Surface charge density of a conducting plate I know that surface charge density is inversely proportional to the radius of curvature of a surface, but since the radius of curvature of a conducting sheet is ∞, and I know its charge density is not 0, is this an exception or is there something wrong in the way I understood the relation?
 A: Considering a charge $Q$ on a spherical surface of radius $r$, due to $Q= \int \sigma df $, where $\sigma$ is surface charge density and $df$ the surface element one gets indeed
$$\sigma = \frac{Q}{4\pi r^2}$$
Now if the radius $r$ is varied and the charge $Q$ is kept constant one can indeed observe that the surface charge density changes like $\sim \frac{1}{r^2}$. To give a more intuitive picture, one either expands or contracts the sphere where the charge is put on and the surface charge density shows this behaviour.
If a charge is put on a flat conductor one can apply the a similar picture comparing it with a spherical surface with a $r=\infty$. For a flat conductor contracting or expanding the spherical surface (and keeping this way the charge $Q$ constant and varying $r$ ) corresponds to a finite perpendicular shift (whose value we will call $a$)   of the conductor. However, $r$ is not changed by such a shift. Or in mathematical terms:  $r +a = \infty +a= \infty$. So the $\sigma$, the surface charge density does not change.
