Question

A nonsinusoidal electromagnetic wave like that described in Section 32.2 has uniform electric and magnetic fields. The magnitude of the Poynting vector for this wave is 11.0 W/m2. Find the magnitude of the electric field.

Answer 1

The magnitude of the electric field will be 64.38 V/m. Mathematically, the electric field is found as the ratio of the voltage per unit length.

What is electric field?

A physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field.

The magnitude of the electric field in terms of the pointing vector for this wave  is found as;

$\rm E_{MAX} = \sqrt{S \times c \times \mu_0 }$

Where,

E is the electric field

c is the speed of light

$\mu_0$ is the permeability

On Putting the given data ;

$\rm E_{MAX} = \sqrt{11.0 \times 3 \times 10^8 \times 4 \times \pi \times 10^{-7}} \\\\\ \rm E_{MAX} =\sqrt{4144.8} \\\\ E_{MAX} =64.38 \ V/m$

Hence, the magnitude of the electric field will be 64.38 V/m.

To learn more about the electric field refer to the link;

https://brainly.com/question/26690770

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