1. **Problem statement:** A point charge $q = -2.5$ C is moved from point B to point A without acceleration, gaining 1 J in potential energy. We need to find the potential difference $V_A - V_B$. 2. **Formula used:** The change in potential energy $\Delta U$ when moving a charge $q$ through a potential difference $\Delta V$ is given by: $$\Delta U = q \Delta V$$ where $\Delta V = V_A - V_B$. 3. **Rearranging the formula to find $\Delta V$:** $$\Delta V = \frac{\Delta U}{q}$$ 4. **Substitute the known values:** $$\Delta V = \frac{1}{-2.5}$$ 5. **Simplify the fraction:** $$\Delta V = \frac{1}{\cancel{2.5}} \times \frac{\cancel{1}}{-1} = -0.4$$ 6. **Interpretation:** The potential difference $V_A - V_B$ is $-0.4$ volts. 7. **Answer choice:** The correct answer is C: $-0.4$ V. This means point A is at a lower potential than point B by 0.4 volts for the negative charge to gain potential energy moving from B to A.