Question

The average value of electric energy density in an electromagnetic wave is: [where \( E_0 \) is the peak value]

• A. \( \frac{\varepsilon_0 E_{\text{rms}}^2}{4} \)
• B. \( \frac{1}{2} \varepsilon_0 E_0^2 \)
• C. \( \frac{1}{2} \varepsilon_0 E_0 \)
• D. \( \frac{1}{4} \varepsilon_0 E_0^2 \)

Hint:

The average energy density of an electric field in an electromagnetic wave is derived using the squared sinusoidal function, which has an average value of \( \frac{E_0^2}{2} \). Apply this to the standard energy density formula \( u_E = \frac{1}{2} \varepsilon_0 E^2 \) to obtain the correct expression.

Answer 1

The Correct Option is D

Step 1: Understanding the Electric Energy Density Formula The energy density of the electric field in an electromagnetic wave is given by: \[ u_E = \frac{1}{2} \varepsilon_0 E^2 \] where: - \( u_E \) is the instantaneous electric energy density, - \( \varepsilon_0 \) is the permittivity of free space, - \( E \) is the electric field at a given instant. Since the electric field in an electromagnetic wave varies sinusoidally, we need to compute its average value over a complete cycle. 
Step 2: Finding the Average Electric Energy Density The time-averaged value of \( E^2 \) for a sinusoidal wave is given by: \[ \langle E^2 \rangle = \frac{E_0^2}{2} \] where \( E_0 \) is the peak value of the electric field. Substituting this into the energy density formula: \[ \langle u_E \rangle = \frac{1}{2} \varepsilon_0 \times \frac{E_0^2}{2} \] \[ \langle u_E \rangle = \frac{1}{4} \varepsilon_0 E_0^2 \] 
Step 3: Conclusion Thus, the average value of electric energy density in an electromagnetic wave is \( \frac{1}{4} \varepsilon_0 E_0^2 \), which matches option (4).