Three identical conducting spheres initially have the following charges: sphere A, 4Q; sphere B, -9Q; and sphere C, 0. Spheres A and B are fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere C is touched to sphere A and then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere C is touched to sphere B and then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1?
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To find the ratio of the electrostatic force between spheres A and B at the end of experiment 2 to that at the end of experiment 1, we need to consider how the charges redistribute between the spheres.
In experiment 1:
1. Sphere C is touched to sphere A. When two conductors touch, charge is redistributed between them until their potentials are equal. Since sphere A initially has a greater positive charge, it will transfer some of its charge to sphere C until they have equal charge.
- Sphere A: Charge = 4Q - x (some charge x is transferred to sphere C)
- Sphere B: Charge = -9Q
- Sphere C: Charge = x
2. Sphere C is then touched to sphere B. Again, charge is redistributed until their potentials are equal. Since sphere B initially has a greater negative charge, it will transfer some of its charge to sphere C until they have equal charge.
- Sphere A: Charge = 4Q - x
- Sphere B: Charge = -9Q - x (some charge x is transferred to sphere C)
- Sphere C: Charge = 2x
In experiment 2:
1. Sphere C is touched to sphere B. Since sphere B has a greater negative charge, it will transfer some of its charge to sphere C until they have equal charge.
- Sphere A: Charge = 4Q
- Sphere B: Charge = -9Q - y (some charge y is transferred to sphere C)
- Sphere C: Charge = y
2. Sphere C is then touched to sphere A. Since sphere A has a greater positive charge, it will transfer some of its charge to sphere C until they have equal charge.
- Sphere A: Charge = 4Q - z (some charge z is transferred to sphere C)
- Sphere B: Charge = -9Q - y
- Sphere C: Charge = y + z
Now, let's analyze the forces between the spheres:
In experiment 1:
The electrostatic force between A and B is given by Coulomb's law:
Force(AB1) = k * ((4Q - x) * (-9Q)) / (r^2) ---- (1)
In experiment 2:
The electrostatic force between A and B is given by Coulomb's law:
Force(AB2) = k * ((4Q - z) * (-9Q - y)) / (r^2) ---- (2)
To find the ratio, we can divide equation (2) by equation (1):
Force(AB2) / Force(AB1) = [(4Q - z) * (-9Q - y)] / [(4Q - x) * (-9Q)] ---- (3)
In general, the forces are proportional to the product of the charges involved. Since the spheres have identical charges, the ratio can be simplified as follows:
Force(AB2) / Force(AB1) = [(-9Q - y) * (4 - z)] / [(-9) * (4 - x)] ---- (4)
Hence, the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1 is given by equation (4).
To find the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1, we need to consider the charges distributed on the spheres after each experiment.
In experiment 1:
- When sphere C is touched to sphere A, they will share charges. Since sphere C has a charge of 0, it will acquire a charge of 4Q from sphere A. Now both sphere A and sphere C have a charge of 4Q.
- When sphere C is touched to sphere B, they will also share charges. Sphere C with a charge of 4Q will transfer 4Q of charge to sphere B. Thus, sphere B will now have a charge of -9Q + 4Q = -5Q, while sphere C will have a charge of 4Q - 4Q = 0.
- At the end of experiment 1, sphere A has a charge of 4Q, sphere B has a charge of -5Q, and sphere C has a charge of 0.
In experiment 2 (reversed procedure):
- When sphere C is touched to sphere B, they will share charges. Since sphere C has a charge of 0, it will acquire a charge of -5Q from sphere B. Now both sphere B and sphere C have a charge of -5Q.
- When sphere C is touched to sphere A, they will also share charges. Sphere C with a charge of -5Q transfers -5Q to sphere A. Thus, sphere A will now have a charge of 4Q - 5Q = -Q, while sphere C will have a charge of -5Q - (-5Q) = 0.
- At the end of experiment 2, sphere A has a charge of -Q, sphere B has a charge of -5Q, and sphere C has a charge of 0.
Now, the electrostatic force between two point charges is given by Coulomb's Law: F = (k * q₁ * q₂) / r², where F is the force, k is the electrostatic constant, q₁ and q₂ are the charges, and r is the distance between the charges.
Since the spheres are identical and their separations are much larger than the spheres, the distances between the charges in both experiments (r) can be considered the same.
The ratio of the forces can be found by simply taking the ratio of the products of the charges:
Ratio = (Force in Experiment 2) / (Force in Experiment 1)
= [(k * (-Q) * (-5Q)) / r²] / [(k * 4Q * (-5Q)) / r²]
= (-5Q²) / (-20Q²)
= 1/4
Therefore, the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1 is 1/4.