Two charges, Q1 and Q2, are held a fixed distance apart. Q1 is exactly twice Q2. You do an experiment: Part a) you double Q1 and measure the force between the charges. Part b) You return Q1 to the original charge, but double Q2 and measure the force between the two charges. How would the force measured in Part a) compare to the force measured in Part b)?
1. The force measured in Part a) would be the same as the force measured in Part b).
2. The force measured in Part a) would be greater than the force measured in Part b).
3. The force measured in Part a) would be smaller than the force measured in Part b).
4. There is no way to tell how the forces in Part a) and Part b) compare since the original force between Q1 and Q2 was never measured.
k Q1 Q2 / d^2
2Q1 Q2 = k Q1 2Q2
To determine the answer, let's analyze the situation using Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Given that Q1 is exactly twice Q2, we can denote the charges as Q1 = 2Q and Q2 = Q, where Q is a constant. The distance between the charges remains fixed.
In Part a), when Q1 is doubled, the new charge becomes 2(2Q) = 4Q. Let's denote the force measured in Part a) as Fa.
In Part b), Q1 is returned to its original charge (2Q), but Q2 is doubled to 2Q. Let's denote the force measured in Part b) as Fb.
According to Coulomb's Law, the force can be expressed as:
Fa = k * (4Q * Q) / d^2
Fb = k * (2Q * 2Q) / d^2
Here, k is Coulomb's constant, and d represents the distance between the charges. We can see that the only difference between Fa and Fb lies in the charges involved.
Simplifying the equations, we get:
Fa = 16Q^2 * k / d^2
Fb = 8Q^2 * k / d^2
Since 16Q^2 > 8Q^2, we can conclude that the force measured in Part a) (Fa) would be greater than the force measured in Part b) (Fb).
Therefore, the correct answer is 2. The force measured in Part a) would be greater than the force measured in Part b).