which change increases the electric force between objects?\ntwo neutral objects are moved closer…

Explanation:

Step1: Recall Coulomb's law

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

Step2: Analyze each option

• Option 1: Neutral objects have $q_1 = q_2=0$, so $F = 0$ regardless of distance.
• Option 2: Adding electrons to negatively - charged objects increases $|q_1|$ and $|q_2|$. Since $F = k\frac{q_1q_2}{r^2}$, an increase in $|q_1|$ and $|q_2|$ will increase $F$.
• Option 3: Moving oppositely - charged objects farther apart increases $r$. As $F = k\frac{q_1q_2}{r^2}$, an increase in $r$ will decrease $F$.
• Option 4: Reducing the charge on positively - charged objects decreases $q_1$ and $q_2$. Since $F = k\frac{q_1q_2}{r^2}$, a decrease in $q_1$ and $q_2$ will decrease $F$.

Answer:

Electrons are added to two negatively charged objects.