Answer:
a) 8*10^-8C/m²
b) +9.04*10^3N/C
c) = -9.04*10^3N/C
Explanation:
Given
Side length, L = 50cm = 0.5m
Charge on the plate, Q = 4*10^-8C
Surface charge density, σ = Q/A
The surface charge density of each part is then half of the total charge density of the plate. Thus,
σ(face) = 1/2σ
σ(face) = Q/2A
σ(face) = Q/2L²
Now we plug in, since we have Q and L
σ(face) = 4*10^-8 / 2*0.5²
σ(face) = 4*10^-8 / 0.5
σ(face) = 8*10^-8C/m²
Magnitude of electric field above the plate is,
E = σ(face) / E•
E = 8*10^-8 / 8.85*10^-12
E = 9.04*10^3 N/C
If we assume this plate lies on the side of the "xy" plane, the electric field is directed in the positive "z" direction. As such,
E = +9.04*10^3N/C
Electric field below the plate has the same magnitude, but different direction. So, E = -9.04*10^3N/C
