the table shows the charges and the distance between five different pairs of objects.\n| charge 1 | charge 2…

Explanation:

Step1: Recall Coulomb's law

The force 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 a constant. For the first - pair with $q_1 = q$, $q_2 = q$ and $r = d$, $F=k\frac{q\times q}{d^{2}}$.

Step2: Calculate $W$

For the second - pair with $q_1 = q$, $q_2 = 2q$ and $r = d$, $W = k\frac{q\times2q}{d^{2}}=2k\frac{q^{2}}{d^{2}} = 2F$.

Step3: Calculate $X$

For the third - pair with $q_1 = q$, $q_2 = q$ and $r = 2d$, $X=k\frac{q\times q}{(2d)^{2}}=\frac{1}{4}k\frac{q^{2}}{d^{2}}=\frac{1}{4}F$.

Step4: Calculate $Y$

For the fourth - pair with $q_1 = 3q$, $q_2 = q$ and $r = d$, $Y = k\frac{3q\times q}{d^{2}}=3k\frac{q^{2}}{d^{2}} = 3F$.

Step5: Calculate $Z$

For the fifth - pair with $q_1 = q$, $q_2 = q$ and $r = 3d$, $Z=k\frac{q\times q}{(3d)^{2}}=\frac{1}{9}k\frac{q^{2}}{d^{2}}=\frac{1}{9}F$.

Step6: Compare the forces

Comparing the values of $W = 2F$, $X=\frac{1}{4}F$, $Y = 3F$, $Z=\frac{1}{9}F$, we have $Y>W>X>Z$.

Answer:

Y, W, X, Z