Answer:
-4.1μC is the final charge on the third sphere
Explanation:
From the given data, q1 and q2 are brought into contact as they are both conductors, as such there will be evenly distribution of charges.
a) charge on each sphere(Q) = q1 + q2 / 2
= +3.8 μC + (- 2.6 μC) / 2 = 1.2μC/2 = 0.6μC
b) Now, one of those two spheres is brought into contact with the third sphere ; Q is brought into contact with q3 = Q + q3 / 2
= 0.6μC - 8.8 μC /2 = -8.2 μC/2
= -4.1μC is the final charge on the third sphere.