If we have any simple object like sheet,rod or sphere with uniform charge at them, is electric field at surface always finite?
For sphere I read that it is finite because as test approaches the surface the area patch which can cause infinite field at a point is also getting smaller hence charge on patch is also getting smaller. So it is applicable for all case and if not why not?
If you look at the field of an infinite sheet of charge (with charge density $\sigma$), the field is:
$$ \vec E(x) = \frac{\sigma}{2\epsilon_0} \frac{\vec x}{|x|} $$
It's uniform. At a distance $x$ from the sheet, the total charge at radius $r\pm dr/2 $, with $r \gg x $, contributing to the field is:
$$ \rho = \sigma rdr$$
The transverse component of the field cancel, leaving a longitudinal component:
$$ E_x \propto E\times \frac x r $$
where:
$$ E \propto \frac{\rho}{r^2} \propto \frac {\sigma dr} r$$
So contributions from distant charge falls of as:
$$ dE(r) \propto \frac x {r^2} $$
So you for well behaved charge distributions, you can ignore contributions from $r \gg x $.
If if you get close enough to a surface ($x \ll R_c$) where $R_c$ is the smallest radius of curvature, it looks flat, and
$$ E = c_0 \times \frac{\sigma}{\epsilon_0} $$
For an infinite sheet: $c_0 = \frac 1 2$, and for a conducting sphere: $c_0 = 1$.
