Answer:
S = (√(ε₀/μ₀)(Eo²Sin²(kx- ωt)i
Step-by-step 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.