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

We need to find the ratio of the average electric energy density to the total average energy density of an electromagnetic wave.

An electromagnetic wave carries energy in both its electric and magnetic field components, and the instantaneous energy densities are given by $$u_E = \frac{1}{2}\epsilon_0 E^2$$ for the electric field and $$u_B = \frac{B^2}{2\mu_0}$$ for the magnetic field.

Since in an electromagnetic wave the electric and magnetic fields are related by $$E = cB \quad \text{where } c = \frac{1}{\sqrt{\mu_0 \epsilon_0}},$$ substituting $$B = E/c$$ into the magnetic energy density yields $$u_B = \frac{B^2}{2\mu_0} = \frac{E^2}{2\mu_0 c^2} = \frac{E^2}{2\mu_0 \cdot \frac{1}{\mu_0 \epsilon_0}} = \frac{\epsilon_0 E^2}{2} = u_E.$$ This shows that in an electromagnetic wave, the electric and magnetic energy densities are equal at every instant.

From this it follows that the total energy density, being the sum of the electric and magnetic contributions, is $$u_{total} = u_E + u_B = u_E + u_E = 2u_E,$$ and hence the required ratio is $$\frac{u_E}{u_{total}} = \frac{u_E}{2u_E} = \frac{1}{2}.$$ Since this holds at every instant, it also holds for the averages.

The correct answer is Option (4): $$\frac{1}{2}$$.