PX275 - E2b - dirac delta

considering a sphere
for a point charge, $Q$ , at the origin, a distance, $r$ , from $Q$ , the electric field is given by:

\vec{E} = \frac{Q}{4 π ε_{0} r^{2}} {\hat{e}}_{r}

for the divergence theorem:

\iint_{S} \vec{E} \cdot d \vec{s} = ∭_{B} (\vec{\nabla} \cdot \vec{E}) d V

the volume integral

∭_{V} (\vec{\nabla} \cdot \vec{E}) d V = \frac{Q}{4 π ε_{0}} ∭_{V} \frac{1}{r^{2}} \underset{\frac{\partial}{\partial r} (1) = 0}{\underset{⏟}{\frac{\partial}{\partial r} (r^{2} \frac{1}{r^{2}})}} d V = 0

the surface integral

\vec{E} = \frac{Q}{4 π ε_{0} r^{2}} {\hat{e}}_{r}

\begin{aligned} \iint_{S} \vec{E} \cdot \vec{d} s & = \frac{Q}{4 π ε_{0}} \iint_{S} \frac{1}{r^{2}} r^{2} \sin θ d θ d ϕ \\ = \frac{Q}{4 π ϵ_{0}} \int_{0}^{2 π} \int_{0}^{π} \sin θ d θ d ϕ \\ = \frac{Q}{4 π ϵ_{0}} 2 π [- \cos θ]_{0}^{π} \\ ∴ \iint_{S} \vec{E} \cdot \vec{d} s & = \frac{Q}{ϵ_{0}} \end{aligned}

these results are not consistent

the dirac delta

in order to define $\vec{\nabla} \cdot \vec{E}$ across all space including $r = 0 :$

\vec{\nabla} \cdot \vec{E} = \frac{Q}{ϵ_{0}} δ (r)

where, $δ (r)$ is the dirac delta

in 3D, it is written as $δ^{3} (r)$
the function is a distribution of infinite height and zero width

\int_{- x_{1}}^{x_{2}} δ (x) d x = 1

from the divergence , if $Q$ is within a volume, $V$ , then:

∭_{V} \frac{Q}{ϵ_{0}} δ (\vec{r}) d V = \frac{Q}{ϵ_{0}} \iint_{V} δ (\vec{r}) d V = \frac{Q}{ϵ_{0}}

more generally, if the charge, $Q$ , is at a position $\vec{a}$ , then define:

δ (\vec{r - a}) = 0, \forall r \neq 0