PHYSICS HIGH SCHOOL

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.

Answers

Answer 1

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

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