Question

The RMS value of the electric field of an electromagnetic wave emitted by a source is \( 660 \) N/C. The average energy density of the electromagnetic wave is:

• A. \( 1.75 \times 10^{-6} \) J/m\(^3\)
• B. \( 2.75 \times 10^{-6} \) J/m\(^3\)
• C. \( 4.85 \times 10^{-6} \) J/m\(^3\)
• D. \( 3.85 \times 10^{-6} \) J/m\(^3\)

Hint:

The energy density of an electromagnetic wave is given by \( u = \frac{\varepsilon_0 E_{\text{rms}}^2}{2} \).

Answer 1

The Correct Option is D

Step 1: Apply Energy Density Formula The average energy density is given by: \[ u = \frac{\varepsilon_0 E_{\text{rms}}^2}{2} \] Step 2: Compute Energy Density \[ u = \frac{(8.85 \times 10^{-12}) (660)^2}{2} \] \[ u = 3.85 \times 10^{-6} \text{ J/m}^3 \] Thus, the correct answer is \( 3.85 \times 10^{-6} \) J/m³.

Answer 2

Given:
- RMS value of the electric field, \( E_{rms} = 660 \, \text{N/C} \)
- We need to find the average energy density \( u \) of the electromagnetic wave.

Step 1: Relation between electric field and energy density:
The average energy density of an electromagnetic wave is given by:
\[ u = \frac{\varepsilon_0 E_{rms}^2}{2} \] where \( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \) (permittivity of free space).

Step 2: Substitute the values:
\[ u = \frac{8.854 \times 10^{-12} \times (660)^2}{2} = \frac{8.854 \times 10^{-12} \times 435,600}{2} \] \[ = \frac{3.854 \times 10^{-6}}{2} = 3.854 \times 10^{-6} \, \text{J/m}^3 \]

Step 3: Rounding the answer:
\[ u \approx 3.85 \times 10^{-6} \, \text{J/m}^3 \]

Therefore, the average energy density of the electromagnetic wave is:
\[ \boxed{3.85 \times 10^{-6} \, \text{J/m}^3} \]