# Solve a problem without using separation of variables

There's an additional exercise from Introduction to Electrodynamics by Griffith.

Problem 4.34 A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius $R$ and dielectric constant $K$). Find the electric potential inside and outside the sphere.

Now this problem can be solved by using separation of variables method, but I want to develop somewhat straightforward method other than separation of variables.

So my idea is like this: As the medium is linear dielectric, $$P=\epsilon_0(K-1)E$$ and $$E=E_p + E_D$$ where $E_p$ is electric field by the point dipole and $E_D$ is by polarized dielectric.

Now there's no net charge other than the point dipole and the surface charge of the sphere, let's consider the surface charge density of the sphere. $$\sigma(r')=P \cdot \hat{r'}$$ $$=\epsilon_0(K-1)E \cdot \hat{r'}$$ $$=\epsilon_0(K-1)(E_p + E_D) \cdot \hat{r'}$$ but since $E_p$ is given as $$E_p (r') = {1 \over 4\pi \epsilon_0} {[3(p \cdot \hat{r'})\hat{r'}-p]\over r'^3},$$ $$E_p (r') \cdot \hat{r'} = {1 \over 4\pi \epsilon_0} {2p \cdot \hat{r'}\over R^3}$$ $$={2V_p(r')\over R}$$ where $V_p$ is electric potential by the point dipole. So $\sigma$ can be rewritten as $$\sigma(r')=\epsilon_0(K-1)(2V_p(r')/R + E_D \cdot \hat{r'})$$. So now we can calcaulate electric potential $V_D$ by dielectric. $$V_D={1 \over 4\pi \epsilon_0}\oint_S{\sigma dA'\over \eta}$$ (where $\eta$ is separation vector given as $\eta = r - r'$ (or its magnitude by context)) $$={\epsilon_0(K-1) \over 4\pi \epsilon_0}\oint_S{(2V_p(r')/R + E_D \cdot \hat{r'}) dA'\over \eta}$$ Now let's focus on the second term of the integrand. $$\oint_S {E_D \cdot \hat{r'}\over \eta} dA' = \oint_S {E_D \over \eta} \cdot (\hat{r'} dA')$$ $$=\oint_S {E_D \over \eta} \cdot d\vec{A'}$$ (which means the scalar $dA'$ is now turned into the vector $d\vec{A'}$. Let me omit the vector notation for convenience after this.) But since $E_D=-\nabla' V_D$, $$\oint_S {E_D \over \eta} \cdot d\vec{A'}=-\oint_S {\nabla' V_D \over \eta} \cdot d\vec{A'}$$ (By Gauss's theorem,) $$=-\int_V \nabla' \cdot {\nabla' V_D \over \eta} d\tau'$$ (By the formula $\nabla fg = f \nabla g + g \nabla f, \nabla'^2V_D = 0$ in $V$) $$=-\int_V \nabla' ({1 \over \eta}) \cdot (\nabla' V_D ) d\tau'$$ (Since $(\nabla f) \cdot A = -\nabla \cdot (fA) + f \nabla \cdot A$ ,) $$=\int_V V_D \nabla'^2 ({1 \over \eta}) d\tau' - \oint_S V_D \nabla'({1 \over \eta}) \cdot dA'$$ But as we know that $$\nabla'^2({1\over \eta})=-4\pi\delta^3(\eta),$$ $$\int_V V_D \nabla'^2 ({1 \over \eta}) d\tau'=-4\pi\int_V V_D \delta^3(\eta) d\tau'$$ $$=-4\pi V_D(r)$$(if r is in V) $$=0$$(if r is out of V)

And the second term is $$\oint_S V_D \nabla'({1 \over \eta}) \cdot dA' = \oint_S V_D ({\hat{\eta} \over \eta^2}) \cdot dA' = \oint_S V_D d\Omega' = 4\pi\bar{V_D}$$ which means that the integral is $(4\pi)\cdot$(Average of $V_D$ over solid angle measured from $r$) So $V_D$, the potential that we wanted to find, is $$V_D={(K-1) \over 4\pi }\oint_S{(2V_p(r')/R) dA'\over \eta}+(K-1)(- V_D(r)+{\bar{V_D}})$$ in the case where $r$ is in V. In turn, $$K \cdot V_D(r) - (K-1) {\bar{V_D}}(r)={(K-1) \over 4\pi R }\oint_S{2V_p(r') dA'\over \eta}$$

Now, this is the point where I'm stuck. My problem will be solved if:

$${\bar{V_D}}(r) = {2\over 3}V_D(r)$$ $$\oint_S{2V_p(r') dA'\over \eta}={2\over 3}{r^3\over R^2}V_p(r)$$