1. **Problem:** A point charge has a charge of $6.00 \times 10^{-11}$ C. At what distance from the point charge is the electric potential 12.0 V? 2. **Formula:** The electric potential $V$ due to a point charge $q$ at a distance $r$ is given by: $$V = \frac{kq}{r}$$ where $k = 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2$ is Coulomb's constant. 3. **Rearrange to find $r$:** $$r = \frac{kq}{V}$$ 4. **Substitute values:** $$r = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-11}}{12.0}$$ 5. **Calculate numerator:** $$8.99 \times 10^9 \times 6.00 \times 10^{-11} = 8.99 \times 6.00 \times 10^{-2} = 53.94 \times 10^{-2} = 0.5394$$ 6. **Calculate $r$:** $$r = \frac{0.5394}{12.0} = 0.04495 \ \text{m}$$ 7. **Answer:** The distance from the point charge where the potential is 12.0 V is approximately $0.0449$ m.