1. **Problem statement:** We want to find how the magnitude of the electric force between two point charges changes when each charge is doubled and the distance between them is halved. 2. **Formula used:** The electric force between two point charges is given by Coulomb's law: $$F = k \frac{q_1 q_2}{r^2}$$ where $F$ is the force, $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. 3. **Initial force:** Let the original charges be $q_1$ and $q_2$, and the original distance be $r$. Then the original force is: $$F = k \frac{q_1 q_2}{r^2}$$ 4. **New charges and distance:** Each charge is doubled, so new charges are $2q_1$ and $2q_2$. The distance is halved, so new distance is $\frac{r}{2}$. 5. **New force:** Substitute new values into Coulomb's law: $$F_{new} = k \frac{(2q_1)(2q_2)}{\left(\frac{r}{2}\right)^2} = k \frac{4 q_1 q_2}{\frac{r^2}{4}}$$ 6. **Simplify denominator:** $$F_{new} = k \frac{4 q_1 q_2}{\frac{r^2}{4}} = k \frac{4 q_1 q_2}{1} \times \frac{4}{r^2} = k \frac{16 q_1 q_2}{r^2}$$ 7. **Compare new force to original force:** $$\frac{F_{new}}{F} = \frac{k \frac{16 q_1 q_2}{r^2}}{k \frac{q_1 q_2}{r^2}} = 16$$ 8. **Conclusion:** The new force is 16 times the original force. **Final answer:** c) becomes 16 times the original one.