Appendix for ED formulas

This document supplements some rather complex proof processes within ED Formulas and ED formulas (2)

1 $∭_{V} E (x) d^{3} x = - \frac{p}{3 ϵ_{0}}$

Let $V$ be a spherical volume of radius $R$ that completely encloses all charges, meaning the charge density $ρ (x^{'})$ vanishes for all $r^{'} > R$ . Without loss of generality, we place the coordinate origin at the center of the sphere $V$ . According to Coulomb's law, the electric field $E (x)$ at any field point $x$ inside the sphere is given by the volume integral over the source coordinates $x^{'}$ :

E (x) = \frac{1}{4 π ε_{0}} \int_{V} ρ (x^{'}) \frac{x - x^{'}}{| x - x^{'} |^{3}} d^{3} x^{'}

We want to evaluate the volume integral of $E (x)$ over the entire sphere $V$ :

\int_{V} E (x) d^{3} x = \frac{1}{4 π ε_{0}} \int_{V} [\int_{V} ρ (x^{'}) \frac{x - x^{'}}{| x - x^{'} |^{3}} d^{3} x^{'}] d^{3} x

Since the region of integration is bounded and the charge distribution is well-behaved, we can apply Fubini's theorem to switch the order of the volume integrals over $d^{3} x$ and $d^{3} x^{'}$ :

\int_{V} E (x) d^{3} x = \frac{1}{4 π ε_{0}} \int_{V} ρ (x^{'}) [\int_{V} \frac{x - x^{'}}{| x - x^{'} |^{3}} d^{3} x] d^{3} x^{'}

Let us denote the inner vector integral as $I (x^{'})$ :

I (x^{'}) = \int_{V} \frac{x - x^{'}}{| x - x^{'} |^{3}} d^{3} x

We recognize that the integrand can be expressed as the negative gradient with respect to the field point $x$ :

\frac{x - x^{'}}{| x - x^{'} |^{3}} = - \nabla (\frac{1}{| x - x^{'} |})

Substituting this identity into $I (x^{'})$ and applying a corollary of the divergence theorem (converting a volume integral of a gradient into a closed surface integral), we obtain:

I (x^{'}) = - \int_{V} \nabla (\frac{1}{| x - x^{'} |}) d^{3} x = - \oint_{\partial V} \frac{1}{| x - x^{'} |} \hat{n} d A

where $\partial V$ represents the bounding spherical surface at $r = R$ , and $\hat{n}$ is the outward unit normal vector. On this spherical surface, the field point is $x = R \hat{x}$ (where $\hat{x}$ is the radial unit vector $\hat{r}$ ), and the area element is $d A = R^{2} d Ω$ . Thus, the integral becomes

I (x^{'}) = - R^{2} \oint_{r = R} \frac{1}{| x - x^{'} |} \hat{x} d Ω

Because all charges are enclosed within the sphere, the source point always lies inside ( $r^{'} < R$ ), while the field point in the surface integral lies exactly on the boundary ( $r = R$ ). Since $r^{'} < r$ , we can expand the reciprocal distance $1 / | x - x^{'} |$ into a series of Legendre polynomials:

\frac{1}{| x - x^{'} |} = \sum_{l = 0}^{\infty} \frac{(r^{'})^{l}}{R^{l + 1}} P_{l} (\cos γ)

where $γ$ is the angle between $x$ and $x^{'}$ .

To evaluate the solid angle integral, we temporarily align the $z$ -axis along the direction of $x^{'}$ , making $γ = θ$ (the standard polar angle). The unit vector $\hat{x}$ expands in Cartesian components as:

\hat{x} = \sin θ \cos ϕ \hat{i} + \sin θ \sin ϕ \hat{j} + \cos θ \hat{k}

Integrating over the azimuthal angle $ϕ$ from $0$ to $2 π$ , the terms containing $\cos ϕ$ and $\sin ϕ$ vanish identically due to periodicity. The only surviving component is along the $\hat{k}$ direction (which is the direction of ${\hat{x}}^{'}$ ):

\oint_{r = R} \frac{1}{| x - x^{'} |} \hat{x} d Ω = {\hat{x}}^{'} \int_{0}^{2 π} d ϕ \int_{0}^{π} [\sum_{l = 0}^{\infty} \frac{(r^{'})^{l}}{R^{l + 1}} P_{l} (\cos θ)] \cos θ \sin θ d θ

Noting that $\cos θ = P_{1} (\cos θ)$ , we invoke the orthogonality condition for Legendre polynomials:

\int_{0}^{π} P_{l} (\cos θ) P_{1} (\cos θ) \sin θ d θ = \frac{2}{2 (1) + 1} δ_{l, 1} = \frac{2}{3} δ_{l, 1}

Hence, only the $l = 1$ term yields a non-zero contribution to the infinite series:

\oint_{r = R} \frac{1}{| x - x^{'} |} \hat{x} d Ω = {\hat{x}}^{'} \cdot 2 π \cdot \frac{r^{'}}{R^{2}} \cdot \frac{2}{3} = \frac{4 π}{3 R^{2}} (r^{'} {\hat{x}}^{'}) = \frac{4 π}{3 R^{2}} x^{'}

Substituting this result back into the expression for $I (x^{'})$ , the radius $R^{2}$ cancels out smoothly:

I (x^{'}) = - R^{2} (\frac{4 π}{3 R^{2}} x^{'}) = - \frac{4 π}{3} x^{'}

We now feed $I (x^{'})$ back into our original total electric field volume integral:

\int_{V} E (x) d^{3} x = - \frac{1}{3 ε_{0}} \int_{V} ρ (x^{'}) x^{'} d^{3} x^{'}

By definition, the total electric dipole moment $p$ of the localized charge distribution with respect to our chosen origin is:

p = \int_{V} ρ (x^{'}) x^{'} d^{3} x^{'}

Replacing the integral with $p$ , we arrive at:

\int_{V} E (x) d^{3} x = - \frac{p}{3 ε_{0}}