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

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.