Respuesta :
This is a problem dealing with the surface charge density. Based on the given problem above, your charge is 8nc on a square plate of 22 cm x22 cm given that E= (surface charge density)/2ε₀ so answer would be 8nc/22/22/2ε₀. Therefore, the magnitude of the electric field to the right (x>0) of the right face of the slab would be 8nc/22/22/2ε₀.
Answer:
E = 9338.4 N/C
Explanation:
As we know that electric field due to a charged sheet is given by the formula
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
here we know that
[tex]\sigma[/tex] = charge density
[tex]\sigma = \frac{Q}{A}[/tex]
[tex]\sigma = \frac{Q}{L^2}[/tex]
[tex]\sigma = \frac{8\times 10^{-9}}{0.22^2}[/tex]
[tex]\sigma = 1.65 \times 10^{-7}C/m^2[/tex]
now we have
[tex]E = \frac{1.65 \times 10^{-7}}{2(8.85 \times 10^{-12}}[/tex]
[tex]E = 9338.4 N/C[/tex]