We know that the electric field at the surface of a conductor only have a normal component equal to $\rho/\varepsilon$ (finite number).

But let’s consider the point $\text{P}$ (at the surface of a conductor). Assume that there is a charge at an infinitesimal distance from the point $\text{p}$. We can obtain the field at the $\text{P}$ by the formula $E=Kq/r$. Obviously, $E\sim1/r$. So the normal component of the field is infinite. Now if we add the field due to other charges, it will remain infinite. So where could I be possibly wrong?

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    $\begingroup$ Nowhere! Field is not defined on the surface of conductor, it is $\infty$. $\endgroup$ – ABC Apr 30 '13 at 8:34
  • $\begingroup$ Are you sure? Can't you approximate the electric field at the surface of the conductor to be the midpoint between zero and the magnitude of the jumpy discontinuity? Specifically, as $E_{surface} = \frac{\sigma}{2\epsilon_0}$. $\endgroup$ – Loonuh Jan 11 '16 at 4:45

Implicit in the derivation of $E=\rho/2\epsilon_0$ (you were off by a factor of two by the way) is the assumption that the charge can be approximated by a continuous charge distribution. You correctly note that as you approach a real sheet of charge composed of point charges this treatment breaks down. Let's calculate the $z$-component of the electric field of a square array of equal point charges $q$ located on the $x-y$ plane at lattice sites $(a i, a j, 0)$, where $a$ is the lattice spacing and $i,j$ are integers. Using Coulomb's law you get:

$$ E_{z}=\frac{q}{4\pi\epsilon_{0}}\sum_{i,j=-\infty}^{\infty}\frac{z}{\left(a^{2}i^{2}+a^{2}j^{2}+z^{2}\right)^{\frac{3}{2}}}. $$

This can't apparently be written in closed form (tried Mathematica). Here is an approximate numerical calculation of this function. $z$ is measured in units of $a$ but the vertical scale is arbitrary (actually $E_z \to 2\pi$ on this scale):

enter image description here

Note that as $z\to 0$ the field diverges as $z^{-2}$, as you note.

We can get the large distance limit by taking $a\to 0$ and $q\to 0$ such that $q/a^2 = \rho$ is the finite constant charge density. We replace $ a i = x $ and $ a j = y $ and note that, since for large $z$ the summand is a smooth function of $i,j$ the sum can be replaced by an integral:

$$ a^{2}\sum_{i,j=-\infty}^{\infty} \to \int\mathrm{d}x\mathrm{d}y. $$

The result is

$$ E_{z}=\frac{\rho}{4\pi\epsilon_{0}}\int\mathrm{d}x\mathrm{d}y\ \frac{z}{\left(x^{2}+y^{2}+z^{2}\right)^{\frac{3}{2}}}, $$

which straightforwardly evaluates to $E_z = \rho/2\epsilon_0$. By passing into the continuum limit you lose all information about the discreteness at short distances.

  • $\begingroup$ what you mean is that the sum of infinite electrical fields does not add to infinite but it converges to ρ/2ϵ0 ? and this is proven mathematically? $\endgroup$ – user23835 May 1 '13 at 6:51

The conductor would be influenced by the point charge: additional charge would build up at the conductor surface until the tangential component of the field has vanished (as usual).

The Wikipedia page about mirror charges (see picture on the right) illustrates the final configuration. Imagine the infinitesimal distance by moving the charge closer to the surface: the infinitesimal distance would lead to a dipole, where the charge with opposite sign is the 'virtual' mirror charge in the conductor.


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