In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10`Hz and amplitude `48Vm^(-1)`. The total energy density of the electromagnetic field of the wave is :

Options

A. `2.0xx10^(-7)Jm^(-3)`
B. `2.0xx10^(-8) Jm^(-3)`
C. `2.0xx10^(-9)Jm^(-3)`
D. `2.0xx10^(-10)Jm^(-3)`

Explanation

To find the total energy density of the electromagnetic field of the wave, we can follow these steps:

### Step 1: Understand the formula for energy density
The total energy density \( u \) of an electromagnetic wave is given by the sum of the energy densities of the electric field and the magnetic field:
\[ u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \]
Where:
- \( \epsilon_0 \) is the permittivity of free space (\( \approx 8.85 \times 10^{-12} \, \text{F/m} \))
- \( E \) is the amplitude of the electric field
- \( B \) is the amplitude of the magnetic field
- \( \mu_0 \) is the permeability of free space (\( \approx 4\pi \times 10^{-7} \, \text{T m/A} \))

### Step 2: Relate electric and magnetic fields
In an electromagnetic wave, the electric field \( E \) and the magnetic field \( B \) are related by the equation:
\[ B = \frac{E}{c} \]
Where \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

### Step 3: Calculate the magnetic field amplitude
Given the electric field amplitude \( E = 48 \, \text{V/m} \):
\[ B = \frac{E}{c} = \frac{48}{3 \times 10^8} \approx 1.6 \times 10^{-7} \, \text{T} \]

### Step 4: Calculate the energy density due to the electric field
Using the formula for the energy density due to the electric field:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
Substituting the values:
\[ u_E = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (48)^2 \]
Calculating \( 48^2 \):
\[ 48^2 = 2304 \]
Now substituting:
\[ u_E = \frac{1}{2} \times (8.85 \times 10^{-12}) \times 2304 \]
Calculating:
\[ u_E \approx \frac{1}{2} \times (8.85 \times 10^{-12} \times 2304) \approx 1.02 \times 10^{-8} \, \text{J/m}^3 \]

### Step 5: Calculate the energy density due to the magnetic field
Using the formula for the energy density due to the magnetic field:
\[ u_B = \frac{1}{2} \frac{B^2}{\mu_0} \]
Substituting the values:
\[ u_B = \frac{1}{2} \times \frac{(1.6 \times 10^{-7})^2}{4\pi \times 10^{-7}} \]
Calculating \( (1.6 \times 10^{-7})^2 \):
\[ (1.6 \times 10^{-7})^2 = 2.56 \times 10^{-14} \]
Now substituting:
\[ u_B = \frac{1}{2} \times \frac{2.56 \times 10^{-14}}{4\pi \times 10^{-7}} \approx \frac{2.56 \times 10^{-14}}{2.513 \times 10^{-6}} \approx 1.02 \times 10^{-8} \, \text{J/m}^3 \]

### Step 6: Total energy density
Since \( u_E \) and \( u_B \) are equal:
\[ u = u_E + u_B = 1.02 \times 10^{-8} + 1.02 \times 10^{-8} = 2.04 \times 10^{-8} \, \text{J/m}^3 \]
This can be approximated as:
\[ u \approx 2 \times 10^{-8} \, \text{J/m}^3 \]

### Final Answer
The total energy density of the electromagnetic field of the wave is:
\[ \boxed{2 \times 10^{-8} \, \text{J/m}^3} \]