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4 years ago

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A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave

. Then, determine the amount of energy U that flows through area of 0.0287 m 2 over an interval of 13.7 s, assuming that the area is perpendicular to the direction of wave propagation. S

1 answer:

7nadin3 [17]4 years ago

5 0

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

$I = \frac{1}{2} C{ {\varepsilon _0}E_m^2$

Here, $\varepsilon _0$ is the permittivity of the free space,

$E_m$ is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10 −12 C² /N⋅m² for ${\varepsilon _0}$

and 93.3 V/m for ${E_{\rm{m}$

$I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2$

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

$E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J$

The energy flowing through the given area is 4.55 J

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3 years ago

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Answer:

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Explanation:

Potential V is given as;

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