We first know the relationship between the electric field and electric potential on the surface of a charged conducting sphere. Consider a sphere of radius \(R\) and total charge \(q\). If you enclose the conducting sphere by a Gaussian sphere of radius \(r\) (\(r > R\)) and apply Gauss's law, you will find the electric field on the Gaussian surface as:

\[\begin{align*} EA=&\frac{q}{{{\epsilon }_{0}}} \\ \text{or,}\quad E=&\frac{kq}{{{r}^{2}}} \end{align*}\]

Which is the same as the electric field of a point charge. So, the total charge \(q\) on the sphere acts as if all the charge were concentrated at the centre of the sphere. And the electric potential is also the same as that of a point charge which is

\[V=\frac{kq}{{{r}^{{}}}}\]

On the surface of the conductor (\(r=R\)), the electric field is \(E=kq/R^2\) and the electric potential is \(V=kq/R\). So,

\[{{V}_{\text{s}}}=R{{E}_{\text{s}}}\]

The subscript \(\text{s}\) in the above equation represents the *surface*. If you keep the radius \(R\) constant and increase the charge in the conducting sphere, the electric potential also increases. But there is a limit on how much charge you can add in the sphere for a certain radius. If you keep on adding the charge on the sphere, the electric potential becomes maximum at a certain point at which the air is ionized and becomes conductive. And you keep on adding the charge any further, the electric discharge through the air takes place. So the maximum potential to the conducting sphere of a given radius is limited, that is any further adding of charge results the electric discharge. The maximum electric field at which the air is ionized is about \(3.0\times {{10}^{6}}\text{V/m}\)

The maximum potential of the sphere decreases by decreasing the radius of the sphere and even adding the small amount of charge can reach the limit and further adding causes electric discharge. Therefore the conducting sphere can withstand large electric potential if the radius of the conducting sphere is large enough; it means you can add even more charge on it until the electric discharge through the air. We have the relationship \(E_s = V_s/R\), and therefore the sharp edges or sharp point in a conductor has very less radius of curvature and even small electric potential can cause sufficiently large electric field which can cause the ionization of the air and then results the electric discharge through the air also called corona discharge. The blue glow through the sharp edges are called *corona*.