Question

The electric field in an electromagnetic wave is given by \( E = (50 \, NC^{-1}) \sin \omega (t - x/c) \). The energy contained in a cylinder of volume V is \( 5.5 \times 10^{-12} \, J \). The value of V is \(\dots\dots\dots\) \( cm^3 \).
(given \( \epsilon_0 = 8.8 \times 10^{-12} \, C^2 N^{-1} m^{-2} \))

Hint:

In electromagnetic waves, the total average energy density is \( \epsilon_0 E_{rms}^2 \) or \( \frac{1}{2} \epsilon_0 E_0^2 \). This accounts for both the electric and magnetic components which contribute equally on average.

Answer 1

Correct Answer: 500

Step 1: Understanding the Concept:
The average energy density \( u_{avg} \) of an electromagnetic wave is the total energy per unit volume, which is the sum of the average electric and magnetic energy densities.
Step 2: Key Formula or Approach:
1. Average Energy Density: \( u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 \).
2. Total Energy: \( U = u_{avg} \times V \).
Step 3: Detailed Explanation:
Given:
- Peak Electric Field \( E_0 = 50 \, NC^{-1} \).
- Total Energy \( U = 5.5 \times 10^{-12} \, J \).
- \( \epsilon_0 = 8.8 \times 10^{-12} \, SI \, units \).
1. Calculate Average Energy Density:
\[ u_{avg} = \frac{1}{2} \times 8.8 \times 10^{-12} \times (50)^2 \]
\[ u_{avg} = 4.4 \times 10^{-12} \times 2500 = 11000 \times 10^{-12} = 1.1 \times 10^{-8} \, J/m^3 \]
2. Calculate Volume V in \( m^3 \):
\[ V = \frac{U}{u_{avg}} = \frac{5.5 \times 10^{-12}}{1.1 \times 10^{-8}} \]
\[ V = 5 \times 10^{-4} \, m^3 \]
3. Convert to \( cm^3 \):
Since \( 1 \, m^3 = 10^6 \, cm^3 \):
\[ V = 5 \times 10^{-4} \times 10^6 = 500 \, cm^3 \]
Step 4: Final Answer:
The value of V is 500 \( cm^3 \).