Understanding the Dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \)

The question asks for the dimension of the quantity \( \dfrac{1}{\mu_0 \varepsilon_0} \), where \( \mu_0 \) is the vacuum permeability and \( \varepsilon_0 \) is the vacuum permittivity. These constants appear together in fundamental relationships in electromagnetism.

Relating \( \mu_0 \) and \( \varepsilon_0 \) to the Speed of Light

A key relationship in physics connects the vacuum permeability \( \mu_0 \), the vacuum permittivity \( \varepsilon_0 \), and the speed of light in vacuum, denoted by \( c \). This relationship is given by the formula:

\( c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \)

This equation shows that the speed of light \( c \) is directly related to the constants \( \mu_0 \) and \( \varepsilon_0 \). The expression \( \dfrac{1}{\mu_0 \varepsilon_0} \) is therefore related to \( c^2 \).

Calculating the Dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \)

From the relationship \( c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}} \), we can square both sides to find the expression whose dimension we need:

\( c^2 = \left(\dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}\right)^2 \)

\( c^2 = \dfrac{1}{\mu_0 \varepsilon_0} \)

So, the dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \) is the same as the dimension of \( c^2 \).

Determining the Dimension of Speed

Speed is defined as distance traveled per unit time. Its dimension is length divided by time.

• Dimension of Length is \( [L] \)
• Dimension of Time is \( [T] \)
• Dimension of Speed, \( [c] \), is \( \dfrac{[L]}{[T]} = [LT^{-1}] \)

Finding the Dimension of \( c^2 \)

Since the dimension of speed \( c \) is \( [LT^{-1}] \), the dimension of \( c^2 \) will be the square of the dimension of \( c \).

\( [c^2] = [LT^{-1}]^2 \)

Applying the exponent to both parts inside the bracket:

\( [c^2] = [L^2 T^{-2}] \)

Final Dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \)

As we established that the dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \) is the same as the dimension of \( c^2 \), we find that the dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \) is \( [L^2 T^{-2}] \). This result aligns with one of the given options.

Understanding the fundamental constants like vacuum permeability \( \mu_0 \) and vacuum permittivity \( \varepsilon_0 \) and their relation to the speed of light is crucial for determining the dimension of \( \dfrac{1}{\mu_0 \varepsilon_0} \).