In an electromagnetic wave, the amplitude of electric field is `1 V//m`. The frequency of wave is `5xx10^(14)Hz`. The wave is propagating along `z`-axis. The average energy density of electric field, in `"joule"//m^(3)`,will be

Options

A. `2.2 xx 10^(-12)J//m^(3)`
B. `2.2 xx 10^(-10)J//m^(3)`
C. `2.2 xx 10^(-11)J//m^(3)`
D. `2.2 xx 10^(-13)J//m^(3)`

Answer

To find the average energy density of the electric field in an electromagnetic wave, we can follow these steps:

### Step 1: Understand the formula for average energy density
The average energy density (u_e) of the electric field in an electromagnetic wave is given by the formula:
\[ u_e = \frac{1}{2} \epsilon_0 E_{\text{rms}}^2 \]
where \( E_{\text{rms}} \) is the root mean square (RMS) value of the electric field, and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Relate the peak electric field to RMS
The RMS value of the electric field can be calculated from the peak electric field (\( E_0 \)) using the relationship:
\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \]
Given that the amplitude (peak value) of the electric field \( E_0 = 1 \, \text{V/m} \), we can substitute this value into the equation.

### Step 3: Calculate \( E_{\text{rms}} \)
Substituting \( E_0 \) into the RMS formula:
\[ E_{\text{rms}} = \frac{1 \, \text{V/m}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, \text{V/m} \]

### Step 4: Substitute \( E_{\text{rms}} \) into the energy density formula
Now, substitute \( E_{\text{rms}} \) back into the energy density formula:
\[ u_e = \frac{1}{2} \epsilon_0 \left( \frac{1}{\sqrt{2}} \right)^2 \]
This simplifies to:
\[ u_e = \frac{1}{2} \epsilon_0 \cdot \frac{1}{2} = \frac{1}{4} \epsilon_0 \]

### Step 5: Substitute the value of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) is approximately:
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \]
Now substituting this value into the equation:
\[ u_e = \frac{1}{4} \times (8.85 \times 10^{-12}) \]

### Step 6: Calculate the average energy density
Now, performing the calculation:
\[ u_e = \frac{8.85 \times 10^{-12}}{4} = 2.2125 \times 10^{-12} \, \text{J/m}^3 \]
Rounding it off, we get:
\[ u_e \approx 2.21 \times 10^{-12} \, \text{J/m}^3 \]

### Final Answer
Thus, the average energy density of the electric field is:
\[ \boxed{2.21 \times 10^{-12} \, \text{J/m}^3} \]