Dimensions of `1/(mu_0 in_0)` , where symbols have their usual meaning are

Options

A. `L^(-1)T`
B. `L^2T^2`
C. `L^2T^(-2)`
D. `LT^(-1)`

Answer

To find the dimensions of \( \frac{1}{\mu_0 \epsilon_0} \), we can follow these steps:

### Step 1: Understand the relationship between \( \mu_0 \), \( \epsilon_0 \), and the speed of light \( c \)
The speed of light in a vacuum is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
This implies that:
\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]

### Step 2: Find the dimensions of \( c \)
The speed of light \( c \) has dimensions of length per time:
\[ [c] = L T^{-1} \]

### Step 3: Square the dimensions of \( c \)
Now, we square the dimensions of \( c \):
\[ [c^2] = (L T^{-1})^2 = L^2 T^{-2} \]

### Step 4: Relate the dimensions of \( \mu_0 \epsilon_0 \) to \( c^2 \)
From the relationship \( \mu_0 \epsilon_0 = \frac{1}{c^2} \), we can write:
\[ [\mu_0 \epsilon_0] = \frac{1}{[c^2]} = \frac{1}{L^2 T^{-2}} = L^{-2} T^{2} \]

### Step 5: Find the dimensions of \( \frac{1}{\mu_0 \epsilon_0} \)
To find the dimensions of \( \frac{1}{\mu_0 \epsilon_0} \), we take the reciprocal of the dimensions we found in Step 4:
\[ \left[\frac{1}{\mu_0 \epsilon_0}\right] = \frac{1}{L^{-2} T^{2}} = L^{2} T^{-2} \]

### Final Answer
The dimensions of \( \frac{1}{\mu_0 \epsilon_0} \) are:
\[ L^{2} T^{-2} \]

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