Question

A sinusoidal electromagnetic wave has an average intensity of 100 W/m2W/m2. By what factor would the electric-field amplitude of the wave have to be increased in order for the wave to have an average intensity of 8100 W/m2W/m2

Answer 1

Answer:

The electric field is increase to 80 $\frac{V}{m}$  so average intensity became 8100 $\frac{W}{m^{2} }$.

Explanation:

Given :

Initial average intensity  $I_{1} = 100 \frac{W}{m^{2} }$

Final average intensity   $I_{2} = 8100 \frac{W}{m^{2} }$

The average intensity is proportional to the square of the electric field.

∴  $I = \frac{1}{2}c\epsilon_{o} E_{o}^{2}$

    $I$ ∝ $E_{o^{2} }$

⇒  $\sqrt{I_{1} } = E_{1}$

  $\sqrt{100 } = E_{1}$

So initial electric field is $E_{1} = 10 \frac{V}{m}$

Now intensity become $8100\frac{W}{m^{2} }$ so electric field is,

  $E_{2} = \sqrt{8100}$

So final electric field  $E_{2} = 90 \frac{V}{m}$

So electric field increase by 80 $\frac{V}{m}$