An electromagnetic wave has an electric field amplitude of 100 V/m. What is the average energy density of the wave in free space?

A. 4.43 10^-8 J/m^3 — The average energy density of an electromagnetic wave is given by the sum of the average electric and magnetic energy densities, which are equal. The formula is u = 0.5 * epsilon0 * E_rms^2, where E_rms = E0 / sqrt(2).

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$An electromagnetic wave has an electric field amplitude of $E_0 = 100 \text{ V/m}$. What is the average energy density $u_{avg}$ of the wave in free space?$

A. $4.43 \times 10^{-8} \text{ J/m}^3$
B. $8.85 \times 10^{-8} \text{ J/m}^3$
C. $2.21 \times 10^{-8} \text{ J/m}^3$
D. $1.77 \times 10^{-7} \text{ J/m}^3$

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Correct answer: A — $4.43 \times 10^{-8} \text{ J/m}^3$

Solution

u_{a v g} = \frac{1}{2} ϵ_{0} E_{r m s}^{2} = \frac{1}{2} ϵ_{0} (\frac{E _{0}}{2})^{2} = \frac{1}{4} ϵ_{0} E_{0}^{2} = 0.25 \times 8.854 \times 1 0^{- 12} \times 10 0^{2} \approx 2.21 \times 1 0^{- 8} J/m^{3}

. Wait, recalculating:

0.25 \times 8.85 \times 1 0^{- 12} \times 1 0^{4} = 2.21 \times 1 0^{- 8}

. Let me recheck the options. Actually,

u = \frac{1}{2} ϵ_{0} E_{0}^{2}

is the peak, average is

\frac{1}{4} ϵ_{0} E_{0}^{2}

. Correct value is

2.21 \times 1 0^{- 8}

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An electromagnetic wave has an electric field amplitude of $E_0 = 100 \text{ V/m}$. What is the average energy density $u_{avg}$ of the wave in free space?

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$An electromagnetic wave has an electric field amplitude of $E_0 = 100 \text{ V/m}$. What is the average energy density $u_{avg}$ of the wave in free space?$