The total average energy density (u) of an electromagnetic wave is given by the sum of the average energy densities of the electric field (uE) and the magnetic field (uB). However, for an electromagnetic wave, these two quantities are equal, i.e., uE = uB. Therefore, the total energy density is given by u = uE + uB = 2uE.

The average energy density of the electric field is given by uE = 1/2 ε0 E^2, where ε0 is the permittivity of free space and E is the amplitude of the electric field.

Given that ε0 = 8.85×10−12 C^2/Nm^2 and E = 50 V/m, we can substitute these values into the equation to find uE:

uE = 1/2 * 8.85×10−12 C^2/Nm^2 * (50 V/m)^2
= 1/2 * 8.85×10−12 C^2/Nm^2 * 2500 V^2/m^2
= 1.11×10−8 J/m^3

Therefore, the total average energy density of the electromagnetic field of the wave is u = 2uE = 2 * 1.11×10−8 J/m^3 = 2.22×10−8 J/m^3.

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The total average energy density (u) of an electromagnetic wave is given by the sum of the average energy densities of the electric field (uE) and the magnetic field (uB). However, for an electromagnetic wave, these two quantities are equal, i.e., uE = uB. Therefore, the total energy density is given by u = uE + uB = 2uE.

The average energy density of the electric field is given by uE = 1/2 ε0 E^2, where ε0 is the permittivity of free space and E is the amplitude of the electric field.

Given that ε0 = 8.85×10−12 C^2/Nm^2 and E = 50 V/m, we can substitute these values into the equation to find uE:

uE = 1/2 * 8.85×10−12 C^2/Nm^2 * (50 V/m)^2
   = 1/2 * 8.85×10−12 C^2/Nm^2 * 2500 V^2/m^2
   = 1.11×10−8 J/m^3

Therefore, the total average energy density of the electromagnetic field of the wave is u = 2uE = 2 * 1.11×10−8 J/m^3 = 2.22×10−8 J/m^3.

Knowee AI · Accepted Answer

The total average energy density (u) of an electromagnetic wave is given by the sum of the average energy densities of the electric field (uE) and the magnetic field (uB). However, for an electromagnetic wave, these two quantities are equal, i.e., uE = uB. Therefore, the total energy density is given by u = uE + uB = 2uE.

The average energy density of the electric field is given by uE = 1/2 ε0 E^2, where ε0 is the permittivity of free space and E is the amplitude of the electric field.

Given that ε0 = 8.85×10−12 C^2/Nm^2 and E = 50 V/m, we can substitute these values into the equation to find uE:

uE = 1/2 * 8.85×10−12 C^2/Nm^2 * (50 V/m)^2
   = 1/2 * 8.85×10−12 C^2/Nm^2 * 2500 V^2/m^2
   = 1.11×10−8 J/m^3

Therefore, the total average energy density of the electromagnetic field of the wave is u = 2uE = 2 * 1.11×10−8 J/m^3 = 2.22×10−8 J/m^3.

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010 Hz and an amplitude of 50Vm−1. The total average energy density of the electromagnetic field of the wave is : [Use ε0=8.85×10−12C2/Nm2 ]

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