Let $[\varepsilon_{0}]$ denote the dimensional formula of the permittivity of vacuum. If $\mathrm{M}=$ mass, $\mathrm{L}=$ length, $\mathrm{T}=$ time and $\mathrm{A}=$ electric current, then:

(A) $[\varepsilon_{0}]=[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}]$

(B) $[\varepsilon_{0}]=[\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \mathrm{~A}^{-2}]$

(C) $[\varepsilon_{0}]=[\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-1} \mathrm{~A}]$

(D) $[\varepsilon_{0}]=[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{2} \mathrm{~A}]$

JEE Main · 2013 · 07 Apr 2013 · Offline · Units & Measurement

Let

[ε_{0}]

denote the dimensional formula of the permittivity of vacuum. If

M =

mass,

L =

length,

T =

time and

A =

electric current, then:

Answer: (A)

Coulomb's law,

F = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{r ^{2}}

, lets us solve for the permittivity as

ε_{0} = \frac{q ^{2}}{4 π F r ^{2}}

, and the dimensions follow by substitution. Charge is

[A T]

, so

q^{2}

[A^{2} T^{2}]

, force is

[M L T^{- 2}]

and

r^{2}

[L^{2}]

. Combining,

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

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option