Coulomb's Law PhET Simulation: Investigation of How Magnitude (size) of Charge Affects Electrostatic Force Part 1

Question

For this part, we will be adjusting the size of CHARGE 1 ONLY.
Set the distance between the two charges for 4 cm
Set charge 2 for 8 µC
Given the charge values and the distances, determine the electrostatic force between the charged particles and if they are attracted, repelled, or neither.

**If two charges are attracted, their electrostatic force is negative** **If the two charges are repelled, their electrostatic force is positive**

Answer 1

First, let's state Coulomb's Law:

F = k * |q1 * q2| / r^2

where,
F is the electrostatic force between the charged particles,
k = Coulomb's constant (8.9875 x 10^9 N·m²/C²),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.

Now, you provided the following information:

r = 4 cm (we need to convert it to meters: 0.04 m),
q2 = 8 µC (we need to convert it to coulombs: 8 x 10^-6 C).

Now, let's start investigating the effect of the magnitude (size) of charge 1 (q1) on the electrostatic force and the behavior between the two charged particles. Since there is no specific value given for q1, we can assume a few different scenarios and analyze the results:

Scenario 1 - The magnitude of charge 1 is equal to the magnitude of charge 2 (q1 = 8 µC):
q1 = 8 x 10^-6 C

F = (8.9875 x 10^9 N·m²/C²) * (8 x 10^-6 C * 8 x 10^-6 C) / (0. 04 m)^2
F = 1.44 N (positive)

In this case, the electrostatic force is positive, which means the two charges repel each other.

Scenario 2 - The magnitude of charge 1 is smaller than the magnitude of charge 2 (q1 = 4 µC):
q1 = 4 x 10^-6 C

F = (8.9875 x 10^9 N·m²/C²) * (4 x 10^-6 C * 8 x 10^-6 C) / (0.04 m)^2
F = 0.72 N (positive)

In this case, the electrostatic force is also positive, which means the two charges repel each other, but with less force as q1 decreases.

Scenario 3 - The magnitude of charge 1 is larger than the magnitude of charge 2 (q1 = 12 µC):
q1 = 12 x 10^-6 C

F = (8.9875 x 10^9 N·m²/C²) * (12 x 10^-6 C * 8 x 10^-6 C) / (0.04 m)^2
F = 2.16 N (positive)

In this case, the electrostatic force is still positive, which means the two charges repel each other with increased force as q1 increases.

To conclude, when charge 1 is positive, and the distance and charge 2 are kept constant, the two charges always repel each other. The magnitude of the repulsion (electrostatic force) increases as the magnitude of charge 1 increases.