What is the maximum value of the electric field in an electromagnetic wave whose average intensity is 7.05 w/m^2?

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22-08-2022
Physics

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What is the maximum value of the electric field in an electromagnetic wave whose average intensity is 7.05 w/m^2?

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smecloudbird18 smecloudbird18 · Answer 1 · 2022-09-05T11:59:18-04:00

The maximum value of the electric field in an electromagnetic wave is

[tex]E[/tex] = 2.3×[tex]10^{2}[/tex] N/C

Electric field is the space surrounding an electric charge q in which another charge [tex]q_{0}[/tex], experiences a force of attraction or repulsion.

Given, the average intensity is

I = 7.O5 w/[tex]m^{2}[/tex]

The expression of average intensity can be written as ,

I = [tex]\frac{1}{2}[/tex]∈[tex]_{0}[/tex][tex]E^{2}[/tex]

[tex]E^{2}[/tex] = 2I/∈[tex]_{0}[/tex]C

Here , ∈[tex]_{0}[/tex] is the permittivity of free space , E is the required value of the electric field and c is the speed of light.

Putting the given values in above equation ,

[tex]E^{2}[/tex] = [2 (7.05w/[tex]m^{2}[/tex])}/ (8.85×[tex]10^{-12}[/tex][tex]c^{2}/N.m^{2}[/tex])(3×[tex]10^{8} m/s[/tex])

[tex]E^{2}[/tex] = 5.31×[tex]10^{2}[/tex] N/C

[tex]E[/tex] = 2.3×[tex]10^{2}[/tex] N/C

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