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How to state mathematically rigorously the fact that when the volume of a charged object of any shape approaches $0$ the electric field on the surface of a spherical region centered at the object is constant in magnitude and normal to the surface in direction? I'm not quite satisfied with the explanation that it becomes a point charge.

Mathematically speaking, I wish to know why it is that $$ \lim_{V\to0^+}\mathbf{E}(x,y,z;V)=\frac{Q}{\epsilon_0\unicode{x222F}_{\partial\Omega}1\operatorname{d}\mathbf{A}}\hat{n}=\frac{Q}{4\epsilon_0\pi r^2}\hat{n}\quad\forall x,y,z $$ where $\mathbf{E}$ dependes on $V$, the volume of the charged object, shrinking in scale.

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  • $\begingroup$ You need also not only the volume shrink to 0, but the whole object became small (contained in a ball of radius $\epsilon$), otherwise you may get a thin wire or a plane $\endgroup$ – patta Apr 10 at 11:25
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If I move a finite charge by a small amount, I expect to affect the far-away electric field by a small amount. Approaching zero movement, will approach zero change in the far field.

Now imagine your object, shrinked to small size. Make it rotate around its centre; the charges move a little, hence the far field changes little. It can rotate in all directions, that implies a spherical symmetry, that is the field of a point charge (or a spherical distribution of charges).

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I would frame the question like this. Let $\rho_1\left(\mathbf{r}\right)$ be some charge density that vanishes outside the sphere of radius $\mathbf{R}$. I will now use a unitless constant $\alpha>0 \in \mathbb{R}$ to define:

$\rho_\alpha\left(\mathbf{r}\right)=\rho_1\left(\mathbf{r}\cdot \alpha\right)$ clearly, letting $\alpha \gg 1$ squeezes the charge density.

The scalar potential due to this charge density is (using electrostatics):

$\phi_\alpha\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\int d^3 r' \frac{\rho_\alpha\left(\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}= \frac{1}{4\pi\epsilon_0}\int d^3 r' \frac{\rho_1\left(\alpha\cdot\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}$

We can change the integration variable to $\boldsymbol{\zeta}=\alpha\mathbf{r}'$:

$\phi_\alpha\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\int d^3 \zeta \frac{\rho_1\left(\boldsymbol{\zeta}\right)}{\left|\alpha\cdot\mathbf{r}-\boldsymbol{\zeta}\right|}$

Now, assuming $\alpha r\gg R $:

$\phi_\alpha\left(\mathbf{r}\right) \approx \frac{1}{4\pi\epsilon_0}\left[\frac{1}{\alpha r}\int d^3 \zeta\,\, \rho_1\left(\boldsymbol{\zeta}\right) + \frac{1}{\alpha^2 r^2}\int d^3 \zeta\,\, \left(\mathbf{\hat{r}}.\boldsymbol{\zeta}\right)\rho_1\left(\boldsymbol{\zeta}\right)+\dots\right]$

So in the limit of $\alpha\to \infty$ (small volume for $\rho$) or $r\to\infty$ (observer far away), one gets:

$\phi_\alpha\left(\mathbf{r}\right) \approx \frac{1}{4\pi\epsilon_0}\left[\frac{1}{\alpha r}\int d^3 \zeta\,\, \rho_1\left(\boldsymbol{\zeta}\right)\right]$

which is independent of orientation of $\mathbf{r}$, i.e. spherically symmetric potential -> uniform radial field.

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  • $\begingroup$ I think I need a little bit of explanation from $\phi_\alpha\left(\mathbf{r}\right) = \frac{1}{4\pi\epsilon_0}\int d^3 \zeta \frac{\rho_1\left(\boldsymbol{\zeta}\right)}{\left|\alpha\cdot\mathbf{r}-\boldsymbol{\zeta}\right|}$ to $\phi_\alpha\left(\mathbf{r}\right) \approx \frac{1}{4\pi\epsilon_0}\left[\frac{1}{\alpha r}\int d^3 \zeta\,\, \rho_1\left(\boldsymbol{\zeta}\right) + \frac{1}{\alpha^2 r^2}\int d^3 \zeta\,\, \left(\mathbf{\hat{r}}.\boldsymbol{\zeta}\right)\rho_1\left(\boldsymbol{\zeta}\right)+\dots\right]$. Also do you mean $r>\frac{R}{\alpha}$? $\endgroup$ – Sayako Hoshimiya Apr 11 at 12:24
  • $\begingroup$ It's Taylor expansion. In general $f=\frac{1}{\left|\mathbf{a}-\boldsymbol{\zeta}\right|}=\frac{1}{a}\cdot\left(1+\left(\frac{\zeta}{a}\right)^2-2\left(\frac{\zeta}{a}\right)\left(\mathbf{\hat{a}.\boldsymbol{\hat{\zeta}}}\right)\right)^{-1/2}$. Now expand for small $\left(\frac{\zeta}{a}\right)$, keep only first two terms, then $a\to r\alpha$ $\endgroup$ – Cryo Apr 12 at 4:22
  • $\begingroup$ I corrected my error with $\alpha r\gg R$. Thanks $\endgroup$ – Cryo Apr 12 at 4:25

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