Charged Solid metal sphere surrounded by a Concentric spherical Shell - Electric Field in different regions

 

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.

Electric field E vs distance r — solid sphere (+Q) inside thick conducting shell