Let `[varepsilon_0]` denote the dimensional formula of the permittivity of vacuum. If M =mass , L=length, T=time and I= electric current Then

Options

A. `[varepsilon_0]=[M^(-1)L^(-3)T^2I]`
B. `[varepsilon_0]-[M^(-1)L^(-3)T^4I^2]`
C. `[varepsilon_0]=[M^(-1)L^(2)T^(-1)I^(-2)]`
D. `[varepsilon_0]=[M^(-1)L^(2)T^(-1)I]`

Answer

To find the dimensional formula of the permittivity of vacuum, denoted as \(\varepsilon_0\), we will use Coulomb's law and the definitions of charge and force. Here’s a step-by-step solution:

### Step 1: Write down Coulomb's Law
Coulomb's law states that the electric force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(R\) is given by:

\[ F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{R^2} \]

### Step 2: Rearrange to express \(\varepsilon_0\)
From the equation above, we can rearrange it to express \(\varepsilon_0\):

\[ \varepsilon_0 = \frac{q_1 q_2}{F R^2} \]

### Step 3: Substitute the definitions of charge and force
We know that charge \(q\) is defined as:

\[ q = I \cdot t \]

where \(I\) is the electric current and \(t\) is time. Thus, for \(q_1\) and \(q_2\):

\[ q_1 = I t \quad \text{and} \quad q_2 = I t \]

So, we can substitute these into the equation for \(\varepsilon_0\):

\[ \varepsilon_0 = \frac{(I t)(I t)}{F R^2} = \frac{I^2 t^2}{F R^2} \]

### Step 4: Substitute the definition of force
Force \(F\) is defined as:

\[ F = m a \]

where \(a\) is acceleration. Acceleration can be expressed as:

\[ a = \frac{d}{t^2} \]

where \(d\) is distance. Therefore, we can write:

\[ F = m \frac{d}{t^2} = m \frac{L}{T^2} \]

Substituting this into our equation for \(\varepsilon_0\):

\[ \varepsilon_0 = \frac{I^2 t^2}{\left(m \frac{L}{T^2}\right) R^2} \]

### Step 5: Substitute \(R\) with \(L\)
Since \(R\) is also a length, we can write \(R^2 = L^2\). Thus, we have:

\[ \varepsilon_0 = \frac{I^2 t^2}{m \frac{L}{T^2} L^2} = \frac{I^2 t^2}{m L^3 T^{-2}} = \frac{I^2 t^2 T^2}{m L^3} \]

### Step 6: Simplify the expression
Now we can simplify the expression:

\[ \varepsilon_0 = \frac{I^2 T^4}{m L^3} \]

### Step 7: Write the dimensional formula
Thus, the dimensional formula of \(\varepsilon_0\) can be expressed as:

\[ \varepsilon_0 = M^{-1} L^{-3} T^{4} I^{2} \]

### Final Answer
The dimensional formula of the permittivity of vacuum \(\varepsilon_0\) is:

\[ \varepsilon_0 = [M^{-1} L^{-3} T^{4} I^{2}] \]

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