one charge is decreased to one - third of its original value, and a second charge is decreased to one - half…

Explanation:

Step1: Recall Coulomb's law

The electrical force between two charges is given by $F = k\frac{q_1q_2}{r^2}$, where $k$ is a constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. Let the original charges be $q_1$ and $q_2$, so the original force $F_1=k\frac{q_1q_2}{r^2}$.

Step2: Calculate new - charges

The new first - charge $q_1'=\frac{1}{3}q_1$ and the new second - charge $q_2'=\frac{1}{2}q_2$.

Step3: Calculate new force

The new force $F_2 = k\frac{q_1'q_2'}{r^2}=k\frac{\frac{1}{3}q_1\times\frac{1}{2}q_2}{r^2}=\frac{1}{6}k\frac{q_1q_2}{r^2}$.

Step4: Compare new and original forces

Since $F_1 = k\frac{q_1q_2}{r^2}$ and $F_2=\frac{1}{6}k\frac{q_1q_2}{r^2}$, we can see that $F_2=\frac{1}{6}F_1$.

Answer:

It will decrease to one - sixth the original force.