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Your whole derivation is correct. Even in the presence of two different dieletric materials, the $\mathbf{D}$ field will not be affected, but for the free charge density that you already dealt with. So the field will be $$\mathbf{D}=\begin{cases}\dfrac{1}{5}\beta r^3\hat{\mathbf{r}},&\text{if $r<a$},\\ \dfrac{1}{5}\beta ...


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We start with the integral $$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$ Since we are integrating over $\vec{r}$, we can without loss of generality, arrange for $\vec{r}'$to lie along the $+Z$ axis, so that $\vec{r}'=r'\hat{z}$. Then the angle between $\vec{r}$ and $\vec{r}'$ is the standard angle (in spherical coordinates) ...


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To solve the integral, I came across the multipole expansion $$\frac{1}{\left|\vec{r}-\vec{r}_{0}\right|}=\sum_{l=0}^{\infty}\frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}\left(\cos\sphericalangle\left(\vec{r},\vec{r}_{0}\right)\right),\;\; r_{<}=\min\left(r,r_{0}\right),\; r_{>}=\max\left(r,r_{0}\right)$$ With $\vec{r}_0=\vec{r}^{\prime}$ one has ...


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The covariant formulation of EM is precisely this. The formulation as a gauge theory also does this. ($c = 1$ in the following) Given the $E$- and $B$-fields as spatial three-vectors in some frame, we construct the antisymmetric field strength tensor as (roman indices are spatial indices, summation over repeated indices implied) $$ F^{0i} := E^i \; ...



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