force f acts between two charges, $q_1$ and $q_2$, separated by a distance d. if $q_1$ is increased to twice…

force f acts between two charges, $q_1$ and $q_2$, separated by a distance d. if $q_1$ is increased to twice its original value and the distance between the charges is also doubled, what is the new force acting between the charges in terms of f? $\frac{1}{4}f$ $\frac{1}{2}f$ f 2f

force f acts between two charges, $q_1$ and $q_2$, separated by a distance d. if $q_1$ is increased to twice its original value and the distance between the charges is also doubled, what is the new force acting between the charges in terms of f? $\frac{1}{4}f$ $\frac{1}{2}f$ f 2f

Answer

Answer:

B. $\frac{1}{2}F$

Explanation:

Step1: Recall Coulomb's law

$F = k\frac{q_1q_2}{d^{2}}$

Step2: Find new - charge and distance values

New $q_1'=2q_1$, new $d' = 2d$

Step3: Calculate new force $F'$

$F'=k\frac{q_1'q_2}{d'^{2}}=k\frac{(2q_1)q_2}{(2d)^{2}}=k\frac{2q_1q_2}{4d^{2}}=\frac{1}{2}k\frac{q_1q_2}{d^{2}}$ Since $F = k\frac{q_1q_2}{d^{2}}$, then $F'=\frac{1}{2}F$.