Question

question 13 of 25 which of the following changes would double the force between two charged particles? a. doubling the amount of charge on each particle b. increasing the distance between the particles by a factor of 2 c. doubling the amount of charge on one of the particles d. decreasing the distance between the particles by a factor of 2

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