Consider a solid conducting sphere with a radius 0.7 cm and charge ?7.3 pC on it. There is a conducting spherical shell concentric to the sphere. The shell has an inner radius 2.8 cm(with 2.8 cm > 0.7 cm) and outer radius 5 cm and a net charge 27.9 pC on the shell. Denote the charge on the inner surface of the shell by Q?2 and that on the outer surface of the shell by Q??2. Find the charge Q??2. Answer in units of pC

To find the charge on the outer surface of the conducting spherical shell, we need to apply the principles of electrostatics, particularly Gauss's law and the behavior of conductors in electrostatic equilibrium. Let's break this down step by step.

Understanding the System

We have a solid conducting sphere with a radius of 0.7 cm and a charge of -7.3 pC. Surrounding this sphere is a conducting spherical shell with an inner radius of 2.8 cm and an outer radius of 5 cm, which has a total charge of +27.9 pC. The charges on the inner and outer surfaces of the shell will be denoted as Q_inner and Q_outer, respectively.

Applying Gauss's Law

According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero. This means that any excess charge resides on the outer surface of the conductor. The charge distribution will adjust itself so that the electric field inside the conducting material is zero.

Charge Distribution

• The charge on the solid sphere is -7.3 pC.
• To maintain electrostatic equilibrium, the inner surface of the shell must have a charge that exactly cancels the electric field produced by the solid sphere. Therefore, the charge on the inner surface of the shell, Q_inner, will be +7.3 pC.
• Now, since the total charge on the shell is +27.9 pC, we can find the charge on the outer surface of the shell.

Calculating Q_outer

The total charge on the shell is the sum of the charges on its inner and outer surfaces:

Q_total = Q_inner + Q_outer

Substituting the known values:

27.9 pC = 7.3 pC + Q_outer

To find Q_outer, we rearrange the equation:

Q_outer = 27.9 pC - 7.3 pC

Calculating this gives:

Q_outer = 20.6 pC

Final Result

Thus, the charge on the outer surface of the conducting spherical shell, Q_outer, is 20.6 pC.