Let `[in_(0)]` denote the dimensional formula of the permittivity of vacuum. If M= mass, L=length, T=Time and A= electric current, then:

Options

A. `[epsilon_(0)]=[M^(-1)L^(-3)T^(4)A^(2)]`
B. `[epsilon_(0)]=[M^(-1)L^(2)T^(-1)A^(-2)]`
C. `[epsilon_(0)]=[M^(-1)L^(2)T^(-1)A]`
D. `[epsilon_(0)]=[M^(-1)L^(-3)T^(2)A]`

Answer

To find the dimensional formula of the permittivity of vacuum (ε₀), we can start from the equation that relates the Coulomb's force to ε₀. The equation is given as:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

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.

### Step 1: Rearranging the equation
From the equation, we can express ε₀ as:

\[ \epsilon_0 = \frac{1}{4\pi} \frac{q_1 q_2}{F r^2} \]

### Step 2: Identifying units
The SI unit of force (F) is Newton (N), which can be expressed in terms of fundamental dimensions as:

\[ 1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2 \]

The unit of charge (Coulomb, C) can be expressed in terms of current (A) and time (s):

\[ 1 \, \text{C} = 1 \, \text{A} \cdot \text{s} \]

### Step 3: Substituting units into ε₀
Now substituting the units into the expression for ε₀:

\[ \epsilon_0 = \frac{1}{4\pi} \frac{(C)(C)}{(N)(m^2)} \]

Substituting the units of C and N:

\[ \epsilon_0 = \frac{(A \cdot s)(A \cdot s)}{(kg \cdot m/s^2)(m^2)} \]

### Step 4: Simplifying the expression
This simplifies to:

\[ \epsilon_0 = \frac{A^2 \cdot s^2}{kg \cdot m \cdot m^2/s^2} \]

\[ \epsilon_0 = \frac{A^2 \cdot s^2 \cdot s^2}{kg \cdot m^3} \]

### Step 5: Final dimensional formula
Now, we can express the dimensional formula of ε₀ as:

\[ [\epsilon_0] = \frac{A^2 \cdot T^4}{M \cdot L^3} \]

This means:

- Mass (M) has a power of -1,
- Length (L) has a power of -3,
- Time (T) has a power of 4,
- Electric current (A) has a power of 2.

Thus, the dimensional formula of permittivity of vacuum is:

\[ [\epsilon_0] = M^{-1} L^{-3} T^4 A^2 \]

### Conclusion
The correct option based on the dimensional analysis is:

\[ [\epsilon_0] = M^{-1} L^{-3} T^4 A^2 \]