To solve this problem it is necessary to apply the concepts related to electric field due to non conducting wall. This is defined as
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
We have that
[tex]\sigma[/tex] is the charge per unit area of the wall and [tex]\epsilon_0[/tex] the vacuum permittivity
First we need to convert to SI Units,
[tex]\sigma= 10.81 \mu C/cm^2 (\frac{100^2cm^2}{1m^2})(\frac{10^{-6} C}{1 \mu C})[/tex]
[tex]\sigma = 0.1081C/m^3[/tex]
Now replacing we have,
[tex]E = \frac{0.1081C/m^3}{2(8.85*10^{-12}C^2/Nm^2)}[/tex]
[tex]E = 6.1073*10^9N/C[/tex]
Therefore the electric field is [tex]6.1073*10^9N/C[/tex]
