Question 55: EM waves

Question

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $$5 	imes 10^{10} \mathrm{~Hz}$$ and an amplitude of $$50 \mathrm{~Vm}^{-1}$$. The total average energy density of the electromagnetic field of the wave is : [Use $$\varepsilon_0=8.85 	imes 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$$ ]

Accepted Answer

The average energy density of an electromagnetic wave is given by the formula:
$$U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$$
For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:
$$U_E = \frac{1}{2} \varepsilon_0 E^2$$
Given that the electric field amplitude ($E$) is $50 \, \mathrm{Vm}^{-1}$ and the permittivity of free space ($\varepsilon_0$) is $8.85 	imes 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$, we can calculate the energy density due to the electric field as follows:
$$U_E = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes (50)^2 = 1.10625 	imes 10^{-8} \, \mathrm{J/m^3}$$
Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.

Answer

$$4.425 	imes 10^{-8} \mathrm{Jm}^{-3}$$

Answer

$$2.212 	imes 10^{-10} \mathrm{Jm}^{-3}$$

Answer

$$2.212 	imes 10^{-8} \mathrm{Jm}^{-3}$$

Answer

$$1.106 	imes 10^{-8} \mathrm{Jm}^{-3}$$

EM waves

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