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Maksim231197 [3]
2 years ago
12

A square metal plate of edge length 12 cm and negligible thickness has a total charge of 5.6 × 10-6 C. (a) Estimate the magnitud

e E of the electric field just off the center of the plate (at, say, a distance of 0.49 mm from the center) by assuming that the charge is spread uniformly over the two faces of the plate. (b) Estimate E at a distance of 29 m (large relative to the plate size) by assuming that the plate is a charged particle.
Physics
1 answer:
hodyreva [135]2 years ago
8 0

Answer:

a) 2.2*10^7 N/C

b) 60 N/C

Explanation:

To start with, we say that the

Area charge of the plate, σ = q/2A

σ = 5.6*10^-6 / 2(12*10^-2)²

σ = 5.6*10^-6 / 0.0288

σ = 1.944*10^-4 C²/m

Next, we find the electric field

Electric field, E = σ/Eo

Electric field, E = 1.944*10^-4 / 8.85*10^-12

Electric field, E = 2.2*10^7 N/C

b)

E = kq/r²

E = [8.99*10^9 * 5.6*10^-6] / 29²

E = 50344 / 841

E = 60 N/C

This means that the E at a distance of 29 m is 60 N/C

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a) <em>473.33 nm </em>

<em>b) 568 nm</em><em> and </em><em>406 nm</em>

<em>c) </em>bluish green and blue

Explanation:

a) As the light traverses the layer of oil it first reflects at the front surface of the oil. Here the index of refraction increases from that of air to that of the oil , so a phase change occurs.  The light then reflects from the rear surface of oil. The index of refraction increases from that of the oil to that of the glass , so again a phase change occurs.  Thus two phase changes occur.

In thin-film interference with 0 or 2 phase changes, condition for constructive interference is:

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So:

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<em><u>For m=2</u></em>

λ=710 nm

<em><u>For m=3</u></em>

λ=473.33 nm

<em><u>For m=4</u></em>

λ=355 nm

<em>Thus the only wavelength in the visible spectrum </em><em>(400 - 700 nm)</em><em> that will give constructive interference is </em><em>473.33 nm </em>

b)

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2t=(m+1/2)λ/n=(2m+1)*λ/2n

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λ=4tn/(2m+1)

<em><u>For m=1</u></em>

λ=946.667 nm

<em><u>For m=2</u></em>

λ=568 nm

<em><u>For m=3</u></em>

λ=405.33 nm

<em><u>For m=4</u></em>

λ=315.56 nm

<em>Thus the wavelengths in the visible spectrum (</em><em>400 to 700 nm)</em><em> that will give destructive interference are </em><em>568 nm</em><em> and </em><em>406 nm</em>

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<em />

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