A 2.55x10^-4 C charge is held 0.350 m from a second 7.62x10^-6 C charge. The second one is released. When it reaches a distance of 1.55 m from the other charge, how much has the electric potential energy changed?
(Unit=J)

The change in the electric potential energy at the final given distance is 38.69 J.

Given;

To find:

Change in the potential energy

The initial potential energy between the charges is calculated as;

The final potential energy between the charges is calculated as;

The change in the potential energy is calculated as;

ΔV = V₁ - V₂

     = 49.97 J  -  11.28 J

     = 38.69 J

Therefore, the electric potential energy changed by 38.69 J

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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.