The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
[tex]U=qV[/tex]
The potential is given by the magnitude of the electric field, E, times the distance, d:
[tex]V=Ed[/tex]
So we have
[tex]U=qEd[/tex] (1)
However, the electric field is equal to the electrical force F divided by the charge q:
[tex]E= \frac{F}{q} [/tex]
Therefore (1) becomes
[tex]U=Fd[/tex]
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
[tex]U=Fd=(3.6 \cdot 10^{-4}N)(9.8 \cdot 10^{-5} m)=3.53 \cdot 10^{-8} J[/tex]
