Dimension of Permittivity is :

The correct answer is Option (4) → \( [\mathrm{M^{-1}L^{-3}T^{4}A^{2}}] \)

The permittivity \( \varepsilon_0 \) appears in Coulomb's law, so its dimensions follow from the force between two charges.

    \( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} \;\Rightarrow\; \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^{2}} \)

    \( [q] = [\mathrm{AT}] \)
    \( [F] = [\mathrm{MLT^{-2}}] \)
    \( [r^{2}] = [\mathrm{L^{2}}] \)

    \( \displaystyle [\varepsilon_0] = \frac{[\mathrm{A^{2}T^{2}}]}{[\mathrm{MLT^{-2}}][\mathrm{L^{2}}]} = [\mathrm{M^{-1}L^{-3}T^{4}A^{2}}] \)