# Charged object shrinking in scale to volume $0$ causes electric field becoming constant

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.

• 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 – patta Apr 10 at 11:25

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).

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.

• 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 $r>\frac{R}{\alpha}$? – Sayako Hoshimiya Apr 11 at 12:24
• 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 $a\to r\alpha$ – Cryo Apr 12 at 4:22
• I corrected my error with $\alpha r\gg R$. Thanks – Cryo Apr 12 at 4:25