Answer:
[tex]\sigma_{max}=3*10^{-5}\frac{C}{m^2}[/tex]
Explanation:
The electric field on the surface of a conductor is given by:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
Here [tex]\sigma[/tex] is the surface charge density and [tex]\epsilon_0[/tex] the permittivity of free space. Thus, the highest surface charge density that can exist in a conductor is given by the value of the dielectric breakdown of the air multiplied by the permittivity of free space:
[tex]\sigma_{max}=\epsilon_oE_{breakdown}\\\sigma_{max}=(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(3*10^6\frac{V}{m})\\\sigma_{max}=3*10^{-5}\frac{C}{m^2}[/tex]
