"Oppositely charged parallel plates are separated by T 5.33 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?"

Explanation:

Given that,

Plate separation

d = 5.33mm = 0.00533m

Potential difference

V = 600V

A. Magnitude of electric field?

The magnitude of electric field is given as

E = |-∆V/∆x|

E = ∆V/∆x

E = 600/0.00533

E = 112,570.36 N/C

E = 1.126 × 10^5 N/C

B. Force on an electron between plate?

Force in an electric field is determine by using the formula

F = qE

Where q is an electron of charge

q = e = 1.609×10^-19C

F = eE

F = 1.609×10^-19 × 1.126×10^5

F = 1.81 × 10^-14 N

C. Work done to move the negative plate from 5.33mm to 2.90mm

Work done is given as

W = - F∆x Cosθ

The angle between the force and the displacement is 0°

∆x = x2 - x1

∆x = 2.90 — 5.33

∆x = -2.43mm

∆x = -0.00243m

So, W = -F∆xCosθ

W = - 1.81×10^-14 × -0.00243Cos0

NOTE: -×- =+, Cos0 =1

W = 4.398 × 10^-17 J

W ≈ 4.4 × 10^-17 J

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.

answer: a. same unit

explanation: