decide how the sketches below would be listed, if they were listed in order of decreasing force between the…

Explanation:

Step1: Recall Coulomb's law

The force between two charges $F = k\frac{q_1q_2}{r^2}$, where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between them.

Step2: Analyze the first - type of sketches

For the sketches with charges of magnitude $+ 1$ and $+1$: The force is inversely proportional to the square of the distance between the charges. The closer the charges, the stronger the force.

Step3: Compare the distances

Let's assume the side - length of each small square in the grid is $d$.

• In the first sketch (top - left), if we assume the distance between the two $+1$ charges is $r_1 = 3d$.
• In the second sketch (top - right), the distance between the two $+1$ charges is $r_2=d\sqrt{(1)^2+(1)^2}=\sqrt{2}d$.
• In the third sketch (bottom - left), the distance between the two $+1$ charges is $r_3 = 2d$.
• In the fourth sketch (bottom - right), using Coulomb's law with $q_1 = 1$ and $q_2 = 2$, and assume the distance $r_4=\sqrt{(1)^2+(1)^2}=\sqrt{2}d$. The force $F_4=k\frac{1\times2}{(\sqrt{2}d)^2}=k\frac{2}{2d^2}=k\frac{1}{d^2}$. For the first three sketches with $q_1 = q_2=1$:
• $F_1 = k\frac{1\times1}{(3d)^2}=k\frac{1}{9d^2}$
• $F_2=k\frac{1\times1}{(\sqrt{2}d)^2}=k\frac{1}{2d^2}$
• $F_3=k\frac{1\times1}{(2d)^2}=k\frac{1}{4d^2}$

Step4: Rank the forces

Comparing the magnitudes of the forces: $F_4>F_2>F_3>F_1$.

Answer:

Bottom - right: 1 (strongest) Top - right: 2 Bottom - left: 3 Top - left: 4 (weakest)