Answer:
[tex]E=158.19\frac{kN}{m}[/tex]
Explanation:
Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]
The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]
Recall that the surface charge density is defined as:
[tex]\sigma=\frac{q}{S}[/tex]
Now, we get the electric field strength:
[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]
