In the Atomic Scale setting of the simulation, what happens to the force between q1=−2e and q2=4e when the distance between them is doubled?
The force weakens by a factor of 2.
The force weakens by a factor of 4.
The force strengthens by a factor of 2.
The force strengthens by a factor of 4
To determine what happens to the force between two charges, q1 and q2, as the distance between them is doubled, we need to apply Coulomb's Law.
Coulomb's Law states that the force (F) between two charged particles is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them.
Mathematically, Coulomb's Law can be expressed as:
F = k * (q1 * q2) / r^2
where k is the electrostatic constant.
In our scenario, q1 = -2e and q2 = 4e, where e is the elementary charge. Let's assume e = 1 (for simplicity).
Now, if we double the distance between the charges, the new distance (2r) will be twice the original distance (r). Plugging this into Coulomb's Law, we have:
F' = k * (q1 * q2) / (2r)^2
Simplifying this equation, we get:
F' = k * (q1 * q2) / (4 * r^2)
Comparing this with the original force (F), we can see that the force (F') is equal to F divided by 4. Therefore, we can conclude that the force weakens by a factor of 4 when the distance between the charges is doubled.
Finally, based on the given answer choices, the correct option would be:
The force weakens by a factor of 4.