The dimensions of `epsilon_(0)mu_(0)` are

Options

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

Explanation

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

### Step 1: Understand the relationship between \( c \), \( \epsilon_0 \), and \( \mu_0 \)
We know from electromagnetic theory that the speed of light \( c \) is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]

### Step 2: Square both sides of the equation
To eliminate the square root, we square both sides:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]

### Step 3: Rearrange the equation
Rearranging the equation gives us:
\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]

### Step 4: Substitute the dimensions of \( c \)
The speed of light \( c \) has dimensions of length per time, which can be represented as:
\[ [c] = L T^{-1} \]
Squaring this gives:
\[ [c^2] = L^2 T^{-2} \]

### Step 5: Substitute back into the equation
Now substituting \( c^2 \) into our rearranged equation:
\[ \mu_0 \epsilon_0 = \frac{1}{L^2 T^{-2}} = L^{-2} T^{2} \]

### Step 6: Conclusion
Thus, the dimensions of \( \epsilon_0 \mu_0 \) are:
\[ [\epsilon_0 \mu_0] = L^{-2} T^{2} \]

### Final Answer
The dimensions of \( \epsilon_0 \mu_0 \) are \( L^{-2} T^{2} \).

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