Respuesta :
Answer:
[tex]Q_A=\frac{3}{8} Q[/tex]
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:
[tex]Q_A=Q_B=\frac{Q}{2}[/tex]
- Now when sphere A is brought in contact with sphere C then the charges on A & C will be:
[tex]Q_A=Q_C=\frac{Q}{4}[/tex]
- Now when again sphere A is brought in contact with sphere B then the charges will be as:
[tex]Q_A=Q_B=(\frac{Q}{4} +\frac{Q}{2} )\div 2[/tex]
[tex]Q_A=Q_B=\frac{3}{8} Q[/tex]
Solution: The final charge on sphere A is Q/4 ( in units of electric charge)
The Energy conservation principle implies that when spheres A and B are in contact the original charge on sphere A will redistribute between the two spheres, as the spheres are identical, the charge will redistribute an equal amount over the two spheres, or sphere A after the separation will keep half of the charge (Q/2) and sphere B ( Q/2).
Now sphere A has Q/2 as charge; we place sphere C and sphere A in contact, and again the charge will redistribute, in that case as charge in sphere A is Q/2 and again the spheres are identical, after the separation spheres A will keep Q/4.