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2 years ago

At what distance from a 27 mw point source of electromagnetic waves is the electric field amplitude 0. 060 v/m ?

1 answer:

n200080 [17]2 years ago

5 0

The distance from a 27 mw point source of electromagnetic waves where the electric field amplitude 0. 060 v/m will be 21.21 m .

Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.

The highest point of a wave is known as 'crest' , whereas the lowest point is known as 'trough'. Electromagnetic waves can be split into a range of frequencies. This is known as the electromagnetic spectrum.

c = 3 * $10^{8}$ m/s

∈ = 8.85 * $10^{- 12}$ $C^{2}$ / N/ $m^{2}$

E = 0. 060 v/m

I = P / 4π $r^{2}$

Also , I = c ∈ $E^{2}$ /2

$r^{2}$ = P / 4π I equation 1

substituting the value of I in equation 1

$r^{2}$ = 2 P / 4π (c ∈ $E^{2}$ )

$r^{2}$ = 2 * 27 * $10^{-3}$ / 4 * 3.14 * 3 * $10^{8}$ * 8.85 * $10^{- 12}$ * $(0.06)^{2}$

r = 21.21 m

To learn more about Electromagnetic waves here

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3 years ago

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

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$E_{total}=4.82*10^6N/C$

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We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

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<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

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3 years ago

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Answer:

$V(t)= 240V* H(t-5)$

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Thus we define a function f:

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Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)