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saul85 [17]
3 years ago
7

A solid insulating sphere of radius R = 1.0 m that carries a positive charge Q1 = 1.0 mC uniformly distributed over it is concen

tric with a very thin insulating shell of radius 2R that carries the charge Q2 = -1.0 mC. The charge Q2 is distributed uniformly over the shell.
Use Gauss's law to obtain the magnitude of the electric field:

a. E1 inside the sphere, r < R

b. E2 between the sphere and the shell, R < r < 2R

c. E3 outside the shell, r > 2R

d. Plot the electric field in the radial direction on a graph. Be sure to label the actual magnitude of the electric field at r = 0, r= R, and r = 2R.

Physics
1 answer:
RoseWind [281]3 years ago
3 0

Answer: see attach file

Explanation: In order to solve this problem we have used the Gaussian law in the different regions. Details are included in the attach.

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Explanation:

Question 1

1.The wires are made of conducting materials e.g copper

2. A light bulb is made of filament material e.g tungsten with gases enclosed in a glass material

3.The lens is made of transparent glass material

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Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a wa
creativ13 [48]

Answer:

in oil film        λ = 303.57 10⁻⁹ m

in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

             c = λ₀ f

we divide the two expression

            c / v = λ₀ / λ

the refractive index is

             

              n = c / v

              n = λ₀ /λ

              λ = λ₀ / n

let's calculate

in oil film

            λ = 425 10⁻⁹ / 1.40

            λ = 303.57 10⁻⁹ m

in the water film

            λ = 425 10⁻⁹ / 1.33

            λ = 319.55 10⁻⁹

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