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

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 \( \text{NC}^{-1} \).