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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.