A uniform electric field of 10 N C^ -1 exists in the vertically...

The relationship between electric field and potential difference is given by $\Delta V = -\vec{E} \cdot \Delta \vec{r}$.

Given that the electric field is in the vertically downward direction and the displacement is in the upward direction, the angle between $\vec{E}$ and $\Delta \vec{r}$ is $180^{{}^{{}^{\circ}}}$.

Magnitude of electric field, $E = 10 \text{ N C}^{-1}$.

Displacement, $\Delta r = 50 \text{ cm} = 0.5 \text{ m}$.

Increase in electric potential, $\Delta V = -E \Delta r \cos(180^{{}^{{}^{\circ}}}) = E \Delta r$.

Substituting the values, $\Delta V = 10 \times 0.5 = 5 \text{ V}$.