A solid conductor sphere of radius ‘a’ has a positive net charge 2Q as shown in Figure A. A conductor spherical shell of inner radius ‘b’ and external radius ‘c’ has the same center with the solid sphere and has a net charge -Q. Using Gauss' Law determine the electric field in the regions denoted as (1), (2), (3) & (4) in the Figure B, and the charge distribution over the shell when all the system is in electrostatic equilibrium.

To determine the electric field in the different regions and the charge distribution over the shell in this system, we can use Gauss' Law. Gauss' Law states that the net electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of the medium.

For region (1):
First, we draw a Gaussian surface enclosing the solid conductor sphere. The electric field within a conductor in electrostatic equilibrium is zero. Therefore, the electric field in region (1) is zero.

For region (2):
We again draw a Gaussian surface enclosing the solid conductor sphere. Since the electric field within a conductor in electrostatic equilibrium is zero, the electric field in region (2) is also zero.

For region (3):
Now, we draw a Gaussian surface enclosing the conductor spherical shell. The electric field within a conductor in electrostatic equilibrium is again zero. Therefore, the electric field in region (3) is zero.

For region (4):
We draw a Gaussian surface inside the conductor shell but outside the solid conductor sphere. Due to the spherical symmetry of the system, the electric field at any point on this Gaussian surface will be the same and directed radially outwards or inwards. Let's consider it as directed outwards.
The Gaussian surface encloses the conductor shell, which has a net charge of -Q. Therefore, the net charge enclosed by the Gaussian surface is -Q.
The net electric flux through the Gaussian surface is related to the electric field and the enclosed charge by Gauss' Law as follows:
Net flux = Electric field * Surface area = (Electric field)(4πr^2)
Since the electric field is radially outward, the direction of the surface area vector is also radially outward. Thus, the angle between the electric field and the surface area is 0 degrees, and the cosine of the angle is 1.
Therefore, Net flux = (Electric field)(4πr^2) * 1 = 4πr^2(Electric field)

According to Gauss' Law, the net electric flux through the closed Gaussian surface is equal to the net charge enclosed divided by the permittivity of the medium. Therefore, we have:
4πr^2(Electric field) = (-Q) / ε₀ (where ε₀ is the permittivity of free space)

Simplifying this equation, we find:
Electric field = (-Q) / (4πε₀r^2)

This equation gives us the magnitude of the electric field in region (4). The negative sign indicates that the electric field is directed inwards into the Gaussian surface.

Regarding the charge distribution over the shell when the system is in electrostatic equilibrium, it can be determined by considering the fact that in equilibrium, the electric field inside a conductor is zero. Since the solid conductor sphere has a positive net charge of 2Q and the conductor spherical shell has a net charge of -Q, the charges on the inner and outer surfaces of the shell will redistribute themselves such that the net electric field inside the shell is zero.

For the charge distribution, the positive net charge on the solid sphere induces an equal amount of negative charge on the inner surface of the shell. This negative charge distribution on the inner surface of the shell cancels out the positive charge inside the shell, resulting in zero electric field within the shell.

Therefore, the charge distribution on the inner surface of the shell will be -Q, while the outer surface will have no charge.

Note: It's important to remember that Gauss' Law is based on the assumption of a closed surface, ideal conductors, and electrostatic equilibrium.