The uniformly charged ball: interior and exterior fields, continuous potential and finite energy

Setup: the density is bounded and supported in ${\bar{B}}_a$, hence in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$; its total charge is $\int \rho =\frac{3Q}{4\pi a^3}\frac{4\pi a^3}{3}=Q$ by [F1]; the sphere is null, so the value of $\rho$ there is immaterial ([F1]). Writing $r=\mathrm{\mid }x\mathrm{\mid }$ and $h(r):=\frac{Q}{4\pi {\epsilon }_0{a}^3}$ for $r<a$ and $h(r):=\frac{Q}{4\pi {\epsilon }_0{r}^3}$ for $r\ge a$, the field is $E(x)=h(r)x$ and the potential is the radial function $\phi (x)=f(r)$ with $f(r)=\frac{Q(3a^2-r^2)}{8\pi {\epsilon }_0{a}^3}$ for $r\le a$ and $f(r)=\frac{Q}{4\pi {\epsilon }_{0}r}$ for $r\ge a$. Both branches are smooth on their open regions, $h$ is continuous at $r=a$ with common value $\frac{Q}{4\pi {\epsilon }_0{a}^3}$, and $f$ is continuous with $f(a)=\frac{Q}{4\pi {\epsilon }_{0}a}$ from both branches; moreover $f^{\mathrm{\prime }}(r)=-rh(r)$ on each side, with common value $f^{\mathrm{\prime }}(a)=-\frac{Q}{4\pi {\epsilon }_0{a}^2}$, so the one-sided limits of $\mathrm{\nabla }\phi =f^{\mathrm{\prime }}(r)x\mathrm{/}r$ agree on $S_a$ ([F2]).

Clause (a): on each coordinate line the sphere has at most two intersection points. On every smooth subinterval the derivative of $\phi$ is the corresponding component of $-E$, with matching limits at the intersection points by step 1.1. Apply the FTC [F4] on those subintervals and pass their endpoints to an intersection; the resulting increment formula is the integral of the continuous component $-E_i$ along the whole short line segment. Dividing by the segment length proves the same partial derivative at the intersection. Thus all partial derivatives exist and are continuous, so the continuous-partial-derivatives characterization gives total differentiability at every point with $\mathrm{\nabla }\phi =-E$; $E$ is continuous because $h$ is, and both $E$ and $\phi$ are smooth on each open side ([F2]).

Clause (b): for $\mathrm{\mid }x\mathrm{\mid }<a$, $E(x)=\frac{Q}{4\pi {\epsilon }_0{a}^3}x$ is affine, so $\mathrm{div}E=\frac{3Q}{4\pi {\epsilon }_0{a}^3}=\rho (x)\mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$; for $\mathrm{\mid }x\mathrm{\mid }>a$, $E(x)=\frac{Qx}{4\pi {\epsilon }_0\mathrm{\mid }x{\mathrm{\mid }}^3}$ is radial of the Coulomb shape, and the quotient rule with ${\mathrm{\partial }}_i\mathrm{\mid }x\mathrm{\mid }=x_i\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ gives $\mathrm{div}E=\frac{Q}{4\pi {\epsilon }_0}{\sum }_i(\frac{1}{\mathrm{\mid }x{\mathrm{\mid }}^3}-\frac{3x_i^2}{\mathrm{\mid }x{\mathrm{\mid }}^5})=0=\rho (x)\mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$ componentwise, since for all indices $j,k$, ${\mathrm{\partial }}_j{E}_k=\frac{Q}{4\pi {\epsilon }_0}({\delta }_{jk}\mathrm{\mid }x{\mathrm{\mid }}^{-3}-3x_j{x}_k\mathrm{\mid }x{\mathrm{\mid }}^{-5})$ is symmetric in $j,k$ ([F2]).

Clause (c): the sphere $S_a$ is a compact embedded smooth two-sided hypersurface with outward normal $n(x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$; the field $E$ is $C^1$ on $\{\mathrm{\mid }x\mathrm{\mid }<a\}$ and on $\{\mathrm{\mid }x\mathrm{\mid }>a\}$ with one-sided $C^1$ traces on $S_a$: inside, ${\mathrm{\partial }}_j{E}_k=\frac{Q}{4\pi {\epsilon }_0{a}^3}{\delta }_{jk}$; outside the derivative formula in step 3.1 extends continuously to $\mathrm{\mid }x\mathrm{\mid }=a>0$. By step 2.1 both traces are the continuous field $E$, so $[E]=E_{+}-E_{-}=0$ on $S_a$ ([F3]). The interface lemma [F3] therefore gives $\mathrm{div}E=\{\mathrm{div}E{\}}_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}+n\cdot [E]{\delta }_{S_a}$ and $\mathrm{curl}E=\{\mathrm{curl}E{\}}_{\mathrm{p}\mathrm{i}\mathrm{e}\mathrm{c}\mathrm{e}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}}+n\times [E]{\delta }_{S_a}$; the jump terms vanish, and the piecewise expressions equal $\rho \mathrm{/}{\epsilon }_0$ and $0$ respectively as locally integrable functions by step 3.1 and the nullity of the sphere ([F1], [F3]). Hence $\mathrm{div}E=\rho \mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^3)$; by the static-to-distributional reading these are the weak equations for the source $(\rho ,0)$ with zero singular part ([F4]).

