2)/(Coulomb Second^2) = -86.28 Volts