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Step-by-Step Solution
Step 1: Recall Coulomb’s Law
Coulomb’s law for the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by
$$F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 \, q_2}{r^2}.$$
Step 2: Express $\epsilon_0$ from Coulomb’s Law
Rearrange to isolate the vacuum permittivity $\epsilon_0$:
$$\epsilon_0 = \frac{1}{4\pi} \cdot \frac{q_1 \, q_2}{F \, r^2}.$$
Step 3: Identify the Dimensional Formula of Each Quantity
Charge $(q)$: In SI units, charge has dimension $[A\,T]$ (current $\times$ time).
Force $(F)$: Dimension is $[M\,L\,T^{-2}]$ (mass $\times$ acceleration).
Distance $(r)$: Dimension is $[L]$.
Step 4: Assign Dimensions to the Terms in the Expression
Substitute the dimensional forms into
$$\epsilon_0 = \frac{q_1 \, q_2}{F \, r^2}.$$
Ignoring the constant $4\pi$, which is dimensionless, we get:
$$[\epsilon_0] = \frac{[A\,T] \times [A\,T]}{[M\,L\,T^{-2}] \times [L^2]}.$$
Step 5: Simplify the Expression
Combine the exponents step by step:
\[
[A \, T] \times [A \, T] = [A^2 \, T^2],
\]
\[
F \, r^2 = [M\,L\,T^{-2}] \times [L^2] = [M\,L^3\,T^{-2}].
\]
Thus,
\[
[\epsilon_0]
= \frac{[A^2 \, T^2]}{[M \, L^3 \, T^{-2}]}
= [M^{-1} \, L^{-3} \, T^{4} \, A^{2}].
\]
Step 6: State the Final Result
Therefore, the dimensional formula for the permittivity of free space is
$$[\epsilon_0] = [M^{-1} \, L^{-3} \, T^{4} \, A^{2}].$$
