Question

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 69.1 V/m . Calculate the intensity ???? of this wave. Then, determine the amount of energy ???? that flows through area of 0.0247 m^2 over an interval of 13.1 s, assuming that the area is perpendicular to the direction of wave propagation.

Answer 1

Answer:

The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.

Explanation:

Given that,

Electric field amplitude E₀= 69.1 V/m

Area A= 0.0247 m²

Time t= 13.1 s

We need to calculate the intensity

Using formula of intensity

$S=\dfrac{1}{2}c\epsilon_{0}E_{0}^2$

Where, c = speed of light

Put the value into the formula

$S=\dfrac{1}{2}\times3\times10^{8}\times8.85\times10^{-12}\times(69.1)^2$

$S=6.3385\ N/m^2$

(b). We need to calculate the energy

Using formula of energy

$E=SAt$

Where, A = area

t = time

S = intensity

Put the value into the formula

$E =6.3385\times0.0247\times13.1$

$E =2.0509\ J$

Hence, The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.