Three identical conducting spheres A, B, and C are labeled with initial charges A: q, B: -q/2, C: 0. If C is touched to B, then C is removed and touched to A, what is the final charge on each sphere?

• A. 3q/8 on each sphere
• B. q/2 on each sphere
• C. 3q/4 on each sphere
• D. 0 on each sphere

Answer: (A)

When two identical conducting spheres touch, charges flow between them until both have the same electric potential. For identical spheres, that means they share the total charge they carried when they were connected, giving equal amounts of charge to each.

First, C and B touch. Their combined charge is 0 + (-q/2) = -q/2. Since the spheres are identical, they split this equally, so each of B and C ends up with -q/4.

Next, C (now with -q/4) touches A (with q). The combined charge is q + (-q/4) = 3q/4. Again, identical spheres share equally, so A and C each get (3q/4)/2 = 3q/8.

Thus, after both steps, A has 3q/8, C has 3q/8, and B remains at -q/4. The total charge is 3q/8 + (-q/4) + 3q/8 = q/2, which matches the initial total.