The electric field on the surface of an irregularly shaped conductor varies from 56 kN $C^{-1}$ to 112 kN$C^{-1}$ . If local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is $62x nC{m}^{-2}$  .Find  $x$

$(Given{\in }_0=8.85\times {10}^{-12}{C}^{2}N{m}^{-2})$ 

The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is the most intense where the radius of curvature is smallest and vice-versa. The local charge density and the electric field intensity are related by 
$E=\frac{\sigma }{{\in }_0}$
$\Rightarrow \sigma =E{\in }_0$
At the point on the surface where the radius of curvature is the greatest, we have 
$\sigma ={\in }_0{E}_{\min }=(8.85\times {10}^{-12}{C}^{2}N{m}^{-2})(5.6\times {10}^{4}N{C}^{-1})$
$\Rightarrow \sigma =496nC{m}^{-2}$