The ratio of average electric energy density and total average energy density of electromagnetic wave is:

Options

A. `2`
B. `3`
C. `1/2`
D. `1`

Answer

To solve the problem of finding the ratio of the average electric energy density to the total average energy density of an electromagnetic wave, we can follow these steps:

### Step 1: Understand the Definitions
- The average electric energy density \( u_E \) in an electromagnetic wave is given by:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
where \( \epsilon_0 \) is the permittivity of free space and \( E \) is the electric field strength.

### Step 2: Calculate the Average Magnetic Energy Density
- The average magnetic energy density \( u_B \) is given by:
\[ u_B = \frac{1}{2} \frac{B^2}{\mu_0} \]
where \( \mu_0 \) is the permeability of free space and \( B \) is the magnetic field strength.

### Step 3: Relate Electric and Magnetic Fields
- In an electromagnetic wave, the electric field \( E \) and the magnetic field \( B \) are related by:
\[ B = \frac{E}{c} \]
where \( c \) is the speed of light in vacuum.

### Step 4: Substitute \( B \) in the Magnetic Energy Density
- Substituting \( B \) in the expression for \( u_B \):
\[ u_B = \frac{1}{2} \frac{(E/c)^2}{\mu_0} = \frac{1}{2} \frac{E^2}{\mu_0 c^2} \]
- Since \( c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \), we can write:
\[ \mu_0 c^2 = \frac{1}{\epsilon_0} \]
- Thus, we can simplify \( u_B \):
\[ u_B = \frac{1}{2} \epsilon_0 E^2 \]

### Step 5: Calculate Total Average Energy Density
- The total average energy density \( u \) is the sum of the electric and magnetic energy densities:
\[ u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \]
- Substituting \( u_B \):
\[ u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2 = \epsilon_0 E^2 \]

### Step 6: Find the Ratio
- Now, we can find the ratio of the average electric energy density to the total average energy density:
\[ \text{Ratio} = \frac{u_E}{u} = \frac{\frac{1}{2} \epsilon_0 E^2}{\epsilon_0 E^2} = \frac{1}{2} \]

### Final Answer
Thus, the ratio of the average electric energy density to the total average energy density of an electromagnetic wave is:
\[ \frac{1}{2} \]