Dipole in a sphere

The general solution of the Laplace equation ${\nabla}^2\phi = 0$ (because of absence of any free charge density) in spherical coordinates using principle of superposition (adding the extra potential term due to dipole ) is given in terms of spherical coordinates as follows:
$\phi\left(r, \theta, \phi\right) = \begin{cases} \phi_1\left(r, \theta\right) = \displaystyle\sum_{l = 0}^{\infty}\left[\left(A_lr^l + \dfrac{B_l}{r^{l+1}}\right)P_l(\cos\theta)\right] + \dfrac{p\cos\theta}{4\pi\epsilon_{0}r^2} & r < R \\ \phi_2\left(r, \theta\right) = \displaystyle\sum_{l = 0}^{\infty}\left[\left(D_lr^l + \dfrac{C_l}{r^{l+1}}\right)P_l(\cos\theta)\right] & r > R \\ \end{cases}$
Note that the dipole term is only added in $\phi_1$ , since it is account for in $\phi_2$ in the constant $C_1$ . Also,the solution is independent of the azimuthal angle $\phi$ because of azimuthal symmetry. Due to the same reason, the solution is expressed in associated Legendre polynomials instead of spherical harmonics .

Now, the solution should be indicative of the following facts: - The potential should go to zero at an infinitely long distance from the dipole.
- The potential must be finite in the absence of the dipole at $r = 0$ .

For abiding by point 1, $D_l$ should be set to zero. For abiding by point 2, $B_l$ should be set to zero (since dipole term is accounted for separately). So our solution now looks like: $\phi\left(r, \theta, \phi\right) = \begin{cases} \phi_1\left(r, \theta\right) = \displaystyle\sum_{l = 0}^{\infty}\left[\left(A_lr^l\right)P_l(\cos\theta)\right] + \dfrac{p\cos\theta}{4\pi\epsilon_{0}r^2} & r < R \\ \phi_2\left(r, \theta\right) = \displaystyle\sum_{l = 0}^{\infty}\left[\left(\dfrac{C_l}{r^{l+1}}\right)P_l(\cos\theta)\right] & r > R \\ \end{cases}$

Now, we apply the dielectric boundary conditions:
1) $\epsilon_{0}\epsilon_r E_{1r} = \epsilon_{0}E_{2r}$ at $r=R$ due to absence of a surface charge density of the sphere $r = R$ .
2) $E_{1t} = E_{2t}$ at $r=R$ .
where $E_{1r}$ and $E_{2r}$ are the electric field in radial direction for $r < R$ and $r > R$ respectively and $E_{1t}$ and $E_{2t}$ are the electric field in tangential direction for $r < R$ and $r > R$ respectively.

Using the first boundary condition, we get:
$\begin{aligned} \epsilon_{0}\epsilon_r E_{1r} &= \epsilon_{0}E_{2r}\\ \epsilon_r \left(-\dfrac{\partial\phi_1}{\partial r}\right)_{r=R} & = \left(-\dfrac{\partial\phi_2}{\partial r}\right)_{r=R}\\ \epsilon_r \left(\displaystyle \sum_{l=0}^{\infty}\left[l A_lr^{l-1}P_l(\cos\theta)\right] - \dfrac{p\cos\theta}{2\pi\epsilon_0 R^3}\right) & = \displaystyle \sum_{l=0}^{\infty} \left[-(l+1)\dfrac{C_l}{R^{l+2}}P_l(\cos\theta)\right]\\ \\ \text{For } l\neq 1\\ \epsilon_r\left(l A_l R^{l-1}\right) & = -(l+1)\dfrac{C_l}{R^{l+2}}\\ C_l & = \dfrac{-l}{l+1}\epsilon_r R^{2l+1}A_l \longrightarrow \boxed{1}\\ \\ \text{For } l = 1 & \\ \epsilon_r\left(A_1\cos\theta - \dfrac{p\cos\theta}{2\pi\epsilon_0 R^3}\right) & = \dfrac{-2C_1\cos\theta}{R^3}\\ C_1 & = \epsilon_r\left(\dfrac{p}{4\pi\epsilon_0} - \dfrac{A_1R^3}{2}\right) \longrightarrow \boxed{2}\\ \end{aligned}$

Using the second boundary condition, we get: $\begin{aligned} E_{1\theta} &= E_{2\theta} \text{(only in }\theta\text{ direction because of azimuthal symmetry)}\\ \left(-\dfrac{\partial\phi_1}{\partial \theta}\right)_{r=R} & = \left(-\dfrac{\partial\phi_2}{\partial \theta}\right)_{r=R}\\ \displaystyle \sum_{l=0}^{\infty}\left[A_l R^l \dfrac{d\left(P_l(\cos\theta)\right)}{d\theta}\right] - \dfrac{p \sin\theta}{4\pi\epsilon_0 R^2} & = \displaystyle \sum_{l=0}^{\infty}\left[\dfrac{C_l}{R^{l+1}} \dfrac{d\left(P_l(\cos\theta)\right)}{d\theta}\right]\\ \\ \text{For } l\neq 1\\ A_l R^l &= \dfrac{C_l}{R^{l+1}} \longrightarrow \boxed{3}\\ \\ \text{For } l = 1\\ A_1R + \dfrac{p}{4\pi\epsilon_0R^2} & = \dfrac{C_1}{R^2}\\ C_1 & = A_1 R^3 + \dfrac{p}{4\pi\epsilon_0} \longrightarrow \boxed{4}\\ \end{aligned}$ From equations $\boxed{1}$ and $\boxed{3}$ , $A_l = C_l = 0 \ \forall \ l\neq 1$ .
From equations $\boxed{2}$ and $\boxed{4}$ , $A_1 = \dfrac{p\left(\epsilon_r - 1\right)}{2\pi\epsilon_0 R^3 \left(\epsilon_r + 2\right)}$ and $C_1 = \dfrac{3 p \epsilon_r}{4\pi\epsilon_0\left(\epsilon_r + 2\right)}$ .

Using the above results and substituting in the solution form of potential, we get: $\phi\left(r, \theta, \phi\right) = \begin{cases} \phi_1\left(r, \theta\right) = \dfrac{p\cos\theta}{4\pi\epsilon_0 r^2}\left[1 + 2\dfrac{r^3}{R^3}\dfrac{\left(\epsilon_r - 1\right)}{\left(\epsilon_r + 2\right)}\right] & r<R \\ \phi_2\left(r, \theta\right) = \dfrac{p\cos\theta}{4\pi\epsilon_0 r^2}\left[\dfrac{3}{\epsilon_r + 2}\right] & r > R \\ \end{cases}$

Since we want to find the potential at a point inside the sphere, we substitute the values of the given parameters into $\phi_1$ with $r = 5\sqrt{2}$ and $\theta = \dfrac{\pi}{4}$ , we get $\phi_1 \approx \color{#69047E}{\boxed{2.8 \times {10}^{12} V}}$ .