Dimension of `(1/(mu_0)(varepsilon_0))` is

Options

A. `L/T`
B. `T/L `
C. `L^2/T^2`
D. `T^2/L^2`

Answer

To find the dimension of \( \frac{1}{\mu_0 \varepsilon_0} \), we can use the relationship between the speed of light \( c \) and these constants.

### Step-by-step Solution:

1. **Understanding the relationship**: We know that the speed of light \( c \) is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \]
This can be rearranged to express \( \mu_0 \varepsilon_0 \):
\[ c^2 = \frac{1}{\mu_0 \varepsilon_0} \]
Therefore, we can write:
\[ \mu_0 \varepsilon_0 = \frac{1}{c^2} \]

2. **Finding the dimensions of \( c^2 \)**: The speed of light \( c \) has dimensions of length per time:
\[ [c] = \frac{L}{T} \]
Thus, the dimensions of \( c^2 \) are:
\[ [c^2] = \left(\frac{L}{T}\right)^2 = \frac{L^2}{T^2} \]

3. **Finding the dimensions of \( \mu_0 \varepsilon_0 \)**: Since we have established that \( \mu_0 \varepsilon_0 = \frac{1}{c^2} \), we can find the dimensions of \( \mu_0 \varepsilon_0 \):
\[ [\mu_0 \varepsilon_0] = \frac{1}{[c^2]} = \frac{1}{\frac{L^2}{T^2}} = \frac{T^2}{L^2} \]

4. **Finding the dimensions of \( \frac{1}{\mu_0 \varepsilon_0} \)**: Now, we can find the dimensions of \( \frac{1}{\mu_0 \varepsilon_0} \):
\[ \left[\frac{1}{\mu_0 \varepsilon_0}\right] = \frac{1}{\frac{T^2}{L^2}} = \frac{L^2}{T^2} \]

5. **Conclusion**: Therefore, the dimension of \( \frac{1}{\mu_0 \varepsilon_0} \) is:
\[ \frac{L^2}{T^2} \]

### Final Answer:
The dimension of \( \frac{1}{\mu_0 \varepsilon_0} \) is \( \frac{L^2}{T^2} \).

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