2016:
The potential energy of the particle is given by:
$$U = qV$$
where $q$ is the charge of the particle and $V$ is the electric potential at the position of the particle.
The electric potential at a point in an electric field is given by:
$$V = \frac{kQ}{r}$$
where $k$ is the Coulomb constant, $Q$ is the charge creating the electric field, and $r$ is the distance from the charge to the point.
In this case, the electric field is uniform, so the electric potential is constant throughout the field. The electric potential at the origin is therefore:
$$V = \frac{kQ}{r} = \frac{100 V/m \cdot 10 \mu C}{10^{-2} m} = 10 V$$
The potential energy of the particle is therefore:
$$U = qV = 10 \mu C \cdot 10 V = 100 \mu J$$
2017:
The electric potential at a point a/2 from the
