JEE MAINS 2013 Measurement

Let [ε_o] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length and T = time and A = electric current then,

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(A) [ ε_o ] = [ M⁻¹ L⁻³ T⁴ A² ]                          

(B) [ ε _o] = [ M⁻¹ L² T⁻¹ A⁻² ]

(C) [ ε_o ] = [ M⁻¹ L² T⁻¹ A ]

(D) [ ε_o ] = [ M⁻¹ L⁻³ T² A ]

 Let’s consider Coulomb’s law of electrostatic force.

F = { 1 / ( 4πε_o ) }  { ( q₁q₂ ) / r²  }

Therefore,

ε_o = { 1 / (4πF ) } { ( q₁q₂ ) / r² }

Dimensions of ‘ε_o’ = Dimensions of { 1 / ( 4πF ) } 

                                                X  Dimensions of { ( q₁q₂ ) / r² }

_ _ _ _ _ _ _ _ _ _ (1)

Here,

4 and π are numbers without any dimension

And

[ F ]= [ M¹,L¹,S^-2 ]

For [ q ]

We know that current is rate of flow of charge.

Therefore,

                                                    I = q / t

i.e.

                                                   q = I t

Therefore,

                                               [ q ] = [ I ] X [ t ]

       [ q ] =   [ M⁰, L⁰, S^-1, A¹ ]

And r is radius

     [ r ] = [ M⁰, L¹, S⁰, A⁰ ]

Putting this value in equation ( 1 ) we get,

        [ ε_o ] = [ 1 / ( F ) ]  X [ ( q₁q₂ ) / r² ]

        [ ε_o ] = { 1 / [ M¹, L¹, S^-2 ] } X [ M⁰, L⁰, S^-1, A¹ ] [ M⁰, L⁰, S^-1, A¹ ]

 / [ M⁰, L², S⁰ , A⁰ ]  }

        [ε_o]=[M^-1,L^-1,S²]  X {[M⁰,L⁰,S^-2 ,A²] / [M⁰,L²,S⁰ ,A⁰] }

Dimensions of ‘ε_o’= [M^-1,L^-3,S⁴, A² ]