Concept: Notice that the charge Q is given only in the solid metal
sphere. Not to the outer spherical shell. However -Q is induced in the
inner surface of the shell and +Q induced in the outer surface of the
shell, due to the charge Q in the solid metal sphere.
Apply Gauss law in different regions to arrive at the Electric Field in
that region. \(\phi_E=\dfrac{q_{encl.}}{\epsilon_{o}}\)
0 < r < R, $\hspace{0.5 cm}q_{encl.}=0, \hspace{0.5cm}\phi_E=0,
\hspace{0.5cm}E=0$ as all charges will be in the outer most
surface for a solid metal sphere.
R < r < a, $\hspace{0.5cm}q_{encl.}=Q,
\hspace{0.5cm}\phi_E=\dfrac{Q}{\epsilon_{o}},
\hspace{0.5cm}E=k\dfrac{Q}{r^2}$ Here we see E is
inversely proportional to $r^2$
a < r < b, $\hspace{0.5cm}q_{encl.}=(Q)+(-Q)=0,
\hspace{0.5cm}\phi_E=0, \hspace{0.5cm}E=0$ note that the Gaussian surface
in this region will enclose only the metal sphere Q and inner surface of
the shell -Q. Not the outer shell +Q. For non-conducting shell with a dielectric constant K, the E is not equal to zero, Instead it will be Eext. - Eint.
b < r < $\infty\hspace{0.5cm}$$q_{encl.}=Q+(-Q)+(+Q)=Q, \hspace{0.5cm}\phi_E=\dfrac{Q}{\epsilon_{o}},
\hspace{0.5cm}E=k\dfrac{Q}{r^2}$ here the Gaussian surface will enclose all
the three charge regions.