Question

A solid conducting sphere of radius a is at the center of a hollow conducting sphere of inner radius b and outer radius c. The solid sphere carries a charge q > 0, the outer sphere carries an excess charge of -3q on its outer surface. derive expressions for the magnitude of the electric field in the following regions: Final answers not given. 

Answer 1

The charge distributions in both the sphere and the shell are characterized by having spherical symmetry around a common center, to determine the electric field at several distances r from this center we will use a spherical Gaussian surface in each of the regions.

(a)

To determine inside the solid sphere, consider a Gaussian surface of radius . As inside a conductor in electrostatic equilibrium it is not possible to have charge, we see that ; consequently, by Gauss's law and by the symmetry for .

(b)

In the region between the surface of the solid sphere and the inner surface of the shell, we construct a spherical Gaussian surface of radius , where . Where the charge inside this surface is , which is the charge of the solid sphere. Due to the spherical symmetry, the electric field lines must be directed radially outward and of constant magnitude on the Gaussian surface.

is parallel to in each of the points. Therefore, and Gauss's law gives us

By symmetry is constant anywhere on the surface, so it can come out of the integral. So,

Where we have used the fact that the area of the surface of the sphere is . Now we solve depending on the electric field:

(c)

In this region the electric field of be equal to zero (), since it is a spherical shell is also a condutor in equilibrium ().

(d)

In this region, where , the spherical Gaussian surface that we will use surrounds a total charge equal to . Therefore, the application of Gauss's law to this surface gives us