if the distance between two charges is increased to three times the original distance, how will the…

Answer:

D. It will decrease to one - ninth the original force.

Explanation:

Step1: Recall Coulomb's law

The electrical force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = k\frac{q_1q_2}{r^{2}}$, where $k$ is the electrostatic constant.

Step2: Consider the new distance

Let the original distance be $r_1 = r$ and the new distance be $r_2=3r$. The original force $F_1 = k\frac{q_1q_2}{r^{2}}$, and the new force $F_2=k\frac{q_1q_2}{(3r)^{2}}$.

Step3: Simplify the new - force expression

$F_2 = k\frac{q_1q_2}{9r^{2}}=\frac{1}{9}\times k\frac{q_1q_2}{r^{2}}$. Since $F_1 = k\frac{q_1q_2}{r^{2}}$, we have $F_2=\frac{1}{9}F_1$. So the force decreases to one - ninth the original force.