Question

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First, sphere A is brought into contact with sphere B and then separated from it. After that, sphere A is brought into contact with sphere C and then separated from it. Finally, sphere A is brought into contact with sphere B again, and then separated from it

What is final charge of sphere A?

Answer 1

The final charge on sphere A, after it is brought into contact with spheres B and C and then brought into contact with sphere B again, is Q/8.

Step-by-step explanation:

Charge Distribution on Metallic Spheres

When two identical conducting spheres come into contact, the total charge is redistributed evenly between them. This is because as conductors, the charge is free to move and will do so until the system reaches a state of electrostatic equilibrium, where the charges have no net force acting upon them. The spheres will then each have half of the total initial charge once they are separated.

For the situation with spheres A and B: sphere A with an initial charge Q is brought into contact with neutral sphere B, resulting in each sphere having a charge of Q/2 after separation. When sphere A, now with charge Q/2, is subsequently brought into contact with another neutral sphere C, the charge Q/2 is then shared equally, leaving both spheres with a charge of Q/4. Finally, when sphere A with charge Q/4 is brought into contact with sphere B, which also has charge Q/4 at this point, the resulting charge on each sphere after separation will be Q/8.

Therefore, the final charge on sphere A is Q/8 after undergoing these contacts and separations.

Answer 2

`Q_A=(3)/(8) Q`

Step-by-step explanation:

Given that sphere A has Q charge on it.

Spheres B & C are neutral.

After the sphere A & B are brought in contact, the charges flow till both the metallic spheres attain the same quantity of charge.

So,

• Charge on the each of the metallic spheres A & B after initially brought in contact:

`Q_A=Q_B=(Q)/(2)`

• Now when sphere A is brought in contact with sphere C then the charges on A & C will be:

`Q_A=Q_C=(Q)/(4)`

• Now when again sphere A is brought in contact with sphere B then the charges will be as:

`Q_A=Q_B=((Q)/(4) +(Q)/(2) )/ 2`

`Q_A=Q_B=(3)/(8) Q`