a. A: [tex]\frac{3}{8}q[/tex], B: [tex]-\frac{q}{4}[/tex], C: [tex]\frac{3}{8}q[/tex]
When two conducting spheres touch, the charge on them is redistributed such that the potential on the two spheres is the same:
[tex]V_1 = V_2\\\frac{Q_1}{C_1}=\frac{Q_2}{C_2}[/tex]
where Q refers to the charge and C to the capacitance of the sphere. For identical spheres, the capacitance is the same, so the previous equation becomes
[tex]Q_1 = Q_2[/tex]
which means that the charge distributes equally on the two spheres.
Here initially we have:
Sphere A: charge of q
Sphere B: charge of -q/2
Sphere C: charge of 0
At first, sphere C is touched to sphere B. Since the total charge of the two sphere was
[tex]-\frac{q}{2}+0=-\frac{q}{2}[/tex]
After touching each sphere will have a charge half of this value:
[tex]q_B = q_C = \frac{1}{2}(-\frac{q}{2})=-\frac{q}{4}[/tex]
Then sphere C (charge of -q/4) touches sphere A (charge of +q). So the total charge is
[tex]-\frac{q}{4}+q=+\frac{3}{4}q[/tex]
Since the charge distributes equally, each sphere will receive 1/2 of this charge:
[tex]q_A = q_C = \frac{1}{2}(+\frac{3}{4}q)=\frac{3}{8}q[/tex]
So the final charge on the 3 spheres will be
A: [tex]\frac{3}{8}q[/tex], B: [tex]-\frac{q}{4}[/tex], C: [tex]\frac{3}{8}q[/tex]
b. A: [tex]\frac{1}{2}[/tex], B: [tex]0[/tex], C: [tex]0[/tex]
At first, sphere C is touched to sphere A. Since the total charge of the two sphere was
[tex]q+0=q[/tex]
After touching each sphere will have a charge half of this value:
[tex]q_A = q_C = \frac{1}{2}(q)=\frac{q}{2}[/tex]
Then sphere C (charge of q/2) touches sphere B (charge of -q/2). So the total charge is
[tex]-\frac{q}{2}+q/2=0[/tex]
Since the charge distributes equally, each sphere will receive 1/2 of this charge, which simply means a charge of zero:
[tex]q_B = q_C = 0[/tex]
So the final charge on the 3 spheres will be
A: [tex]\frac{1}{2}[/tex], B: [tex]0[/tex], C: [tex]0[/tex]
