The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:
![U=qV](https://img.qammunity.org/2019/formulas/physics/middle-school/myr8otw67bq4rxlmrfawqalqcl1igcht9i.png)
The potential is given by the magnitude of the electric field, E, times the distance, d:
![V=Ed](https://img.qammunity.org/2019/formulas/physics/college/vsdvzd7s7k1ee4cgpltsgax4tq1luxplvl.png)
So we have
![U=qEd](https://img.qammunity.org/2019/formulas/physics/college/bgba1yfgvcm5zi80d6266owri7z990ulqt.png)
(1)
However, the electric field is equal to the electrical force F divided by the charge q:
![E= (F)/(q)](https://img.qammunity.org/2019/formulas/physics/college/ayaegoc251d1g0wf7v7ozdxm8c7feawij9.png)
Therefore (1) becomes
![U=Fd](https://img.qammunity.org/2019/formulas/physics/college/4q5wnvk7o8yaxkuen1j0gf2k5ge64hqqbm.png)
And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
![U=Fd=(3.6 \cdot 10^(-4)N)(9.8 \cdot 10^(-5) m)=3.53 \cdot 10^(-8) J](https://img.qammunity.org/2019/formulas/physics/college/zbi21ef5d421wh6qpmw2lj2funp4htl1dk.png)