1. **State the problem:** A positive charge $q=3$ C is moved from point A to point B by the electrostatic force of charge $Q$ at point O. The work done by this force is $1.5$ J. We need to find the potential difference $V_B - V_A$. 2. **Recall the relationship between work and potential difference:** The work done by the electrostatic force when moving a charge $q$ from A to B is related to the potential difference by $$W = q(V_A - V_B)$$ Note that the work done by the force is positive if the force moves the charge in the direction of decreasing potential energy. 3. **Rearrange to find $V_B - V_A$:** $$V_B - V_A = -\frac{W}{q}$$ 4. **Substitute the known values:** $$V_B - V_A = -\frac{1.5}{3} = -0.5$$ 5. **Interpret the result:** The potential at B is $0.5$ volts lower than at A, so the potential difference $V_B - V_A = -0.5$ V. **Final answer:** $\boxed{-0.5\text{ V}}$ which corresponds to choice B.