The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:

The potential is given by the magnitude of the electric field, E, times the distance, d:

So we have

(1)
However, the electric field is equal to the electrical force F divided by the charge q:

Therefore (1) becomes

And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
Answer:
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To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

Then the potential energy is

PART A) The force is excepted to be along x-axis.
Therefore we take a derivative of U with respect to x.



This expression is the resultant magnitude of the Force F.
PART B) The magnitude of loss in potential energy as the particle falls to the center

According to conservation of energy,

