Find the electric force experienced by a charge reduced to a unit area of an arbitary conductor if the surface density of the charge equals `sigma`.

Explanation

Near the conductor `E = E_(n) = (sigma)/(epsilon_(0))`
This field can be written as the sum of two parts `E_(1)` and `E_(2), E_(1)` is the electric field due to an infinitesinal area `dS`.
Very near it `E_(1) = +- (sigma)/(2epsilon_(0))`
The remaining part contributes `E_(2) = (sigma)/(2 epsilon_(0))` on both sides. In calculating the force on the element `dS` we drop `E_(1)` (because it is a self-force) Thus
`(dF)/(dS) = sigma . (sigma)/(2 epsilon_(0)) = (sigma^(2))/(2 epsilon_(0))`