Physics, 25.12.2019 23:31 pascente5

An electric field is set up between two parallel plates, each of area 2.0 m2, by putting 1.0 μc of charge on one plate and −1.0 μc of charge on the other. the plates are separated by 4.0 mm with their centers opposite each other, and the charges are distributed uniformly over the surface of the plates. what is the magnitude of the electric field between the plates at a distance of 1.0 mm from the positive plate, but not near the edges of the plates? (ϵ0=8.85×10−12 c2/n · m2)

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Answer from: nany1982

The system in the question is a capacitor. So,

$C = \epsilon_0 \frac{A}{d} = (8.85\times 10^{-12}) \times \frac{2}{4\times 10^{-6}} = 4.425 \times 10^{-6} F$

$V = \frac{Q}{C} = \frac{1 \times 10^{-6}}{4.425 \times 10^{-6}} = 0.225 V\\$

$E = V/d = 225/(4\times 10^{-6}) = 56.25 \times 10^6 ~ V/m$

E is constant between the plates, so the E-field at a distance of 1mm from the positive plate is $56.25 \times 10^6~ V/m.$

Explanation:

According to the information given in the question, we can find the capacitance of this system. However, we should find a way to calculate E-field from capacitance. The way to do that is to find the potential difference between the plates and then calculate E using their relationship with V.

Normally, we should differentiate V in order to find E. Since E is constant between the plates, $V = \int{E(r)} \, dr = \int {E} \, dr = Er$ . This potential difference is calculated according to the distance between the plates, so we should divide V by 4mm to find E.

E is constant between parallel plate capacitors. This can be proven by the Gauss’ Law, however it is not necessary for this question.

Answer from: Quest

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explanation:

Answer from: Quest

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(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.

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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.