Find the potential `varphi` at the edge of a thin disc of radius `R` carrying the uniformly distributed charge with surface density `sigma`.

Explanation

By definition, the potentail in the case fo a surface charge distribution is defined by intergal `varphi = (1)/(4pi epsilon_(0)) int (sigma dS)/(r )`, In order to simply intergation we shall choose the area element `dS` in the from of a part of the ring of radius `r` width `dr` in (Fig). Then `dS = 2theta r dr, r = 2R cos theta` and `dr = -2R sin theta d theta`. After subsituating these expressions into intergal
`varphi = (1)/(4pi epsilon_(0)) int (sigma dS)/(r )`, we obtain the expression for for `varphi` at the point `O`:
`varphi = - (sigma R)/(pi epsilon_(0)) int_(pi//2)^(0) theta sin theta d theta`
We intergate bty parts,
denoting `theta = u` and `sin theta d theta = dv` ,
`int theta sin theta d theta = -theta cos theta `
`+int cos theta d theta = -theta cos theta + sin theta`
which gives`-1` after substuting the limits of interfation. As a result, we obtain
`varphi = sigmaR//pi epsilon_(0)`