Question

The ratio of average electric energy density and total average energy density of electromagnetic wave is:

• A. 2
• B. 3
• C. $\frac{1}{2}$
• D. 1

Hint:

In an electromagnetic wave, the electric and magnetic fields contribute equally to the total energy density.

Answer 1

The Correct Option is C

To determine the ratio of the average electric energy density to the total average energy density of an electromagnetic wave, let's first understand the relevant physical concepts and equations involved in electromagnetic waves.

In electromagnetic waves, electric and magnetic fields oscillate perpendicular to each other and to the direction of wave propagation. The energy in an electromagnetic wave is uniformly distributed between the electric and magnetic fields.

The average electric energy density (\(u_e\)) is given by:

\(u_e = \frac{1}{2}\epsilon E^2\)

The average magnetic energy density (\(u_m\)) is given by:

\(u_m = \frac{1}{2}\mu H^2\)

For electromagnetic waves in a medium where there are electric fields \(E\) and magnetic fields \(H\), the permeability \(\mu\) and permittivity \(\epsilon\) are related, and we have:

\(E = cB\), where \(c\) is the speed of light, and \(B\) is the magnetic field.

The total average energy density (\(u_{total}\)) of the electromagnetic wave is:

\(u_{total} = u_e + u_m\)

Since both energy densities are equal, we have \(u_e = u_m\), which simplifies to:

\(u_{total} = \frac{1}{2}\epsilon E^2 + \frac{1}{2}\mu H^2\)

But \(u_e = u_m\), hence:

\(u_{total} = 2u_e\)

Therefore, the ratio of the average electric energy density to the total average energy density is:

\(\frac{u_e}{u_{total}} = \frac{u_e}{2u_e} = \frac{1}{2}\)

Thus, the correct answer is: \(\frac{1}{2}\).