A thin conducting plate 50.0 cm on a side lies in the xy plane. If a total charge of 4.00 ? 10 ?8 C is placed on the plate, find (a) the charge density on the plate, (b) the electric field just above the plate, and (c) the electric field just below the plate.

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