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Assoli18 [71]
1 year ago
11

The microwave radiation left over from the big bang explosion of the universe has an average energy density of 4. 08 × 10–14 j/m

3. what is the rms value of the electric field of this radiation?
Physics
1 answer:
julsineya [31]1 year ago
8 0

The rms value of electric field will be 6.52*10^-3V/m.

To find the answer, we have to know about the energy density.

<h3>How to find the rms value of electric field of radiation?</h3>
  • We have the expression for average energy density as,

                    U=\frac{\epsilon_0E_0^2}{2}

  • It is given that, the average energy density of 4. 08 × 10^–14 j/m3, Thus, the maximum value of electric field is,

               E_0=\sqrt{\frac{2U}{\epsilon_0} }=\sqrt{\frac{2*4.08*10^{-14}}{8.85*10^{-12}}} =9.22*10^{-3}V/m

  • Thus, the rms value of electric field will be,

                        E_{rms}=\frac{E_0}{\sqrt2} =6.52*10^{-3}V/m

thus, we can conclude that, the rms value of electric field will be 6.52*10^-3V/m.

Learn more about the energy density here:

brainly.com/question/14453487

#SPJ4

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Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

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