Question

1. observe: now slowly increase q₂ above 0.0 × 10⁶ c. what do you observe, and what does this tell you about the electrostatic force?

Explanation:

Step1: Recall Coulomb's law

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

Step2: Analyze the effect of increasing $q_2$

When $q_2$ is increased above $0.0\times 10^{ - 6}\text{C}$ (assuming $q_1$ and $r$ are constant), according to the formula, the value of the force $F$ will increase. If the charges have the same sign, the repulsive force between them will get stronger. If the charges have opposite signs, the attractive force between them will get stronger.

Answer:

As $q_2$ is increased above $0.0\times 10^{ - 6}\text{C}$, the magnitude of the electrostatic force between the charges (assuming other factors constant) increases. If the charges are like - charged, the repulsive force increases; if they are opposite - charged, the attractive force increases.