To solve the problem, we need to use Coulomb's law, which states that the force [tex]\( F \)[/tex] between two charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( d \)[/tex] is given by:
[tex]\[ F = k \frac{q_1 q_2}{d^2} \][/tex]
where [tex]\( k \)[/tex] is Coulomb's constant. Let's start by calculating the forces in each scenario given in the table.
1. For charges [tex]\( q \)[/tex] and [tex]\( q \)[/tex] separated by distance [tex]\( d \)[/tex]:
[tex]\[
F = k \frac{q \cdot q}{d^2} = k \frac{q^2}{d^2}
\][/tex]
This force is denoted as [tex]\( F \)[/tex].
2. For charges [tex]\( q \)[/tex] and [tex]\( 2q \)[/tex] separated by distance [tex]\( d \)[/tex]:
[tex]\[
W = k \frac{q \cdot 2q}{d^2} = k \frac{2q^2}{d^2} = 2F
\][/tex]
3. For charges [tex]\( q \)[/tex] and [tex]\( q \)[/tex] separated by distance [tex]\( 2d \)[/tex]:
[tex]\[
X = k \frac{q \cdot q}{(2d)^2} = k \frac{q^2}{4d^2} = \frac{F}{4}
\][/tex]
4. For charges [tex]\( 3q \)[/tex] and [tex]\( q \)[/tex] separated by distance [tex]\( d \)[/tex]:
[tex]\[
Y = k \frac{3q \cdot q}{d^2} = k \frac{3q^2}{d^2} = 3F
\][/tex]
5. For charges [tex]\( q \)[/tex] and [tex]\( q \)[/tex] separated by distance [tex]\( 3d \)[/tex]:
[tex]\[
Z = k \frac{q \cdot q}{(3d)^2} = k \frac{q^2}{9d^2} = \frac{F}{9}
\][/tex]
Now, let's rank these forces from greatest to least:
- [tex]\( 3F \)[/tex] (Y)
- [tex]\( 2F \)[/tex] (W)
- [tex]\( \frac{F}{4} \)[/tex] (X)
- [tex]\( \frac{F}{9} \)[/tex] (Z)
So, the list of forces from greatest to least is:
[tex]\[ Y, W, X, Z \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{Y, W, X, Z} \][/tex]
