A thin spherical shell of radius 7.6 cm carries a uniform surface charge density of 6.7 times 10-9 C/m2. The magnitude of the electric field (in N/C) at r = 7.8 cm is approximately:.

 E= 1.968 10¹¹  N/C

Explanation:

To calculate the electric field of a sphere we can use Gaussian law

        Ф = ∫ E. dA = / ε₀

In this case we must use a Gaussian surface of spherical shape, since the radii of the sphere and the electric field line have been parallel, consequently the scalar product is reduced to an algebraic product.

         The error of a sphere is

          A = 4π π r²

         E (4π r³) = / εo                           (1)

to find the charge inside the sera we use the concept of density

         ρ = quint / V

         qint = rho V

the volume of a sphere is

          V = 4/3 pi r³

we substitute in 1

    E 4π r² = ρ 4/3 π r³

     E = ρ / 3ε₀

now let's change the charge density to the charge

      E = Q 3 / 4π r³    r / 3ε₀

       E = (1 / 4πε₀) Q / r²

       E = k Q /r²

calculate

        E = ρr / 3ε₀

       E = 6.7 10⁻⁹  7.8  10⁻² / 3 8.85 10⁻¹²

         E= 1.968 10¹¹  N/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.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.