Electromagnetic waves transport energy. This problem shows you which parts of the energy are stored in the electric and magnetic fields, respectively, and also makes a useful connection between the energy density of a plane electromagnetic wave and the Poynting vector. In this problem, we explore the properties of a plane electromagnetic wave traveling at the speed of light calong the x axis through vacuum. Its electric and magnetic field vectors are as follows: E= EoSin(kx- ωt)j
B= BoSin (kx- ωt)k

Throughout, use these variables (E, B, Eo, Bo, k, x, and w) in your answers. You will also need the permittivity of free space εo and the permeability of free space μo.

Required:
What is the instantaneous energy μE(t) in the electric field of the wave?

S = (√(ε₀/μ₀)(Eo²Sin²(kx- ωt)i

Explanation:

The instantaneous energy of the wave is given by the Poynting vector, S = 1/μ₀(E × B) where μ₀ = permeability of free space

So, substituting the variables into the equation with E= EoSin(kx- ωt)j and B= BoSin (kx- ωt)k

So,  S = 1/μ₀(E × B)

S = 1/μ₀(EoSin(kx- ωt)j × BoSin (kx- ωt)k)

S = 1/μ₀(EoSin(kx- ωt) × BoSin (kx- ωt) (j × k)

S = 1/μ₀(EoBoSin²(kx- ωt)i  (i = j × k)

Now,

Eo/Bo = c where c = speed of light

So, Bo = Eo/c

Thus

S = 1/μ₀(EoBoSin²(kx- ωt)i

S = 1/μ₀(Eo(Eo/c)Sin²(kx- ωt)i

S = 1/cμ₀(Eo²Sin²(kx- ωt)i

Also,c = 1/√μ₀ε₀ where ε₀ =permittivity of free space

So,

S = 1/cμ₀(Eo²Sin²(kx- ωt)i

S = 1/(1/√μ₀ε₀)μ₀(Eo²Sin²(kx- ωt)i

S = 1/(1/√(ε₀/μ₀)(Eo²Sin²(kx- ωt)i

S = (√(ε₀/μ₀)(Eo²Sin²(kx- ωt)i

which is the instantaneous energy in the electric field.

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