Two infinite planes of charge lie parallel to each other and to the yz plane. one is at x--1 m and has a surface charge density of σ =-2.0 pc/m2. the other is at x = 5 m and has a surface charge density of σ 5.8 με/m2. find the electric field for the following locations. a)x-1 m n/c (b)-1m

    Ea = 3.2 10⁵ N / C

Explanation:

To calculate the electric field of each plane we will use Gauss's law, we create a Gaussian surface that is a cylinder that has the axis perpendicular to the plane, in this case the flow line between the cylinder walls and the surface is zero and all the flow is perpendicular to the base of the cylinder.

We apply the law of gauss flow to each side is the value of the electric field (E) for the area of the cylinder (A); whereby the flow in the two directions is 2 E A

                 Φ = 2E A = ρₙt / εo

Where ρₙ is the charge inside the cylinder, as the charge density gives us, σ = Q / A

                 ρₙ = σ A

     By which we can clear the electric field

                E = σ A / 2εo

where it is worth 8.85 10⁻¹² C² / N² m². Let's calculate with this equation in the field for each plane

1  plane    σ= -2.0 pC / m²

               E1 = -2. 10⁻¹² / 2 8.85 10⁻¹²

               E1 = -0.113 N / C

The field line is directed to the plane

2 Plane    σ = 5.8 mC / m2

               E2 = 5.8 10-6 / 2 8.85 10-12

               E2 = 3,277 10 5 N / A

Field lines leave the plane

As we have the values of each field in the whole space. Let's calculate in the field at the point x = 1 m

   To do this we must add the fields at the selected point vectorally, for this distance the point is between the two planes, so the field of plane 1 points to the left and the point of plane 2 also points to the left, consequently field adds

               Ea = E1 + E2

               Ea = 0.113 + 3.277 10⁵

               Ea = 3.2 10⁵ N / C

At point X = -1 m in this case the point is on plane 1, so this plane does not generate any field, it is an equipotential surface, the total field is equal to field 2

              Ea = E1 = 3.2 10⁵ N / C

Note that there is no difference in numbers values by the difference between the load between each plane

0.55x745.7j and a calculator

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.