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

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 $f$?\n$\\frac{1}{4}f$\n$\\frac{1}{2}f$\n$f$\n$2f$

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 $f$?\n$\\frac{1}{4}f$\n$\\frac{1}{2}f$\n$f$\n$2f$

Answer

Explanation:

Step1: Recall Coulomb's law

The force between two charges is given by $F = k\frac{q_1q_2}{d^2}$, where $k$ is the electrostatic - constant.

Step2: Find the new force

The new $q_1'=2q_1$ and the new $d' = 2d$. The new force $F'=k\frac{q_1'q_2}{d'^2}=k\frac{(2q_1)q_2}{(2d)^2}$.

Step3: Simplify the new - force expression

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

Answer:

$\frac{1}{2}F$