The dimensional formula of permittivity `(epsilon_0)` of free space is :

Options

A. `M^(-1)L^(-3) T^4A^2`
B. `M^(-1)L^(-2)T^2 A`
C. `M^(-1)L^(-2)T^(-2)A`
D. `M^(-1)L^(-2)T^(-2) A^(-2)`

Answer

To find the dimensional formula of permittivity of free space \((\epsilon_0)\), we can start from the formula for the electrostatic force given by Coulomb's law:

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

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 formula
From the equation, we can express \(\epsilon_0\) as:

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

### Step 2: Identifying dimensions
Now we need to find the dimensions of each term in the equation:

1. **Charge \(q\)**: The dimension of electric charge is given by:
\[ [q] = A \cdot T \]
where \(A\) is the dimension of current and \(T\) is the dimension of time.

2. **Force \(F\)**: The dimension of force is:
\[ [F] = M \cdot L \cdot T^{-2} \]
where \(M\) is mass, \(L\) is length, and \(T\) is time.

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

### Step 3: Substituting dimensions into the formula
Substituting these dimensions into the expression for \(\epsilon_0\):

\[ \epsilon_0 = \frac{(q_1 q_2)}{F \cdot r^2} = \frac{(A \cdot T)(A \cdot T)}{M \cdot L \cdot T^{-2} \cdot L^2} \]

This simplifies to:

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

### Step 4: Final dimensional formula
Thus, the dimensional formula of permittivity \((\epsilon_0)\) is:

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

### Conclusion
The correct answer is:

\[ \epsilon_0 \text{ has the dimensional formula } M^{-1} L^{-3} A^2 T^4 \]