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Q.

Let $[ε_{0}]$ denote the dimensional formula of the perimitivity of vacuum. If M = mass, L = length, T = Time and A = electric current, then

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a

$[ε_{0}] = [M^{2} L^{2} T^{- 1} A^{- 2}]$

b

$[ε_{0}] = [M^{- 1} L^{- 3} T^{4} A^{2}]$

c

$[ε_{0}] = [M^{- 1} L^{- 3} T^{2} A]$

d

$[ε_{0}] = [M^{- 1} L^{2} T^{- 1} A^{2}]$

answer is B.

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Detailed Solution

From Coulomb’s law $F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{R^{2}}$

$ε_{0} = \frac{q_{1} q_{2}}{4 π F R^{2}}$

Substituting the units

Hence, $ε_{0} = \frac{{C^{2}}_{2}}{N - m^{2}}$

$= \frac{{[A T]}^{2}}{[M L T^{- 2}] [L^{2}]}$

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

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Let [ε0] denote the dimensional formula of the perimitivity of vacuum. If M = mass, L = length, T = Time and A = electric current, then