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

Step 1: Recall the Energy Densities

The average electric energy density (\( \langle u_E \rangle \)) and the average magnetic energy density (\( \langle u_B \rangle \)) of an electromagnetic wave are equal: \[ \langle u_E \rangle = \langle u_B \rangle \] The total average energy density (\( \langle u_{\text{total}} \rangle \)) is the sum of the electric and magnetic energy densities: \[ \langle u_{\text{total}} \rangle = \langle u_E \rangle + \langle u_B \rangle \]

Step 2: Find the Ratio

Since \( \langle u_E \rangle = \langle u_B \rangle \), we have: \[ \langle u_{\text{total}} \rangle = \langle u_E \rangle + \langle u_E \rangle = 2 \langle u_E \rangle \] Thus: \[ \langle u_E \rangle = \frac{1}{2} \langle u_{\text{total}} \rangle \] The ratio of average electric energy density to the total average energy density is: \[ \frac{\langle u_E \rangle}{\langle u_{\text{total}} \rangle} = \frac{1}{2} \]

Conclusion: The ratio is \( \frac{1}{2} \) (Option 3).