In a plane electromagnetic wave, the time-averaged energy flux per unit area is which of the following?

In a plane electromagnetic wave, the time-averaged energy flux per unit area is which of the following?

• A. S_avg = μ0 c E0^2 / 2
• B. S_avg = ε0 c E0^2 / 2
• C. S_avg = E0^2 /(μ0 c)
• D. S_avg = E0^2 /(2 μ0 c)

Answer: (D)

The main idea is the energy flow carried by a plane electromagnetic wave, described by the Poynting vector. In vacuum, the energy flux magnitude is S = (1/μ0) E × B, and for a plane wave E and B are perpendicular with B = E/c. If E(t) = E0 cos(ωt) and B(t) = (E0/c) cos(ωt), then the instantaneous flux is S(t) = (1/μ0) E B = (E0^2/(μ0 c)) cos^2(ωt). Averaging over time, ⟨cos^2⟩ = 1/2, so the time-averaged energy flux is S_avg = E0^2/(2 μ0 c). This is also equal to (1/2) ε0 c E0^2, since ε0 c = 1/(μ0 c). You can also think of it as the average energy density u_avg = ε0 E0^2/2 transported at speed c, giving S_avg = u_avg c.

The main idea is the energy flow carried by a plane electromagnetic wave, described by the Poynting vector. In vacuum, the energy flux magnitude is S = (1/μ0) E × B, and for a plane wave E and B are perpendicular with B = E/c.

If E(t) = E0 cos(ωt) and B(t) = (E0/c) cos(ωt), then the instantaneous flux is S(t) = (1/μ0) E B = (E0^2/(μ0 c)) cos^2(ωt). Averaging over time, ⟨cos^2⟩ = 1/2, so the time-averaged energy flux is S_avg = E0^2/(2 μ0 c). This is also equal to (1/2) ε0 c E0^2, since ε0 c = 1/(μ0 c). You can also think of it as the average energy density u_avg = ε0 E0^2/2 transported at speed c, giving S_avg = u_avg c.