Clause (d), field energy: $\mathrm{\mid }E{\mathrm{\mid }}^2=h(r{)}^2{r}^2$, so the polar formula gives ${\int }_{{\mathbb{R}}^3}\mathrm{\mid }E{\mathrm{\mid }}^{2}d{\lambda }_3=4\pi ({\int }_0^a\frac{Q^2{r}^2}{16{\pi }^2{\epsilon }_0^2{a}^6}{r}^{2}dr+{\int }_a^{\mathrm{\infty }}\frac{Q^2}{16{\pi }^2{\epsilon }_0^2{r}^4}{r}^{2}dr)=\frac{Q^2}{4\pi {\epsilon }_0^2}(\frac{a^5}{5a^6}+\frac{1}{a})=\frac{3Q^2}{10\pi {\epsilon }_0^{2}a},$ using ${\int }_0^a{r}^{4}dr=a^5\mathrm{/}5$ and ${\int }_a^{\mathrm{\infty }}{r}^{-2}dr=1\mathrm{/}a$ ([F1], [F2]); multiplying by ${\epsilon }_0\mathrm{/}2$ gives $\frac{{\epsilon }_0}{2}\int \mathrm{\mid }E{\mathrm{\mid }}^2=\frac{3Q^2}{20\pi {\epsilon }_{0}a}$, which is finite.

Clause (d), assembly integral: since $\rho$ vanishes off $\{\mathrm{\mid }x\mathrm{\mid }<a\}$, $\frac{1}{2}{\int }_{{\mathbb{R}}^3}\rho \phi d{\lambda }_3=\frac{1}{2}\cdot \frac{3Q}{4\pi a^3}\cdot \frac{Q}{8\pi {\epsilon }_0{a}^3}{\int }_{\mathrm{\mid }x\mathrm{\mid }<a}(3a^2-\mathrm{\mid }x{\mathrm{\mid }}^2)d{\lambda }_3,$ and the polar formula gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }<a}(3a^2-\mathrm{\mid }x{\mathrm{\mid }}^2)d{\lambda }_3=4\pi {\int }_0^a(3a^2{r}^2-r^4)dr=4\pi (a^5-\frac{a^5}{5})=\frac{16\pi a^5}{5}$ ([F1]); hence $\frac{1}{2}\int \rho \phi =\frac{1}{2}\cdot \frac{3Q^2}{32{\pi }^2{\epsilon }_0{a}^6}\cdot \frac{16\pi a^5}{5}=\frac{3Q^2}{20\pi {\epsilon }_{0}a}$, which equals the field energy of step 5.1; the same value as the smooth-source identity of [F4] is thus obtained although $\rho$ is only piecewise continuous, the computation being valid because the finiteness and equalities were verified directly for this density with a null interface ([F1]).

Clause (e): for $\mathrm{\mid }x\mathrm{\mid }>a$ the second branch of $E$ is exactly the point-charge field $\frac{Qx}{4\pi {\epsilon }_0\mathrm{\mid }x{\mathrm{\mid }}^3}$ of [F4] with $x_0=0$, so the exterior field is the Coulomb field of the total charge $Q$ even though the source is extended; for $\mathrm{\mid }x\mathrm{\mid }<a$ the field is the affine field $\frac{Q}{4\pi {\epsilon }_0{a}^3}x$; the potential is continuous at $\mathrm{\mid }x\mathrm{\mid }=a$ and decays as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$ by its exterior branch; and $\mathrm{\mid }E(x)\mathrm{\mid }\le \frac{\mathrm{\mid }Q\mathrm{\mid }}{4\pi {\epsilon }_0{a}^2}$ on all of ${\mathbb{R}}^3$, the maximum being attained at $\mathrm{\mid }x\mathrm{\mid }=a$, so the field energy is finite and no divergence occurs ([F1], [F2], [F4]).

Clause (f): the verification produced the displayed candidate, its pointwise equations off the sphere, its distributional equations with zero surface-charge jump, and its exact finite energy; no uniqueness is claimed, since uniqueness would require boundary data and an infinity condition and the density is not continuous, and the pair $(E,(\rho ,0))$ is read through the weak predicate and the static-to-distributional dictionary only. No empirical content is asserted, and the sphere carries no surface charge because $[E]=0$; this is exactly the scope stated.