Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -6q. Sphere B carries a charge of -5q. Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it.
To understand the final charge distribution on the spheres after the described sequence of events, let's break down each step and analyze the charge transfers that occur.
Step 1: Spheres A and B are touched together and then separated.
When spheres A and B are touched together, electrons can move from one sphere to another until both reach equilibrium. Since sphere A carries a charge of -6q and sphere B carries a charge of -5q, they have an excess of electrons. When they are touched together, electrons flow from the more negatively charged sphere (A) to the less negatively charged sphere (B) until they both have an equal negative charge. Let's calculate the final charge on spheres A and B after they are separated.
The total charge before the touch is -6q + (-5q) = -11q.
When spheres A and B are separated, the charge is divided equally between the two spheres because they are identical. Therefore, each sphere will have a charge of (-11q)/2 = -5.5q.
Step 2: Sphere C is touched to sphere A and separated from it.
Since sphere A has a charge of -5.5q after the previous step, when sphere C is touched to it, both spheres will equalize their charge. Electrons will flow from the more negatively charged sphere (A) to the initially neutral sphere (C) until they both have an equal negative charge.
As a result, sphere A will lose electrons and its charge will decrease, and sphere C will gain electrons and acquire a negative charge. However, since sphere A and C were initially identical, touching them together will divide the final charge equally between the two spheres.
Thus, both sphere A and sphere C will have a charge of -5.5q/2 = -2.75q.
Step 3: Sphere C is touched to sphere B and separated from it.
In this step, sphere C (which now carries a charge of -2.75q) is touched to sphere B (which still carries a charge of -5.5q).
When they are touched, electrons will redistribute between them until they reach an equal charge. Since sphere B has a lower charge (-5.5q) than sphere C (-2.75q), electrons will flow from sphere C to sphere B.
After they are separated, the final charge distribution will depend on the ratio of charges initially on spheres C and B. The new charges will be proportional to the initial charges as they are being divided yet again.
Using the initial charges of -2.75q for sphere C and -5.5q for sphere B, we find that the ratio is -2.75q / -5.5q = 0.5.
Therefore, in this step, sphere B will gain half of the charge from sphere C, while sphere C will lose half of its charge. As a result:
- Sphere B will have a final charge of -5.5q + (0.5 * 2.75q) = -5.5q + 1.375q = -4.125q.
- Sphere C will have a final charge of -2.75q - (0.5 * 2.75q) = -2.75q - 1.375q = -4.125q.
In summary, after the described sequence of events:
- Sphere A will have a final charge of -2.75q.
- Sphere B will have a final charge of -4.125q.
- Sphere C will have a final charge of -4.125q.
Please note that the above calculations assume that there is no loss of charge during the transfer process. In real-world scenarios, some charge may be lost due to factors such as leakage or environmental influences.