Answer :
Explanation:
As the given spheres are connected by a thin wire so, the potential on the spheres are the same.
[tex]\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}[/tex] ......... (1)
Hence, total charge will be as follows.
[tex]q_{1} + q_{2}[/tex] = Q = -95.5 nC .......... (2)
Using the above two equations, the final equation will be as follows.
[tex]q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}[/tex]
and, [tex]q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}[/tex]
Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.
[tex]q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}[/tex]
= [tex]\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}[/tex]
= 82.714 nC
Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.