Question 24.7: A Sphere Inside a Spherical Shell A solid insulating sphere ......

Conceptualize Notice how this problem differs from Example 24.3. The charged sphere in Figure 24.10 appears in Active Figure 24.17, but it is now surrounded by a shell carrying a charge -2 Q.

Categorize The charge is distributed uniformly throughout the sphere, and we know that the charge on the conducting shell distributes itself uniformly on the surfaces. Therefore, the system has spherical symmetry and we can apply Gauss’s law to find the electric field in the various regions.

Analyze In region ②–between the surface of the solid sphere and the inner surface of the shell-we construct a spherical gaussian surface of radius r, where a<r<b, noting that the charge inside this surface is +Q (the charge on the solid sphere). Because of the spherical symmetry, the electric field lines must be directed radially outward and be constant in magnitude on the gaussian surface.

The charge on the conducting shell creates zero electric field in the region r<b, so the shell has no effect on the field due to the sphere. Therefore, write an expression for the field in region ② as that due to the sphere from part (A) of Example 24.3:

E_{2}=k_{e} \frac{Q}{r^{2}} \quad(\text { for } a<r<b)

Because the conducting shell creates zero field inside itself, it also has no effect on the field inside the sphere. Therefore, write an expression for the field in region ① as that due to the sphere from part (B) of Example 24.3:

E_{1}=k_{e} \frac{Q}{a^{3}} r \quad(\text { for } r<a)

In region ④, where r>c, construct a spherical gaussian surface; this surface surrounds a total charge q_{\text {in }}=Q+ (-2 Q)=-Q. Therefore, model the charge distribution as a sphere with charge -Q and write an expression for the field in region ④ from part (A) of Example 24.3:

E_{4}=-k_{e} \frac{Q}{r^{2}} \quad(\text { for } r>c)

In region ③, the electric field must be zero because the spherical shell is a conductor in equilibrium:

E_{3}=0 \quad(\text { for } b<r<c)

Construct a gaussian surface of radius r, where b<r<c, and note that q_{\text {in }} must be zero because E_{3}=0. Find the amount of charge q_{\text {inner }} on the inner surface of the shell:

\begin{aligned}& q_{\text {in }}=q_{\text {sphere }}+q_{\text {inner }} \\& q_{\text {inner }}=q_{\text {in }}-q_{\text {sphere }}=0-Q=-Q\end{aligned}

Finalize The charge on the inner surface of the spherical shell must be -Q to cancel the charge +Q on the solid sphere and give zero electric field in the material of the shell. Because the net charge on the shell is -2 Q, its outer surface must carry a charge -Q