question 13 of 25 which of the following changes would double the force between two charged particles? a…

Explanation:

Step1: Recall Coulomb's law

The force between two charged particles is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is the Coulomb's constant, $q_1$ and $q_2$ are the charges of the two particles, and $r$ is the distance between them.

Step2: Analyze option A

If we double the charge on each particle, $q_1$ becomes $2q_1$ and $q_2$ becomes $2q_2$. Then the new force $F'=k\frac{(2q_1)(2q_2)}{r^2}=4k\frac{q_1q_2}{r^2}=4F$.

Step3: Analyze option B

If we increase the distance between the particles by a factor of 2, $r$ becomes $2r$. Then the new force $F' = k\frac{q_1q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1q_2}{r^2}=\frac{1}{4}F$.

Step4: Analyze option C

If we double the charge on one of the particles, say $q_1$ becomes $2q_1$, then the new force $F'=k\frac{(2q_1)q_2}{r^2}=2k\frac{q_1q_2}{r^2}=2F$.

Step5: Analyze option D

If we decrease the distance between the particles by a factor of 2, $r$ becomes $\frac{r}{2}$. Then the new force $F'=k\frac{q_1q_2}{(\frac{r}{2})^2}=4k\frac{q_1q_2}{r^2}=4F$.

Answer:

C. Doubling the amount of charge on one of the particles