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, we can use Coulomb's law, which states:

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

Where:
- \( F \) is the force between two charges,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges,
- \( \varepsilon_0 \) is the permittivity of vacuum.

### Step 1: Rearranging Coulomb's Law
We can rearrange the equation to isolate \( \varepsilon_0 \):

\[ \varepsilon_0 = \frac{q_1 q_2}{F r^2} \cdot 4\pi \]

Since \( 4\pi \) is a dimensionless constant, we can ignore it for dimensional analysis:

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

### Step 2: Finding Dimensions of Each Term
1. **Dimensions of Charge (\( q_1 \) and \( q_2 \))**:
The dimension of electric charge is given by:
\[ [q] = [I][T] \quad \text{(where \( I \) is electric current and \( T \) is time)} \]
Therefore, for \( q_1 \) and \( q_2 \):
\[ [q_1] = [q_2] = [I][T] \]

2. **Dimensions of Force (\( F \))**:
The dimension of force is given by Newton's second law:
\[ [F] = [M][L][T^{-2}] \]

3. **Dimensions of Distance (\( r \))**:
The dimension of distance is:
\[ [r] = [L] \]

### Step 3: Substituting Dimensions into the Equation
Now, we substitute the dimensions into the equation for \( \varepsilon_0 \):

\[ [\varepsilon_0] = \frac{[q_1][q_2]}{[F][r^2]} = \frac{([I][T])([I][T])}{[M][L][T^{-2}][L^2]} \]

### Step 4: Simplifying the Expression
Now we simplify the expression:

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

This can be rewritten as:

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

### Final Result
Thus, the dimensional formula for the permittivity of vacuum \( \varepsilon_0 \) is:

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

### Conclusion
The correct option for the dimensional formula of the permittivity of vacuum is:

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

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