The positively charged particle is released from the rest in a uniform electric field. The electric potential energy of the charge
a. decreases because the charge moves along the electric field.
b. decreases because the charge moves opposite the electric field.
c. remains constant because the electric field is uniform.
d. increases because the charge moves along the electric field.

When a positively charged particle is released from rest in a uniform electric field, it experiences an electric force $\vec{F} = q\vec{E}$ in the direction of the electric field.

Thus, the particle moves along the direction of the electric field.

The work done by the electric field on the positive charge is positive. Since the change in electric potential energy is the negative of the work done by the conservative electric force ($\Delta U = -W$), the electric potential energy of the charge decreases.

Alternatively, electric potential $V$ decreases in the direction of the electric field. Since $U = qV$ and $q$ is positive, the electric potential energy decreases as the charge moves along the electric field.

Answer: a