Question

decide how the sketches below would be listed, if they were listed in order of decreasing force between the charges. that is, select \1\ beside the sketch with the strongest force between the charges, select \2\ beside the sketch with the next strongest force between the charges, and so on. note for advanced students: since the units of each charge are not written, you may assume any convenient and reasonable unit, for example coulombs or multiples of e.

Explanation:

Step1: Recall Coulomb's law

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

Step2: Analyze each sketch

Let's assume the grid - squares have side - length $a$.

• For the first sketch: $q_1 = 2$, $q_2=1$, and $r = 2a$. So $F_1=k\frac{2\times1}{(2a)^2}=k\frac{2}{4a^2}=k\frac{1}{2a^2}$.
• For the second sketch: $q_1 = 2$, $q_2 = 1$, and $r = 3a$. So $F_2=k\frac{2\times1}{(3a)^2}=k\frac{2}{9a^2}$.
• For the third sketch: $q_1 = 1$, $q_2 = 1$, and $r = 3a$. So $F_3=k\frac{1\times1}{(3a)^2}=k\frac{1}{9a^2}$.
• For the fourth sketch: $q_1 = 1$, $q_2 = 2$, and $r=\sqrt{(2a)^2+(1a)^2}=\sqrt{4a^2 + a^2}=\sqrt{5}a$. So $F_4=k\frac{1\times2}{(\sqrt{5}a)^2}=k\frac{2}{5a^2}$.

Step3: Compare the magnitudes of the forces

We have $F_1 = k\frac{1}{2a^2}\approx k\frac{4.5}{9a^2}$, $F_2=k\frac{2}{9a^2}$, $F_3=k\frac{1}{9a^2}$, $F_4=k\frac{2}{5a^2}\approx k\frac{3.6}{9a^2}$.
Since $\frac{4.5}{9a^2}>\frac{3.6}{9a^2}>\frac{2}{9a^2}>\frac{1}{9a^2}$, we have $F_1 > F_4>F_2>F_3$.

Answer:

First sketch: 1
Second sketch: 3
Third sketch: 4
Fourth sketch: 2