Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will be

Options

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

Explanation

To find the dimensions of permittivity (\( \epsilon_0 \)), we start with the formula for the Coulomb force:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

Where:
- \( F \) is the force,
- \( Q_1 \) and \( Q_2 \) are the charges,
- \( r \) is the distance between the charges,
- \( \epsilon_0 \) is the permittivity of free space.

### Step 1: Rearranging the Formula
We can rearrange the formula to express \( \epsilon_0 \):

\[ \epsilon_0 = \frac{Q_1 Q_2}{4 \pi F r^2} \]

### Step 2: Finding Dimensions of Each Quantity
Next, we need to find the dimensions of each quantity involved:

1. **Charge (\( Q \))**: The dimension of charge can be expressed in terms of current and time:
\[ [Q] = [I][T] = A \cdot T \]
Therefore, for \( Q_1 \) and \( Q_2 \):
\[ [Q_1 Q_2] = [Q_1][Q_2] = (A \cdot T)(A \cdot T) = A^2 T^2 \]

2. **Force (\( F \))**: The dimension of force is given by:
\[ [F] = [M][L][T^{-2}] = M L T^{-2} \]

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

### Step 3: Substituting Dimensions into the Rearranged Formula
Now substituting these dimensions into the rearranged formula for \( \epsilon_0 \):

\[ \epsilon_0 = \frac{[Q_1 Q_2]}{[F][r^2]} = \frac{A^2 T^2}{(M L T^{-2})(L^2)} \]

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

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

### Step 5: Writing the Final Dimensions
Thus, the dimensions of permittivity (\( \epsilon_0 \)) are:

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

### Conclusion
The final answer for the dimensions of permittivity is:

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