Answer:
[tex]E = 6.77 *10^{3} N/C[/tex]
Explanation:
THE GIVEN sheet can be taken as two horizontal force with surface charge density is
at one surface is ∈_1 = [tex]95*10^ {-9} C/mm2[/tex]
at oher surface is ∈_2= [tex]-25*10^{-9} C/mm2[/tex]
the magnitude of electric field due to surface charge is given as[tex]\frac{∈}{2∈_O}[/tex]
So, electric field at P (2 CM below from surface is) = E_1 +E_2
[tex]E = \frac{∈_1}{2∈_O} +\frac{∈_2}{2∈_O}[/tex]
[tex]E = \frac{95*10^{-9} +25*10^{-9}}{2*8.85*10^{-12}}[/tex]
[tex]E = 6.77 *10^{3} N/C[/tex]