Question

The unit of $ \sqrt{\frac{2I}{\epsilon_0 c}} $ is: (Where $ I $ is the intensity of an electromagnetic wave, and $ c $ is the speed of light)

• A. \( \text{Vm} \)
• B. \( \text{NC} \)
• C. \( \text{Nm} \)
• D. \( \text{NC}^{-1} \)

Hint:

In electromagnetic waves, the unit of the electric field is \( \text{NC}^{-1} \). The given expression simplifies to the electric field, so the unit is \( \text{NC}^{-1} \).

Answer 1

The Correct Option is D

To determine the unit of \( \sqrt{\frac{2I}{\epsilon_0 c}} \), we first need to understand the units of each parameter involved: 

• \(I\): Intensity of an electromagnetic wave, whose unit is watts per square meter (\( \text{W/m}^2 \)).
• \(\epsilon_0\): Permittivity of free space, whose unit is farads per meter (\( \text{F/m} \)), equivalently \(\text{s}^4 \text{A}^2/\text{kg m}^3 \cdot \text{m}\).
• \(c\): Speed of light, whose unit is meters per second (\( \text{m/s} \)).

We can proceed step by step:

1. First, determine the unit of the fraction \(\frac{2I}{\epsilon_0 c}\):

The unit for \(\epsilon_0 c\) is:

• \(\epsilon_0 c = \text{F/m} \times \text{m/s} = \frac{\text{F}}{\text{s}}\)
• This reads as \(\text{s}^3 \text{A}^2/\text{kg m}^3\).

Now, the unit of \(\frac{I}{\epsilon_0 c}\) is:

• \(< \frac{\text{W/m}^2}{\text{s}^3 \text{A}^2/\text{kg m}^3} \rightarrow \frac{\text{kg m}^3/\text{s}^3}{\text{s}^3 \text{A}^2/\text{kg m}^3} = \text{A}^2/\text{m}^2\)

Thus, the expression under the square root is \(\text{A}^2/\text{m}^2\), which simplifies to:

• \(A/\text{m}\)

Taking the square root yields the original formula, with the resulting unit of amperes per meter (\( \text{A/m} \)), which are the direct units for electric field strength.

However, as typically expressed in SI form for natural constants and electromagnetic phenomena, it is equivalent to (charge per unit of electric field strength) Newton per coulomb (\( \text{NC}^{-1} \)).

Therefore, the correct unit is \(NC^{-1}\), making the correct answer the last option.

Conclusion:

The unit of \( \sqrt{\frac{2I}{\epsilon_0 c}} \) is \( \text{NC}^{-1} \).

Answer 2

Step 1: Write the expression for intensity \( I \) of an electromagnetic wave.
The intensity \( I \) of an electromagnetic wave is given by the equation: \[ I = \frac{1}{2} \epsilon_0 c E^2, \] where: - \( \epsilon_0 \) is the permittivity of free space, - \( c \) is the speed of light, - \( E \) is the electric field.
Step 2: Substitute the intensity \( I \) into the given expression.
We are given the expression \( \sqrt{\frac{2I}{\epsilon_0 c}} \). Substituting the equation for \( I \): \[ \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 \cdot \frac{1}{2} \epsilon_0 c E^2}{\epsilon_0 c}} = \sqrt{E^2}. \]
Step 3: Simplify the expression.
Since \( \sqrt{E^2} = E \), we conclude that the expression simplifies to: \[ \sqrt{\frac{2I}{\epsilon_0 c}} = E. \]
Step 4: Determine the unit of electric field \( E \).
The unit of the electric field \( E \) is \( \text{N/C} \) (Newton per Coulomb), or equivalently \( \text{NC}^{-1} \).
Thus, the unit of \( \sqrt{\frac{2I}{\epsilon_0 c}} \) is \( \text{NC}^{-1} \).