force f acts between a pair of charges, q1 and q2, separated by a distance d. for each of the statements…

force f acts between a pair of charges, q1 and q2, separated by a distance d. for each of the statements, use the drop - down menus to express the new force in terms of f. q1 is halved, q2 is doubled, but the distance between the charges remains d. q1 and q2 are unchanged. the distance between the charges is doubled to 2d. q1 is doubled and q2 is tripled. the distance between the charges remains d.

Answer

Explanation:

Step1: Recall Coulomb's Law

$F = k\frac{q_1q_2}{d^2}$, where $k$ is a constant.

Step2: Analyze first - case

New $q_1'=\frac{q_1}{2}$, $q_2' = 2q_2$, $d'=d$. Then $F'=k\frac{\frac{q_1}{2}\times2q_2}{d^2}=k\frac{q_1q_2}{d^2}=F$.

Step3: Analyze second - case

$q_1' = q_1$, $q_2' = q_2$, $d' = 2d$. Then $F'=k\frac{q_1q_2}{(2d)^2}=k\frac{q_1q_2}{4d^2}=\frac{F}{4}$.

Step4: Analyze third - case

$q_1' = 2q_1$, $q_2' = 3q_2$, $d' = d$. Then $F'=k\frac{2q_1\times3q_2}{d^2}=6k\frac{q_1q_2}{d^2}=6F$.