Question

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm and a net charge of − 95.5 nC, whereas sphere B, which is initially neutral, has a radius of 74.4 cm. The two spheres are then connected by a thin metal wire. What is the charge on sphere B after equilibrium has been reached?

Answer 1

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          $\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

              $q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          $q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and,    $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          $q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

                     = $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.