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

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

Step-by-step 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=(1)/(2)c\epsilon_(0)E_(0)^2`

Where, c = speed of light

Put the value into the formula

`S=(1)/(2)*3*10^(8)*8.85*10^(-12)*(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*0.0247*13.1`

`E =2.0509\ J`

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