An uncharged conducting sphere of radius ais coated with a thick

insulating shell (dielectric constant ${\mathit{\epsilon }}_{\mathbf{r}}$) out to radius b.This object is now placed in an otherwise uniform electric field ${\overrightarrow{\mathbf{E}}}_{\mathbf{0}}$. Find the electric field in the insulator.

The external electric field is ${\overrightarrow{E}}_0$.

The radius of conducting sphere is $a$.

The dielectric constant of the insulating shell coated on the sphere up to radius $b$is ${\epsilon }_r$.

The potential inside a conductor is zero.

The potential in free space in the presence of an electric field ${\overrightarrow{E}}_0$is

$\mathbf{V}\mathbf{=}\mathbf{-}{\mathbf{E}}_{\mathbf{0}}\mathbf{rcos\theta }\mathbf{+}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{r}}^{\frac{{\mathbf{B}}_{\mathbf{I}}}{\mathbf{I}\mathbf{+}\mathbf{1}}}{\mathbf{P}}_{\mathbf{1}}\left(\mathrm{cos\theta }\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, $r$and $\theta$are spherical polar coordinates.

The potential inside a dielectric is represented in general as

$\mathbf{V}\mathbf{=}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\left(A_I{r}^I+r^{\frac{B_1}{I+1}}\right){\mathbf{P}}_{\mathbf{I}}\left(\mathrm{cos\theta }\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here, $P_I$is the Lagrange polynomial and $A_I$and $B_I$are its coefficients.

Following equation (1), the potential outside the sphere is

$V_{out}=-E_{0}r\cos \theta +\sum _{I-0}^{\infty }{r}^{\frac{B_I}{I+1}}{P}_I\left(\cos \theta \right)$

Following equation (2), the potential inside the dielectric is

$V_{mind}=\sum _{i=0}^{\infty }\left(A_I{r}^i+\frac{B_I}{r^I+1}\right)P_I\left(\cos \theta \right)$

The potential inside the sphere is

$V_{in}=0$

The boundary conditions are

$$\begin{gathered} V_{out}\left(r=b\right)=V_{mid}\left(r=b\right).....\left(3\right) \\ {\epsilon }_0\frac{\partial V_{out}}{\partial r}\left(r=b\right)={\epsilon }_r\frac{\partial V_{mid}}{\partial r}\left(r=b\right).....\left(4\right) \\ {V}_{mid}\left(r=a\right)=V_{in}\left(r=a\right).....\left(5\right) \end{gathered}$$

From equation (3) it follows

$-E_{0}b\cos \theta +\sum _{I-0}^{\infty }{b}^{\frac{B_I}{I+1}}{P}_I\left(\cos \theta \right)=\sum _{I-0}^{\infty }\left(A_I{b}^I+\frac{B_I}{b^{I+1}}\right)P_I\left(\cos \theta \right).....\left(6\right)$

From equation (4) it follows

localid="1657603467848" $-E_0\cos \theta -\left(I+1\right)\sum _{I-0}^{\infty }{b}^{\frac{B_I}{I+2}}{P}_I\left(\cos \theta \right)={\epsilon }_0\sum _{I-0}^{\infty }\left(I{A}_I{b}^{I-1}-\left(I+1\right)\frac{B_I}{b^{I-2}}\right)P_I\left(\cos \theta \right)...\left(7\right)$

From equation (5) it follows

$\sum _{I-0}^{\infty }\left(A_I{a}^I+\frac{B_I}{a^{I+1}}\right)P_I\left(\cos \theta \right)=0.....\left(8\right)$

From equation (6) and for$I\ne 1$

$B_I=A_I\left(b^{2I+1}-a^{2I+1}\right).....\left(9\right)$

From equation (7) and for$I\ne 1$

$$\begin{gathered} A_I=B_I \\ =0.....\left(10\right) \end{gathered}$$

From equation (6) and for $I=1$

$B_1-E_0{b}^3=A_1\left(b^3-a^3\right).....\left(11\right)$

From equation (7) and for $I=1$

$-2B_1-E_0{b}^3={\epsilon }_r{A}_1\left(b^3+2a^3\right).....\left(12\right)$

From equations (11) and (12)

$A_1=\frac{-3E_0}{2\left[1+{\left({\displaystyle \frac{a}{b}}\right)}^3\right]+{\epsilon }_r\left[1+2{\left({\displaystyle \frac{a}{b}}\right)}^3\right]}$

Substitute this and equation (10) into the potential for the mid section

$V_{mid}=\frac{-3E_0}{2\left[1+{\left({\displaystyle \frac{a}{b}}\right)}^3\right]+{\epsilon }_r\left[1+2{\left({\displaystyle \frac{a}{b}}\right)}^3\right]}\left(r-\frac{a^3}{r^2}\right)\cos \theta$

The expression for the electric field in the mid section is

localid="1657603377788" $$\begin{gathered} \overrightarrow{E}=-\nabla V_{mid} \\ =\frac{3E_0}{2\left[1+{\left({\displaystyle \frac{a}{b}}\right)}^3\right]+{\epsilon }_r\left[1+2{\left({\displaystyle \frac{a}{b}}\right)}^3\right]}\left\{\left(1+2\frac{a^3}{r^3}\right)\cos \theta \stackrel{⏜}{r}-\left(1-\frac{a^3}{r^3}\right)\sin \theta \stackrel{⏜}{r}\right\} \end{gathered}$$

Thus, the electric field is$=\frac{3E_0}{2\left[1+{\left({\displaystyle \frac{a}{b}}\right)}^3\right]+{\epsilon }_r\left[1+2{\left({\displaystyle \frac{a}{b}}\right)}^3\right]}\left\{\left(1+2\frac{a^3}{r^3}\right)\cos \theta \stackrel{⏜}{r}-\left(1-\frac{a^3}{r^3}\right)\sin \theta \stackrel{⏜}{r}\right\}$.