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

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${MLT}^{- 1} A^{- 1}$

${MLT}^{- 2} A^{- 2}$

$M^{- 1} L^{- 3} T^{+ 4} A^{2}$

$M^{2} L^{- 2} T^{- 2} A^{2}$

answer is C.

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Detailed Solution

$Force, F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}} \Rightarrow ε_{0} = \frac{q_{1} \cdot q_{2}}{4 π {Fr}^{2}} So dimension of ε_{0} = \frac{[AT]^{2}}{[{MLT}^{- 2}] [L^{2}]} = [M^{- 1} L^{- 3} T^{4} A^{2}]$

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